Middle School Math 8: Linear Equations, Functions & the Pythagorean Theorem

The number that filled a counter

On 1 December 2014, YouTube posted a short notice about one music video. Psy's "Gangnam Style" had been watched more than 2,147,483,647 times. That was a problem, because the old counter under every video could not count any higher. BBC News explained why three days later: the counter stored each view count as a 32-bit integer, and the largest number a 32-bit integer can hold is exactly 2,147,483,647. YouTube's engineers had seen it coming and moved every video to a 64-bit counter.

The number looks like a random string of digits, but it is not. It is 2³¹ − 1: thirty-one 2s multiplied together, minus one. In this lesson you will see why a computer's counter stops there, and you will check the number by hand in five short steps, using rules for exponents that you will need all year.

A word about where you are. This course is the eighth-grade year of mathematics. It follows Middle School Math 7, where you worked with negative numbers, proportional relationships written as y = kx, two-step equations, angles, circles and probability. It ends at the door of Algebra I, which assumes you can already handle exponents, slope, the equation of a line, systems of two equations, functions and the Pythagorean theorem. Everything in between is in these thirty-three lessons, and it starts with powers.

This lesson covers Common Core standard 8.EE.A.1, which asks you to know and apply the properties of integer exponents to write equivalent numerical expressions. Here every exponent is a positive whole number. The next lesson adds zero and negative exponents.

Why thirty-one switches stop at 2,147,483,647

A computer stores a whole number as a row of tiny switches called bits. Each bit is either off or on, written 0 or 1. So the question "how big a number can 32 bits hold?" is really a counting question: how many different on-off patterns can a row of switches make?

Count them for short rows. One switch has 2 patterns: off, on. Two switches have 4: off-off, off-on, on-off, on-on. A third switch doubles the count again, because each of the 4 old patterns can be followed by off or by on, which gives 8. Every switch you add doubles the number of patterns.

The view counter had 32 switches, but it kept the first one to mark whether a number is negative. Views are never negative, but the counter was built to hold signed numbers, so only 31 switches were left for the size of the count. That gives 2³¹ patterns. One of them, all switches off, stands for 0, so the patterns cover 0, 1, 2 and so on up to 2³¹ − 1.

A car's odometer works the same way with digits instead of switches. A three-digit odometer has 10 × 10 × 10 = 10³ = 1,000 readings, from 000 to 999, so its biggest reading is 10³ − 1. When counting starts at zero, the largest number is one less than the number of patterns.

Base, exponent and power

In 2³¹, the 2 is the base, the number being multiplied. The raised 31 is the exponent, and it counts how many factors of the base there are. The whole expression is a power, read "two to the thirty-first power".

So 2⁵ = 2 × 2 × 2 × 2 × 2 = 32. It is not 2 × 5 = 10. Two exponents have their own names. 7² is "seven squared", because a square 7 units on a side has an area of 7² = 49 square units. 4³ is "four cubed", because a cube 4 units along each edge has a volume of 4³ = 64 cubic units. A base with an exponent of 1 is just the base: 9¹ = 9.

Multiplying powers with the same base

To check 2³¹ without a calculator, you need a way to multiply powers. Write a small case out in full and count:

2³ × 2⁴ = (2 × 2 × 2) × (2 × 2 × 2 × 2).

That is three 2s followed by four more, seven 2s in all, so 2³ × 2⁴ = 2⁷. Check with the values: 8 × 16 = 128, and 2⁷ = 128. The exponents were added, 3 + 4 = 7, because counting factors is adding. Nothing about 3 and 4 was special. For any base a and any positive whole numbers m and n,

a^m × aⁿ = a^{m + n}.

Key idea: When you multiply powers of the same base, keep the base and add the exponents. You are counting how many factors there are altogether.

Now the check. Start from 2⁵ = 32.

Step 1. 2¹⁰ = 2⁵ × 2⁵ = 32 × 32 = 1,024.

Step 2. 2²⁰ = 2¹⁰ × 2¹⁰ = 1,024 × 1,024 = 1,048,576.

Step 3. 2³⁰ = 2²⁰ × 2¹⁰ = 1,048,576 × 1,024 = 1,073,741,824.

Step 4. 2³¹ = 2³⁰ × 2¹ = 1,073,741,824 × 2 = 2,147,483,648.

Step 5. 2³¹ − 1 = 2,147,483,647.

That is the number in the BBC's report, digit for digit. Each step added exponents: 5 + 5 = 10, then 10 + 10 = 20, then 20 + 10 = 30, then 30 + 1 = 31.

Dividing powers with the same base

Division runs the other way. Write 2⁷ ÷ 2³ as a fraction with seven 2s on top and three underneath. Each 2 on the bottom cancels one on top, since 2 ÷ 2 = 1, and four 2s are left. So 2⁷ ÷ 2³ = 2⁴ = 16. Check: 128 ÷ 8 = 16. The exponents were subtracted, 7 − 3 = 4, so for any base a that is not zero,

a^m ÷ aⁿ = a^{m − n}.

For now, keep m bigger than n so that some factors are left over. What happens when m equals n, or is smaller, is the whole of the next lesson.

Here is the rule at work on the new counter. With 64 switches, again keeping one for the sign, the largest count is 2⁶³ − 1 = 9,223,372,036,854,775,807, about 9.2 quintillion, which the BBC rounded to 9.22 quintillion. How many times as many patterns does the new counter have as the old one? Divide: 2⁶³ ÷ 2³¹ = 2^{63 − 31} = 2³² = 4,294,967,296. The new counter has more than four billion times as much room.

A power of a power

Computer memory is measured in powers of 2 as well. The table of binary prefixes kept by the National Institute of Standards and Technology (NIST) gives kibi as 2¹⁰ = 1,024 and builds the bigger prefixes from it: gibi is written there as (2¹⁰)³, three factors of 2¹⁰. How do you simplify a power raised to a power?

Write a small one out. (2³)⁴ means four factors of 2³:

(2³)⁴ = 2³ × 2³ × 2³ × 2³ = 2^{3 + 3 + 3 + 3} = 2¹².

Adding four 3s is the same as 4 × 3, so the exponents multiply:

(a^m)ⁿ = a^{m × n}.

So (2¹⁰)³ = 2³⁰, and NIST's table agrees: one gibibyte is 2³⁰ = 1,073,741,824 bytes, the same number you reached in Step 3 above.

A power of a product or a quotient

What about a product raised to a power, such as (2 × 5)³? Write it out and regroup the factors:

(2 × 5)³ = (2 × 5) × (2 × 5) × (2 × 5) = (2 × 2 × 2) × (5 × 5 × 5) = 2³ × 5³.

Check: 10³ = 1,000, and 2³ × 5³ = 8 × 125 = 1,000. The exponent goes to each factor:

(a × b)ⁿ = aⁿ × bⁿ, and in the same way (a ÷ b)ⁿ = aⁿ ÷ bⁿ.

The rule is handy read backwards. 2⁶ × 5⁶ looks like a lot of work, but it equals (2 × 5)⁶ = 10⁶ = 1,000,000. And for a fraction, (2/3)⁴ = 2⁴/3⁴ = 16/81.

Key idea: All four rules come from writing powers out as repeated multiplication and counting factors. If you ever forget one, write a small example in full and count.

Negative bases and the order of operations

From seventh grade you know that a product with an even number of negative factors is positive and one with an odd number is negative. Powers of negative numbers follow that rule: (−2)⁴ = (−2) × (−2) × (−2) × (−2) = 16, but (−2)⁵ = −32.

Brackets matter. In −2⁴ the exponent belongs only to the 2, so −2⁴ = −(2 × 2 × 2 × 2) = −16. In the same way, powers are worked out before multiplication: 3 × 2⁴ = 3 × 16 = 48, while (3 × 2)⁴ = 6⁴ = 1,296.

Worked examples

Worked example 1. Write 5⁴ × 5³ as a single power and evaluate it.

Step 1. Same base, so add the exponents: 5^{4 + 3} = 5⁷.

Step 2. 5⁷ = 78,125.

Check: 5⁴ = 625 and 5³ = 125, and 625 × 125 = 78,125.

Worked example 2. Simplify 3⁸ ÷ 3⁵.

Step 1. Same base, so subtract the exponents: 3^{8 − 5} = 3³.

Step 2. 3³ = 27.

Check: 3⁸ = 6,561 and 3⁵ = 243, and 6,561 ÷ 243 = 27.

Worked example 3. Simplify (10²)³.

Step 1. A power of a power, so multiply the exponents: 10^{2 × 3} = 10⁶ = 1,000,000.

Check: 10² = 100, and 100 × 100 × 100 = 1,000,000. The tempting answer 10⁵ adds the exponents, which is the rule for multiplying two powers, not for raising a power to a power.

Worked example 4. Write 4³ × 2⁵ as a single power of 2.

Step 1. The bases differ, so no rule applies yet. But 4 = 2², so 4³ = (2²)³ = 2⁶.

Step 2. Now the bases match: 2⁶ × 2⁵ = 2¹¹ = 2,048.

Check: 4³ = 64 and 2⁵ = 32, and 64 × 32 = 2,048.

Worked example 5. Simplify (9 × 2⁷) ÷ (3 × 2⁴).

Step 1. Divide the plain numbers and the powers separately: (9 ÷ 3) × (2⁷ ÷ 2⁴).

Step 2. 9 ÷ 3 = 3, and 2⁷ ÷ 2⁴ = 2³ = 8.

Step 3. 3 × 8 = 24.

Check: 9 × 128 = 1,152 and 3 × 16 = 48, and 1,152 ÷ 48 = 24.

Worked example 6. Simplify (2³ × 3²)².

Step 1. The power of a product sends the outside exponent to each factor: (2³)² × (3²)².

Step 2. The power of a power multiplies exponents: 2⁶ × 3⁴ = 64 × 81 = 5,184.

Check: inside the brackets, 2³ × 3² = 8 × 9 = 72, and 72² = 5,184.

Worked example 7. A phone game gives 2ⁿ coins for clearing level n. How many times as many coins does level 18 give as level 11?

Step 1. "How many times as many" means divide: 2¹⁸ ÷ 2¹¹.

Step 2. Subtract the exponents: 2⁷ = 128. Level 18 gives 128 times as many coins.

Check: 2¹⁸ = 262,144 and 2¹¹ = 2,048, and 2,048 × 128 = 262,144.

Common misconceptions

"2³ × 2⁴ = 2¹², because you multiply the exponents." Multiplying exponents is the rule for a power of a power. For a product of two powers you add them: three factors and four factors make seven, so the answer is 2⁷ = 128, while 2¹² = 4,096.

"2³ × 2⁴ = 4⁷." The base stays the same. Every factor in the product is a 2, so the answer is a power of 2. 4⁷ is 16,384, more than a hundred times too big.

"2³ + 2⁴ = 2⁷." There is no rule for adding powers. Work each one out: 8 + 16 = 24. The add-the-exponents rule belongs to multiplication only.

"2³ × 3⁴ = 6⁷." Different bases with different exponents cannot be combined into one power. The value is 8 × 81 = 648, while 6⁷ = 279,936.

"(2 + 3)² = 2² + 3²." The exponent shares out over a product, not over a sum. (2 + 3)² = 5² = 25, but 2² + 3² = 4 + 9 = 13.

"−3⁴ = 81." Without brackets the exponent belongs to the 3 alone, so −3⁴ = −81. Only (−3)⁴ is 81.

Where this leaves us

• An exponent counts factors of the base: 2³¹ is thirty-one 2s multiplied together.
• A row of n switches makes 2ⁿ patterns. Counting from zero, the largest number it can show is one less than that, which is why 31 bits stop at 2³¹ − 1 = 2,147,483,647.
• Same base, multiplying: add the exponents. Same base, dividing: subtract them.
• A power of a power: multiply the exponents. A power of a product: give the exponent to each factor.
• There is no rule for adding powers, and different bases combine only after you rewrite one base as a power of the other.

The quotient rule raises a question you can already ask. If 2⁷ ÷ 2³ = 2⁴, what is 2³ ÷ 2³, which the rule says is 2⁰? And what is 2³ ÷ 2⁷, which the rule says is 2⁻⁴? The next lesson answers both.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.A.1). thecorestandards.org
2. BBC News. (2014, December 4). Gangnam Style music video 'broke' YouTube view limit. BBC. bbc.com
3. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 10.2, Use multiplication properties of exponents. Prealgebra 2e. OpenStax. openstax.org
4. Illustrative Mathematics. (2019). Grade 8, Unit 7, Lesson 8: Combining bases. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. National Institute of Standards and Technology. (n.d.). Definitions of the SI units: The binary prefixes. NIST Reference on Constants, Units, and Uncertainty. physics.nist.gov

Two answers that look right

Here are two answers from a quiz, written by a student who knew the last lesson well: 2⁰ = 0 and 2⁻³ = −8.

The reasoning behind each one sounds sensible. For the first: "An exponent counts how many 2s you multiply. Zero 2s means there is nothing there, and nothing is 0." For the second: "The exponent is negative, so the answer must be negative. 2³ is 8, so 2⁻³ is −8."

Both answers are wrong. This lesson finds the exact step where each argument breaks. Once you see it, you will also see what a zero or negative exponent has to mean, and why there was never really a choice about it.

This lesson finishes Common Core standard 8.EE.A.1, which asks you to know and apply the properties of integer exponents. The integers include zero and the negative whole numbers, and the standard's own example is 3² × 3⁻⁵ = 3⁻³ = 1/3³ = 1/27. By the end of the lesson you will be able to explain every step of it.

Where "2⁰ = 0" breaks

The first argument fails at the words "nothing is 0". When you multiply, the starting point is not 0. Think about how a calculator works out 2³. It starts at 1 and multiplies by 2 three times: 1 × 2 × 2 × 2 = 8. If it started at 0, every power would come out as 0, since 0 × 2 = 0. So 2⁰ means "start at 1 and multiply by 2 no times", and that leaves the 1 untouched.

That is one argument. Here are four checks, and they all agree with it.

Check 1: the pattern. Read the powers of 2 from the top down.

Each time the exponent drops by 1, one factor of 2 is taken away, which means dividing by 2. The next value down is 2 ÷ 2 = 1. For the pattern to carry on, 2⁰ must be 1, not 0.

Check 2: the quotient rule. Any number other than zero divided by itself is 1, so 2³ ÷ 2³ = 8 ÷ 8 = 1. The quotient rule from the last lesson says 2³ ÷ 2³ = 2^{3 − 3} = 2⁰. Both statements are true only if 2⁰ = 1.

Check 3: the product rule. The product rule says 2⁰ × 2⁵ = 2^{0 + 5} = 2⁵ = 32. So 2⁰ is a number that leaves 32 unchanged when you multiply by it, and the only such number is 1. If 2⁰ were 0, then 2⁰ × 2⁵ would be 0, and the product rule would be broken.

Check 4: a tournament. A knockout tournament that starts with 64 teams, which is 2⁶, halves the field every round. After round one 2⁵ = 32 teams are left, then 2⁴, 2³, 2² and 2¹. After the sixth round 2⁰ teams are left. That is the champion: one team, not zero.

Nothing in these checks depended on the base being 2. For any base a that is not zero, a⁰ = 1. So 7⁰ = 1, (−5)⁰ = 1 and (3/4)⁰ = 1. The rule needs a to be non-zero because Check 2 divides a number by itself, and 0 ÷ 0 has no value. Eighth-grade work leaves 0⁰ alone.

Remember: Any non-zero number raised to the power 0 is 1. A zero exponent means "multiply by the base no times", which leaves the starting 1.

Where "2⁻³ = −8" breaks

The second argument fails at "the exponent is negative, so the answer must be negative". The sign of the exponent and the sign of the value are two different things. To see what a negative exponent does, keep the pattern going below 2⁰. Each step down still divides by 2.

Halving a positive number gives a smaller positive number, never a negative one. So 2⁻³ = 1/8. Look at the 8: the student's 2³ = 8 was the right number in the wrong place. It belongs under a fraction bar, not behind a minus sign.

The quotient rule says the same. Work out 2³ ÷ 2⁶ by cancelling. There are three 2s on top and six underneath. The three on top cancel three of the six, leaving 1 on top and three 2s underneath:

2³ ÷ 2⁶ = (2 × 2 × 2)/(2 × 2 × 2 × 2 × 2 × 2) = 1/(2 × 2 × 2) = 1/2³ = 1/8.

The rule gives 2^{3 − 6} = 2⁻³. Both are true only if 2⁻³ = 1/2³. In general, for any base a that is not zero and any positive whole number n,

a⁻ⁿ = 1/aⁿ.

A negative exponent means "divide by the base this many times". Another way to say it: a⁻ⁿ is the reciprocal of aⁿ, the number you multiply it by to get 1, because a⁻ⁿ × aⁿ = a⁰ = 1.

Remember: A negative exponent never makes a number negative. It moves the power to the other side of a fraction bar.

The rules still work, so use them

The reason for defining zero and negative exponents this way is that all four rules from the last lesson keep working for every integer exponent. The checks above were built to make sure of it. That turns the standard's example into a two-line calculation:

3² × 3⁻⁵ = 3^{2 + (−5)} = 3⁻³ = 1/3³ = 1/27.

Check it the long way. 3² = 9 and 3⁻⁵ = 1/3⁵ = 1/243, so the product is 9/243. Divide the top and bottom by 9 and you get 1/27. It matches.

Powers of ten and the metric prefixes

Negative powers of ten are where you will meet negative exponents most often. The metric prefixes, as NIST lists them, are powers of ten:

To write 10⁻ⁿ as a decimal, put the 1 in the nth place after the decimal point. 10⁻³ = 1/1,000 = 0.001, with the 1 in the third decimal place and two zeros in front of it. Count decimal places, not zeros.

Negative exponents on fractions

A fraction raised to a negative power flips over. Take (1/2)⁻³. By the definition it is 1/(1/2)³ = 1 ÷ (1/8), and dividing 1 by 1/8 gives 8. So (1/2)⁻³ = 8 = 2³. In the same way, (2/3)⁻² = (3/2)² = 9/4: flip the fraction, then make the exponent positive.

The same move works across a fraction bar. A factor with a negative exponent on the bottom can move to the top with a positive exponent: 1/2⁻³ = 2³ = 8, because dividing by 1/8 is the same as multiplying by 8.

Worked examples

Worked example 1. Evaluate 5⁰ + 5⁻¹.

Step 1. 5⁰ = 1.

Step 2. 5⁻¹ = 1/5¹ = 1/5.

Step 3. 1 + 1/5 = 6/5 = 1.2.

Check: 5⁻¹ × 5¹ should equal 5⁰ = 1, and (1/5) × 5 = 1.

Worked example 2. Write 4⁻³ as a fraction and a decimal.

Step 1. 4⁻³ = 1/4³ = 1/64.

Step 2. 1 ÷ 64 = 0.015625.

Check: (1/64) × 4³ = (1/64) × 64 = 1, as a power times its reciprocal should be.

Worked example 3. Simplify 10⁴ × 10⁻⁷.

Step 1. Same base, so add the exponents: 4 + (−7) = −3.

Step 2. 10⁻³ = 0.001.

Check: 10⁴ = 10,000 and 10⁻⁷ = 0.0000001, and 10,000 × 0.0000001 = 0.001.

Worked example 4. Simplify 2⁵ ÷ 2⁸.

Step 1. Subtract the exponents, top minus bottom: 5 − 8 = −3.

Step 2. 2⁻³ = 1/8 = 0.125.

Check: 32 ÷ 256 = 0.125.

Worked example 5. Simplify (3⁻²)⁻¹.

Step 1. A power of a power, so multiply the exponents: (−2) × (−1) = 2.

Step 2. 3² = 9.

Check: 3⁻² = 1/9, and raising a number to the power −1 flips it, so (1/9)⁻¹ = 9.

Worked example 6. Simplify 6⁻² × 6⁵ ÷ 6³.

Step 1. Multiply first: 6^{−2 + 5} = 6³.

Step 2. Then divide: 6³ ÷ 6³ = 6⁰ = 1.

Check: a number divided by itself is 1.

Worked example 7. Suppose a ball always bounces back to half the height it fell from, and you drop it from 2 m. After n bounces it rises to 2 × (1/2)ⁿ = 2¹ × 2⁻ⁿ metres. How high does it rise after the fourth bounce?

Step 1. Put in n = 4: 2¹ × 2⁻⁴.

Step 2. Add the exponents: 2^{1 + (−4)} = 2⁻³ = 1/8 m = 0.125 m, which is 12.5 cm.

Check by halving four times: 2 m, then 1 m, 0.5 m, 0.25 m and 0.125 m.

Common misconceptions

"A negative exponent makes the number negative." It makes a reciprocal: 2⁻³ = 1/8, a positive number less than 1. A value is negative only when the base is negative and the power has an odd number of factors, as in (−2)³ = −8.

"Anything to the power 0 is 0." It is 1, as all four checks showed. The pattern of halving reaches 1, not 0, and a tournament ends with one champion.

"2⁻³ = −6." This mixes up a power with multiplication, the same mistake as writing 2³ = 6. The exponent counts factors; here it counts how many times you divide by 2.

"−5⁰ = 1." Without brackets the exponent belongs to the 5 only, so −5⁰ = −(5⁰) = −1. With brackets, (−5)⁰ = 1.

"10⁻⁴ = 0.00001, because it needs four zeros." 10⁻⁴ = 1/10,000 = 0.0001. The 1 sits in the fourth decimal place, so there are only three zeros between the decimal point and the 1.

The short version

• a⁰ = 1 for any non-zero base a. The halving pattern, the quotient rule and the product rule all force it.
• a⁻ⁿ = 1/aⁿ. A negative exponent means divide by the base n times, so it gives a reciprocal, not a negative number.
• With these meanings, all four exponent rules work for every integer exponent, which is why 3² × 3⁻⁵ = 3⁻³ = 1/27.
• 10⁻ⁿ is a decimal with the 1 in the nth place after the point, and the metric prefixes milli, micro and nano are 10⁻³, 10⁻⁶ and 10⁻⁹.
• A fraction to a negative power flips: (a/b)⁻ⁿ = (b/a)ⁿ.

With powers of ten that can go up or down, you can now write any number, the width of a hair or the population of the planet, as a short decimal times a power of ten. That is scientific notation, and it is the next lesson.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.A.1). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 10.4, Divide monomials (zero exponents). Prealgebra 2e. OpenStax. openstax.org
3. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 10.5, Integer exponents and scientific notation. Prealgebra 2e. OpenStax. openstax.org
4. Illustrative Mathematics. (2019). Grade 8, Unit 7, Lesson 6: What about other bases? IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. National Institute of Standards and Technology. (n.d.). Metric (SI) prefixes. Office of Weights and Measures. nist.gov

A country inside a planet

On 1 April 2020, the U.S. census counted 331,449,281 people living in the fifty states and the District of Columbia. On 15 November 2022, the United Nations marked the day it had projected the world's population would reach 8 billion. Here is the question for this lesson: about how many times as many people live on Earth as live in the United States?

Try it before you read on. Dividing 8,000,000,000 by 331,449,281 on paper is a long job, and it is easy to lose track of the zeros. There is a faster way that you can do in your head, and it rests on the powers of ten from the last two lessons.

This lesson covers Common Core standard 8.EE.A.3: using numbers written as a single digit times a power of 10 to estimate very large or very small quantities, and to say how many times as much one is as the other. The standard's own example compares these same two populations, using the figures of 2010. You will also start on 8.EE.A.4 by reading and writing scientific notation. The next lesson calculates with it.

One digit and a power of ten

Round each number to a single non-zero digit, then write it as that digit times a power of ten.

Step 1. 331,449,281 is about 300,000,000. That is 3 followed by eight zeros, so it is 3 × 10⁸.

Step 2. 8,000,000,000 is 8 followed by nine zeros, so it is 8 × 10⁹.

Step 3. Divide the digits and the powers of ten separately: (8 × 10⁹) ÷ (3 × 10⁸) = (8 ÷ 3) × (10⁹ ÷ 10⁸) ≈ 2.67 × 10¹, which is about 27.

So the world has roughly 27 times as many people as the United States. How good is that? A calculator gives 8,000,000,000 ÷ 331,449,281 ≈ 24.1. The estimate came out a little high because rounding 331 million down to 300 million made the number you divide by smaller. Still, "between 20 and 30 times" is right, and you got it from 8 ÷ 3 and one subtraction of exponents. The two counts come from different years, 2020 and 2022. That is fine for a "how many times" estimate, but it would not be fine for an exact answer.

The point: Rounding to one digit times a power of ten turns a hard division into an easy one. Divide the digits, then subtract the exponents.

Scientific notation

Those estimates are a rough form of scientific notation, the way scientists and calculators write very large and very small numbers. A number is in scientific notation when it is written as

a × 10ⁿ, where 1 ≤ a < 10 and n is an integer.

The first part, a, is a number from 1 up to but not including 10, and it can have as many decimal places as the measurement deserves. The power of ten carries the size. The census count is exactly 3.31449281 × 10⁸, and rounded to three digits it is 3.31 × 10⁸. NASA's fact sheet for Earth gives the planet's mass the same way: the table lists 5.9722 in a column headed "10²⁴ kg", which means 5.9722 × 10²⁴ kg.

Large numbers: count the places the point moves

To write 331,449,281 in scientific notation, put the decimal point just after the first digit, which gives 3.31449281, and count how many places the point moved to get there. Its home is at the end of the number, after the final 1, so it moved eight places to the left. Moving the point eight places left divides a number by 10⁸, so to keep the value the same you multiply by 10⁸: 331,449,281 = 3.31449281 × 10⁸.

Going back is the reverse. 4.5 × 10⁶ means move the point in 4.5 six places to the right, filling the empty places with zeros: 4,500,000.

Small numbers: negative exponents

The U.S. Environmental Protection Agency (EPA) gives a sense of scale for air pollution. The fine particles called PM_2.5 have diameters of 2.5 micrometres or less, and the average human hair is about 70 micrometres across. A micrometre is 10⁻⁶ m, so a hair is about 70 × 10⁻⁶ m = 0.00007 m wide, and a fine particle is at most 2.5 × 10⁻⁶ m = 0.0000025 m wide.

70 × 10⁻⁶ is not yet in scientific notation, because 70 is not less than 10. Write 70 as 7 × 10¹. Then 70 × 10⁻⁶ = 7 × 10¹ × 10⁻⁶ = 7 × 10⁻⁵ m. Check by moving the point: in 0.00007 the point has to move five places to the right to reach 7, so the exponent is −5.

How many times as wide as the largest fine particle is a hair?

(7 × 10⁻⁵) ÷ (2.5 × 10⁻⁶) = (7 ÷ 2.5) × 10^{−5 − (−6)} = 2.8 × 10¹ = 28.

The EPA rounds this to "30 times larger". Look at the exponents: −5 − (−6) = −5 + 6 = 1. The seventh-grade rule for subtracting a negative number is doing real work here.

The point: In scientific notation the exponent carries the size and the first number carries the detail. A positive exponent means a large number, a negative exponent means a small positive number, and the exponent counts how far the decimal point moves.

Comparing numbers in scientific notation

Which is bigger, 4.2 × 10⁻³ or 9.8 × 10⁻⁴? The 9.8 catches the eye, but look at the exponents first. 10⁻³ = 0.001 and 10⁻⁴ = 0.0001, so the first number is 0.0042 and the second is 0.00098. The first is more than four times as big.

When two numbers are both in scientific notation, the one with the larger exponent is larger, whatever their leading numbers are. Only when the exponents match do you compare the leading numbers. This works because the leading number is always at least 1 and less than 10, so no leading number can make up for a whole power of ten.

Be careful with negative exponents: −3 is larger than −4, because it is further right on the number line. So 10⁻³ is ten times as large as 10⁻⁴.

Worked examples

Worked example 1. Write 5,280,000 in scientific notation.

Step 1. Put the point after the first digit: 5.28.

Step 2. The point moved from the end of 5,280,000 six places to the left, so the exponent is 6: 5.28 × 10⁶.

Check: moving the point in 5.28 six places right gives 5,280,000.

Worked example 2. Write 0.00046 in scientific notation.

Step 1. The first non-zero digit is 4, so the leading number is 4.6.

Step 2. Getting from 0.00046 to 4.6 moves the point four places to the right. The number is small, so the exponent is negative: 4.6 × 10⁻⁴.

Check: 4.6 × 0.0001 = 0.00046.

Worked example 3. Write 9.07 × 10⁻³ in standard form.

Step 1. A negative exponent means a small number, so move the point three places to the left.

Step 2. 9.07 becomes 0.907, then 0.0907, then 0.00907.

Check: 0.00907 has its first non-zero digit in the third decimal place, matching 10⁻³.

Worked example 4. Rewrite 45 × 10⁷ and 0.6 × 10⁻³ in scientific notation.

Step 1. 45 = 4.5 × 10¹, so 45 × 10⁷ = 4.5 × 10¹ × 10⁷ = 4.5 × 10⁸.

Step 2. 0.6 = 6 × 10⁻¹, so 0.6 × 10⁻³ = 6 × 10⁻¹ × 10⁻³ = 6 × 10⁻⁴.

Check: when the leading number gets 10 times smaller, the exponent goes up by 1, and when it gets 10 times bigger, the exponent goes down by 1, so the value never changes.

Worked example 5. Put 3.2 × 10⁵, 9.9 × 10⁴, 1.1 × 10⁶ and 4.5 × 10⁵ in order from smallest to largest.

Step 1. Sort by exponent: 10⁴ first, then the two 10⁵ numbers, then 10⁶.

Step 2. Between the two 10⁵ numbers, 3.2 is smaller than 4.5.

Step 3. The order is 9.9 × 10⁴, 3.2 × 10⁵, 4.5 × 10⁵, 1.1 × 10⁶.

Check in standard form: 99,000, 320,000, 450,000, 1,100,000.

Worked example 6. One song has 2 × 10⁹ plays and another has 5 × 10⁶. How many times as many plays does the first song have?

Step 1. (2 × 10⁹) ÷ (5 × 10⁶) = (2 ÷ 5) × 10^{9 − 6} = 0.4 × 10³.

Step 2. 0.4 × 10³ = 400. (In scientific notation that is 4 × 10².)

Check: 5,000,000 × 400 = 2,000,000,000.

Common misconceptions

"The exponent is the number of zeros." That is true only when the leading number is a single digit. 3.31 × 10⁸ = 331,000,000 has six zeros, not eight. The exponent counts how many places the decimal point moves.

"A negative exponent means a negative number." 7 × 10⁻⁵ is 0.00007, the width of a hair: small, but positive. A negative number in scientific notation has its minus sign in front, as in −7 × 10⁵.

"45 × 10⁷ is in scientific notation." It has the right value, but the leading number must be at least 1 and less than 10. The scientific form is 4.5 × 10⁸.

"9.8 × 10⁻⁴ is bigger than 4.2 × 10⁻³, because 9.8 is bigger than 4.2." Compare the exponents first. −3 is larger than −4, so 4.2 × 10⁻³ is the larger number.

"10⁹ is only a bit more than 10⁸." It is ten times as much. Each step up in the exponent multiplies by 10, which is why the world's population, at about 10⁹ scale, holds the U.S. population, at about 10⁸ scale, more than twenty times over.

Putting it together

• To estimate, round to a single digit times a power of ten. To compare two estimates, divide the digits and subtract the exponents.
• Scientific notation writes a number as a × 10ⁿ with 1 ≤ a < 10. The exponent counts how far the decimal point moves: positive for large numbers, negative for small ones.
• To fix a leading number that is too big or too small, move its decimal point and change the exponent the opposite way, so the value stays the same.
• To compare two numbers in scientific notation, compare exponents first, then leading numbers.
• Estimates are only as good as their rounding. 27 against a true 24 is fine for "how many times"; it would not be fine for a census.

So far you have only divided numbers in scientific notation. The next lesson multiplies, adds and subtracts them too, finds how long sunlight takes to reach you, and reads the strange "E" numbers a calculator prints.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.A.3, 8.EE.A.4). thecorestandards.org
2. Epstein, B., and Lofquist, D. (2021, April 26). First 2020 Census data release shows U.S. resident population of 331,449,281. U.S. Census Bureau. census.gov
3. United Nations. (2022). Day of Eight Billion. un.org
4. U.S. Environmental Protection Agency. (2026, September 18). Particulate matter (PM) basics. epa.gov
5. Williams, D. R. (2024, November 15). Earth fact sheet. NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov
6. Illustrative Mathematics. (2019). Grade 8, Unit 7, Lesson 13: Definition of scientific notation. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

How old is the sunlight on your face?

NASA's fact sheet for Earth gives the average distance from the Earth to the Sun as 149.598 million kilometres. Light covers 299,792,458 metres every second, a speed the National Institute of Standards and Technology lists as exact. So when sunlight lands on your face, how long ago did it leave the Sun?

Time is distance divided by speed. The two numbers are clumsy to divide as they stand, so the plan is to write both in scientific notation, in the same units, and let the exponent rules do the heavy lifting. You will get the answer to the nearest second. The same method then finds how far light goes in a year, how heavy the Earth and Moon are together, and how long the whole world would need to fill YouTube's new view counter.

This lesson covers Common Core standard 8.EE.A.4: performing operations with numbers in scientific notation, including problems that mix decimals and scientific notation; choosing units of a sensible size for very large or very small quantities; and interpreting scientific notation produced by technology.

Multiplying and dividing: two jobs, then tidy up

To multiply or divide numbers in scientific notation, split each one into its leading number and its power of ten, and deal with the two parts separately. That is allowed because multiplication can be done in any order, and dividing is multiplying by a reciprocal. Here is the sunlight problem from start to finish.

Step 1. Put the distance in metres. 149.598 million km = 149,598,000 km = 1.49598 × 10⁸ km. A kilometre is 10³ m, so the distance is 1.49598 × 10⁸ × 10³ = 1.49598 × 10¹¹ m.

Step 2. Write the speed in scientific notation: 299,792,458 m/s = 2.99792458 × 10⁸ m/s.

Step 3. Divide the leading numbers and subtract the exponents:

(1.49598 × 10¹¹) ÷ (2.99792458 × 10⁸) = (1.49598 ÷ 2.99792458) × 10^{11 − 8} = 0.4990 × 10³ seconds.

Step 4. Tidy up. 0.4990 is less than 1, so it is not a proper leading number. Move its point one place right to get 4.990, and take 1 off the exponent to keep the value the same: 4.990 × 10² s = 499.0 s.

Step 5. Change to minutes: 499 ÷ 60 = 8 remainder 19, so 8 minutes 19 seconds, or about 8.3 minutes.

Check: NASA's Space Place site, working in miles, also gives about 8.3 minutes. A rough check with one-digit estimates agrees too: (1.5 × 10¹¹) ÷ (3 × 10⁸) = 0.5 × 10³ = 500 s. The sunlight on your face left the Sun a little over eight minutes ago.

Why this matters: In scientific notation a hard multiplication or division becomes one easy calculation with ordinary numbers and one easy calculation with exponents. The only extra job is tidying the leading number at the end.

When the tidy-up changes the exponent

Here is the same method where a small change in the input flips the last step. Compare two products:

(2 × 10⁴) × (4 × 10⁵) = (2 × 4) × 10^{4 + 5} = 8 × 10⁹. Nothing to tidy: 8 is already between 1 and 10.

(3 × 10⁴) × (4 × 10⁵) = (3 × 4) × 10^{4 + 5} = 12 × 10⁹. Now the leading number has passed 10. Write 12 as 1.2 × 10¹, so the answer is 1.2 × 10¹ × 10⁹ = 1.2 × 10¹⁰.

The rule for tidying: if the leading number comes out at 10 or more, move its point one place left and add 1 to the exponent. If it comes out below 1, move its point one place right and subtract 1 from the exponent. Either way the value does not change.

Multiplying works the same way on a bigger scale. How far does light travel in a year? A year has about 365.25 days, so it has about 365.25 × 24 × 3,600 = 31,557,600 = 3.15576 × 10⁷ seconds. Then

(2.99792458 × 10⁸ m/s) × (3.15576 × 10⁷ s) = (2.99792458 × 3.15576) × 10^{8 + 7} ≈ 9.46 × 10¹⁵ m.

That is 9.46 × 10¹² km, or about 9 trillion kilometres, the distance NASA calls a light-year.

Adding and subtracting: match the powers first

NASA lists the Earth's mass as 5.9722 × 10²⁴ kg and the Moon's as 0.07346 × 10²⁴ kg, which is 7.346 × 10²² kg in scientific notation. What do the two weigh together?

You cannot add 5.9722 and 7.346 here, because they count different things: one counts units of 10²⁴ kg and the other counts units of 10²² kg. It is like adding 5 dollars and 7 cents and calling the answer 12 dollars. First write both with the same power of ten. Moving the Moon's point two places left takes 2 off the size of the leading number, so add 2 to the exponent:

7.346 × 10²² = 0.07346 × 10²⁴.

Now the leading numbers count the same thing, and you add them:

5.9722 × 10²⁴ + 0.07346 × 10²⁴ = (5.9722 + 0.07346) × 10²⁴ = 6.04566 × 10²⁴ ≈ 6.05 × 10²⁴ kg.

The Moon adds only about one percent. Subtraction works the same way. NASA gives the Earth's closest distance to the Sun, its perihelion, as 147.095 × 10⁶ km, and its farthest, its aphelion, as 152.100 × 10⁶ km. Both already share the power 10⁶, so

152.100 × 10⁶ − 147.095 × 10⁶ = 5.005 × 10⁶ km.

Over a year the distance to the Sun changes by about 5 million kilometres.

Why this matters: Multiplying and dividing never need matching powers, but adding and subtracting always do, for the same reason that you line up the decimal points before adding decimals.

Mixing decimals and scientific notation

Real problems often hand you one number in each form. How many hairs, laid side by side, make 1 centimetre, if a hair is 7 × 10⁻⁵ m wide, the EPA's figure from the last lesson?

Step 1. Write the decimal in scientific notation: 1 cm = 0.01 m = 1 × 10⁻² m.

Step 2. Divide: (1 × 10⁻²) ÷ (7 × 10⁻⁵) = (1 ÷ 7) × 10^{−2 − (−5)} ≈ 0.143 × 10³ = 143.

Check: 143 × 0.00007 m = 0.01001 m, which is just over 1 cm. About 143 hairs fit across a centimetre.

Choosing a sensible unit

The U.S. Geological Survey gives the spreading rate of the Mid-Atlantic Ridge, the seam down the middle of the Atlantic Ocean floor, as about 2.5 centimetres per year, or 25 km in a million years. You could convert that to metres per second:

0.025 m ÷ (3.15576 × 10⁷ s) ≈ 7.9 × 10⁻¹⁰ m/s.

That is correct, and almost useless: nobody can picture 7.9 × 10⁻¹⁰. In millimetres per year the same rate is 25 mm a year, a length you can mark with a ruler. The standard uses exactly this example: millimetres per year is a good unit for seafloor spreading. The same goes the other way: the distance to the Sun is 1.49598 × 10¹⁴ mm, which is true and silly, while 1.5 × 10⁸ km is easy to read.

A good unit makes the numbers easy to picture and compare. Scientific notation is for the cases where no ordinary unit does that.

Reading a calculator's E

Calculators and spreadsheets print scientific notation with a capital E. Microsoft's help page for Excel explains that in this format E+n multiplies the number before it by 10 to the nth power, and gives the example of 12345678901 displayed as 1.23E+10, which is 1.23 × 10¹⁰. Excel switches to this format by itself for numbers with 12 or more digits. Read the E as "times ten to the":

The minus sign in 7.9E-10 belongs to the exponent, so the number is tiny and positive. The E has nothing to do with subtraction.

Worked examples

Worked example 1. Find (6 × 10⁵) × (4 × 10⁻²).

Step 1. Leading numbers: 6 × 4 = 24. Exponents: 5 + (−2) = 3. So 24 × 10³.

Step 2. Tidy: 24 = 2.4 × 10¹, so the answer is 2.4 × 10⁴.

Check: 600,000 × 0.04 = 24,000.

Worked example 2. Find (9.6 × 10⁷) ÷ (3.2 × 10⁻³).

Step 1. Leading numbers: 9.6 ÷ 3.2 = 3. Exponents: 7 − (−3) = 7 + 3 = 10.

Step 2. The answer is 3 × 10¹⁰. No tidying needed.

Check: dividing by a small number should give a big answer, and it does.

Worked example 3. Find 4.1 × 10⁶ + 3.5 × 10⁵.

Step 1. Match the powers: 3.5 × 10⁵ = 0.35 × 10⁶.

Step 2. Add the leading numbers: 4.1 + 0.35 = 4.45. The answer is 4.45 × 10⁶.

Check: 4,100,000 + 350,000 = 4,450,000.

Worked example 4. Find 2.3 × 10⁻⁴ − 8 × 10⁻⁵.

Step 1. Match the powers: 8 × 10⁻⁵ = 0.8 × 10⁻⁴.

Step 2. Subtract: 2.3 − 0.8 = 1.5. The answer is 1.5 × 10⁻⁴.

Check: 0.00023 − 0.00008 = 0.00015.

Worked example 5. YouTube's new counter tops out at about 9.2 × 10¹⁸ views. If all 8 × 10⁹ people on Earth watched one video once every second, without stopping, how long would the counter take to fill?

Step 1. Time = views ÷ views per second = (9.2 × 10¹⁸) ÷ (8 × 10⁹) = 1.15 × 10⁹ seconds.

Step 2. Change to years: (1.15 × 10⁹) ÷ (3.15576 × 10⁷) ≈ 0.365 × 10² = 36.5 years.

Check: the units work out, views divided by views per second gives seconds. The engineers built in more than three decades of the whole planet watching nonstop.

Worked example 6. A spreadsheet shows 7.9E-10. Write it in scientific notation and in standard form.

Step 1. E means "times ten to the", so it is 7.9 × 10⁻¹⁰.

Step 2. Move the point ten places left: 0.00000000079.

Check: the first non-zero digit sits in the tenth decimal place, matching the exponent −10.

Common misconceptions

"To multiply, multiply the exponents." The powers of ten are being multiplied, so their exponents add: 10⁵ × 10⁻² = 10³. Multiplying exponents is the power of a power rule, which is a different situation.

"To add, add the exponents." 3 × 10⁴ + 2 × 10⁴ is 5 × 10⁴, not 5 × 10⁸. When the powers match, they stay the same, just as 3 tens plus 2 tens is 5 tens.

"Add the leading numbers whatever the powers." 5 × 10⁶ + 3 × 10⁴ is not 8 × 10⁶. Match the powers first: 5 × 10⁶ + 0.03 × 10⁶ = 5.03 × 10⁶.

"12 × 10⁹ is a finished answer." The value is right, but the leading number must be less than 10. Tidy it to 1.2 × 10¹⁰.

"4.99E2 means 4.99 × 2." E stands for "times ten to the", so 4.99E2 is 4.99 × 10² = 499.

"A smaller unit is always better because it is more exact." Changing the unit does not change how precise a measurement is. Seafloor spreading in metres per second is no more exact than 25 mm a year, only harder to read.

What to carry forward

• Multiply or divide: work the leading numbers and the powers of ten separately, adding exponents to multiply and subtracting to divide.
• Add or subtract: rewrite the numbers with the same power of ten first, then add or subtract the leading numbers.
• Tidy at the end: a leading number of 10 or more, or below 1, gets its point moved and its exponent changed the opposite way.
• Choose units that make the numbers easy to picture, such as millimetres per year for seafloor spreading, and keep scientific notation for sizes no ordinary unit handles.
• A calculator's E means "times ten to the": 7.9E-10 is 7.9 × 10⁻¹⁰.

The exponent 2 has turned up again and again as "squared". The next module runs it backwards: if a square has an area of 144 square feet, how long is each side? That question leads to square roots, cube roots, and numbers that can never be written as fractions.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.A.4). thecorestandards.org
2. Williams, D. R. (2024, November 15). Earth fact sheet. NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov
3. Williams, D. R. (2024, January 11). Moon fact sheet. NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov
4. National Institute of Standards and Technology. (2022). CODATA value: Speed of light in vacuum. NIST Reference on Constants, Units, and Uncertainty. physics.nist.gov
5. NASA Space Place. (n.d.). What is a light-year? NASA Science. spaceplace.nasa.gov
6. Kious, W. J., and Tilling, R. I. (1996). Understanding plate motions. In This dynamic Earth: The story of plate tectonics (online edition). U.S. Geological Survey. pubs.usgs.gov
7. Microsoft. (n.d.). Display numbers in scientific (exponential) notation. Microsoft Support. support.microsoft.com

A room with 144 square feet of floor

An apartment listing describes a square bedroom with 144 square feet of floor. You want to run a strip of LED lights along one wall, and the strips are sold in 10-foot and 16-foot lengths. Which one fits?

The floor is a square, so its area is side × side. You need a number that gives 144 when it is multiplied by itself. Try a few. 10 × 10 = 100, which is too small. 13 × 13 = 169, which is too big. 12 × 12 = 144. Each wall is 12 feet long, so the 10-foot strip fits and the 16-foot strip does not.

You have just found a square root, the exponent 2 run backwards. This lesson covers Common Core standard 8.EE.A.2: using square root and cube root symbols to represent the solutions of equations x² = p and x³ = p, where p is a positive rational number, and evaluating the roots of small perfect squares and perfect cubes. The last part of that standard, knowing that √2 is irrational, gets a lesson of its own in Lesson 7.

The square root symbol

A square root of a number p is a number that gives p when you multiply it by itself. The symbol √ means the positive square root, so √144 = 12. Squaring and taking a square root undo each other, the way adding 5 and subtracting 5 undo each other: 12² = 144, and √144 = 12.

Numbers such as 144, whose square roots are whole numbers, are called perfect squares. They come up often enough to be worth knowing by heart.

Read the table either way. Across the top to the bottom row is squaring: 13² = 169. From the bottom row back up is taking a square root: √169 = 13.

Two solutions to x² = 144

Now a question with no room in it: which numbers x make x² = 144 true? 12 works. So does −12, because (−12) × (−12) = 144, a negative times a negative being positive. The equation has two solutions:

x = 12 or x = −12, written x = ±√144 = ±12.

The sign ± is read "plus or minus". The symbol √144 on its own still means only 12; it is the ± in front that brings back the negative solution. In the bedroom, a wall cannot be −12 feet long, so the situation throws that solution away. In a problem with no situation, keep both.

What about x² = −9? No number works. A positive number times itself is positive, a negative number times itself is positive, and 0 times itself is 0. Squares are never negative, so in this course a negative number has no square root.

In short: √p is one positive number, but the equation x² = p has two solutions, √p and −√p, when p is positive. Let the situation decide whether the negative one makes sense.

A falling ball: the positive solution is the one that happened

OpenStax's College Physics gives the distance y that an object falls from rest as y = ½gt², when air resistance is small enough to ignore, with g = 9.80 m/s² near the Earth's surface. Half of 9.80 is 4.9, so the formula becomes y = 4.9t², with y in metres and t in seconds after the drop. How long does a ball take to fall 19.6 m?

Step 1. Put in the distance: 4.9t² = 19.6.

Step 2. Divide both sides by 4.9: t² = 4.

Step 3. Take square roots: t = ±√4 = ±2.

Step 4. A time after the drop cannot be negative, so the ball falls for 2 seconds.

Check: 4.9 × 2² = 4.9 × 4 = 19.6. The textbook makes the same choice in its own worked example of a rock dropped from a cliff: the equation hands back a plus and a minus answer, and only the positive one describes what happened.

Roots of fractions and decimals

A fraction's square root is the square root of its top over the square root of its bottom, as long as both are perfect squares: √(9/16) = 3/4, because (3/4)² = 9/16. Decimals work through the perfect squares you already know. √0.25 = 0.5, because 0.5 × 0.5 = 0.25. √0.04 = 0.2, because 0.2 × 0.2 = 0.04. √1.44 = 1.2, because 1.2 × 1.2 = 1.44.

Count decimal places to stay safe. Squaring a number with one decimal place gives two decimal places, so a perfect-square decimal like 0.04 or 1.44 has an even number of decimal places, and its root has half as many.

Square roots that are not whole numbers

A square garden has an area of 50 m². How long is each side? It is √50, and 50 is not in the table. You can still trap it.

Step 1. 7² = 49 and 8² = 64. Since 50 lies between 49 and 64, √50 lies between 7 and 8, and much nearer to 7.

Step 2. Try tenths: 7.1² = 50.41, a little too big. So √50 is between 7.0 and 7.1.

Step 3. Try hundredths: 7.07² = 49.9849, just under 50, and 7.08² = 50.1264, over. So √50 is between 7.07 and 7.08.

Each side of the garden is about 7.07 m. You could keep going to more decimal places, and the digits would never stop or settle into a repeating pattern. Lesson 7 shows why some square roots behave like that.

Cube roots

NIST's list of units used with the SI defines the litre as 1 L = 1 dm³ = 10³ cm³, so a litre is 1,000 cubic centimetres. Suppose a box in the shape of a cube holds exactly one litre. How long is each edge?

The volume of a cube is edge × edge × edge, so the edge e satisfies e³ = 1,000. Since 10 × 10 × 10 = 1,000, the edge is 10 cm. You have found a cube root: a number that gives p when it is used as a factor three times. It is written with a small 3 tucked into the root sign, ³√1,000 = 10.

Cubes behave differently from squares in one important way. The equation x³ = 64 has only one solution, x = 4, because (−4)³ = (−4)(−4)(−4) = −64. Three negative factors make a negative product, so a negative number cannot be the cube root of a positive one. No ± is needed. The other side of the same fact is that negative numbers do have cube roots: ³√(−64) = −4.

Fractions and decimals work as before: ³√(8/27) = 2/3, because (2/3)³ = 8/27, and ³√0.001 = 0.1, because 0.1 × 0.1 × 0.1 = 0.001.

Worked examples

Worked example 1. Solve x² = 81.

Step 1. 9² = 81, so √81 = 9.

Step 2. The equation has two solutions: x = 9 or x = −9.

Check: 9 × 9 = 81 and (−9) × (−9) = 81.

Worked example 2. Solve x² = 49/100.

Step 1. √49 = 7 and √100 = 10, so √(49/100) = 7/10.

Step 2. x = 7/10 or x = −7/10, which is x = ±0.7.

Check: (7/10)² = 49/100.

Worked example 3. A square map in a video game is made of 4,096 square tiles. How many tiles run along each edge?

Step 1. 60² = 3,600 and 70² = 4,900, so the answer is between 60 and 70.

Step 2. 4,096 ends in 6. A square ends in 6 only when the number being squared ends in 4 or 6, so try 64 and 66.

Step 3. 64² = 4,096. Each edge has 64 tiles.

Check: 64 × 64 = 64 × 60 + 64 × 4 = 3,840 + 256 = 4,096.

Worked example 4. Solve x³ = 343.

Step 1. From the table, 7³ = 343.

Step 2. The only solution is x = 7.

Check: 7 × 7 = 49 and 49 × 7 = 343.

Worked example 5. A cube-shaped storage box has a volume of 0.125 m³. How long is each edge?

Step 1. The edge is ³√0.125. Since 5³ = 125, try 0.5.

Step 2. 0.5 × 0.5 × 0.5 = 0.125, so each edge is 0.5 m, which is 50 cm.

Worked example 6. Estimate √20 to two decimal places.

Step 1. 4² = 16 and 5² = 25, so √20 is between 4 and 5.

Step 2. 4.4² = 19.36 and 4.5² = 20.25, so it is between 4.4 and 4.5.

Step 3. 4.47² = 19.9809 and 4.48² = 20.0704, so √20 ≈ 4.47.

Worked example 7. Using y = 4.9t², how long does a ball take to fall 44.1 m?

Step 1. 4.9t² = 44.1, so t² = 44.1 ÷ 4.9 = 9.

Step 2. t = ±3, and only the positive time makes sense: 3 seconds.

Check: 4.9 × 9 = 44.1.

Common misconceptions

"√144 = 72, half of 144." A square root is not half. Half of 144 is 72, but 72 × 72 = 5,184. The square root is the number that multiplies by itself to give 144, which is 12.

"√16 = ±4." The symbol √ means the positive root only, so √16 = 4. The equation x² = 16 has two solutions, ±4, and the ± has to be written in.

"√(9 + 16) = 3 + 4 = 7." Square roots do not split over addition. √(9 + 16) = √25 = 5. They do split over multiplication: √(9 × 16) = √144 = 12 = 3 × 4.

"√0.4 = 0.2." 0.2 × 0.2 = 0.04, not 0.4. √0.4 is about 0.63, because 0.63 × 0.63 is about 0.40.

"The cube root of 27 is 9, because cube root means divide by 3." 27 ÷ 3 = 9, but 9 × 9 × 9 = 729. The cube root of 27 is 3, because 3 × 3 × 3 = 27.

"x³ = 64 has two solutions, ±4." (−4)³ = −64, so −4 is not a solution. The ± belongs to square roots, not cube roots.

What you now know

• A square root of p is a number whose square is p, and √p means the positive one. A cube root of p is a number whose cube is p.
• x² = p has two solutions, x = ±√p, when p is positive, and none when p is negative. A situation such as a length or a time may rule out the negative solution.
• x³ = p has exactly one solution, ³√p, for any p, positive or negative.
• Roots of fractions come from roots of the top and the bottom: √(9/16) = 3/4 and ³√(8/27) = 2/3.
• A square root that is not a whole number can be trapped between consecutive whole numbers, then tenths, then hundredths.

The trapping of √50 never seemed to end. Before finding out why, the next lesson looks at decimals that never end but do repeat, like 0.333..., and asks a question that starts arguments in every eighth-grade class: is 0.999... the same number as 1?

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.A.2). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.7, Simplify and use square roots. Prealgebra 2e. OpenStax. openstax.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 13: Cube roots. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Urone, P. P., and Hinrichs, R. (2022). Section 2.7, Falling objects. College Physics 2e. OpenStax. openstax.org
5. Newell, D. B., and Tiesinga, E. (Eds.). (2019). The International System of Units (SI) (NIST Special Publication 330), Section 4: Non-SI units that are accepted for use with the SI (the litre). National Institute of Standards and Technology. nist.gov

A one-line question for university students

In 1978, David Tall and Rolph Schwarzenberger of the University of Warwick reported what happened when they asked first-year mathematics students, fresh from school, a single question: is 0.999..., with the nines going on forever, equal to one, or just less than one? Most of the students said it was just less than one. Some said the difference was "infinitely small". Others called it "the nearest you can get to one" without actually being one.

These were strong students, and their answer is not silly. This lesson sets out the case on each side, then settles the question, and on the way shows you how to turn any repeating decimal into a fraction. The argument that settles it and the method for fractions turn out to be the same piece of algebra.

This lesson covers Common Core standard 8.NS.A.1: understanding informally that every number has a decimal expansion, showing that the decimal expansion of a rational number repeats eventually, and converting a decimal that repeats eventually into a rational number.

The case that 0.999... is less than 1

Write the decimal out a few digits at a time and measure the gap to 1.

Every one of these is less than 1. The gap shrinks by a factor of 10 with each new digit, but it never becomes zero, and adding one more 9 never closes it. So, the argument runs, 0.999... creeps closer and closer to 1 without ever getting there, and it must be just a little less than 1.

The evidence here is real. Every decimal in the table, and every string of nines you could ever finish writing, really is less than 1. Keep that in mind, because the whole question is whether 0.999... is one of those decimals.

The case that 0.999... equals 1

Argument 1: thirds. Long division gives 1/3 = 0.333..., with the threes going on forever. Multiply both sides by 3. On the left, 3 × 1/3 = 1. On the right, every 3 becomes a 9, so 3 × 0.333... = 0.999.... If 1/3 really equals 0.333..., then 1 equals 0.999.... Tall and Schwarzenberger list this argument, along with a cousin of it: 1/9 = 0.111... and 2/9 = 0.222..., and so on up to 8/9 = 0.888..., so 9/9 ought to be 0.999.... But 9/9 = 1.

Argument 2: shift and subtract. Call the number x, so x = 0.999.... Multiplying by 10 moves every digit one place to the left, so 10x = 9.999.... The two numbers have exactly the same endless tail of nines after the decimal point, so subtracting one from the other removes the tail completely:

10x − x = 9.999... − 0.999... = 9, so 9x = 9, and x = 1.

Argument 3: nothing fits between. If two numbers are different, there is room between them; their average, for one, sits halfway between. What decimal could sit between 0.999... and 1? To be bigger than 0.999..., it would need a digit bigger than 9 somewhere, and there is no such digit. Nothing fits between them, so they are not two different numbers.

What settles it

Both sides agree about every decimal you can finish writing: 0.9, 0.99, 0.999 and every longer string of nines falls short of 1. What they disagree about is what the three dots mean.

The first case treats 0.999... as a very long string of nines that ends somewhere far away. But the dots say the nines never end. So 0.999... is not any of the decimals in the table. It names the single number that those decimals close in on, the number every one of them falls short of by less and less. Read the dots that way and the gap in the first case is not "infinitely small". There is no gap at all, because there is no last nine to leave one. That is also why Argument 2 works: the tail of nines in 10x and the tail in x are identical, with no final digit sticking out at the end of either.

So 0.999... = 1, and mathematicians agree about it. It is one number written two ways, the way 1/2 and 2/4 are one number. Other numbers have two decimal spellings too: 0.5 = 0.4999... and 2.37 = 2.36999.... Tall and Schwarzenberger's students were not careless. School had shown them plenty of decimals that end, and few that do not, so the dots meant something vague to them.

The upshot: A repeating decimal names one exact number, the value that its finished pieces close in on. 0.999... is not a little less than 1. It is 1.

Why every fraction's decimal ends or repeats

In seventh grade you saw that long division of a fraction either stops or repeats. Here is why nothing else can happen. Divide 1 by 7 and keep track of the remainders.

When you divide by 7, the remainder can only be 0, 1, 2, 3, 4, 5 or 6. If it is ever 0, the division stops and the decimal ends. If it is never 0, there are only six other remainders to go round, so within a few steps one of them must come back. From that moment the division repeats itself exactly, digits and all. So 1/7 = 0.142857142857..., with the block 142857 repeating forever.

The same reasoning works for any fraction. Dividing by a whole number n leaves remainders from 0 up to n − 1, so either a remainder of 0 ends the decimal, or a remainder returns and the digits repeat from there. A fraction can never produce a decimal that goes on forever without repeating. Textbooks often show the repeating block with a bar over it; OpenStax writes 1.333... as 1.3 with a line over the 3. This course writes the dots.

From a repeating decimal back to a fraction

Argument 2 is really a method, and it works on any repeating decimal. The trick is to multiply by the power of ten that shifts the decimal by exactly one repeating block, so that the two tails line up and cancel.

Worked example 1. Write 0.777... as a fraction.

Step 1. Let x = 0.777.... The block is one digit long, so multiply by 10: 10x = 7.777....

Step 2. Subtract: 10x − x = 7.777... − 0.777... = 7, so 9x = 7.

Step 3. x = 7/9.

Check: 7 ÷ 9 = 0.777..., because each step of the division is 70 ÷ 9 = 7 remainder 7.

Worked example 2. Write 0.363636... as a fraction.

Step 1. Let x = 0.3636.... The block, 36, is two digits long, so multiply by 100: 100x = 36.3636....

Step 2. Subtract: 100x − x = 36, so 99x = 36.

Step 3. x = 36/99 = 4/11, dividing the top and bottom by 9.

Check: 4 ÷ 11 gives 40 ÷ 11 = 3 remainder 7, then 70 ÷ 11 = 6 remainder 4, and the remainder 4 starts the pattern again: 0.3636....

Worked example 3. Write 0.142857142857... as a fraction.

Step 1. The block has six digits, so multiply by 10⁶: 1,000,000x = 142,857.142857....

Step 2. Subtract: 999,999x = 142,857, so x = 142,857/999,999.

Step 3. Since 142,857 × 7 = 999,999, the fraction simplifies to 1/7, as the long division promised.

Worked example 4. Write 0.1666..., where only the 6 repeats, as a fraction.

Step 1. Let x = 0.1666.... Here the repeating starts one place late, so first shift the non-repeating 1 to the left of the point: 10x = 1.666....

Step 2. Shift once more by one block: 100x = 16.666....

Step 3. Subtract the two numbers with matching tails: 100x − 10x = 16.666... − 1.666... = 15, so 90x = 15.

Step 4. x = 15/90 = 1/6.

Check: 1 ÷ 6 = 0.1666..., since after the first step the remainder is 4 every time and 40 ÷ 6 = 6 remainder 4.

Worked example 5. Write 2.454545... as a fraction.

Step 1. Let x = 2.4545.... The block is two digits, so 100x = 245.4545....

Step 2. Subtract: 100x − x = 245.4545... − 2.4545... = 243, so 99x = 243.

Step 3. x = 243/99 = 27/11, which is 2 and 5/11.

Check: 27 ÷ 11 = 2 remainder 5, then 50 ÷ 11 = 4 remainder 6, then 60 ÷ 11 = 5 remainder 5, and the 5 repeats the pattern: 2.4545....

Worked example 6. Decimals that end are rational too. 0.375 = 375/1,000 = 3/8. By the reasoning of the dispute, it also has a second spelling, 0.374999....

The upshot: Every decimal that ends or repeats is a fraction, and every fraction gives a decimal that ends or repeats. Multiplying by the right power of ten and subtracting gets you from the decimal back to the fraction.

Common misconceptions

"0.999... is the largest number below 1." There is no largest number below 1. Whatever number below 1 you name, the number halfway between it and 1 is bigger, and still below 1.

"A repeating decimal is only an approximation." 0.333... is exactly 1/3. It is a decimal cut off after a few digits, such as 0.333, that is an approximation.

"A decimal that never ends must repeat." Fractions always give decimals that end or repeat, but some numbers are not fractions. The next lesson meets √2, whose decimal never ends and never repeats.

"For 0.3636..., multiply by 10 and subtract." That gives 10x − x = 3.6363... − 0.3636... = 3.2727..., which still has a tail. The block has two digits, so multiply by 100.

"0.1666... = 16/99." That treats 16 as the repeating block, which describes 0.1616.... When the repeating starts late, shift the non-repeating part out of the way first, as in Worked example 4. The answer is 1/6.

Looking back

• Every fraction's decimal ends or repeats, because long division by n has only n possible remainders, and a repeated remainder repeats the digits.
• To turn a repeating decimal into a fraction, let x be the decimal, multiply by the power of ten that shifts it by one repeating block, and subtract so the tails cancel. Shift any non-repeating digits out of the way first.
• The dots in 0.999... mean the nines never end, so the number has no gap below 1. 0.999... = 1, one number with two decimal spellings.
• Decimals that end or repeat are exactly the rational numbers, the numbers that can be written as fractions.

That leaves a question hanging. Are there decimals that go on forever without ever repeating? If so, they cannot be fractions. The next lesson finds one in a clay tablet nearly four thousand years old and in a square drawn by Plato, and proves that the diagonal of a square is such a number.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, The Number System (8.NS.A.1). thecorestandards.org
2. Tall, D. O., and Schwarzenberger, R. L. E. (1978). Conflicts in the learning of real numbers and limits. Mathematics Teaching, 82, 44-49. University of Warwick repository. wrap.warwick.ac.uk
3. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.3, Decimals and fractions (repeating decimals). Prealgebra 2e. OpenStax. openstax.org
4. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 14: Decimal representations of rational numbers. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 15: Infinite decimal expansions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

A square drawn in clay

Somewhere in southern Mesopotamia, at some time between about 1900 and 1600 BCE, someone pressed a square and its two diagonals into a small clay tablet. Along one side they wrote 30. Along a diagonal they wrote two numbers in the base-60 system that Babylonian scribes used. The tablet is now in the Yale Babylonian Collection, catalogued as YBC 7289.

The first number on the diagonal is 1;24,51,10, which means 1 + 24/60 + 51/60² + 10/60³. Work it out and you get 1.41421296.... The second number, 42;25,35, is 30 times the first, about 42.43. The scribe was giving the length of a square's diagonal: about 1.41421296 when the side is 1, and so about 42.43 when the side is 30.

The true value is 1.41421356..., so the scribe's number is off by less than a millionth. This lesson is about that number, the diagonal of a square whose sides are 1, and about why no scribe and no computer will ever write it down exactly.

This lesson covers Common Core standards 8.NS.A.1, knowing that numbers that are not rational are called irrational; 8.NS.A.2, using rational approximations to compare irrational numbers, locate them on a number line and estimate expressions such as π²; and the last part of 8.EE.A.2, knowing that √2 is irrational.

A square twice as big

More than a thousand years after the tablet was made, in Plato's dialogue Meno, Socrates draws a square with sides of 2 feet, so its area is 4 square feet. He asks a boy how long the side must be for a square with twice the area, 8 square feet. The boy says 4 feet. Socrates shows him that a 4-foot square holds four of the small squares, an area of 16, which is four times as big, not twice. The boy tries 3 feet, but 3 × 3 = 9, which is still not 8. He is stuck.

Then Socrates draws the answer. You can build it yourself on a grid. Take four 1-by-1 squares and put them together into a 2-by-2 square with corners at (0, 0), (2, 0), (2, 2) and (0, 2). Now join the midpoints of its four sides: (1, 0), (2, 1), (1, 2) and (0, 1). That gives a tilted square sitting inside the big one. Each side of the tilted square is a diagonal of one of the small squares, and each of those diagonals cuts its small square exactly in half.

So the tilted square is made of four half-squares. Four halves make 2 whole small squares, so its area is 2. In Socrates' version every length is doubled, the small squares have area 4, and the tilted square has area 8: exactly the doubled square the boy was looking for. Its side is the diagonal of the original square, which Plato has Socrates call by the name the learned give it, the diagonal.

Back to side 1. The tilted square has area 2, so if its side, the diagonal d, is squared, you get 2: d² = 2. A length is positive, so d = √2. The diagonal of a square with side 1 is √2, the number the Babylonian scribe approximated as 1.41421296.

Is √2 a fraction?

The scribe's number is itself a fraction: 1;24,51,10 = 305,470/216,000. Square it and you get about 1.9999983, very close to 2 but not 2. Is there some better fraction whose square is exactly 2? You could hunt for one forever. Instead, here is an argument that settles it for good. It works by supposing there is such a fraction and showing that the supposition leads to something impossible.

Step 1. Suppose √2 = a/b, where a and b are whole numbers and the fraction is in lowest terms, so a and b have no common factor. In particular, they are not both even.

Step 2. Square both sides: 2 = a²/b², so a² = 2b². That makes a² an even number.

Step 3. An odd number times itself is odd: 3 × 3 = 9, 5 × 5 = 25, and so on. Since a² is even, a cannot be odd. So a is even, and a = 2c for some whole number c.

Step 4. Put 2c in place of a: (2c)² = 2b², so 4c² = 2b², and dividing by 2, b² = 2c². Now b² is even, so by the same reasoning b is even.

Step 5. Both a and b are even. But Step 1 said they are not both even. The supposition contradicts itself, so it was false: there is no fraction a/b equal to √2.

A number that cannot be written as a fraction of two integers is called irrational. √2 is irrational. The same is true of the square root of every whole number that is not a perfect square, so √3, √5, √7 and √8 are all irrational too; the proofs follow the same idea with more work.

Worth holding on to: √2 is a perfectly real length, the diagonal of a square you can draw, but it is not a fraction. Every decimal that ends, and every fraction, falls just short of it or just past it.

Rational, irrational and real

The last lesson showed that every fraction has a decimal that ends or repeats, and every decimal that ends or repeats is a fraction. So the decimal of an irrational number can do neither. It goes on forever without ever settling into a repeating block. Illustrative Mathematics' lesson summary on decimal expansions uses the two irrational numbers you will meet most often as its examples: √2 and π.

The last example has a pattern, one more 0 each time, but no block of digits repeats forever, so it is irrational. A pattern is not the same as a repeat.

Rational and irrational numbers together make up the real numbers. Every point on the number line is a real number, and every real number has a point on the line. √2 has its point: lay the diagonal of a 1-by-1 square along the number line from 0, and its far end marks √2.

Trapping √2 between decimals

You cannot write √2 exactly, but you can pin it down as closely as you like. The standard describes the method: truncate the decimal, and square the two candidates on either side.

Each row traps √2 in an interval ten times narrower than the row before. To go on, try 1.4141, 1.4142 and so on until the square passes 2, and keep the last one that stays below. The process never ends, because no decimal that ends squares to exactly 2, but each step gives you one more correct digit. The scribe's 1.41421296 sits inside every one of these intervals.

Comparing and placing irrational numbers

To compare an irrational number with a decimal, compare their squares, or trap the irrational number first.

√7 or 2.6? 2.6² = 6.76, which is less than 7, and 2.7² = 7.29, which is more. So 2.6 < √7 < 2.7, and √7 is the larger of the two.

π or √10? π is about 3.1416, and 3.1416² is about 9.87, less than 10. So π < √10. (√10 is about 3.16.)

Estimating π². π is between 3.14 and 3.15. Then π² is between 3.14² = 9.8596 and 3.15² = 9.9225. A closer value, 3.1416² ≈ 9.8697, gives π² ≈ 9.87.

To place numbers on a number line, trap each one between tenths, then mark it in the right gap:

Check one: 2.2² = 4.84 and 2.3² = 5.29, so √5 is between 2.2 and 2.3, and −√5 is between −2.3 and −2.2.

Worked examples

Worked example 1. Decide whether each number is rational or irrational: √36, √(9/25), √15, 0.2525..., 1.010010001....

√36 = 6: rational. √(9/25) = 3/5: rational. 15 is not a perfect square, so √15 is irrational. 0.2525... repeats, so it is rational; in fact it is 25/99. 1.010010001... has growing gaps of zeros and no repeating block: irrational.

Worked example 2. Trap √30 between two hundredths.

Step 1. 5² = 25 and 6² = 36, so √30 is between 5 and 6.

Step 2. 5.4² = 29.16 and 5.5² = 30.25, so it is between 5.4 and 5.5.

Step 3. 5.47² = 29.9209 and 5.48² = 30.0304, so it is between 5.47 and 5.48, and closer to 5.48.

Worked example 3. Put 2π, √40 and 6.3 in order from smallest to largest.

Step 1. 2π ≈ 2 × 3.1416 = 6.2832.

Step 2. 6.3² = 39.69 and 6.4² = 40.96, so √40 is between 6.3 and 6.4; and 6.32² = 39.9424 and 6.33² = 40.0689, so √40 ≈ 6.32.

Step 3. The order is 2π (about 6.28), then 6.3, then √40 (about 6.32). Three numbers within 0.05 of one another, sorted without a calculator.

Worked example 4. How long is the diagonal of a square tile with 10 cm sides?

Step 1. Scaling the unit square up by 10 scales its diagonal by 10, so the diagonal is 10√2 cm.

Step 2. 10 × 1.414 ≈ 14.14 cm. A ruler on a real tile reads just over 14 cm.

Worked example 5. How close is the scribe's value? Square 1.41421296.

Step 1. 1.41421296² ≈ 1.9999983.

Step 2. The square misses 2 by about 0.0000017, less than two millionths. That is accurate enough to build anything the Babylonians could build, and still not exactly 2, because no fraction can be.

Common misconceptions

"√2 = 1.414." 1.414 is an approximation: 1.414² = 1.999396. No decimal that ends equals √2, because every such decimal is a fraction.

"π = 22/7." 22/7 = 3.142857..., a repeating decimal and so a rational number. π = 3.14159... is irrational. 22/7 is a handy approximation, close but not equal.

"Irrational numbers are not real, or cannot be placed on a number line." √2 is the length of a diagonal you can draw, and it has an exact point on the number line. Irrational numbers are real numbers; they just have no fraction or ending decimal to name them.

"A decimal with a pattern repeats." 0.101001000100001... has a pattern, but no block of digits repeats forever, so it is irrational.

"Every square root is irrational." √49 = 7 and √(4/9) = 2/3 are rational. A square root of a whole number is irrational exactly when the number is not a perfect square.

Pulling it together

• The diagonal of a square with side 1 is √2, because the square built on that diagonal has area 2, as Plato's doubled square shows.
• √2 is irrational: supposing it equals a fraction in lowest terms forces both the top and the bottom to be even, which is impossible.
• Irrational numbers have decimals that never end and never repeat. Together with the rational numbers they make up the real numbers, which fill the number line.
• To trap an irrational square root, square candidates on either side and narrow the interval one decimal place at a time.
• To compare or place irrational numbers, use rational approximations: √7 > 2.6 because 2.6² < 7, and π² ≈ 9.87.

That finishes the number system for eighth grade. The next module leaves numbers for shapes: you will slide, flip and turn figures on a grid, and find out what it really means for two shapes to be the same.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, The Number System (8.NS.A.1, 8.NS.A.2). thecorestandards.org
2. Cuneiform Digital Library Initiative. (2024, November 6). MCT 042 YBC 07289 artifact entry (P255048), with photographs from the Yale Babylonian Collection. cdli.earth
3. Institute for the Study of the Ancient World. (2010). YBC 7289. In Before Pythagoras: The culture of Old Babylonian mathematics [Exhibition]. New York University. isaw.nyu.edu
4. Plato. (1871/n.d.). Meno (B. Jowett, Trans.), 82b-85b. The Internet Classics Archive, MIT. classics.mit.edu
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 7.1, Rational and irrational numbers. Prealgebra 2e. OpenStax. openstax.org
6. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 15: Infinite decimal expansions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Seven pieces, not five

Tetris drops pieces made of four squares joined edge to edge. MathWorld, the Wolfram encyclopedia of mathematics, counts how many such shapes there are: five, if you are allowed to flip a shape over, but seven if you may only slide it and turn it, because then a shape and its mirror image count as different. Tetris uses seven pieces. As a piece falls, you can slide it sideways and rotate it, but nothing in the game flips it over.

That is why the game has both an S piece and a Z piece. Picture the S: two squares side by side on the bottom row, and two more on the row above, shifted one square to the right. The Z is the same picture shifted one square to the left. Hold one up to a mirror and you get the other. Turn the S as much as you like: you get an S lying on its side, or an S standing up, but never a Z. The J and L pieces are another mirror pair.

Slides, turns and flips are the three rigid motions, ways of moving a shape without stretching, shrinking or bending it. This lesson compares them side by side and gives each a coordinate rule. It covers Common Core standards 8.G.A.1, verifying by experiment that rotations, reflections and translations keep lengths, angles and parallel lines, and 8.G.A.3, describing their effects using coordinates.

The three motions, defined

A translation slides every point of a figure the same distance in the same direction. To describe one you give the direction and the distance, such as "3 units right and 2 units down".

A reflection flips a figure across a line, called the line of reflection. Each point moves straight across the line, along a path at right angles to it, and lands the same distance away on the other side. Points on the line itself do not move. To describe a reflection you name the line.

A rotation turns a figure about a fixed point, called the centre of rotation, through an angle. To describe one you give the centre, the angle and the direction: clockwise, the way a clock's hands move, or counterclockwise. The centre does not move, and every other point travels round it on a circle.

The figure you start with is called the pre-image, and the figure you end with is its image. Labels carry over with a prime mark: the image of point A is A', read "A prime".

The three side by side

Read the table down its columns and you have each motion's full description. Read it across its rows and you see what makes reflections the odd one out: they are the only motion that reverses the order of the vertices. That single row is the whole reason Tetris needs seven pieces instead of five.

What matters here: All three motions move a figure without changing its size or shape. What separates them is what you must specify, which points stay put, and whether the figure's orientation is kept or reversed.

The coordinate rules, worked on one triangle

Take the right triangle with vertices A(1, 1), B(4, 1) and C(1, 3). Its side AB runs 3 units along a grid line, its side AC runs 2 units up a grid line, and the right angle is at A.

Translation 3 right and 2 down. The rule is (x, y) → (x + 3, y − 2). A(1, 1) goes to A'(4, −1), B(4, 1) to B'(7, −1), and C(1, 3) to C'(4, 1).

Reflection in the x-axis. Each point moves straight up or down across the x-axis, so its distance from the axis stays the same and only the sign of y changes: (x, y) → (x, −y). A'(1, −1), B'(4, −1), C'(1, −3).

Reflection in the y-axis. Now the points move straight left or right, so only the sign of x changes: (x, y) → (−x, y). A'(−1, 1), B'(−4, 1), C'(−1, 3).

Rotation 90° counterclockwise about the origin. The rule is (x, y) → (−y, x). Here is why. Think of the path from the origin to B(4, 1) as "4 right, then 1 up". Turn that whole path a quarter turn counterclockwise: "right" becomes "up" and "up" becomes "left". So the path becomes "4 up, then 1 left", which ends at (−1, 4). The old y-coordinate, with its sign changed, became the new x, and the old x became the new y. So A'(−1, 1), B'(−1, 4), C'(−3, 1).

Rotation 180° about the origin. Two quarter turns: (x, y) → (−y, x) → (−x, −y). Both signs change. A'(−1, −1), B'(−4, −1), C'(−1, −3).

What survives the move

The standard asks you to check by experiment that rigid motions take lines to lines, keep the lengths of segments and the sizes of angles, and take parallel lines to parallel lines. You can check all of it in the table.

Lengths. AB is 3 units long. In every image, A' and B' are still 3 apart: (4, −1) to (7, −1) is 3 units along a grid line, and in the rotated image (−1, 1) to (−1, 4) is 3 units up a grid line. AC is 2 units, and so is every A'C'. The slanted side BC does not lie along a grid line, so the grid does not measure it directly. Trace the triangle on thin paper, move the paper the way the motion says, and the traced BC lands exactly on B'C'. In Lesson 28 you will measure slanted lengths with coordinates too.

Angles. The right angle at A stays a right angle. In the 90° rotation, A'B' runs straight up and A'C' runs straight left, which still meet square.

Parallel lines. Take the parallelogram P(0, 0), Q(3, 0), R(4, 2), S(1, 2). Sides PQ and SR are both flat, and sides PS and QR both go "1 right, 2 up", so opposite sides are parallel. Rotate it 90° counterclockwise: P'(0, 0), Q'(0, 3), R'(−2, 4), S'(−2, 1). Now P'Q' and S'R' both run straight up, and P'S' and Q'R' both go "2 left, 1 up". Opposite sides are still parallel.

What matters here: A rigid motion can change where a figure is and which way it faces. It never changes a length, an angle or a pair of parallel lines.

Telling a flip from a turn

Go round the original triangle in the order A, B, C. From A(1, 1) you go right to B(4, 1), then up and left to C(1, 3), then down to A: that is counterclockwise. Now go round the image after the reflection in the x-axis in the same letter order: A'(1, −1) right to B'(4, −1), then down and left to C'(1, −3), then up to A'. That is clockwise. The reflection reversed the orientation. Do the same for the rotated image and you will find the order is still counterclockwise. Translations and rotations keep orientation; reflections reverse it. That is the check that tells you an S has become a Z.

Worked examples

Worked example 1. Find the image of P(−2, 5) under each motion: a translation 4 left and 1 up; a reflection in the y-axis; a rotation of 90° clockwise about the origin.

Step 1. Translation: (x − 4, y + 1) gives (−6, 6).

Step 2. Reflection in the y-axis: change the sign of x, giving (2, 5).

Step 3. 90° clockwise: (x, y) → (y, −x) gives (5, 2).

Check Step 3 with the path idea: the path to P is "2 left, 5 up". A quarter turn clockwise turns "left" into "up" and "up" into "right", so the path becomes "2 up, 5 right", which ends at (5, 2).

Worked example 2. Name the motion that takes (2, 5) to each point: (−2, 5); (2, −5); (−2, −5); (−5, 2).

(−2, 5): only x changed sign, so a reflection in the y-axis. (2, −5): only y changed sign, so a reflection in the x-axis. (−2, −5): both signs changed, so a rotation of 180° about the origin. (−5, 2): the coordinates swapped and the new x is the old y with its sign changed, which is the rule (−y, x): a rotation of 90° counterclockwise.

Worked example 3. After a translation (x, y) → (x − 2, y + 5), a point's image is (3, −4). Where was the point?

Step 1. Undo the slide: add 2 to x and subtract 5 from y.

Step 2. The original point was (5, −9).

Check: 5 − 2 = 3 and −9 + 5 = −4.

Worked example 4. A rectangle has vertices (1, 2), (5, 2), (5, 4) and (1, 4). Reflect it in the x-axis and check its side lengths.

Step 1. Change the sign of each y: (1, −2), (5, −2), (5, −4), (1, −4).

Step 2. The original sides are 4 and 2 units long. In the image, (1, −2) to (5, −2) is 4 units and (5, −2) to (5, −4) is 2 units. The rectangle is unchanged in size and shape; it has only moved below the axis.

Worked example 5. In a game, a piece sits on the squares (1, 1), (2, 1), (2, 2) and (3, 2), an S shape. Can a rotation of 90° counterclockwise about the origin turn it into the Z shape (1, 2), (2, 2), (2, 1), (3, 1)?

Step 1. Rotate: (−1, 1), (−1, 2), (−2, 2), (−2, 3). That is an S standing on its end.

Step 2. Check orientation instead of trying every rotation. The Z is the mirror image of the S, so it has the opposite orientation, and no rotation or translation can reverse orientation. Only a reflection could. So no sequence of slides and turns makes S into Z, and Tetris needs both pieces.

Common misconceptions

"A reflection in the x-axis changes the x-coordinate." It moves points up and down across the x-axis, so it is the y-coordinate that changes sign: (3, 4) goes to (3, −4). The coordinate that stays the same is the one measured along the mirror line.

"A 90° rotation just swaps the coordinates." Swapping, (x, y) → (y, x), is a reflection in the line y = x. The rotation also changes a sign: (−y, x) for counterclockwise, (y, −x) for clockwise. Check with (4, 1): a quarter turn counterclockwise sends it to (−1, 4), not (1, 4).

"A rotation always turns a shape about its own middle." A rotation can be about any point, inside the shape or outside it, and the answer depends on which one. Always say where the centre is.

"A 180° rotation is a kind of reflection." A 180° rotation keeps orientation, and a reflection reverses it, so the two are different. (Doing two reflections in a row, in the x-axis and then in the y-axis, does produce the 180° rotation, because two reversals cancel.)

"Moving a shape far away makes it smaller." It may look smaller on a screen that zooms out, but a translation by any distance keeps every length the same.

Summing up

• Translations slide, reflections flip across a line, and rotations turn about a centre. A translation needs a direction and distance, a reflection needs a line, and a rotation needs a centre, an angle and a direction.
• Coordinate rules: translation (x + a, y + b); reflection in the x-axis (x, −y), in the y-axis (−x, y); rotation about the origin by 90° counterclockwise (−y, x), by 180° (−x, −y), by 90° clockwise (y, −x).
• All three keep lengths, angle measures and parallel lines, which you can check with coordinates or tracing paper.
• Translations and rotations keep a figure's orientation; reflections reverse it. That is why mirror-image pieces such as S and Z cannot be turned into each other.

If one figure can be carried exactly onto another by rigid motions, the two figures are the same size and shape. The next lesson turns that idea into the mathematical meaning of "congruent" and shows how to find the moves.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.A.1, 8.G.A.3). thecorestandards.org
2. Weisstein, E. W. (n.d.). Tetromino. MathWorld, a Wolfram Web Resource. mathworld.wolfram.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 5: Coordinate moves. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 7: No bending or stretching. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Kirk, D., et al. (2023). Section 10.5, Tessellations (rigid motions). Contemporary Mathematics. OpenStax. openstax.org
6. Wikipedia contributors. (n.d.). Tetris (gameplay section, background only). Wikipedia. en.wikipedia.org

Two right triangles on a grid

Triangle ABC has corners at A(1, 1), B(4, 1) and C(1, 5). Triangle PQR has corners at P(−2, 3), Q(−2, 6) and R(−6, 3). Each has a right angle, and each has one leg 3 units long and one leg 4 units long, but they sit in different places and face different ways. Are they the same?

In everyday speech, "congruent" means "the same size and shape". Eighth-grade geometry turns that into something you can check. Two figures are congruent when there is a sequence of rigid motions, translations, reflections and rotations, that carries one figure exactly onto the other. Since rigid motions never change a length or an angle, figures that can be moved onto each other really are the same size and shape. To prove two figures congruent, you find the moves.

The idea is old. Euclid's Elements, written around 300 BCE, lists among its basic assumptions that things which coincide with one another are equal to one another, and its fourth proposition proves a fact about triangles by placing one triangle on top of another. This lesson covers Common Core standard 8.G.A.2: understanding that a figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections and translations, and describing a sequence that shows the congruence. It also continues 8.G.A.3, working with coordinates.

The method, worked end to end

Step 1. Match the corresponding parts. Decide which corner of ABC should land on which corner of PQR. The right angles must match, so A goes to P. AB is the 3-unit leg, running from (1, 1) to (4, 1). In PQR the 3-unit leg is PQ, running from (−2, 3) up to (−2, 6). So B goes to Q. That leaves C going to R, and the 4-unit legs AC and PR match.

Step 2. Translate one matching corner onto the other. To move A(1, 1) to P(−2, 3), slide 3 left and 2 up: (x, y) → (x − 3, y + 2).

A(1, 1) → (−2, 3), which is P. B(4, 1) → (1, 3). C(1, 5) → (−2, 7).

Step 3. Turn about that corner until the sides line up. After the slide, the 3-unit leg runs from P to (1, 3): 3 units to the right. In PQR it runs 3 units up, to Q. A quarter turn counterclockwise turns "right" into "up", so rotate 90° counterclockwise about P. Check each point by its path from P. The image of B is "3 right of P", and after the turn it is "3 up from P", which is (−2, 6) = Q. The image of C is "4 up from P", and after the turn it is "4 left of P", which is (−6, 3) = R.

Step 4. Check every vertex. The sequence sends A to P, B to Q and C to R. Every corner lands where it should, so the whole triangle does.

The answer: translate 3 units left and 2 units up, then rotate 90° counterclockwise about P(−2, 3). The triangles are congruent, and you write ▵ABC ≅ ▵PQR. The order of the letters records which corner goes where: A with P, B with Q, C with R.

The sequence is not the only one. Rotate ABC 90° counterclockwise about the origin first, using (x, y) → (−y, x): A(1, 1) → (−1, 1), B(4, 1) → (−1, 4), C(1, 5) → (−5, 1). Then translate 1 left and 2 up: (−1, 1) → (−2, 3) = P, (−1, 4) → (−2, 6) = Q, (−5, 1) → (−6, 3) = R. A different sequence, the same conclusion. Any correct sequence proves the congruence.

Bottom line: To show two figures are congruent, match their corresponding parts, then slide one corner onto its partner, turn or flip until the rest lines up, and check every vertex.

A second pair, where one change flips the answer

Now take triangle XYZ with X(2, −1), Y(5, −1) and Z(2, −5). It too has a right angle, at X, with legs of 3 and 4. Try the same method.

Step 1. Match: A goes to X, B to Y (the 3-unit legs, both running right), and C to Z.

Step 2. Look at the 4-unit legs. In ABC, AC runs up from A. In XYZ, XZ runs down from X. Meanwhile AB and XY both run to the right. No rotation can turn "up" into "down" while leaving "right" as "right": a turn moves both legs together. The triangles face opposite ways.

Step 3. Check orientation, as in the last lesson. Going A to B to C is counterclockwise: right, then up and to the left. Going X to Y to Z is clockwise: right, then down and to the left. Orientation is reversed, so a reflection is needed.

Step 4. Reflect ABC in the x-axis, (x, y) → (x, −y): A(1, 1) → (1, −1), B(4, 1) → (4, −1), C(1, 5) → (1, −5). Then translate 1 unit right: (1, −1) → (2, −1) = X, (4, −1) → (5, −1) = Y, (1, −5) → (2, −5) = Z.

So ▵ABC ≅ ▵XYZ, by a reflection in the x-axis and then a translation 1 unit right. The only thing that changed from the first pair was which way one leg pointed, and that one change meant a reflection had to be in the sequence. Mirror-image figures are congruent; you just need a flip to show it. Joyce's commentary on Euclid's fourth proposition points out the physical side of this: to move a paper triangle onto its mirror image you have to lift it out of the page and turn it over.

When the answer is no

A rectangle 2 units by 6 units and a rectangle 3 units by 4 units both have an area of 12 square units. Are they congruent? No. Rigid motions keep every length, so any image of the 2-by-6 rectangle still has sides of 2 and 6, and can never land exactly on sides of 3 and 4. Equal area is not enough.

The same reasoning gives a quick test. If you can find a length or an angle in one figure with no matching length or angle in the other, the figures are not congruent, and no sequence of moves will ever work.

Using congruence once you have it

When ▵ABC ≅ ▵PQR, every part of one matches the corresponding part of the other, because the moves carried each part onto its partner without changing it. So AB = PQ, BC = QR and AC = PR, and the angles at A and P are equal, as are the angles at B and Q and at C and R. That is useful: measure a part of one figure and you know the matching part of the other.

Worked examples

Worked example 1. Segment JK runs from J(0, 0) to K(3, 1). Segment LM runs from L(4, 2) to M(3, 5). Find a sequence of rigid motions that takes JK onto LM.

Step 1. JK goes "3 right, 1 up". LM goes "1 left, 3 up". A quarter turn counterclockwise turns "3 right, 1 up" into "3 up, 1 left", which matches.

Step 2. Rotate JK 90° counterclockwise about the origin: J(0, 0) stays at (0, 0) and K(3, 1) → (−1, 3).

Step 3. Translate 4 right and 2 up: (0, 0) → (4, 2) = L and (−1, 3) → (3, 5) = M.

Check: both endpoints land in place, so the segments are congruent.

Worked example 2. Square S has vertices (1, 1), (3, 1), (3, 3), (1, 3). Square T has vertices (−5, 0), (−3, 0), (−3, 2), (−5, 2). Describe a single rigid motion that maps S onto T.

Step 1. Both squares have sides of 2 and face the same way, so a slide should do it.

Step 2. (1, 1) must go to (−5, 0): that is 6 left and 1 down, (x − 6, y − 1).

Check: (3, 1) → (−3, 0), (3, 3) → (−3, 2), (1, 3) → (−5, 2). All four corners land.

Worked example 3. ▵DEF ≅ ▵UVW. Angle E is 37° and side DF is 8 cm. What do you know about ▵UVW?

Step 1. The letter order matches D with U, E with V and F with W.

Step 2. So angle V = 37° and side UW = 8 cm.

Worked example 4. One triangle has sides 5, 5 and 6. Another has sides 5, 6 and 6. Are they congruent?

Step 1. The first has two sides of 5; the second has only one.

Step 2. A rigid motion keeps all three lengths, so no sequence can turn one into the other. They are not congruent, even though they share the lengths 5 and 6.

Worked example 5. In Tetris, are the S and Z pieces congruent?

Step 1. Each is four unit squares, and each is the mirror image of the other.

Step 2. A reflection carries one exactly onto the other, so yes: the S and Z pieces are congruent. The game treats them as different pieces because its moves include only slides and turns, not because their size or shape differs.

Common misconceptions

"Two figures with the same area are congruent." A 2-by-6 rectangle and a 3-by-4 rectangle both have area 12, but their side lengths differ, so no rigid motion maps one onto the other.

"Figures that face different ways are not congruent." Turning or flipping a figure is allowed. PQR faces a different way from ABC and is still congruent to it.

"A mirror image is a different shape." It is a congruent shape with the opposite orientation. A reflection is a rigid motion, so mirror images count as congruent.

"The order of letters in ▵ABC ≅ ▵PQR does not matter." It records which corner matches which. Writing ▵ABC ≅ ▵QPR would claim that A matches Q, and that would be false here.

"There is only one correct sequence." Many sequences can work. ABC reached PQR both by sliding first and by turning first. Any sequence that lands every vertex correctly is a valid proof.

Recap

• Two figures are congruent when a sequence of translations, reflections and rotations carries one exactly onto the other.
• The method: match corresponding parts, translate one matching vertex onto its partner, rotate or reflect until the sides line up, and check every vertex with coordinates.
• If the orientation is reversed, the sequence must include a reflection. Mirror images are congruent.
• If any length or angle has no match, the figures are not congruent. Equal area is not enough.
• In a congruence statement such as ▵ABC ≅ ▵PQR, the letter order gives the matching corners, and matching sides and angles are equal.

Rigid motions keep lengths and angles. The next module asks what happens to angles when lines cross, uses parallel lines to show why every triangle's angles add to 180°, and then allows one more kind of move, a dilation, which changes size but not shape.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.A.2, 8.G.A.3). thecorestandards.org
2. Euclid. (c. 300 BCE/1996). Elements, Book I, common notions (D. E. Joyce, Ed.). Clark University. mathcs.clarku.edu
3. Euclid. (c. 300 BCE/1996). Elements, Book I, Proposition 4, with commentary on the method of superposition (D. E. Joyce, Ed.). Clark University. mathcs.clarku.edu
4. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 12: Congruent polygons. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 13: Congruence. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

A shadow in Alexandria

Around 240 BCE, Eratosthenes of Cyrene, the chief librarian at Alexandria in Egypt, heard about a well in the town of Syene, now Aswan, far up the Nile. At noon on the summer solstice, around 21 June, sunlight lit the whole bottom of the well, so the Sun had to be straight overhead. At noon on the same day in Alexandria, an upright stick did cast a shadow. Eratosthenes measured the angle between the stick and the Sun's rays and found about 7.2 degrees, which is 1/50 of a full circle.

From that one angle and the distance between the two cities, he worked out the size of the whole Earth. His own book on it is lost, and the story comes down through later writers; this lesson follows the summary published by the American Physical Society. The argument rests on a single fact about parallel lines, and the same fact explains why the angles of every triangle add to 180°.

This lesson covers Common Core standard 8.G.A.5: using informal arguments to establish facts about the angles created when parallel lines are cut by a transversal, and about the angle sum and exterior angle of triangles. The last part of that standard, the angle-angle test for similar triangles, comes in Lesson 12.

Two lines that cross

Start with something you met in seventh grade. When two lines cross, they make four angles around the crossing point. The angles opposite each other are called vertical angles, because they share a vertex, and they are equal. Angles next to each other along a line add to 180°. So if one of the four angles is 70°, its two neighbours are each 180° − 70° = 110°, and the angle opposite it is 70°.

Why are vertical angles equal? Each one makes a straight line, 180°, together with the same neighbour. Two angles that both add to 180° with the same angle must be the same size.

Two parallel lines and a transversal

Now draw two parallel lines and a third line crossing both. The crossing line is called a transversal. To have something exact to look at, put the parallel lines on a grid as the horizontal lines y = 0 and y = 4, and let the transversal be the line y = x, which passes through (0, 0), (4, 4) and (6, 6). It crosses the bottom line at (0, 0) and the top line at (4, 4).

At each crossing there are four angles, eight in all. Name them by position: above-right, above-left, below-left and below-right of the crossing point. The line y = x climbs at 45° to the horizontal, so at each crossing the above-right and below-left angles are 45°, and the above-left and below-right angles are 180° − 45° = 135°.

Notice that the two crossings look exactly alike. That is no accident. Slide the bottom crossing up along the transversal, 4 right and 4 up, and it lands on the top crossing. The slide carries the bottom line onto a line through (4, 4) parallel to it, which is the top line, and it carries the transversal along itself. So every angle at the bottom crossing lands on the angle in the same position at the top crossing. This is the fact everything else depends on.

Alternate interior angles are equal because each is vertical to, and so equal to, an angle that corresponds to the other. Same-side interior angles add to 180° because one of them equals the neighbour of the other.

The core of it: When a transversal crosses two parallel lines, the two crossings are copies of each other. Every angle is equal to its partner in the same position, and every pair of angles is either equal or adds to 180°.

Back to Alexandria

Now Eratosthenes' argument makes sense. Picture the Earth as a circle with centre O, and mark Syene and Alexandria on it, with Alexandria to the north. At Syene the Sun is straight overhead, so its ray runs straight down the well, along the line from the Sun through Syene to the centre O. At Alexandria the stick points straight up, along the line from O out through Alexandria. A ray from the Sun strikes the top of the stick at 7.2° to it.

The Sun is so far away that its rays at Syene and at Alexandria are parallel. The line through the stick, running down to the centre O, is a transversal that crosses both parallel rays. The 7.2° angle at the top of the stick and the angle at the centre between the two cities are alternate interior angles, so they are equal. The angle at the Earth's centre between Syene and Alexandria is 7.2°.

7.2° is 1/50 of 360°, so the stretch of the Earth's surface from Syene to Alexandria is 1/50 of the way around. The two cities were reckoned to be about 5,000 stadia apart, so the whole way around is 50 × 5,000 = 250,000 stadia.

How good was that? Nobody knows exactly how long Eratosthenes' stadium was. The APS account says modern scholars have suggested values between 500 and about 600 feet. At 500 feet, 250,000 stadia is 125,000,000 feet, and 125,000,000 ÷ 5,280 ≈ 23,700 miles. At 600 feet it is about 28,400 miles. The Earth is about 24,900 miles around the equator, so he was within about 15 percent either way, using a stick and one angle.

The argument depends on assumptions, and the APS account lists them: that the Sun's rays are parallel, that Alexandria is due north of Syene, and that Syene lies exactly where the Sun is overhead at the solstice. None is perfectly true, but each is close enough for the method to work.

Why a triangle's angles add to 180°

The standard suggests an experiment. Cut out any paper triangle, tear off its three corners, and lay them side by side with their points touching. They always fit together into a straight line, 180°. Parallel lines explain why.

Take the triangle with A(0, 0), C(6, 0) and B(2, 4). Draw the line through B parallel to AC; since AC lies on the x-axis, that is the horizontal line y = 4. At B, three angles now sit side by side along that line: one on the left, the triangle's own angle at B in the middle, and one on the right.

Step 1. Side AB is a transversal crossing the two parallel lines y = 0 and y = 4. The left-hand angle at B and the triangle's angle at A are alternate interior angles, so they are equal.

Step 2. Side BC is another transversal. The right-hand angle at B and the triangle's angle at C are alternate interior angles too, so they are equal.

Step 3. The three angles at B fill a straight line, so they add to 180°. They are the triangle's angles at A, B and C, so A + B + C = 180°.

Nothing in the argument depended on the particular triangle, so it works for every triangle. Euclid proved it this way in Book I, Proposition 32 of the Elements, drawing his parallel line through a different corner.

The exterior angle

Extend one side of a triangle past a corner, say AC past C to a point D. The angle between the extension CD and the side CB is an exterior angle. It sits next to the interior angle C along a straight line, so it equals 180° − C. And since A + B + C = 180°, 180° − C = A + B. An exterior angle equals the sum of the two interior angles opposite it. Euclid states both results together in the same proposition.

Check with numbers: if A = 50° and B = 75°, the exterior angle at C is 50° + 75° = 125°. The other way: C = 180° − 125° = 55°, and 180° − 55° = 125°. Both routes agree.

Worked examples

Worked example 1. Two parallel lines are cut by a transversal, and one of the eight angles is 65°. Find the others.

Step 1. At that crossing, the vertical angle is also 65°, and the two neighbours are 180° − 65° = 115°.

Step 2. The other crossing is a copy of the first, so it has two angles of 65° and two of 115°, in the same positions.

Check: every same-side interior pair is one 65° and one 115°, which add to 180°.

Worked example 2. A triangle has angles of 47° and 68°. Find the third angle.

Step 1. 47° + 68° = 115°.

Step 2. 180° − 115° = 65°.

Check: 47 + 68 + 65 = 180.

Worked example 3. The two interior angles opposite an exterior angle are 38° and 61°. Find the exterior angle.

Step 1. The exterior angle is their sum: 38° + 61° = 99°.

Check: the third interior angle is 180° − 99° = 81°, and 180° − 81° = 99°.

Worked example 4. The angles of a triangle are x°, 2x° and 3x°. Find them.

Step 1. x + 2x + 3x = 180, so 6x = 180 and x = 30.

Step 2. The angles are 30°, 60° and 90°: a right triangle.

Worked example 5. Alternate interior angles formed by parallel lines measure 70° and (3x + 10)°. Find x.

Step 1. Alternate interior angles are equal: 3x + 10 = 70.

Step 2. 3x = 60, so x = 20.

Check: 3 × 20 + 10 = 70.

Worked example 6. Same-side interior angles measure (2x + 40)° and 3x°. Find both.

Step 1. Same-side interior angles add to 180°: 2x + 40 + 3x = 180, so 5x + 40 = 180.

Step 2. 5x = 140, so x = 28. The angles are 2 × 28 + 40 = 96° and 3 × 28 = 84°.

Check: 96 + 84 = 180.

Worked example 7. Eratosthenes' method with made-up modern numbers: two towns 555 km apart, one due north of the other, see the noon Sun 5° apart on the same day. Estimate the Earth's circumference.

Step 1. 5° is 5/360 = 1/72 of a circle.

Step 2. 72 × 555 km = 39,960 km, about 40,000 km. That is close to the real figure: NASA gives the Earth's equatorial radius as 6,378 km, and 2 × π × 6,378 ≈ 40,075 km.

Common misconceptions

"Corresponding angles are always equal." Only when the two lines crossed by the transversal are parallel. If the lines tilt towards each other, the two crossings are no longer copies, and the angles differ.

"A bigger triangle has a bigger angle sum." Every triangle's angles add to 180°. A huge triangle has longer sides, not bigger angles in total.

"An exterior angle equals the interior angle next to it." The two add to 180°. The exterior angle equals the sum of the two interior angles that are not next to it.

"Vertical angles are the ones that point up and down." "Vertical" here comes from "vertex". Vertical angles are the pairs opposite each other at a crossing, whichever way they face.

"Eratosthenes proved that the Earth is round." According to the APS account, most Greeks already believed it was round by around 500 BCE, and Aristotle had given physical evidence, such as the curved shadow of the Earth on the Moon during an eclipse. Eratosthenes measured how big it is.

The takeaway

• When two lines cross, vertical angles are equal and neighbouring angles add to 180°.
• When a transversal crosses two parallel lines, corresponding angles are equal, alternate interior angles are equal, and same-side interior angles add to 180°, because a slide along the transversal carries one crossing onto the other.
• A line through one corner of a triangle, parallel to the opposite side, shows that the three angles add to 180°.
• An exterior angle of a triangle equals the sum of the two interior angles opposite it.
• Eratosthenes used alternate interior angles and parallel sunbeams to turn a 7.2° shadow into the circumference of the Earth.

Rigid motions keep a shape's size. The next lesson adds a move that changes the size and keeps the shape, which is why a photo can be enlarged but a 4-by-6 print cannot fill an 8-by-10 frame without losing part of the picture.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.A.5). thecorestandards.org
2. American Physical Society. (2006, June 1). This month in physics history: June, ca. 240 B.C.: Eratosthenes measures the Earth. APS News. aps.org
3. Euclid. (c. 300 BCE/1996). Elements, Book I, Proposition 32 (D. E. Joyce, Ed.). Clark University. mathcs.clarku.edu
4. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 14: Alternate interior angles. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 1, Lesson 16: Parallel lines and the angles in a triangle. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
6. Williams, D. R. (2024, November 15). Earth fact sheet (equatorial radius). NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov

A photo and a frame

You have a favourite photo printed at 4 inches by 6 inches, and you want an enlargement to fill a frame that is 8 inches by 10 inches. Can the print shop make one that fills the frame exactly, with nothing cut off and no white strips?

Try it. To make the 4-inch side 8 inches, you must double it, and doubling the 6-inch side gives 12 inches, 2 inches too long for the frame. To make the 6-inch side 10 inches, you must multiply by 10/6 = 5/3, and the 4-inch side becomes 4 × 5/3 ≈ 6.67 inches, too narrow. No single enlargement makes both sides fit. The shop has to crop part of the picture.

The reason is that a true enlargement multiplies every length by the same number, and an 8-by-10 rectangle is not the same shape as a 4-by-6 one. This lesson makes that precise. It covers Common Core standards 8.G.A.3, describing the effect of dilations on figures using coordinates, and 8.G.A.4, understanding that figures are similar when one can be obtained from the other by rigid motions and dilations, and describing a sequence that shows it.

What a dilation does

A dilation enlarges or shrinks a figure from a fixed point called the centre of dilation, by a number called the scale factor, k. Every point moves along the straight line from the centre through it, to k times its original distance from the centre. The centre itself does not move.

• If k is greater than 1, every point moves away from the centre and the figure gets bigger.
• If k is between 0 and 1, every point moves towards the centre and the figure gets smaller.
• If k = 1, nothing moves.

When the centre is the origin, the rule is simple: multiply both coordinates by k.

(x, y) → (kx, ky).

Put the photo on a grid with corners at (0, 0), (6, 0), (6, 4) and (0, 4), and dilate it about the origin by 2. The corners go to (0, 0), (12, 0), (12, 8) and (0, 8): a 12-by-8 rectangle, every length doubled. Dilate by 5/3 instead, and they go to (0, 0), (10, 0), (10, 6.67) and (0, 6.67), approximately: 10 by 6.67. Neither image is 10 by 8.

What a dilation keeps and what it changes

Take the right triangle A(2, 1), B(4, 1), C(2, 3) and dilate it by 3 about the origin: A'(6, 3), B'(12, 3), C'(6, 9).

So a dilation keeps every angle and the shape, and multiplies every length by k. It is not a rigid motion, because lengths change. Areas change even more: multiplying two lengths by 3 multiplies the area by 3 × 3 = 9.

So what?: A dilation is the precise version of "the same shape, a different size". Every length is multiplied by one number, and nothing else about the shape changes.

A centre that is not the origin

The rule (kx, ky) only works when the centre is the origin. For any other centre, work with the move from the centre to the point. Dilate P(4, 3) by 2 about the centre C(1, 1).

Step 1. From C to P is 3 right and 2 up.

Step 2. Multiply that move by the scale factor: 6 right and 4 up.

Step 3. Make the new move from C: (1 + 6, 1 + 4) = (7, 5). So P' = (7, 5).

Check: P' lies on the line from C through P, and it is twice as far from C as P is.

Similar figures

Two figures are similar when one can be carried onto the other by a sequence of translations, reflections, rotations and dilations. Rigid motions move and turn and flip; the dilation sets the size. Because none of these moves changes an angle, and only the dilation changes lengths, similar figures have two properties, as OpenStax's Prealgebra states them: corresponding angles are equal, and corresponding sides all have the same ratio.

That gives a test. Divide each side of one figure by the matching side of the other. If every division gives the same number, that number is the scale factor and the figures are similar, provided the angles match too. If two divisions give different numbers, the figures are not similar.

The photo and the frame: 8 ÷ 4 = 2, but 10 ÷ 6 ≈ 1.67. Two different ratios, so an 8-by-10 rectangle is not similar to a 4-by-6 rectangle. That is why the picture has to be cropped. An 8-by-12 print would work: 8 ÷ 4 = 2 and 12 ÷ 6 = 2.

Congruent figures are similar figures with scale factor 1. Every congruent pair is similar, but most similar pairs are not congruent.

Worked examples

Worked example 1. Dilate the square with corners (4, 4), (8, 4), (8, 8), (4, 8) by 1/2 about the origin.

Step 1. Halve every coordinate: (2, 2), (4, 2), (4, 4), (2, 4).

Step 2. The sides were 4 units long and are now 2 units long, half as long, and the corners are still right angles.

Worked example 2. A dilation about the origin sends P(3, −2) to P'(7.5, −5). Find the scale factor.

Step 1. 7.5 ÷ 3 = 2.5 and −5 ÷ −2 = 2.5.

Step 2. Both coordinates were multiplied by the same number, so k = 2.5.

Worked example 3. Are a 3-by-5 rectangle and a 6-by-10 rectangle similar? What about a 3-by-5 and a 5-by-7?

Step 1. 6 ÷ 3 = 2 and 10 ÷ 5 = 2. One scale factor, and all the angles are right angles, so yes, they are similar, with k = 2.

Step 2. 5 ÷ 3 ≈ 1.67 but 7 ÷ 5 = 1.4. The ratios differ, so the second pair is not similar, even though each rectangle is 2 units longer one way.

Worked example 4. Triangle ABC has A(0, 0), B(2, 0), C(0, 1). Triangle DEF has D(1, 1), E(1, 5), F(−1, 1). Show they are similar by finding a sequence of moves.

Step 1. Compare lengths. AB = 2 and DE = 4 (from y = 1 to y = 5); AC = 1 and DF = 2 (from x = 1 to x = −1). Both ratios are 2, so try a dilation by 2.

Step 2. Dilate by 2 about the origin: A(0, 0), B(4, 0), C(0, 2).

Step 3. AB now runs right but DE runs up, so rotate 90° counterclockwise about the origin: (x, y) → (−y, x). A stays at (0, 0), B goes to (0, 4) and C goes to (−2, 0).

Step 4. Translate 1 right and 1 up: (0, 0) → (1, 1) = D, (0, 4) → (1, 5) = E, (−2, 0) → (−1, 1) = F.

A dilation, a rotation and a translation carry ABC onto DEF, so the triangles are similar with scale factor 2.

Worked example 5. A video frame 1,920 pixels wide and 1,080 tall is scaled up to 3,840 by 2,160. Is the new frame a dilation of the old one?

Step 1. 3,840 ÷ 1,920 = 2 and 2,160 ÷ 1,080 = 2.

Step 2. One scale factor, so yes: a dilation by 2. The picture keeps its shape, and it has 2 × 2 = 4 times as many pixels.

Worked example 6. Dilate the segment from Q(3, 5) to R(5, 1) by 1/2 about the centre C(1, 1).

Step 1. From C to Q is 2 right, 4 up; half of that is 1 right, 2 up, so Q' = (2, 3).

Step 2. From C to R is 4 right, 0 up; half of that is 2 right, so R' = (3, 1).

Check: QR goes 2 right and 4 down; Q'R' goes 1 right and 2 down. The image is parallel to the original and half as long.

Common misconceptions

"A dilation by 2 doubles the area." It doubles every length, so a rectangle's width and height both double and its area becomes 2 × 2 = 4 times as big. The 1,920-by-1,080 frame scaled by 2 has four times the pixels.

"Dilations change the angles." Every angle stays the same size. That is exactly what keeps the shape.

"The centre of a dilation must be the origin, or the middle of the shape." It can be any point. The rule (kx, ky) is only the shortcut for the origin; for other centres, scale the move from the centre to each point.

"Any two rectangles are similar, because they all have four right angles." Equal angles are not enough for rectangles. The sides must also share one scale factor, and the 4-by-6 and 8-by-10 rectangles do not.

"Similar means the same as congruent." Similar figures have the same shape and possibly different sizes. Congruent figures have the same shape and the same size.

What to remember

• A dilation with centre C and scale factor k moves every point along the line from C to k times its distance from C. About the origin, (x, y) → (kx, ky).
• k greater than 1 enlarges, k between 0 and 1 shrinks. Angles stay the same, lengths are multiplied by k, and areas by k × k.
• Two figures are similar when a sequence of rigid motions and dilations carries one onto the other. Then corresponding angles are equal and corresponding sides share one ratio.
• To test for similarity, divide matching sides. One ratio throughout, with matching angles, means similar; two different ratios means not similar.
• A 4-by-6 photo is not similar to an 8-by-10 frame, because 8 ÷ 4 = 2 but 10 ÷ 6 ≈ 1.67.

For triangles there is a shortcut: you do not need to check the sides at all. The next lesson starts from a tempting mistake about "growing" a triangle and ends with a way to measure a flagpole from its shadow.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.A.3, 8.G.A.4). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 4: Dilations on a square grid. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 6: Similarity. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 7: Similar polygons. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 9.3, Use properties of angles, triangles, and the Pythagorean theorem (similar triangles). Prealgebra 2e. OpenStax. openstax.org

A triangle "grown" by 2

A student wants a bigger copy of a right triangle drawn on a grid. The original has a horizontal leg of 2 units and a vertical leg of 3 units, with corners at (0, 0), (2, 0) and (2, 3). The student's plan: add 2 units to each leg. The new triangle has legs of 4 and 5, with corners at (0, 0), (4, 0) and (4, 5). Bigger, and still a right triangle. Is it the same shape?

It looks plausible, and it is wrong. This lesson finds exactly where the plan fails, fixes it, and then gives a shortcut that makes checking triangles much quicker: two angles are enough. It finishes with two ways to measure the height of something you cannot climb.

This lesson covers the last part of Common Core standard 8.G.A.5, using informal arguments to establish the angle-angle criterion for similar triangles, and continues 8.G.A.4 on similarity.

Where "add 2 to each side" breaks

Compare how steep the two long sides are. In the original, the slanted side climbs 3 units while it crosses 2, which is 3/2 = 1.5 units up for each unit across. In the student's triangle it climbs 5 while it crosses 4, which is 5/4 = 1.25 up for each unit across. The new slanted side is less steep, so the angle it makes with the bottom leg is smaller. The angles changed, and a triangle with different angles is a different shape.

The ratios of the sides say the same thing. For similar triangles, every side must be multiplied by the same scale factor. Here 4 ÷ 2 = 2 but 5 ÷ 3 ≈ 1.67. Two different factors, so the triangles are not similar.

The fix is to multiply instead of add. Doubling both legs gives 4 and 6, with corners at (0, 0), (4, 0) and (4, 6). The slanted side now climbs 6 for every 4 across, which is 6/4 = 1.5, the same steepness as the original. Same angles, every side doubled: a similar triangle.

Key idea: Shape is about ratios, not differences. Adding the same length to every side changes the ratios between the sides, and so changes the angles. Multiplying every side by the same number keeps the ratios and the angles.

The angle-angle test

Checking all three ratios works, but triangles allow a shortcut. Suppose two triangles have two pairs of equal angles, say 30° and 45° in each. Then their third angles are equal too, because each third angle is 180° − 30° − 45° = 105°. Illustrative Mathematics' lesson on similar triangles shows why two angles are then enough.

Call the triangles ABC and DEF, with the 30° angles at A and D and the 45° angles at B and E.

Step 1. Dilate triangle ABC by the scale factor that makes side AB exactly as long as DE. Dilations do not change angles, so the new triangle still has 30° and 45° at the ends of that side.

Step 2. Move the new triangle with rigid motions so that its side lands on DE, with the 30° corner on D and the 45° corner on E, on the same side as F.

Step 3. From D, the side making 30° with DE points in exactly one direction, and from E the side making 45° points in exactly one direction. Two straight lines cross at only one point, so the third corner must land exactly on F.

So a dilation followed by rigid motions carries ABC onto DEF, and the triangles are similar. This is the angle-angle criterion: if two angles of one triangle are equal to two angles of another, the triangles are similar.

The test is special to triangles. A square and a long thin rectangle both have four right angles, so all their angles match, but they are not similar. For shapes with more sides, angles are not enough; you have to check the sides as well.

Measuring a tree with its shadow

On a sunny day, sunlight arrives in parallel rays, the same fact Eratosthenes relied on. So at a given moment, the Sun's rays make the same angle with flat ground everywhere nearby. Illustrative Mathematics' lesson "The Shadow Knows" sends students outside to use this.

A girl 1.5 m tall casts a shadow 2 m long. At the same moment, a tree casts a shadow 12 m long. How tall is the tree?

Step 1. Each object, its shadow and the sunbeam past its top make a triangle. Each has a right angle where the upright object meets flat ground, and each has the same angle where the sunbeam meets the ground. Two equal angles: by the angle-angle test, the triangles are similar.

Step 2. The tree's shadow is 12 ÷ 2 = 6 times as long as the girl's, so the scale factor is 6.

Step 3. The tree is 6 × 1.5 = 9 m tall.

Check with a ratio inside each triangle: height ÷ shadow is 1.5 ÷ 2 = 0.75 for the girl and 9 ÷ 12 = 0.75 for the tree. In similar triangles, the ratio of two sides within one triangle equals the ratio of the matching sides within the other.

Measuring a tree with a mirror

No sun? Use a small mirror lying flat on the ground. OpenStax's College Physics states the law of reflection: the angle at which light leaves a mirror equals the angle at which it arrives. Stand so that you can just see the top of the tree in the mirror. The light from the treetop comes down to the mirror and bounces up to your eye at the same angle to the ground.

Your eyes are 1.6 m above the ground, you stand 2 m from the mirror, and the mirror is 15 m from the foot of the tree.

Step 1. One triangle is your eye, your feet and the mirror. The other is the treetop, the foot of the tree and the mirror. Each has a right angle at the ground, and each has the same angle at the mirror. By angle-angle, they are similar.

Step 2. The tree's distance is 15 ÷ 2 = 7.5 times yours, so the scale factor is 7.5.

Step 3. The tree is 7.5 × 1.6 = 12 m tall.

Worked examples

Worked example 1. One triangle has angles of 50° and 60°. Another has angles of 50° and 70°. Are they similar?

Step 1. The first triangle's third angle is 180 − 50 − 60 = 70°, so its angles are 50°, 60° and 70°.

Step 2. The second's third angle is 180 − 50 − 70 = 60°, so its angles are also 50°, 60° and 70°.

Step 3. Two pairs of equal angles, so yes, they are similar.

Worked example 2. One triangle has angles of 40° and 60°; another has 40° and 70°. Are they similar?

The first has angles 40°, 60°, 80°; the second has 40°, 70°, 70°. Only one pair of angles matches, so they are not similar.

Worked example 3. Triangle ABC has sides 5, 7 and 9. A similar triangle has its shortest side 15. Find its other sides.

Step 1. The shortest sides correspond, so the scale factor is 15 ÷ 5 = 3.

Step 2. The other sides are 3 × 7 = 21 and 3 × 9 = 27.

Worked example 4. In triangle ABC, point D is on side AB and point E is on side AC, and DE is parallel to BC. AD = 4, AB = 10 and DE = 6. Find BC.

Step 1. Triangles ADE and ABC share the angle at A. Because DE is parallel to BC, side AB is a transversal, so the angles at D and B are corresponding angles and are equal. Two equal angles: the triangles are similar.

Step 2. The scale factor from the small triangle to the large one is AB ÷ AD = 10 ÷ 4 = 2.5.

Step 3. BC = 2.5 × 6 = 15.

Worked example 5. A 2 m pole casts a 3.2 m shadow. At the same moment a building casts a 40 m shadow. How tall is the building?

Step 1. Height ÷ shadow is the same in both triangles: 2 ÷ 3.2 = 0.625.

Step 2. The building is 0.625 × 40 = 25 m tall.

Check with the scale factor: 40 ÷ 3.2 = 12.5, and 12.5 × 2 = 25.

Common misconceptions

"Adding the same amount to every side gives a similar triangle." It changes the ratios between the sides, so it changes the angles. Legs of 2 and 3 grown to 4 and 5 make a less steep triangle. Multiply, do not add.

"You need to check all three angles." Two are enough, because the third is then forced: the three must add to 180°.

"One equal angle is enough." Many different shapes of triangle have a 40° angle. You need two pairs.

"Equal angles make any two shapes similar." Only for triangles. A square and a 1-by-5 rectangle have all their angles equal and are not similar.

"Shadows measured at different times still work." The Sun's angle changes during the day, so the shadow method needs both shadows measured at the same moment.

Summing up

• Similar shapes come from multiplying every length by one scale factor. Adding the same length to every side changes the ratios and the angles.
• Angle-angle criterion: if two angles of one triangle equal two angles of another, the triangles are similar, because a dilation and rigid motions can carry one onto the other.
• In similar triangles, matching sides share one scale factor, and the ratio of two sides within one triangle equals the ratio of the matching sides in the other.
• Shadows at the same moment, or a flat mirror and the law of reflection, make similar right triangles, so you can find a height from lengths on the ground.
• Angles alone decide similarity only for triangles.

The steepness of a slanted side, climb divided by distance across, kept appearing in this lesson. It has a name, slope, and the next module is built around it, starting with the fastest 100 metres ever run.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.A.4, 8.G.A.5). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 8: Similar triangles. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 13: The shadow knows. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 9.3, Use properties of angles, triangles, and the Pythagorean theorem (similar triangles). Prealgebra 2e. OpenStax. openstax.org
5. Urone, P. P., and Hinrichs, R. (2022). Section 25.2, The law of reflection. College Physics 2e. OpenStax. openstax.org

Ten metres every second

On 16 August 2009, at the Olympic Stadium in Berlin, Usain Bolt ran 100 metres in 9.58 seconds. World Athletics still lists it as the fastest time ever run. His average speed was 100 ÷ 9.58 ≈ 10.44 metres per second.

Imagine a runner who kept up exactly that speed from the first instant. After 1 second they would have covered 10.44 m, after 2 seconds 20.88 m, and after t seconds 10.44t metres. Distance would be proportional to time, d = 10.44t, and its graph would be a straight line through (0, 0). That line has a steepness, and its steepness is the speed. This lesson is about that idea: in a proportional relationship, the unit rate is the slope of the graph.

Real sprinters do not run at a steady speed. Bolt took a moment to accelerate out of the blocks, so his real distance-time graph curves at the start before it straightens out. The steady runner is a model, a simpler version that has the same average.

This lesson covers Common Core standard 8.EE.B.5: graphing proportional relationships, interpreting the unit rate as the slope of the graph, and comparing two proportional relationships represented in different ways, such as a distance-time graph and a distance-time equation.

From seventh grade: y = kx

In seventh grade you wrote proportional relationships as y = kx, where k is the constant of proportionality, and you saw that the graph is a straight line through (0, 0) that passes through the point (1, k). The constant k is the unit rate: how much y there is for each 1 unit of x. For the steady runner, k = 10.44 metres per second.

The new idea in eighth grade is a name for how steep the line is. The slope of a line is the change in y divided by the change in x as you move from one point on the line to another, often said as "rise over run". For a proportional relationship, start at (0, 0) and move to (1, k): the rise is k and the run is 1, so the slope is k ÷ 1 = k. The slope of the graph is the unit rate.

Check with two other points on the runner's line, (2, 20.88) and (5, 52.2). The rise is 52.2 − 20.88 = 31.32 and the run is 5 − 2 = 3, and 31.32 ÷ 3 = 10.44. The same slope.

The point: In a proportional relationship y = kx, three things are the same number: the constant of proportionality, the unit rate, and the slope of the graph.

Four cyclists, four descriptions

Four friends ride bikes at steady speeds, and each describes the ride a different way. Distances are in kilometres and times in minutes.

Read the table across a row and you see how to get one number out of each kind of description. Read the last column down and the comparison is easy: Cai is fastest at 0.5 km per minute, Ana is next at 0.4, and Ben and Dee tie at 0.35, even though Ben gave an equation and Dee gave a sentence.

On one set of axes, with time across and distance up, all four rides are straight lines starting at (0, 0). Cai's line is the steepest, because it rises 0.5 for each minute across. Ben's and Dee's are the same line.

The point: To compare proportional relationships given in different forms, turn each into its unit rate. The largest unit rate is the steepest line.

What the slope's units tell you

A slope always carries units: the units of y per unit of x. Cai's slope is 0.5 kilometres per minute. That is the same speed as 30 kilometres per hour, because 0.5 × 60 = 30. If you change the units, the number changes, but the speed does not. When you compare rates, make sure they use the same units first: Dee's "21 km in an hour" could not be compared with the others until it became 0.35 km per minute.

The same reasoning works for prices. Two data plans charge only for what you use, with no monthly fee. Plan X is shown as a graph through (0, 0) and (4, 30), with gigabytes across and dollars up. Plan Y is the equation c = 8g. Plan X's slope is 30 ÷ 4 = 7.50 dollars per gigabyte, and Plan Y's is 8 dollars per gigabyte. Plan X charges less for each gigabyte, so its line is less steep, and for any amount of data it costs less.

Worked examples

Worked example 1. A recipe uses 3 cups of flour for every 2 cups of milk. Write the relationship as an equation and find the slope of its graph, with milk across and flour up.

Step 1. Unit rate: 3 ÷ 2 = 1.5 cups of flour per cup of milk.

Step 2. The equation is f = 1.5m, and the slope is 1.5.

Check: with 4 cups of milk, f = 1.5 × 4 = 6 cups of flour, and 6 ÷ 4 = 1.5.

Worked example 2. A graph of a proportional relationship passes through (6, 15). Find the slope, the equation, and the value of y when x = 10.

Step 1. The line also passes through (0, 0), so the slope is 15 ÷ 6 = 2.5.

Step 2. The equation is y = 2.5x, and when x = 10, y = 25.

Worked example 3. Runner P covers 300 m in 60 seconds. Runner Q's distance is d = 4.5t, in metres and seconds. Who is faster, and by how much?

Step 1. P's unit rate: 300 ÷ 60 = 5 m per second. Q's unit rate: 4.5 m per second.

Step 2. P is faster, by 0.5 m every second.

Worked example 4. Is the relationship in this table proportional? x: 2, 4, 6; y: 7, 14, 22.

Step 1. Divide each y by its x: 7 ÷ 2 = 3.5, 14 ÷ 4 = 3.5, 22 ÷ 6 ≈ 3.67.

Step 2. The ratios are not all the same, so the relationship is not proportional, and its points do not lie on one line through (0, 0).

Worked example 5. At Bolt's average speed of 10.44 m/s, how long would a steady runner take to cover 200 m?

Step 1. d = 10.44t, so 200 = 10.44t.

Step 2. t = 200 ÷ 10.44 ≈ 19.16 seconds.

Check: 10.44 × 19.16 ≈ 200. (World Athletics lists Bolt's 200 m record as 19.19 seconds, close to the steady-runner model.)

Common misconceptions

"Slope is how high the line is." Slope is how steep the line is: how much it rises for each unit across. A line can be high on the page and have a small slope.

"The steeper-looking line is always the faster one." Only when both lines are drawn on the same axes with the same scales. On graphs with different scales, compare the numbers, not the pictures.

"In d = 0.35t, the 0.35 is a distance." It is the rate: 0.35 kilometres for every minute. The distance is d, and it depends on t.

"Every straight-line graph shows a proportional relationship." A proportional relationship's line must pass through (0, 0). A straight line that starts somewhere else is linear but not proportional, as the next two lessons show.

"Slope is run over rise." It is rise over run: the change in y divided by the change in x. Cai's slope is 5 ÷ 10 = 0.5, not 10 ÷ 5 = 2.

What to carry forward

• A proportional relationship y = kx has a straight-line graph through (0, 0) and (1, k).
• Its slope, rise over run, equals its unit rate and its constant of proportionality k, and it carries units of y per unit of x.
• You can find the unit rate from a table (divide y by x in any row), an equation (the number multiplying x), a graph (rise over run, or y ÷ x at any point), or words (rewrite as an amount per one unit).
• To compare relationships in different forms, convert each to a unit rate in the same units. The greater unit rate is the steeper line.

Every slope in this lesson was measured from (0, 0). The next lesson measures slope between any two points, including lines that go downhill or start above zero, and checks whether a wheelchair ramp is legal.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.B.5). thecorestandards.org
2. World Athletics. (n.d.). 100 metres men, senior, all-time top list. worldathletics.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 3, Lesson 2: Graphs of proportional relationships. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 3, Lesson 4: Comparing proportional relationships. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 4.4, Understand slope of a line. Elementary Algebra 2e. OpenStax. openstax.org

Two feet up to the door

A school's side door sits 2 feet, which is 24 inches, above the path outside. The caretaker sketches a ramp that starts 20 feet, or 240 inches, out from the door and climbs steadily to the doorstep. Is it legal?

The U.S. Access Board's guide to the ADA Accessibility Standards, the rules for ramps that wheelchair users can manage, gives the answer in one ratio. Section 405.2 sets the maximum running slope of a ramp at 1:12, and the guide explains that slope means the proportion of vertical rise to horizontal length. A ramp may rise at most 1 inch for every 12 inches it runs across, which the guide also writes as 8.33 percent.

The caretaker's ramp rises 24 inches over 240 inches, a slope of 24 ÷ 240 = 1/10 = 0.1. The limit is 1 ÷ 12 ≈ 0.0833. 0.1 is bigger, so the ramp is too steep. To rise 24 inches at no more than 1 in 12, the ramp must run at least 24 × 12 = 288 inches, which is 24 feet.

That calculation is the whole idea of slope: rise divided by run. This lesson measures it between any two points, for lines that go up, down, flat and straight up, and explains why every pair of points on a line gives the same answer. It covers Common Core standard 8.EE.B.6: using similar triangles to explain why the slope is the same between any two distinct points on a non-vertical line.

Rise over run from any two points

To find the slope between two points, subtract their y-coordinates to get the rise, subtract their x-coordinates in the same order to get the run, and divide. Labelling the points (x₁, y₁) and (x₂, y₂), the small numbers are labels, not exponents: x₁ is read "x one" and means the x-coordinate of the first point. Then

slope = (y₂ − y₁) ÷ (x₂ − x₁).

Take (1, 2) and (4, 8). The rise is 8 − 2 = 6 and the run is 4 − 1 = 3, so the slope is 6 ÷ 3 = 2. Starting from the other point gives (2 − 8) ÷ (1 − 4) = −6 ÷ −3 = 2, the same. What matters is that the top and the bottom are subtracted in the same order.

Uphill, downhill, flat and straight up

The sign tells you the direction and the size tells you the steepness. A slope of −2 is just as steep as a slope of 2, falling 2 for every 1 across instead of rising 2. A slope of 1/12 is gentle, and a slope of 5 is steep. To compare steepness, compare the sizes and ignore the signs.

Why this matters: Slope turns "how steep?" into one number you can compare with a rule. A ramp, a road or a line on a graph is too steep exactly when its slope is bigger than the limit.

Why every pair of points gives the same slope

On the line through (1, 2) and (4, 8), pick a different pair of points, say (2, 4) and (3, 6). The rise is 2 and the run is 1, and the slope is again 2. Is that luck? Similar triangles say no.

Between any two points on a line, draw the slope triangle: go straight across for the run, then straight up for the rise, so that the line itself is the long side. From (1, 2) to (4, 8) the triangle has a run of 3 and a rise of 6; from (2, 4) to (3, 6) it has a run of 1 and a rise of 2.

Step 1. Both triangles have a right angle where the run meets the rise.

Step 2. Both triangles have an angle where the line meets the horizontal run. The two runs are horizontal, so they are parallel, and the line is a transversal crossing them. Those two angles are corresponding angles, so they are equal.

Step 3. Two pairs of equal angles: by the angle-angle test from Lesson 12, the slope triangles are similar.

Step 4. In similar triangles, the ratio of two sides within one triangle equals the ratio of the matching sides within the other. So rise ÷ run is the same for both triangles.

The same argument works for any two slope triangles on any non-vertical line, so the slope is the same no matter which two points you use. That is why it makes sense to talk about "the slope of a line", as one number.

Worked examples

Worked example 1. Find the slope between (−3, 4) and (5, −2).

Step 1. Rise: −2 − 4 = −6. Run: 5 − (−3) = 8.

Step 2. Slope = −6 ÷ 8 = −3/4. The line falls 3 units for every 4 across.

Worked example 2. A ramp must rise 30 inches. Under the 1:12 limit, what is the shortest horizontal run it can have?

Step 1. At the steepest legal slope, every 1 inch of rise needs 12 inches of run.

Step 2. 30 × 12 = 360 inches, which is 30 feet.

Check: 30 ÷ 360 = 1/12, exactly the limit.

Worked example 3. A skateboard ramp rises 3 feet over a run of 8 feet. Another rises 2 feet over 5 feet. Which is steeper?

Step 1. Slopes: 3 ÷ 8 = 0.375 and 2 ÷ 5 = 0.4.

Step 2. 0.4 is larger, so the second ramp is steeper, even though the first one is taller.

Worked example 4. A table gives points on a line: x = 0, 3, 6 with y = 10, 8, 6. Find the slope.

Step 1. From the first row to the second, y changes by 8 − 10 = −2 while x changes by 3.

Step 2. Slope = −2 ÷ 3 = −2/3. Check with the second and third rows: (6 − 8) ÷ (6 − 3) = −2/3.

Worked example 5. A line passes through (2, 1) with slope 3. Find two more points on it.

Step 1. A slope of 3 means up 3 for every 1 across. From (2, 1), go 1 right and 3 up: (3, 4).

Step 2. Go 1 left and 3 down instead: (1, −2).

Check: (4 − 1) ÷ (3 − 2) = 3 and (−2 − 1) ÷ (1 − 2) = 3.

Worked example 6. A path climbs 1 m for every 20 m across. Write its slope as a fraction, a decimal and a percent, and decide whether a ramp that steep would meet the 1:12 limit.

Step 1. 1/20 = 0.05 = 5 percent.

Step 2. 1/12 ≈ 0.0833 = 8.33 percent. 5 percent is less, so it is within the limit.

Common misconceptions

"Slope is run over rise." Slope is rise over run. A ramp rising 2 over a run of 24 has slope 2/24 = 1/12, not 12.

"Subtract in whichever order is easier." The order can be either way, but it must be the same on the top and the bottom. (8 − 2) ÷ (1 − 4) = 6 ÷ −3 = −2 has the wrong sign.

"A vertical line has slope 0." A horizontal line has slope 0, because its rise is 0. A vertical line has a run of 0, and dividing by 0 is impossible, so its slope is undefined.

"A ramp's slope is its rise divided by the length of its surface." The ADA standard uses the horizontal run, the distance across the ground. The sloping surface is a little longer than the run; Lesson 26 shows how to work out exactly how long.

"A negative slope means the line is below the x-axis." It means the line falls as you move to the right. A line with negative slope can be entirely above the x-axis for the part you see, like a slide that ends at a platform.

The takeaway

• Slope = rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁), subtracting in the same order top and bottom.
• Positive slopes rise to the right, negative slopes fall, horizontal lines have slope 0, and vertical lines have no defined slope.
• Slope triangles on the same line are similar by angle-angle, so rise ÷ run is the same between any two points on a non-vertical line.
• Slopes can be written as ratios, decimals or percents. The ADA limit for a ramp's running slope is 1:12, about 8.33 percent: at most 1 inch of rise for every 12 inches of horizontal run.

Knowing a line's slope and one point fixes the whole line. The next lesson turns that into an equation, y = mx + b, and shows how to write it from words, a table, a graph or two points.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.B.6). thecorestandards.org
2. U.S. Access Board. (n.d.). Chapter 4: Ramps and curb ramps (section 405.2, slope). Guide to the ADA Accessibility Standards. access-board.gov
3. Illustrative Mathematics. (2019). Grade 8, Unit 2, Lesson 10: Meet slope. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 3, Lesson 10: Calculating slope. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 4.4, Understand slope of a line. Elementary Algebra 2e. OpenStax. openstax.org

Four dollars before you bowl

A bowling alley charges $4 to rent shoes and $6 for each game. One game costs $10, two games $16, three games $22. The cost goes up by the same $6 for every extra game, so the points (games, cost) lie on a straight line with slope 6. But this is not a proportional relationship: one game costs $10, so cost ÷ games is 10, while two games give 16 ÷ 2 = 8. The line does not pass through (0, 0). With zero games you still pay $4 for the shoes, so the line starts at (0, 4).

The cost is c = 6g + 4: six dollars times the number of games, plus four. Every line that is not vertical has an equation of this shape, y = mx + b. This lesson shows why, and then works through writing the equation from every kind of description you are likely to meet. It covers the second half of Common Core standard 8.EE.B.6: deriving the equation y = mx for a line through the origin and y = mx + b for a line crossing the vertical axis at b.

Deriving y = mx + b, two ways

By sliding a line. Start with the line y = mx through the origin, with slope m. Its points are (x, mx). Translate the whole line b units up. Every point (x, mx) moves to (x, mx + b), so the new line's points satisfy y = mx + b. A translation keeps the line straight and parallel to where it started, so the slope is still m, and the point (0, 0) moves to (0, b). For the bowling alley, slide c = 6g up 4 units and you get c = 6g + 4.

By using slope. Suppose a line has slope m and crosses the y-axis at (0, b). Pick any other point (x, y) on the line. The slope between (0, b) and (x, y) must be m, because slope is the same between any two points on a line:

(y − b) ÷ (x − 0) = m.

Multiply both sides by x: y − b = mx. Add b to both sides: y = mx + b. Every point on the line satisfies this equation.

In y = mx + b, the number m is the slope, and b is the y-intercept, the y-coordinate where the line crosses the y-axis. This way of writing a line is called slope-intercept form. A proportional relationship is the special case b = 0.

Remember: In y = mx + b, m is how much y changes for each 1 that x increases, and b is the value of y when x is 0.

Writing the equation from a table

Table A: x = 0, 1, 2, 3 and y = 5, 8, 11, 14.

Step 1. Each time x goes up 1, y goes up 3, so m = 3.

Step 2. When x = 0, y = 5, so b = 5.

Step 3. y = 3x + 5. Check the last row: 3 × 3 + 5 = 14.

Now a table where a changed input flips the shortcut. Table B: x = 2, 4, 6 and y = 7, 11, 15. Here there is no row with x = 0, so the first y-value is not b.

Step 1. When x goes up 2, y goes up 4, so m = 4 ÷ 2 = 2.

Step 2. Work back to x = 0: going from x = 2 to x = 0 is 2 steps of −1, so y drops by 2 × 2 = 4, from 7 to 3. So b = 3.

Step 3. y = 2x + 3. Check: 2 × 6 + 3 = 15.

Reading b = 7 from the first row would give y = 2x + 7, and the check fails at once: 2 × 2 + 7 = 11, not 7.

Writing the equation from a graph

A line passes through (0, −2) and (3, 4).

Step 1. It crosses the y-axis at (0, −2), so b = −2.

Step 2. From (0, −2) to (3, 4), the rise is 6 and the run is 3, so m = 2.

Step 3. y = 2x − 2. The minus sign belongs to b: b = −2, not 2.

Writing the equation from two points

A line passes through (2, 9) and (6, 1).

Step 1. Slope: (1 − 9) ÷ (6 − 2) = −8 ÷ 4 = −2.

Step 2. The equation looks like y = −2x + b. Put in either point to find b. Using (2, 9): 9 = −2 × 2 + b, so 9 = −4 + b, and b = 13.

Step 3. y = −2x + 13.

Check with the other point: −2 × 6 + 13 = −12 + 13 = 1.

The same method works from a slope and one point. A line with slope 1/2 through (4, 1): 1 = (1/2) × 4 + b = 2 + b, so b = −1 and y = (1/2)x − 1.

Graphing from the equation

To graph y = −3x + 6, start at the y-intercept, (0, 6). The slope −3 means down 3 for every 1 to the right, so step to (1, 3) and then (2, 0). Join the points with a straight line and extend it both ways. Check one point: −3 × 2 + 6 = 0.

Two special lines do not need the full form. A horizontal line through (0, 4) has slope 0, so its equation is y = 0x + 4, which is just y = 4. A vertical line through (3, 0) has undefined slope and cannot be written as y = mx + b at all; every point on it has x = 3, so its equation is x = 3.

What m and b mean in a situation

The slope is a rate: something per something. The intercept is a starting amount: the value when the input is 0. Reading them this way is the skill the rest of the course builds on.

Worked examples

Worked example 1. A gym charges a $25 joining fee and $18 a month. Write the cost c after m months, and find the cost for a year.

Step 1. The rate is $18 per month and the starting amount is $25: c = 18m + 25.

Step 2. For 12 months: c = 18 × 12 + 25 = 216 + 25 = $241.

Worked example 2. Write the equation of the line through (−1, 7) and (3, −1).

Step 1. m = (−1 − 7) ÷ (3 − (−1)) = −8 ÷ 4 = −2.

Step 2. Using (3, −1): −1 = −2 × 3 + b = −6 + b, so b = 5.

Step 3. y = −2x + 5. Check (−1, 7): −2 × (−1) + 5 = 2 + 5 = 7.

Worked example 3. Does the point (4, 10) lie on y = 3x − 2?

Step 1. Put in x = 4: 3 × 4 − 2 = 10.

Step 2. That matches y = 10, so yes. (5, 12) would not: 3 × 5 − 2 = 13, not 12.

Worked example 4. A phone battery reads 100 percent at 8 a.m. and 64 percent at 11 a.m., dropping steadily. Write an equation and say when it will reach 0.

Step 1. Let t be hours after 8 a.m. The points are (0, 100) and (3, 64), so m = (64 − 100) ÷ 3 = −12 and b = 100: B = −12t + 100.

Step 2. B = 0 when 12t = 100, so t = 100 ÷ 12 ≈ 8.3 hours, a little after 4:20 p.m.

Worked example 5. Rewrite y = 5 + 3x and 2y = 4x + 6 in slope-intercept form.

Step 1. y = 5 + 3x is y = 3x + 5: slope 3, intercept 5. The order of the terms does not matter.

Step 2. Divide 2y = 4x + 6 by 2: y = 2x + 3. Slope 2, intercept 3.

Common misconceptions

"In y = 3x + 5, the slope is 5." The slope is the number multiplying x, here 3. The 5 is the y-intercept, the value of y when x = 0.

"Every line goes through the origin." Only lines with b = 0 do. The bowling line starts at (0, 4), because you pay for shoes before you bowl.

"b is where the line crosses the x-axis." b is where it crosses the y-axis. The line y = 2x − 2 crosses the y-axis at −2 and the x-axis at 1.

"The first y-value in a table is always b." Only if that row has x = 0. Otherwise, use the slope to work back to x = 0, as with Table B.

"y = 2x − 2 has y-intercept 2." The intercept carries its sign: y = 2x + (−2), so b = −2.

Looking back

• Every non-vertical line has an equation y = mx + b, where m is the slope and b is the y-intercept. Translating y = mx up by b, or using the slope from (0, b) to any point, both lead to it.
• From words: m is the rate per unit, b is the starting amount. From a table: m is the change in y per 1 in x, and b is y at x = 0. From a graph: read b where the line crosses the y-axis and m as rise over run.
• From two points: find m first, then put one point into y = mx + b to find b, and check with the other point.
• To graph y = mx + b, plot (0, b) and step by the slope. Horizontal lines are y = b; vertical lines are x = a.

Once you can write two prices as lines, you can ask when they are equal. Two phone plans with different fees and different rates cost the same for exactly one amount of data, and finding it means solving an equation with the variable on both sides. That is the next module.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.B.6). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 3, Lesson 8: Translating to y = mx + b. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 3, Lesson 11: Equations of all kinds of lines. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 4.5, Use the slope-intercept form of an equation of a line. Elementary Algebra 2e. OpenStax. openstax.org
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 4.6, Find the equation of a line. Elementary Algebra 2e. OpenStax. openstax.org

Two plans, one price

Two phone plans are on offer. Plan A costs $40 a month plus $5 for each gigabyte of data. Plan B costs $25 a month plus $8 for each gigabyte. Plan B is cheaper if you hardly use data, and Plan A is cheaper if you use a lot. Somewhere in between, they cost exactly the same. Where?

The slow way is a table, gigabyte by gigabyte.

At 5 GB both plans cost $65. Read the bottom row and you can see why: Plan A starts $15 more expensive, but every gigabyte closes the gap by $8 − $5 = $3, and $15 ÷ $3 = 5 gigabytes. After that, Plan B is the dearer one.

The table works, but only because the answer was a small whole number. If the plans cost the same at 4.7 GB, you would never find it this way. The equation does it directly. The plans cost the same when

40 + 5g = 25 + 8g.

This lesson is about solving equations like that, with the variable on both sides. It covers Common Core standard 8.EE.C.7b: solving linear equations with rational number coefficients, including equations whose solutions require expanding expressions with the distributive property and collecting like terms.

Balancing moves

An equation is a balance: the two sides are equal. You may do anything to one side as long as you do exactly the same to the other, and the balance holds. Seventh grade used this for equations like 3x + 4 = 19. With the variable on both sides there is one extra first move: gather the variable terms on one side.

Step 1. Subtract 5g from both sides, so the g-terms are only on the right: 40 = 25 + 3g.

Step 2. Subtract 25 from both sides: 15 = 3g.

Step 3. Divide both sides by 3: g = 5.

Check in the original equation: 40 + 5 × 5 = 65 and 25 + 8 × 5 = 65. Both sides match, so g = 5 is right, and the answer to the question is 5 gigabytes, at which both plans cost $65.

Look at Step 1 and Step 2 again. Subtracting 5g from 8g left 3g, the $3 by which each gigabyte closes the gap. Subtracting 25 from 40 left 15, the starting gap. The algebra did exactly what the table's bottom row did, in three lines.

In short: To solve an equation with the variable on both sides, move all the variable terms to one side and all the plain numbers to the other, using the same operation on both sides each time, then divide by the coefficient.

Methods side by side

Worked examples

Worked example 1. Solve −3x + 7 = 2x − 8.

Step 1. Add 3x to both sides, which keeps the x-term positive: 7 = 5x − 8.

Step 2. Add 8 to both sides: 15 = 5x.

Step 3. Divide by 5: x = 3.

Check: −3 × 3 + 7 = −2 and 2 × 3 − 8 = −2.

Worked example 2. Solve 3(x − 4) = 2(x + 1).

Step 1. Expand both sides: 3x − 12 = 2x + 2.

Step 2. Subtract 2x: x − 12 = 2.

Step 3. Add 12: x = 14.

Check: 3 × (14 − 4) = 30 and 2 × (14 + 1) = 30.

Worked example 3. Solve 5x − 2 + x = 3x + 13.

Step 1. Collect like terms on the left: 5x + x = 6x, so 6x − 2 = 3x + 13.

Step 2. Subtract 3x: 3x − 2 = 13.

Step 3. Add 2: 3x = 15, so x = 5.

Check: 5 × 5 − 2 + 5 = 28 and 3 × 5 + 13 = 28.

Worked example 4. Solve (2/3)x + 4 = (1/2)x + 6.

Step 1. The denominators are 3 and 2, so multiply every term on both sides by 6: 6 × (2/3)x = 4x, 6 × 4 = 24, 6 × (1/2)x = 3x and 6 × 6 = 36. The equation becomes 4x + 24 = 3x + 36.

Step 2. Subtract 3x: x + 24 = 36.

Step 3. Subtract 24: x = 12.

Check: (2/3) × 12 + 4 = 8 + 4 = 12 and (1/2) × 12 + 6 = 6 + 6 = 12.

Worked example 5. Solve 0.2x + 1.5 = 0.05x + 3.

Step 1. Multiply every term by 100: 20x + 150 = 5x + 300.

Step 2. Subtract 5x: 15x + 150 = 300.

Step 3. Subtract 150: 15x = 150, so x = 10.

Check: 0.2 × 10 + 1.5 = 3.5 and 0.05 × 10 + 3 = 3.5.

Worked example 6. Maya has $120 and spends $15 a week. Leo has $30 and saves $15 a week. After how many weeks will they have the same amount?

Step 1. Let w be the number of weeks. Maya has 120 − 15w and Leo has 30 + 15w, so 120 − 15w = 30 + 15w.

Step 2. Add 15w to both sides: 120 = 30 + 30w.

Step 3. Subtract 30: 90 = 30w, so w = 3.

Check: after 3 weeks Maya has 120 − 45 = $75 and Leo has 30 + 45 = $75.

Worked example 7. Solve 4(2x + 1) − 3x = 2(x − 7).

Step 1. Expand: 8x + 4 − 3x = 2x − 14.

Step 2. Collect like terms: 5x + 4 = 2x − 14.

Step 3. Subtract 2x: 3x + 4 = −14. Subtract 4: 3x = −18. Divide by 3: x = −6.

Check: 4 × (−12 + 1) + 18 = −44 + 18 = −26 and 2 × (−13) = −26.

Common misconceptions

"Move a term to the other side and keep its sign." Moving a term is shorthand for subtracting (or adding) it on both sides, which changes its sign on the new side. From 40 + 5g = 25 + 8g, taking 5g away from both sides gives 40 = 25 + 3g, not 40 = 25 + 13g.

"3(x − 4) = 3x − 4." The 3 multiplies every term inside the brackets: 3x − 12. Multiplying only the first term is the most common slip in this lesson.

"5x + 2 = 7x." Only like terms combine. 5x and 2 are different kinds of term, so 5x + 2 stays as it is.

"To clear fractions, multiply just the fractions." Every term on both sides must be multiplied, or the balance breaks. In Worked example 4, the 4 and the 6 became 24 and 36 too.

"The answer is $65." The equation was about g, the number of gigabytes, so its solution is g = 5. The $65 is the cost at that point. Always say what your answer means in the situation.

Where this leaves us

• When two expressions are equal, set them equal and solve: 40 + 5g = 25 + 8g gives g = 5.
• Do the same thing to both sides at every step. Gather variable terms on one side and numbers on the other, then divide by the coefficient.
• Expand brackets with the distributive property, multiplying every term inside, and collect like terms before you move anything.
• Clear fractions by multiplying every term by a common denominator, and clear decimals by multiplying by 10, 100 or 1,000.
• Check every solution in the original equation, and state what it means.

Every equation in this lesson had exactly one solution. But try 2x + 5 = 2x + 9, or 3(x + 2) = 3x + 6, and something strange happens when you subtract 2x or 3x. The next lesson takes those strange cases apart.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.7b). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 2.3, Solve equations with variables and constants on both sides. Elementary Algebra 2e. OpenStax. openstax.org
3. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 2.5, Solve equations with fractions or decimals. Elementary Algebra 2e. OpenStax. openstax.org
4. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 5: Solving any linear equation. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 9: When are they the same? IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Two answers from one homework sheet

A student solved two equations on the same evening. The first was 3(x + 2) = 3x + 6. The student expanded the left side to get 3x + 6 = 3x + 6, subtracted 3x from both sides to get 6 = 6, subtracted 6 to get 0 = 0, and wrote "x = 0".

The second was 2x + 5 = 2x + 9. The student subtracted 2x from both sides, got 5 = 9, decided they had made an arithmetic mistake, and did the whole thing again. And again. Each time it came out 5 = 9.

Neither answer is right, and neither student did any arithmetic wrong. This lesson finds where the reasoning went astray in each case, and shows that every linear equation in one variable has either exactly one solution, no solution, or infinitely many, and that you can tell which by looking at it. It covers Common Core standard 8.EE.C.7a: giving examples of linear equations with one solution, infinitely many solutions or no solutions, and showing which is the case by transforming the equation until it takes the form x = a, a = a, or a = b with a and b different numbers.

Where "0 = 0, so x = 0" goes wrong

Test the student's answer. Put x = 0 into 3(x + 2) = 3x + 6: the left side is 3 × 2 = 6, and the right side is 0 + 6 = 6. It works. But try x = 5: the left side is 3 × 7 = 21 and the right is 15 + 6 = 21. That works too. So does x = −4: 3 × (−2) = −6 and −12 + 6 = −6. Every number you try works.

The mistake was reading 0 = 0 as a statement about x. When the x-terms disappeared, the equation stopped saying anything about x at all. The statement 0 = 0 is simply true, whatever x is. Each step along the way kept the equation balanced, so the original equation is true exactly when 0 = 0 is true, which is always. The equation has infinitely many solutions: every number is a solution.

You can see why without solving at all. Expand the left side: 3(x + 2) = 3x + 6. The two sides are the same expression written two ways, so of course they are equal for every x. An equation like this, true for every value of the variable, is called an identity.

Where "5 = 9 must be a slip" goes wrong

Sometimes 5 = 9 really does mean you made a mistake, so checking was sensible. But here the work was correct. Look at the original equation: 2x + 5 = 2x + 9. Whatever number x is, 2x is the same on both sides, and then one side adds 5 and the other adds 9. The right side is always 4 bigger than the left. They can never be equal.

So 5 = 9 is not an arithmetic slip; it is the equation's honest answer. The original equation is true exactly when 5 = 9 is true, which is never. The equation has no solution. Such an equation is called a contradiction.

Here is a real version. Two streaming plans both cost $10 a month, but one charges a $5 sign-up fee and the other a $9 fee. After m months they have cost 10m + 5 and 10m + 9 dollars. Setting them equal gives 10m + 5 = 10m + 9, which becomes 5 = 9. The plans never cost the same, because the second is always $4 more.

The upshot: When the variable disappears, look at what is left. A true statement such as 6 = 6 means every number is a solution. A false statement such as 5 = 9 means no number is.

The three final forms

Notice how alike the three examples are. Each starts with 2(x + 4) = 2x + 8 on the left. Change the right side's coefficient of x from 2 to 3, and there is one solution. Keep the coefficient at 2 and the constant at 8, and there are infinitely many. Keep the coefficient at 2 but change the constant to 7, and there are none.

Telling in advance

Simplify each side to the form (number)x + (number). Then compare.

• Different coefficients of x: exactly one solution.
• Same coefficient and same constant: infinitely many solutions.
• Same coefficient, different constants: no solution.

The graphs show the same three cases. Think of each side as a line: y = 2x + 8 on one side, and y = 3x + 7, y = 2x + 8 or y = 2x + 7 on the other. Lines with different slopes cross at exactly one point. A line meets itself everywhere. Lines with the same slope and different intercepts are parallel and never meet. Lessons 18 to 21 use this idea for pairs of lines.

Worked examples

Worked example 1. How many solutions does 5x − 3(x − 2) = 2x + 6 have?

Step 1. Expand the left side, taking care with the minus sign: 5x − 3x + 6 = 2x + 6.

Step 2. Collect like terms: 2x + 6 = 2x + 6.

Step 3. Both sides are the same expression, so the equation is an identity with infinitely many solutions. Subtracting 2x gives 6 = 6, which confirms it.

Worked example 2. How many solutions does 4(1 − x) = −4x + 1 have?

Step 1. Expand: 4 − 4x = −4x + 1.

Step 2. Add 4x to both sides: 4 = 1.

Step 3. That is false, so there is no solution.

Worked example 3. How many solutions does 6x − 1 = 2(3x + 1) − x have?

Step 1. Expand and collect on the right: 6x + 2 − x = 5x + 2. The equation is 6x − 1 = 5x + 2.

Step 2. The coefficients, 6 and 5, are different, so there is exactly one solution. Subtract 5x: x − 1 = 2, so x = 3.

Check: 6 × 3 − 1 = 17 and 2 × 10 − 3 = 17.

Worked example 4. Fill in the blank in 4x − 1 = 4x + __ so that the equation has no solution, and then so that it has infinitely many.

Step 1. The coefficients already match. Any constant other than −1 gives no solution, for example 4x − 1 = 4x + 3.

Step 2. The constant −1 gives 4x − 1 = 4x + (−1), the same expression on both sides: infinitely many solutions.

Worked example 5. A number trick: think of a number, double it, add 6, halve the result, then subtract the number you first thought of. Show that you always end with 3.

Step 1. Call the number n. The steps give (2n + 6) ÷ 2 − n.

Step 2. (2n + 6) ÷ 2 = n + 3, and n + 3 − n = 3.

Step 3. So the equation (2n + 6) ÷ 2 − n = 3 is an identity: true for every n. That is why the trick never fails.

Common misconceptions

"0 = 0 means x = 0." When the variable disappears, the statement left over says nothing about x. A true statement means every value of x is a solution, not just 0.

"5 = 9 means I made a mistake." Check your work, because it might. But if the work is right, a false statement means the equation has no solution. The sides of 2x + 5 = 2x + 9 always differ by 4.

"No solution means the answer is 0." Zero is a number, and it either works or it does not. In 2x + 5 = 2x + 9, x = 0 gives 5 = 9, so 0 is not a solution. "No solution" means no number at all works.

"If the x-terms cancel, there are infinitely many solutions." Only if what is left is true. In 4(1 − x) = −4x + 1 the x-terms cancel and leave 4 = 1, which is false: no solution.

"Every equation has exactly one answer." Linear equations in one variable can have one solution, none, or infinitely many. Which one depends on the coefficients and constants.

Recap

• Simplify a linear equation step by step. If you reach x = a, there is exactly one solution. If you reach a true statement such as 8 = 8, every number is a solution. If you reach a false statement such as 8 = 7, there is no solution.
• An identity has the same expression on both sides once simplified. A contradiction has the same coefficient of x on both sides but different constants.
• Different coefficients of x on the two sides always mean exactly one solution.
• In pictures: two lines with different slopes cross once, identical lines meet everywhere, and parallel lines never meet.

That last point is the bridge to the next module. Two plans, two racers or two savings accounts are two lines, and the question "when are they equal?" becomes the question "where do the lines cross?", with two unknowns at once.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.7a). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 2.4, Use a general strategy to solve linear equations (classify equations). Elementary Algebra 2e. OpenStax. openstax.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 7: All, some, or no solutions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 8: How many solutions? IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

A thirty-metre head start

Maya and her younger brother Leo race over 100 metres in the park. Maya runs at a steady 6 metres per second, Leo at 4 metres per second, so Maya gives him a 30-metre head start. Measure every position from Maya's starting line, so the finish is at the 100-metre mark and Leo starts at the 30-metre mark. Does Maya catch him before the finish?

After t seconds, Maya is at d = 6t and Leo is at d = 30 + 4t. Here is the race, second by second:

At 15 seconds they are side by side at the 90-metre mark. Maya then pulls ahead and reaches the finish first: 100 ÷ 6 ≈ 16.7 seconds, against Leo's (100 − 30) ÷ 4 = 17.5 seconds.

Now picture the graph, with time across and distance up. Maya's line starts at (0, 0) and climbs with slope 6. Leo's starts at (0, 30) and climbs with slope 4, more gently. Maya's steeper line closes in on Leo's, and the two cross at (15, 90). That crossing point is the answer to the question.

This lesson covers Common Core standard 8.EE.C.8a, understanding that the solutions of a system of two linear equations correspond to points where their graphs intersect, and part of 8.EE.C.8b, estimating solutions by graphing.

What a system is, and what solves it

A system of linear equations is a set of two (or more) linear equations that are about the same unknowns at the same time. The race is the system

d = 6t and d = 30 + 4t.

A solution of the system is a pair of values that makes both equations true at once. The pair t = 15, d = 90 works in both: 6 × 15 = 90, and 30 + 4 × 15 = 90. Other pairs work in one equation only. The pair t = 10, d = 60 fits Maya's equation but not Leo's, because 30 + 40 = 70, not 60. On the graph, (10, 60) is on Maya's line but not on Leo's.

Every point on a line is a pair that makes that line's equation true, so a pair that makes both equations true must lie on both lines. Two different lines that cross share exactly one point. So the solution of the system is the crossing point.

What matters here: The solution of a system of two equations is the point that lies on both lines. Graphing finds it as the place where the lines cross, and substituting it into both equations checks it.

The same race with a bigger head start

Change one number. This time Leo starts at the 40-metre mark: d = 40 + 4t. Maya's line is unchanged. Solving by looking at the table or the graph, the lines now cross where 6t = 40 + 4t, at t = 20 seconds and d = 120 metres.

But the race ends at 100 metres. The crossing point is past the finish line, so in this race Maya never catches Leo. He finishes in (100 − 40) ÷ 4 = 15 seconds, while she needs 16.7. The mathematics still finds the crossing point; you have to ask whether it falls inside the situation.

Solving a system by graphing

Solve the system y = 2x + 1 and y = −x + 7.

Step 1. Graph the first line: start at its y-intercept (0, 1), and step up 2 for every 1 to the right: (1, 3), (2, 5), (3, 7).

Step 2. Graph the second: start at (0, 7) and step down 1 for every 1 to the right: (1, 6), (2, 5), (3, 4).

Step 3. The lines share the point (2, 5).

Step 4. Check in both equations: 2 × 2 + 1 = 5 and −2 + 7 = 5. The solution is x = 2, y = 5.

When the lines are in a different form, rewrite them first or use their intercepts. For x + y = 6, setting x = 0 gives y = 6 and setting y = 0 gives x = 6, so the line passes through (0, 6) and (6, 0).

When the crossing is between grid points

Solve y = 2x + 1 and y = −x + 5 by graphing.

The first line passes through (1, 3) and (2, 5); the second through (1, 4) and (2, 3). At x = 1 the second line is above; at x = 2 the first line is above. So they cross somewhere between x = 1 and x = 2, about a third of the way along, near (1.3, 3.7).

Check the estimate: at x = 1.3, the first line gives 3.6 and the second gives 3.7, close but not equal. A graph gives a good estimate, not always the exact answer. The exact solution is x = 4/3 and y = 11/3, about (1.33, 3.67), and the next lesson shows how to get it without a graph.

No solution and infinitely many

Some systems have no solution. y = 2x + 1 and y = 2x − 3 have the same slope and different intercepts, so their lines are parallel and never cross. No pair satisfies both, just as in Lesson 17, where 2x + 1 = 2x − 3 simplifies to the false statement 1 = −3.

Some have infinitely many. y = 2x + 1 and 4x − 2y = −2 look different, but rewrite the second: subtract 4x from both sides to get −2y = −4x − 2, then divide by −2 to get y = 2x + 1. It is the same line written another way, so every point on it solves the system.

Worked examples

Worked example 1. Is (3, −1) a solution of the system x + y = 2 and 2x − y = 5?

Step 1. First equation: 3 + (−1) = 2. True.

Step 2. Second equation: 2 × 3 − (−1) = 6 + 1 = 7, not 5. False.

Step 3. The point lies on the first line but not the second, so it is not a solution of the system.

Worked example 2. Plan A costs c = 40 + 5g dollars and Plan B costs c = 25 + 8g for g gigabytes. Where do their lines cross, and what does it mean?

Step 1. From the table in Lesson 16, both cost $65 at 5 GB, so the lines cross at (5, 65).

Step 2. To the left of 5 GB, Plan B's line is lower, so it is cheaper; to the right, Plan A's line is lower. The crossing point is the break-even point.

Worked example 3. How many solutions does the system y = −3x + 4 and y = −3x + 1 have?

Same slope, −3, and different intercepts, 4 and 1: parallel lines, so no solution.

Worked example 4. Solve by graphing: y = x − 2 and y = −2x + 4.

Step 1. Points on the first line: (0, −2), (1, −1), (2, 0), (3, 1).

Step 2. Points on the second: (0, 4), (1, 2), (2, 0), (3, −2).

Step 3. Both contain (2, 0). Check: 2 − 2 = 0 and −4 + 4 = 0. The solution is (2, 0).

Common misconceptions

"A system has two answers, one for each equation." The solution is one pair of values that works in both equations at once, one point on both lines.

"Any point on one of the lines is a solution." (10, 60) is on Maya's line, but Leo is at 70 m at 10 seconds, not 60. A solution must be on both lines.

"The graph always gives the exact answer." When the crossing falls between grid points, a graph gives an estimate. Check by substituting, and use the methods in the next two lessons for exact answers.

"Lines that do not cross on the page are parallel." They may cross off the edge of the page. Compare slopes: only lines with the same slope and different intercepts never meet.

"The crossing point always answers the real question." In the 40-metre head start, the lines cross at 120 m, beyond the finish. Always check that the crossing point makes sense in the situation.

What you now know

• A system of two linear equations asks for a pair of values that makes both equations true.
• That pair is a point on both lines, so a system's solution is where its graphs cross.
• To solve by graphing, draw both lines, read the crossing point, and check it in both equations. A crossing between grid points can only be estimated.
• Different slopes mean one solution; parallel lines mean none; the same line means infinitely many.
• Interpret the crossing in context: it may be a break-even point, a catching-up moment, or a point outside the situation.

Graphs are clear but not always precise. The next lesson finds the exact crossing point of any two lines by substitution, starting with the tickets sold for a school play.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.8a, 8.EE.C.8b). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.1, Solve systems of equations by graphing. Elementary Algebra 2e. OpenStax. openstax.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 11: On both of the lines. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 12: Systems of equations. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

250 tickets and $1,950

The school play sold 250 tickets and took in $1,950. Adult tickets cost $9 and student tickets $6. The box office lost its tally of how many of each it sold. Can you work it out?

Two unknowns need two letters. Let a be the number of adult tickets and s the number of student tickets. The facts give two equations:

a + s = 250 (the number of tickets) and 9a + 6s = 1,950 (the money).

You could graph both lines and look for the crossing point, but the numbers are large and the crossing might not land on a grid point. Substitution finds the exact answer. This lesson covers Common Core standard 8.EE.C.8b: solving systems of two linear equations in two variables algebraically, and solving simple cases by inspection.

The method, worked end to end

Step 1. Solve one equation for one variable. The first equation is easiest, because a has coefficient 1: a = 250 − s.

Step 2. Substitute into the other equation. Wherever a appears in 9a + 6s = 1,950, write (250 − s), with brackets: 9(250 − s) + 6s = 1,950.

Step 3. Solve the equation in one variable. Expand: 2,250 − 9s + 6s = 1,950. Collect: 2,250 − 3s = 1,950. Subtract 2,250: −3s = −300. Divide by −3: s = 100.

Step 4. Substitute back to find the other variable. a = 250 − 100 = 150.

Step 5. Check in both original equations. 150 + 100 = 250, and 9 × 150 + 6 × 100 = 1,350 + 600 = 1,950. Both are true.

The box office sold 150 adult tickets and 100 student tickets. On a graph, the two lines cross at (150, 100) if a is across and s is up.

Why does it work? In Step 1, a = 250 − s says that, in any solution, a and 250 − s are the same number, so you may swap one for the other. That swap leaves one equation with one unknown, which you already know how to solve.

Bottom line: Substitution turns two equations in two unknowns into one equation in one unknown. Solve it, substitute back for the second unknown, and check the pair in both equations.

When the substitution goes strange

Here the same method meets a different input and flips the result. Solve y = 3x + 2 and 6x − 2y = 5.

Step 1. The first equation already gives y.

Step 2. Substitute: 6x − 2(3x + 2) = 5.

Step 3. Expand: 6x − 6x − 4 = 5, which is −4 = 5.

The variable has vanished and left a false statement, exactly as in Lesson 17. No pair of values satisfies both equations, so the system has no solution. Rewrite the second equation to see why: −2y = −6x + 5, so y = 3x − 2.5. Both lines have slope 3 and different intercepts, so they are parallel.

If the substitution ends in a true statement instead, the equations describe the same line. For x + 2y = 4 and 3x + 6y = 12, solve the first for x = 4 − 2y and substitute: 3(4 − 2y) + 6y = 12, so 12 − 6y + 6y = 12, which is 12 = 12. Every point on the line x + 2y = 4 is a solution: infinitely many. Indeed, the second equation is just the first multiplied by 3.

Worked examples

Worked example 1. Solve y = 2x + 1 and y = −x + 5, the system that could only be estimated from a graph in the last lesson.

Step 1. Both equations give y, so set the two expressions equal: 2x + 1 = −x + 5.

Step 2. Add x to both sides: 3x + 1 = 5. Subtract 1: 3x = 4, so x = 4/3.

Step 3. Substitute back: y = 2 × 4/3 + 1 = 8/3 + 3/3 = 11/3.

Step 4. Check in the other equation: −4/3 + 5 = −4/3 + 15/3 = 11/3. The exact solution is (4/3, 11/3), about (1.33, 3.67), just where the graph suggested.

Worked example 2. Solve 2x + 3y = 13 and x − y = −1.

Step 1. The second equation has x with coefficient 1, so solve it for x: x = y − 1.

Step 2. Substitute into the first: 2(y − 1) + 3y = 13.

Step 3. 2y − 2 + 3y = 13, so 5y = 15 and y = 3.

Step 4. x = 3 − 1 = 2.

Step 5. Check: 2 × 2 + 3 × 3 = 13 and 2 − 3 = −1. The solution is (2, 3).

Worked example 3. A jar holds 20 coins, all dimes and quarters, worth $4.25 in total. How many of each?

Step 1. Let d be dimes and q quarters: d + q = 20 and 0.10d + 0.25q = 4.25.

Step 2. d = 20 − q, so 0.10(20 − q) + 0.25q = 4.25.

Step 3. 2 − 0.10q + 0.25q = 4.25, so 0.15q = 2.25 and q = 15.

Step 4. d = 20 − 15 = 5.

Step 5. Check: 5 dimes are $0.50 and 15 quarters are $3.75, a total of $4.25 in 20 coins.

Worked example 4. Sam is three times as old as his sister Kira. In four years he will be twice as old as she will be. How old are they now?

Step 1. Let s be Sam's age and k Kira's. Now: s = 3k. In four years: s + 4 = 2(k + 4).

Step 2. Substitute s = 3k into the second: 3k + 4 = 2k + 8.

Step 3. Subtract 2k: k + 4 = 8, so k = 4. Then s = 3 × 4 = 12.

Step 4. Check: 12 is three times 4, and in four years Sam is 16 and Kira is 8, and 16 is twice 8.

Worked example 5. Solve by inspection: x + y = 7 and x + y = 10.

x + y cannot be 7 and 10 at the same time, so there is no solution. No calculation is needed, which is what the standard means by solving simple cases by inspection.

Common misconceptions

"Substitute back into the equation you rearranged." Putting a = 250 − s back into a + s = 250 gives 250 = 250, which tells you nothing new. Substitute into the other equation.

"9(250 − s) is 2,250 − s." The brackets matter: the 9 multiplies both terms, giving 2,250 − 9s. Writing the substituted expression without brackets is the most common slip.

"Once you find s, you are done." A system's solution is a pair. Substitute back to find the other variable, and give both: 150 adult and 100 student tickets.

"Checking one equation is enough." The pair must satisfy both. A mistake in Step 4 can give a pair that fits the equation you used to find it but not the other one.

"A false statement such as −4 = 5 means you made a mistake." Check your work, but if it is right, the system has no solution: the lines are parallel.

The short version

• Substitution: solve one equation for one variable, substitute that expression, in brackets, into the other equation, solve, substitute back, and check in both equations.
• Choose a variable with coefficient 1 or −1 to rearrange, to avoid fractions.
• If the variable vanishes and leaves a false statement, there is no solution (parallel lines). If it leaves a true statement, there are infinitely many (the same line).
• Simple systems such as x + y = 7 and x + y = 10 can be solved by inspection: here, no solution.

Substitution is clumsy when no variable has coefficient 1, as in 3x + 2y = 13 and 3x + 4y = 17. The next lesson adds or subtracts whole equations to make a variable disappear, starting with two receipts from a snack bar.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.8b). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.2, Solving systems of equations by substitution. Elementary Algebra 2e. OpenStax. openstax.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 13: Solving systems of equations. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 14: Solving more systems. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Two receipts

At a ballgame, your family bought 3 hot dogs and 2 drinks for $13.00. In the second inning you went back for 3 more hot dogs and 4 drinks, and paid $17.00. The receipts do not show the prices of single items. What does one drink cost?

Lay the receipts side by side. The second order has the same 3 hot dogs, plus 2 more drinks, and it cost $4.00 more. Those 2 extra drinks must account for the extra $4.00. So 2 drinks cost $4.00, and one drink costs $2.00. Back on the first receipt, 3 hot dogs and $4.00 of drinks came to $13.00, so the hot dogs cost $9.00, which is $3.00 each.

You just solved a system without rearranging anything. With h for the price of a hot dog and d for a drink, the receipts say 3h + 2d = 13 and 3h + 4d = 17. Subtracting the first receipt from the second took away the hot dogs completely and left 2d = 4. That is the elimination method: add or subtract whole equations so that one variable disappears. This lesson covers more of Common Core standard 8.EE.C.8b, solving systems algebraically, and the standard's own example of solving simple cases by inspection.

Why you may add or subtract equations

If two amounts are equal and you add the same thing to both, they stay equal. An equation is a statement that its two sides are the same number. So if 3h + 4d and 17 are the same number, and 3h + 2d and 13 are the same number, then subtracting the second pair from the first leaves equal results:

(3h + 4d) − (3h + 2d) = 17 − 13, so 2d = 4.

Adding works the same way. The point of choosing to add or to subtract is to make one variable cancel.

Adding when the coefficients are opposites

Solve x + y = 10 and x − y = 4.

Step 1. The y-terms are +y and −y, opposites, so adding the equations cancels them: (x + y) + (x − y) = 10 + 4, which gives 2x = 14.

Step 2. x = 7.

Step 3. Substitute into either equation: 7 + y = 10, so y = 3.

Step 4. Check in both: 7 + 3 = 10 and 7 − 3 = 4. The solution is (7, 3).

The rule for choosing: when one variable has the same coefficient in both equations, subtract. When its coefficients are opposites, add.

Multiplying first

Often neither variable lines up. Solve 2x + 3y = 12 and 5x − y = 13.

Step 1. Multiply every term of the second equation by 3, so the y-terms become opposites: 15x − 3y = 39. Multiplying both sides of a true equation by the same number keeps it true.

Step 2. Add it to the first equation: (2x + 3y) + (15x − 3y) = 12 + 39, so 17x = 51.

Step 3. x = 3.

Step 4. Substitute into 5x − y = 13: 15 − y = 13, so y = 2.

Step 5. Check: 2 × 3 + 3 × 2 = 12 and 5 × 3 − 2 = 13. The solution is (3, 2).

Sometimes both equations need multiplying. Solve 3x + 4y = 10 and 2x + 3y = 7.

Step 1. To make the x-coefficients match, multiply the first equation by 2 and the second by 3: 6x + 8y = 20 and 6x + 9y = 21.

Step 2. Subtract the first new equation from the second: y = 1.

Step 3. Substitute: 3x + 4 = 10, so x = 2.

Step 4. Check: 3 × 2 + 4 × 1 = 10 and 2 × 2 + 3 × 1 = 7. The solution is (2, 1).

Worth holding on to: Elimination makes one variable's coefficients equal or opposite, then subtracts or adds the equations to cancel it. Whatever you do, do it to every term on both sides.

Solving by inspection

The standard gives this example: 3x + 2y = 5 and 3x + 2y = 6 have no solution, because 3x + 2y cannot be 5 and 6 at the same time. Elimination agrees: subtracting gives 0 = 1, a false statement. The two lines are parallel.

Now try 3x + 2y = 5 and 6x + 4y = 10. The second equation is the first multiplied by 2, so they describe the same line, and there are infinitely many solutions. Elimination agrees here too: doubling the first equation and subtracting gives 0 = 0, which is always true.

Which method?

All three methods give the same answer when they work, because they are all finding the one pair that makes both equations true. OpenStax's Elementary Algebra makes the same point: in later courses you will choose the method yourself, so it pays to notice which one the system is asking for.

Worked examples

Worked example 1. At a stationery shop, 2 notebooks and 5 pens cost $8.50, and 2 notebooks and 3 pens cost $6.50. Find each price.

Step 1. 2n + 5p = 8.50 and 2n + 3p = 6.50. The n-terms match, so subtract: 2p = 2.00, and p = 1.00.

Step 2. 2n + 3 = 6.50, so 2n = 3.50 and n = 1.75.

Step 3. Check: 2 × 1.75 + 5 × 1.00 = 8.50. A notebook costs $1.75 and a pen $1.00.

Worked example 2. Solve 4x − 3y = 1 and 2x + 3y = 11.

Step 1. The y-terms are opposites, so add: 6x = 12 and x = 2.

Step 2. 2 × 2 + 3y = 11, so 3y = 7 and y = 7/3.

Step 3. Check: 4 × 2 − 3 × 7/3 = 8 − 7 = 1. The solution is (2, 7/3). Elimination handles fractions in the answer without trouble.

Worked example 3. Solve 5x + 2y = 16 and 3x − 4y = 20.

Step 1. Multiply the first equation by 2: 10x + 4y = 32.

Step 2. Add to the second: 13x = 52, so x = 4.

Step 3. 5 × 4 + 2y = 16, so 2y = −4 and y = −2.

Step 4. Check: 3 × 4 − 4 × (−2) = 12 + 8 = 20. The solution is (4, −2).

Worked example 4. A school club sells 7 hoodies and 3 caps for $273, and 7 hoodies and 1 cap for $245. Find each price.

Step 1. 7h + 3c = 273 and 7h + c = 245. Subtract the second equation from the first: 2c = 28, so a cap costs $14.

Step 2. 7h + 14 = 245, so 7h = 231 and a hoodie costs $33.

Step 3. Check: 7 × 33 + 3 × 14 = 231 + 42 = 273.

Worked example 5. How many solutions does the system 2x − y = 3 and −4x + 2y = −6 have?

Multiply the first equation by −2: −4x + 2y = −6, exactly the second equation. Same line, so infinitely many solutions.

Common misconceptions

"Subtract only the terms you want to cancel." You must subtract whole equations: every term on the left and the numbers on the right. Subtracting 3h from 3h but forgetting to subtract 13 from 17 breaks the balance.

"Multiply just one side of the equation." Multiplying 5x − y = 13 by 3 means 15x − 3y = 39. Writing 15x − 3y = 13 is a different, wrong equation.

"Subtracting a negative is the same as subtracting." Subtracting (2x − 4y) from (5x − y) gives 3x + 3y, because −y − (−4y) = 3y. Many elimination errors are sign errors.

"Adding always cancels the variable with the bigger coefficient." Adding cancels a variable only when its coefficients are opposites, such as 3y and −3y.

"Elimination and substitution can give different answers." If both are done correctly, they give the same pair, because there is only one point on both lines.

What to remember

• Equal amounts added to or subtracted from equal amounts stay equal, so whole equations can be added or subtracted.
• Subtract when a variable's coefficients match; add when they are opposites; multiply one or both equations first when neither is true.
• After eliminating, solve for one variable, substitute to find the other, and check in both original equations.
• A false statement such as 0 = 1 means no solution; a true one such as 0 = 0 means infinitely many. Simple cases, such as 3x + 2y = 5 and 3x + 2y = 6, can be decided by inspection.
• Choose the method the system invites: graphing to see it, substitution when a variable is already isolated, elimination when the equations line up.

You can now solve any system you are given. The harder part is writing the system from a problem in words, and it is harder than it looks: in a famous study, many university students wrote the equation for "six times as many students as professors" backwards. That is the next lesson.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.8b). thecorestandards.org
2. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.3, Solve systems of equations by elimination. Elementary Algebra 2e. OpenStax. openstax.org
3. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.4, Solve applications with systems of equations. Elementary Algebra 2e. OpenStax. openstax.org
4. Khan Academy. (n.d.). 8th grade math: Systems of equations. khanacademy.org

Six times as many students

Here is a one-sentence problem. Write an equation using S for the number of students and P for the number of professors: "There are six times as many students as professors at this university."

Write your answer before reading on.

A great many people write 6S = P. In a study published in 1981, John Clement, Jack Lochhead and George Monk gave this problem to university students, and a 2021 review of the research by Calos Soneira, Sarah Bansilal and Reginald Govender reports that 68 percent of the wrong answers were exactly that equation: 6S = P. The mistake has a name, the reversal error, and it has turned up again and again, from high school students to university students and trainee teachers.

6S = P is wrong, and this lesson traces exactly why, then gives you a check that catches it every time. It covers Common Core standard 8.EE.C.8c: solving real-world and mathematical problems leading to two linear equations in two variables.

Where 6S = P breaks

Test the equation with numbers you can picture. Suppose the university has 10 professors. "Six times as many students" means 60 students. Now put S = 60 and P = 10 into 6S = P: 6 × 60 = 360, which is not 10. The equation says the opposite of the sentence. It claims there are six times as many professors as students.

The correct equation must make 60 and 10 work. Try 6P = S: 6 × 10 = 60. That matches. So the equation is S = 6P: the number of students equals 6 times the number of professors.

Why does 6S = P feel right? The review describes the usual cause as word-order matching: reading the sentence left to right and writing a symbol for each word in turn. "Six times as many students" becomes 6S, "as" becomes =, and "professors" becomes P. The symbols follow the words, but an equation states which numbers are equal, and the words were never in that order.

There is a second cause. Writing 6S can feel like "six students", a label, rather than "6 times the number of students". The letters stand for numbers, not for people. S is the number of students.

So what?: Before you trust an equation built from words, put in numbers you know fit the story. If the equation fails the test, it is wrong, however natural it looked.

Translating relationships

In every row, the test numbers fit the sentence and the correct equation, and fail the tempting version.

From a problem to a system

A problem with two unknowns needs two facts that connect them. Each fact becomes one equation. The steps:

Step 1. Name each unknown as a number, with its unit: "let k be the number of cookies sold".

Step 2. Turn each fact into an equation, and test it with numbers that fit the story.

Step 3. Solve the system by graphing, substitution or elimination.

Step 4. Check the answer against the original words, not just the equations, and state it in a sentence.

Worked examples

Worked example 1. A bake sale sold 150 cookies and brownies for $370. Cookies cost $2 and brownies $3. How many of each were sold?

Step 1. Let k be the number of cookies and r the number of brownies.

Step 2. Count: k + r = 150. Money: 2k + 3r = 370. Test: with 80 cookies and 70 brownies the count would be 150, and the money 160 + 210 = 370.

Step 3. Substitute k = 150 − r: 2(150 − r) + 3r = 370, so 300 + r = 370 and r = 70; then k = 80.

Step 4. 80 cookies and 70 brownies: 150 items and $160 + $210 = $370.

Worked example 2. Tom is 4 years older than twice his cousin's age, and their ages add to 31. How old is each?

Step 1. t = 2c + 4 and t + c = 31.

Step 2. Substitute: 2c + 4 + c = 31, so 3c = 27 and c = 9; then t = 22.

Step 3. Check the words: twice 9 is 18, and 4 more is 22; and 22 + 9 = 31.

Worked example 3. A rectangle's length is 3 cm more than twice its width, and its perimeter is 54 cm. Find its dimensions.

Step 1. l = 2w + 3 and 2l + 2w = 54.

Step 2. Substitute: 2(2w + 3) + 2w = 54, so 6w + 6 = 54 and w = 8; then l = 19.

Step 3. Check: 19 is 3 more than 16, and 2 × 19 + 2 × 8 = 54.

Worked example 4. The standard's own kind of problem: does the line through (0, 2) and (3, 8) meet the line through (1, 1) and (4, 7)?

Step 1. First line: slope (8 − 2) ÷ 3 = 2 and intercept 2, so y = 2x + 2.

Step 2. Second line: slope (7 − 1) ÷ (4 − 1) = 2. Using (1, 1): 1 = 2 + b, so b = −1 and y = 2x − 1.

Step 3. Same slope, different intercepts: the lines are parallel and never meet. The system 2x + 2 = 2x − 1 gives 2 = −1, no solution.

Worked example 5. Does the line through (0, 2) and (3, 8) meet the line through (0, 5) and (5, 0), and where?

Step 1. The first is y = 2x + 2. The second has slope (0 − 5) ÷ 5 = −1 and intercept 5, so y = −x + 5.

Step 2. Different slopes, so they meet once: 2x + 2 = −x + 5 gives 3x = 3, x = 1 and y = 4.

Step 3. Check: (1, 4) is on both, since 2 + 2 = 4 and −1 + 5 = 4.

Worked example 6. A class has 31 students. The number of girls is 5 less than three times the number of boys. How many boys and girls are there?

Step 1. g = 3b − 5 and g + b = 31. Test the first equation: 9 boys would mean 27 − 5 = 22 girls.

Step 2. Substitute: 3b − 5 + b = 31, so 4b = 36 and b = 9; then g = 22.

Step 3. Check: 22 + 9 = 31, and 3 × 9 − 5 = 22.

Common misconceptions

"Six times as many students as professors is 6S = P." This is the reversal error. Test it: with 10 professors and 60 students, 6 × 60 is not 10. The correct equation is S = 6P.

"S means students." S means the number of students. Treating letters as labels, as in "6S means six students", is what makes the reversal feel right.

"Five less than three times b is 5 − 3b." "Less than" reverses the order of the words: five less than 3b is 3b − 5. Test it: with b = 10 you want 25, and 5 − 30 is −25.

"One equation is enough to find two unknowns." k + r = 150 has many solutions: 80 and 70, 100 and 50, and so on. You need a second fact, such as the money taken, to pin down the pair.

"Two lines through different points must cross somewhere." Lines with the same slope are parallel, so they never cross, however long you draw them.

Putting it together

• Name each unknown as a number of something, and write one equation for each fact.
• Test each equation with numbers that fit the story before you solve. The test catches the reversal error: S = 6P, not 6S = P.
• Translate carefully: "5 less than 3b" is 3b − 5, and "4 more than twice c" is 2c + 4.
• Solve the system by the method that suits it, then check the answer against the original words.
• To decide whether the lines through two pairs of points meet, find both slopes: different slopes cross once, equal slopes with different intercepts never meet.

Every equation in these last modules turned an input into an output: games into dollars, seconds into metres. The next module gives that idea its proper name, a function, and asks what makes a rule a function at all.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Expressions and Equations (8.EE.C.8c). thecorestandards.org
2. Soneira, C., Bansilal, S., and Govender, R. (2021). Insights into the reversal error from a study with South African and Spanish prospective primary teachers. Pythagoras, 42(1), article 613. pythagoras.org.za
3. Clement, J., Lochhead, J., and Monk, G. S. (1981). Translation difficulties in learning mathematics. The American Mathematical Monthly, 88(4), 286-290. (The original study; its findings are cited here as reported by Soneira and colleagues.)
4. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 5.4, Solve applications with systems of equations. Elementary Algebra 2e. OpenStax. openstax.org
5. Illustrative Mathematics. (2019). Grade 8, Unit 4, Lesson 15: Writing systems of equations. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Button B4

A vending machine at school has a keypad. Press B4 and a bag of pretzels drops, every time. Press C2 and you always get a granola bar. Two buttons, A1 and A2, both give apple juice, and that is fine: you still know exactly what each button will do.

Now imagine a broken machine where pressing B4 sometimes drops pretzels and sometimes drops chips. You could no longer predict what B4 does. The first machine behaves like a function; the broken one does not.

A function is a rule that assigns to each input exactly one output. The buttons are the inputs and the snacks are the outputs. The working machine gives each button exactly one snack. That is the whole definition, and this lesson tests it on eight relationships side by side. It covers Common Core standard 8.F.A.1: understanding that a function is a rule that assigns to each input exactly one output, and that the graph of a function is the set of ordered pairs made of an input and its output. (The standard notes that function notation is not required in Grade 8. You will meet it in Algebra I.)

Eight relationships, compared

Read down the last column and look for the pattern in the "No" rows. In every one, some input has two different outputs. That is the only thing that stops a relationship from being a function. Two inputs sharing an output, like two buttons giving apple juice, or two students born on the same date when the student is the input, never matters.

Also notice the pairs of rows that swap input and output. Time to height is a function; height to time is not. Student to birthday is a function; date to students is not. Reversing a function does not always give a function.

The core of it: A relationship is a function when no input ever has two different outputs. Outputs may repeat; inputs may not split.

Inputs, outputs and graphs

The input is also called the independent variable, because you choose it, and the output the dependent variable, because it depends on the input. The standard describes a function's graph as the set of ordered pairs (input, output). For the function y = 2x + 3, the pairs include (0, 3), (1, 5), (2, 7) and (−1, 1), and plotting every such pair draws a straight line.

That gives a graph test. If some input had two outputs, the graph would have two points straight above each other, with the same x and different y. So a graph shows a function exactly when no vertical line meets it at more than one point. This is often called the vertical line test.

A circle fails. The circle of points 5 units from the origin includes both (3, 4) and (3, −4), one above the other on the vertical line x = 3, so the input 3 has two outputs. The curve made by the pairs (s, s²), such as (1, 1), (2, 4) and (3, 9), passes: each input appears once.

Checking tables and lists

In a table, look for an input that appears twice with different outputs.

Table A is a function. The output 5 appears twice, for inputs 1 and 3, but no input repeats.

Table B is not a function, because the input 2 has two outputs, 5 and 6.

A list of ordered pairs works the same way. {(1, 2), (3, 4), (1, 5)} is not a function: the input 1 is paired with 2 and with 5. {(1, 2), (3, 2), (5, 2)} is a function, a dull one that gives 2 every time.

Worked examples

Worked example 1. Is the rule "input a whole number, output the number of letters in its English name" a function? What are the outputs for 3, 7 and 11?

Step 1. Each number has one English name with a definite count of letters, so yes.

Step 2. "three" has 5 letters, "seven" has 5, and "eleven" has 6. Two inputs share the output 5, which is allowed.

Worked example 2. Is y² = x a function, with x as the input?

Step 1. Try x = 9: y could be 3 or −3, since both square to 9.

Step 2. One input, two outputs: not a function.

Worked example 3. A phone shop's price list says: under 64 GB of storage, $199; 64 GB to 127 GB, $249; 128 GB or more, $299. With storage as the input, is price a function?

Step 1. Every storage size falls in exactly one band, so it gets exactly one price.

Step 2. Yes, it is a function, even though many inputs share each output and the rule is given in words, not a formula.

Worked example 4. A graph passes through the points (2, 1), (2, 5) and (4, 3). Can it be the graph of a function?

Step 1. The points (2, 1) and (2, 5) are on the vertical line x = 2.

Step 2. The input 2 would have outputs 1 and 5, so no.

Worked example 5. For the function y = 2x + 3, find the output for the input −4, and the input that gives the output 15.

Step 1. Input −4: y = 2 × (−4) + 3 = −5.

Step 2. Output 15: 2x + 3 = 15, so 2x = 12 and x = 6.

Common misconceptions

"A function cannot give the same output twice." It can. In Table A, inputs 1 and 3 both give 5. The rule forbids one input with two outputs, not two inputs with one output.

"A function must have a formula." Tables, lists, graphs and sentences can all describe functions. The vending machine and the phone price list are functions with no formula.

"If the graph is curved, it is not a function." The curve of y = s² passes the vertical line test. Being a function is about one output per input, not about straightness. Straight-line functions are the special ones called linear functions.

"If a relationship is a function, its reverse is too." Time to height is a function, but height to time is not, because the ball reaches most heights twice.

"The test uses horizontal lines." Two points on the same horizontal line just share an output, which is allowed. Two points on the same vertical line share an input, which is not.

Pulling it together

• A function assigns to each input exactly one output. The graph of a function is the set of all its (input, output) pairs.
• A relationship fails to be a function only when some input has two or more different outputs. Repeated outputs are fine.
• Tables and lists: look for a repeated input with different outputs. Graphs: no vertical line may meet the graph twice.
• Functions can be given by words, tables, lists, graphs or equations, and they do not have to be straight lines.
• Swapping input and output can destroy a function, as with time and height for a thrown ball.

The most useful functions in this course are the ones whose graphs are straight lines. The next lesson builds them from just two facts, a starting value and a rate of change, beginning with an 1897 rule that tells the temperature from the chirping of crickets.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Functions (8.F.A.1). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 1: Inputs and outputs. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 2: Introduction to functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., and Honeycutt Mathis, A. (2020). Section 3.5, Relations and functions. Intermediate Algebra 2e. OpenStax. openstax.org

Counting crickets

In November 1897, the physicist Amos Dolbear published a short note in The American Naturalist called "The Cricket as a Thermometer". On warm nights, he wrote, crickets in a field chirp in time with each other, and the rate of their chirping seems to be set entirely by the temperature. At 60°F he counted 80 chirps a minute; at 70°F, 120 a minute. That is, in his words, a change of four chirps a minute for each change of one degree.

Those two facts are enough to build a function: input the temperature, output the chirp rate. Dolbear also turned it round, so that you could input the chirps you hear and get the temperature. This lesson builds linear functions like his from the two numbers that describe every one of them: a rate of change and an initial value. It covers Common Core standard 8.F.B.4, constructing a function to model a linear relationship and interpreting its rate of change and initial value, and 8.F.A.3, interpreting y = mx + b as a linear function whose graph is a straight line.

Building Dolbear's function

Let T be the temperature in °F and N the number of chirps per minute.

Step 1. Two pairs are known: (T, N) = (60, 80) and (70, 120).

Step 2. Rate of change: N rises by 120 − 80 = 40 while T rises by 70 − 60 = 10, so the rate of change is 40 ÷ 10 = 4 chirps per minute for each degree. That matches Dolbear's sentence.

Step 3. Initial value: the function looks like N = 4T + b. Put in (60, 80): 80 = 240 + b, so b = −160.

Step 4. The function is N = 4T − 160. Check with (70, 120): 4 × 70 − 160 = 280 − 160 = 120.

A linear function is a function whose graph is a straight line, y = mx + b. In function language, m is the rate of change, how much the output changes for each increase of 1 in the input, and b is the initial value, the output when the input is 0. They are the slope and the y-intercept from Lesson 15, with names that describe what they do in a situation.

Key idea: Every linear function is fixed by two numbers: how fast the output changes per unit of input, and where it starts when the input is 0.

When the initial value makes no sense

Look hard at b = −160. It says that at 0°F the crickets make −160 chirps a minute. A negative number of chirps is impossible. The rate of change, 4 chirps per degree, means something real; the initial value does not.

That is not a mistake in the algebra. It is a sign that the model only describes part of the world. Dolbear himself wrote that below 50°F a cricket has little energy for music, and his rule gives only 40 chirps a minute there. A linear model is a straight-line description that fits over some range of inputs. Outside that range, you should not trust it. Here the sensible range starts around 50°F; the line's y-intercept, at T = 0, lies far outside it.

Turning it round: temperature from chirps

Dolbear wrote his rule the other way round, with N as the input, as T = 50 + (N − 40)/4. Simplify it: (N − 40)/4 = N/4 − 10, so T = N/4 + 40.

Now the rate of change is 1/4 of a degree per chirp per minute, and the initial value is 40°F. His own example: when the crickets chirp 100 times a minute, T = 100/4 + 40 = 25 + 40 = 65°F. The same relationship, written as a function in the other direction, has its own rate of change and initial value.

Check that the two versions agree: N = 4T − 160 gives 4 × 65 − 160 = 100 chirps at 65°F.

Four ways a linear function can be handed to you

In every case, find the rate of change as the change in output divided by the change in input, and the initial value as the output at input 0, working back from a known point when you have to.

Worked examples

Worked example 1. In the game, how many hits can the character take before running out of health?

Step 1. H = −15h + 250 reaches 0 when 15h = 250, so h ≈ 16.7.

Step 2. Hits come in whole numbers. After 16 hits the character has 250 − 240 = 10 points left, and the 17th hit finishes it.

Worked example 2. For the savings function s = 6w + 19, what do 6 and 19 mean, and when will you reach $100?

Step 1. 6 means you add $6 each week; 19 means you had $19 when you started counting, at week 0.

Step 2. 6w + 19 = 100 gives 6w = 81 and w = 13.5, so you pass $100 during week 14.

Worked example 3. A candle is 12 cm tall after burning for 2 hours and 9 cm tall after 8 hours. Write its height as a function of time and interpret the numbers.

Step 1. Rate of change: (9 − 12) ÷ (8 − 2) = −3 ÷ 6 = −0.5 cm per hour.

Step 2. Initial value: 12 + 2 × 0.5 = 13 cm. Working back 2 hours adds back 1 cm.

Step 3. h = −0.5t + 13. The candle started 13 cm tall and burns 0.5 cm every hour. It burns out when 0.5t = 13, after 26 hours.

Worked example 4. A phone reads 100 percent at 8 a.m. and loses 12 percentage points every hour, as in Lesson 15. On the graph of B = −12t + 100, what does the point where the line crosses the t-axis mean?

Step 1. The line crosses the t-axis where B = 0, at t = 100 ÷ 12 ≈ 8.3 hours.

Step 2. It is the moment the battery dies, a little after 4:20 p.m. The model stops there: the battery cannot go below 0 percent, so the line beyond that point means nothing.

Worked example 5. Use Dolbear's rule to find the temperature when crickets chirp 140 times a minute.

Step 1. T = N/4 + 40 = 140/4 + 40 = 35 + 40 = 75°F.

Check with the other form: 4 × 75 − 160 = 140 chirps.

Common misconceptions

"The initial value is the first number in the table." Only if that row has input 0. The savings data started at week 2, so the initial value had to be worked back from $31 to $19.

"The rate of change is how much the output changes." It is how much the output changes for each 1 unit of input. A change of 40 chirps over 10 degrees is a rate of 4 chirps per degree, not 40.

"A rate of change cannot be negative." When the output falls as the input rises, as with the draining pool or the burning candle, the rate of change is negative.

"A model works for every input." Dolbear's rule gives −160 chirps at 0°F and the battery model gives negative charge after 8.3 hours. Linear models describe a situation over a limited range of inputs.

"Rate of change and initial value are new ideas, different from slope and intercept." For a linear function they are the same two numbers. The new names say what they mean in a situation.

The short version

• A linear function has the form y = mx + b. Its rate of change, m, is the change in output per unit of input; its initial value, b, is the output when the input is 0.
• From two points or a table: rate of change = change in output ÷ change in input, then work back to input 0 for the initial value. From a graph: read the slope and the vertical intercept. From words: the rate is a "per" quantity and the initial value is a starting amount.
• Dolbear's 1897 data, 80 chirps at 60°F and 120 at 70°F, give N = 4T − 160, or T = N/4 + 40: 4 chirps per degree.
• Interpret both numbers in context, and notice when the initial value, or part of the line, lies outside the range where the model makes sense.

Linear functions all grow by the same amount at each step. The next lesson meets functions that do not, and compares functions given in different forms: a table against an equation, a graph against words, and an allowance that adds against one that doubles.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Functions (8.F.A.3, 8.F.B.4). thecorestandards.org
2. Dolbear, A. E. (1897). The cricket as a thermometer. The American Naturalist, 31(371), 970-971. JSTOR Early Journal Content, Internet Archive. archive.org
3. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 8: Linear functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 9: Linear models. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 4.5, Use the slope-intercept form of an equation of a line (interpreting slope and intercept in applications). Elementary Algebra 2e. OpenStax. openstax.org

Two offers for thirty days

Your aunt offers you a summer deal and lets you choose. Offer A: $10 every day for 30 days. Offer B: 1 cent on the first day, 2 cents on the second, 4 cents on the third, doubling every day for 30 days. Which would you take?

Offer A sounds better, and for almost two weeks it is. Here are the running totals, with Offer B converted to dollars:

Offer B's total after d days is 2^d − 1 cents, because 1 + 2 + 4 + ... doubles up to one less than the next power of 2. After 13 days it is 2¹³ − 1 = 8,191 cents, still behind. On day 14 it jumps to 2¹⁴ − 1 = 16,383 cents, $163.83, and passes Offer A for good. After 30 days it is 2³⁰ − 1 cents. You worked out 2³⁰ = 1,073,741,824 in Lesson 1, so Offer B pays $10,737,418.23. Take Offer B.

This lesson is about comparing functions like these, given in different forms, and about the difference between functions that grow by the same amount each step and functions that do not. It covers Common Core standards 8.F.A.2, comparing properties of two functions each represented in a different way, and 8.F.A.3, interpreting y = mx + b as a linear function and giving examples of functions that are not linear.

Constant steps and growing steps

Look at how each total changes from one day to the next. Offer A always adds $10: the same step every time. Offer B adds 1 cent, then 2, then 4, then 8: the steps keep growing. That is the whole difference between a linear and a nonlinear function.

A linear function changes by the same amount for each equal step in the input, so its graph is a straight line with one slope. Offer A's total is T = 10d, a line through (0, 0) with slope 10. A nonlinear function changes by different amounts for equal steps, so its graph curves. Offer B's total, 2^d − 1 cents, curves upwards more and more steeply. It is not linear, and no single rate of change describes it.

The standard gives its own example: the area of a square as a function of its side length, A = s². Its graph contains (1, 1), (2, 4) and (3, 9). From the first point to the second the output rises 3; from the second to the third it rises 5. Equal steps in s, unequal steps in A, so the points do not lie on a straight line, and A = s² is not linear.

The point: In a table with equally spaced inputs, a linear function's outputs change by the same amount at every step. If the change varies, the function is not linear.

Three ways to tell

One caution about tables. If the inputs are not equally spaced, compare rates, not raw steps. For x = 0, 1, 3, 6 with y = 2, 5, 11, 20, the steps in y are 3, 6 and 9, which look unequal, but the steps in x are 1, 2 and 3. The rates are 3 ÷ 1 = 3, 6 ÷ 2 = 3 and 9 ÷ 3 = 3, all the same, so the function is linear: y = 3x + 2.

Comparing two functions given differently

The standard's example asks which of two linear functions has the greater rate of change when one is given as a table and the other as an equation. Turn each into its rate of change and initial value first, then compare.

Function P is the table x = 0, 2, 4 with y = 3, 7, 11. Its rate of change is 4 ÷ 2 = 2, and its initial value is 3.

Function Q is the equation y = 1.5x + 5. Its rate of change is 1.5 and its initial value is 5.

P has the greater rate of change; Q has the greater initial value. Q starts higher, P climbs faster, so P must catch up: 2x + 3 = 1.5x + 5 gives 0.5x = 2, so x = 4, where both equal 11. After that, P is greater.

Two more, one a graph and one in words. Function R is a straight line through (0, 10) and (5, 0): rate of change −10 ÷ 5 = −2, initial value 10. Function S: "a tank starts with 12 litres and drains 1.5 litres a minute": rate of change −1.5, initial value 12. Both decrease, and R decreases faster, because 2 is bigger than 1.5. S starts higher and falls more slowly, so S stays greater for every input from 0 on.

Worked examples

Worked example 1. Is the function in the table x = 1, 2, 3, 4 and y = 2, 5, 10, 17 linear?

Step 1. The x-steps are all 1. The y-steps are 3, 5 and 7.

Step 2. The steps are not equal, so the function is not linear. (It is y = x² + 1.)

Worked example 2. Which is linear: y = 3x − 7, y = 3/x, or y = x³?

Only y = 3x − 7, which has the form y = mx + b. In y = 3/x the input is in the denominator, and in y = x³ it is cubed. Check y = x³ with a table: x = 1, 2, 3 gives 1, 8, 27, with steps 7 and 19.

Worked example 3. Phone plan X is described as "$20 a month plus $3 per gigabyte". Plan Y's graph is a straight line through (0, 14) and (2, 24), with gigabytes across and dollars up. Compare them.

Step 1. X: rate $3 per GB, initial value $20.

Step 2. Y: rate (24 − 14) ÷ 2 = $5 per GB, initial value $14.

Step 3. Y starts cheaper but rises faster. They are equal when 3g + 20 = 5g + 14, so 2g = 6 and g = 3 GB, at $29. Below 3 GB choose Y; above it, X.

Worked example 4. A bacteria colony in a lab dish doubles every hour, starting from 500 cells. Is the number of cells a linear function of time?

Step 1. Hours 0, 1, 2, 3 give 500, 1,000, 2,000, 4,000 cells.

Step 2. The steps are 500, 1,000, 2,000: growing, not equal. The function is nonlinear, like Offer B.

Worked example 5. Is the area of a rectangle 3 cm wide a linear function of its length? Is the area of a square a linear function of its side?

Step 1. A rectangle 3 cm wide has area A = 3L. Each extra centimetre of length adds 3 cm²: linear.

Step 2. A square has A = s². Its side grows in both directions at once, so each extra centimetre adds more area than the last: 1, 4, 9, 16, with steps 3, 5, 7. Not linear.

Common misconceptions

"The doubling offer is better from the start." It is far behind at first: after 10 days it has paid $10.23 against $100. Nonlinear growth can start slow and then overtake any linear function.

"Unequal y-steps always mean nonlinear." Only when the x-steps are equal. With x = 0, 1, 3, 6, the y-steps 3, 6, 9 still give a constant rate of 3.

"y = 5 is not a linear function, because it has no x." It is y = 0x + 5, a horizontal line with rate of change 0.

"To compare rates, compare the biggest numbers you can see." Plan X shows a 20 and Plan Y a 24, but their rates are 3 and 5. Convert each representation to its rate of change before comparing.

"Nonlinear means decreasing." Nonlinear means the rate of change is not constant. Offer B is nonlinear and increasing; the draining tank in Function S is linear and decreasing.

Recap

• A linear function changes by the same amount for each equal step in its input; its graph is a straight line, y = mx + b.
• Nonlinear functions, such as A = s² and a doubling allowance, change by different amounts for equal steps, and their graphs curve.
• Check a table with equally spaced inputs by its steps, and a table with uneven inputs by its rates.
• To compare two functions given in different forms, find each one's rate of change and initial value, then compare them, and find where they are equal if you need to know when one overtakes the other.
• A nonlinear function can start far behind a linear one and still overtake it: one cent doubled for 30 days beats $10 a day.

Some graphs come with no equation at all: a sketch of a phone battery through a day, or of your distance from home on a bike ride. The next lesson reads the story in the shape of a graph, and starts with a mistake almost everybody makes.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Functions (8.F.A.2, 8.F.A.3). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 4: Tables, equations, and graphs of functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 7: Connecting representations of functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., and Honeycutt Mathis, A. (2020). Section 3.6, Graphs of functions. Intermediate Algebra 2e. OpenStax. openstax.org

Speed over a hill

A cyclist rides along a flat road, then up a long hill, over the top, down the other side, and onto flat road again. Sketch a graph of her speed against time.

Many people draw a hill: a line that rises, peaks and falls. It feels right, because the ride was over a hill. It is wrong. The graph is a picture of her speed, not of the road. Going uphill she slows down, so her speed falls. Near the top it is at its lowest. Going downhill she speeds up, so her speed rises, faster than she was going before the hill. Back on the flat, it settles to a steady value. The speed graph looks more like a valley than a hill.

The mistake is common enough to have a name. Lisa Murphy's review of research on how students read graphs calls it the graph-as-picture interpretation: expecting the graph to look like the scene. The review reports it in middle school students, high school students and college physics students, and cites a 1987 study by Lillian McDermott and colleagues in which college physics students struggled to separate the shape of a graph from the path of the motion.

This lesson covers Common Core standard 8.F.B.5: describing the relationship between two quantities qualitatively by analysing a graph, for example where the function is increasing or decreasing, linear or nonlinear, and sketching a graph that shows the features of a function described in words.

Where "it's a hill, so the graph is a hill" breaks

The picture-reading fails at the axes. The horizontal axis is time, not position along the road, and the vertical axis is speed, not height above sea level. Every point on the graph says "at this time, she was going this fast". The road's shape only matters through what it does to her speed.

A second misreading is close behind. Murphy's review names it slope/height confusion: reading the height of a graph when you should read its steepness, or the other way round. On a graph of distance against time, how fast the cyclist is going is shown by how steep the graph is, not by how high it is. A high, flat stretch means she is far from home and not moving. A low, steep stretch means she is near home and moving fast.

Remember: Read a graph through its axes. Ask what the input and output are, then read rising, falling and flat as changes in the output over time, never as the shape of the scene.

The vocabulary of shape

Reading a bike ride

A graph shows Jamal's distance from home, in kilometres, against time in minutes. It is made of straight pieces joining (0, 0), (10, 2), (25, 2) and (35, 0).

From 0 to 10 minutes the graph rises steadily from 0 to 2 km: he rides away from home at 2 ÷ 10 = 0.2 km per minute. From 10 to 25 minutes it is flat at 2 km: he is 2 km from home and not moving, perhaps at a friend's house. From 25 to 35 minutes it falls steadily back to 0: he rides home at 0.2 km per minute.

The flat piece does not mean he rode on flat ground; it means his distance from home did not change. A graph made of straight pieces like this is called a piecewise linear function: each piece is linear, with its own rate of change.

Sketching from a story

To sketch a graph from words, you need the shape, not exact numbers.

Step 1. Decide the input and output and label the axes.

Step 2. Break the story into stages.

Step 3. For each stage, decide whether the output increases, decreases or stays constant, and whether it changes steadily (a straight piece) or speeds up or slows down (a curve).

Step 4. Join the pieces in order, left to right, so that each input has one output.

Story: "Ana walks to school. She sets off slowly, then speeds up when she realises she is late, then stops at the gate." Distance from home against time: starting at 0, the graph rises gently, then rises more and more steeply as she speeds up, which is a curve bending upwards, then goes flat at the school's distance when she stops.

Worked examples

Worked example 1. Describe the graph of a phone's battery charge through a day: charging overnight to 100 percent by 7 a.m., draining slowly through the morning, faster during an hour of gaming at lunch, then charging again from 4 p.m.

Increasing until 7 a.m.; then decreasing gently; then decreasing steeply for the gaming hour; then decreasing gently again until 4 p.m.; then increasing. The steepness of each falling piece shows how fast the charge drops.

Worked example 2. A bathtub fills with the tap running, someone gets in, and later the plug is pulled. Sketch water depth against time.

The depth rises steadily while the tap runs. When the person gets in, the depth jumps up quickly, because their body pushes water up. It stays constant while they soak with the tap off, then falls steadily to 0 after the plug is pulled.

Worked example 3. A ball is thrown straight up and caught at the same height. Describe its height against time.

The height increases quickly at first, then more slowly as the ball slows, reaches a highest point, then decreases, slowly at first and then faster. The graph is a curve, so the function is nonlinear, increasing then decreasing. This time the graph happens to look like the ball's path drawn sideways, but only because height is the output and time runs across.

Worked example 4. On a distance-from-home graph, piece A rises from (0, 0) to (5, 3) and piece B rises from (20, 3) to (22, 4). On which piece is the person moving faster?

Piece A: 3 ÷ 5 = 0.6 km per minute. Piece B: 1 ÷ 2 = 0.5 km per minute. A is faster, even though B is higher on the graph. Speed is steepness, not height.

Worked example 5. Which situation matches a graph that is constant, then increases linearly, then is constant again? (a) A parked car that drives off and then stops. (b) Distance from home for a car parked in the driveway, then driven away at a steady speed, then parked at a shop. (c) Speed of a car that brakes to a stop.

(b): distance from home is constant while it is parked, increases steadily as it drives away, then is constant at the shop. For (a), the speed graph would jump up from 0 and then fall back to 0, and (c) is decreasing.

Common misconceptions

"Riding over a hill gives a hill-shaped graph." That reads the graph as a picture. A speed graph for a ride over a hill dips, because she slows going up and speeds up going down.

"Higher on the graph means faster." On a distance-time graph, height is distance; speed is the steepness. A high, flat piece means far away and stopped.

"A flat piece always means stopped." It means the output is not changing. On a distance graph that is stopped; on a speed graph a flat piece means moving at a steady speed.

"Going down on the graph means going downhill." On a distance-from-home graph it means getting closer to home, whether the road goes up or down.

"A graph can turn back to the left." A graph against time cannot, because each moment has one value. That is the function rule from Lesson 22 at work.

Where this leaves us

• A graph shows how an output changes with an input. Read it through its axes, not as a picture of the scene.
• Describe a graph by its pieces: increasing, decreasing or constant, and linear (straight) or nonlinear (curved). Steeper means a faster rate of change.
• On distance-time graphs, speed is shown by steepness, not by height.
• To sketch from a story, label the axes, split the story into stages, choose the shape of each stage, and join them from left to right.

That completes functions for eighth grade. The next module returns to geometry with one of the most famous facts in mathematics: in a right triangle, the squares on the two shorter sides add up to the square on the longest. A United States congressman found his own proof of it in 1876.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Functions (8.F.B.5). thecorestandards.org
2. Murphy, L. D. (1999). Graphing misinterpretations and microcomputer-based laboratory instruction, with emphasis on kinematics [Review paper]. University of Illinois at Urbana-Champaign. mste.illinois.edu
3. McDermott, L. C., Rosenquist, M. L., and van Zee, E. H. (1987). Student difficulties in connecting graphs and physics: Examples from kinematics. American Journal of Physics, 55(6), 503-513. (Cited as reported in Murphy's review.)
4. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 6: Even more graphs of functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 10: Piecewise linear functions. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

A proof from the House of Representatives

On 7 March 1876, James Garfield, then a member of Congress from Ohio, wrote in his diary that he had shown a mathematics professor a new proof. Later that year the New-England Journal of Education printed it. The editors explained that Garfield had hit on it during mathematical amusements and discussions with other members of Congress, and joked that it was something both houses could agree on without regard to party. Five years later Garfield became president of the United States.

The fact he proved is more than two thousand years old. In any right triangle, call the two sides that meet at the right angle the legs, with lengths a and b, and the side opposite the right angle the hypotenuse, with length c. The hypotenuse is always the longest side. Then

a² + b² = c².

That is the Pythagorean theorem. Euclid proved it as Proposition 47 of Book I of the Elements, and people have kept finding new proofs ever since; Garfield's filled only the bottom third of one column of the journal. This lesson explains two proofs and puts the theorem to work. It covers Common Core standards 8.G.B.6, explaining a proof of the Pythagorean theorem, and 8.G.B.7, applying it to find unknown side lengths in right triangles.

Testing it with numbers first

Take the right triangle with legs 3 and 4. On a grid, draw it with corners at (0, 0), (4, 0) and (0, 3). The theorem says the hypotenuse c satisfies 3² + 4² = c²: 9 + 16 = 25, so c = 5. Measure it with a ruler on squared paper and you get 5 units.

The theorem is about squares in the literal sense. Build a square on each side of the triangle. The square on the 3-unit leg has area 9, the square on the 4-unit leg has area 16, and the square on the hypotenuse has area 25. The two smaller squares together exactly match the big one.

Proof 1: one big square, arranged two ways

Take any right triangle, with legs a and b and hypotenuse c, and make four copies of it. Now draw a big square whose sides are a + b long. You can fit the four triangles into it in two different ways.

Arrangement 1. Put one triangle in each corner of the big square, each with its right angle in the corner, so that each side of the big square is made of one leg a and one leg b. The four hypotenuses then form a tilted square in the middle, with side c. Its corners are a fixed distance along each side of the big square: for a big square with corners (0, 0), (a + b, 0), (a + b, a + b) and (0, a + b), the tilted square's corners are (a, 0), (a + b, a), (b, a + b) and (0, b). The uncovered area is the tilted square: c².

Arrangement 2. Slide the same four triangles around so that they pair up into two rectangles, each a by b, in two opposite corners of the big square. The uncovered area is now two squares, one a by a and one b by b: a² + b².

The big square is the same size both times, and the four triangles cover the same area both times. So the uncovered areas must be equal:

c² = a² + b².

Check with the 3-4-5 triangle. The big square is 7 by 7, area 49. The four triangles each have area ½ × 3 × 4 = 6, so 24 in total. Uncovered area: 49 − 24 = 25. In arrangement 1 that is c² = 5² = 25, and in arrangement 2 it is 3² + 4² = 9 + 16 = 25.

Nothing in the argument used the numbers 3 and 4, so it works for every right triangle.

Proof 2: Garfield's trapezoid

Garfield used half the picture. Take two copies of the right triangle. Stand the first one up with its leg a vertical and its leg b along the ground. Put the second one next to it, turned so that its leg a continues along the ground and its leg b stands vertical. Join the tops of the two vertical legs. The shape you get is a trapezoid: its two vertical sides, a and b, are parallel, and the distance between them, along the ground, is a + b.

Inside the trapezoid are three triangles: the two copies, and between them a third triangle whose two sides are both hypotenuses, length c. The angles where the two copies meet add up with the third triangle's angle to a straight line, and since a right triangle's two smaller angles add to 90°, that third angle is 180° − 90° = 90°. So the middle triangle is a right triangle with legs c and c.

Now find the trapezoid's area two ways.

As a trapezoid: the average of the parallel sides times the distance between them, (a + b)/2 × (a + b) = ½(a + b)².

As three triangles: ½ab + ½ab + ½c² = ab + ½c².

Set them equal and double both sides: (a + b)² = 2ab + c². Expand the left side, using the area of an (a + b)-by-(a + b) square split into an a-by-a square, a b-by-b square and two a-by-b rectangles: a² + 2ab + b² = 2ab + c². Subtract 2ab from both sides: a² + b² = c².

Check with 3, 4, 5: the trapezoid's area is (3 + 4)/2 × 7 = 24.5, and the triangles give 6 + 6 + 12.5 = 24.5.

Why this matters: Both proofs work by computing one area in two ways. Once you have seen why a² + b² = c² must hold, it is a tool you can trust for any right triangle, including ones far too big to measure.

Using the theorem

To find the hypotenuse, add the squares of the legs and take the square root: c = √(a² + b²). To find a leg, subtract: b² = c² − a², so b = √(c² − a²). Lengths are positive, so take the positive square root.

Worked examples

Worked example 1. A right triangle has legs 6 cm and 8 cm. Find the hypotenuse.

Step 1. c² = 6² + 8² = 36 + 64 = 100.

Step 2. c = √100 = 10 cm.

Worked example 2. A right triangle has hypotenuse 13 and one leg 5. Find the other leg.

Step 1. b² = 13² − 5² = 169 − 25 = 144.

Step 2. b = √144 = 12.

Check: 5² + 12² = 25 + 144 = 169 = 13².

Worked example 3: a ladder. OSHA's construction rule for ladders that lean against a wall says the foot should be set out from the wall about one quarter of the ladder's working length, the distance along the ladder from its foot to its top support. A ladder has a working length of 16 feet. How high up the wall does it reach?

Step 1. The foot is 16 ÷ 4 = 4 feet from the wall. The ladder is the hypotenuse, 16 feet; the ground is one leg, 4 feet; the wall is the other leg, h.

Step 2. h² = 16² − 4² = 256 − 16 = 240.

Step 3. h = √240 ≈ 15.5 feet. (15.5² = 240.25, just over 240.)

Worked example 4: a baseball diamond. The Official Baseball Rules lay out the infield as a square with 90 feet between bases. How far is the catcher's throw from home plate to second base?

Step 1. Home, first base and second base form a right triangle with legs 90 and 90; the throw is the hypotenuse.

Step 2. c² = 90² + 90² = 8,100 + 8,100 = 16,200, so c = √16,200 = 90√2 ≈ 127.28 feet.

Step 3. Rule 2.01 tells groundskeepers to measure 127 feet 3 3/8 inches from home plate to place second base. That is 127.28125 feet: the rule writers did this calculation, and their figure is within a fiftieth of an inch of 90√2. The exact answer is irrational, like the diagonal in Lesson 7.

Worked example 5: the ramp. Lesson 14's ramp rises 24 inches over a horizontal run of 288 inches. How long is its sloping surface?

Step 1. s² = 24² + 288² = 576 + 82,944 = 83,520.

Step 2. s = √83,520 ≈ 289.0 inches, only about an inch longer than the run. For a gentle slope, the surface and the run are nearly the same.

Common misconceptions

"c can be any side." c is the hypotenuse: the side opposite the right angle, and always the longest. Putting a leg in place of c gives nonsense, such as a leg longer than the hypotenuse.

"a + b = c." The theorem is about the squares of the sides. 3 + 4 = 7, but the hypotenuse is 5. Two sides of a triangle always add up to more than the third.

"To find a leg, add the squares." Adding gives the hypotenuse. To find a leg, subtract the square of the other leg from the square of the hypotenuse.

"√(a² + b²) = a + b." Square roots do not split over addition: √(36 + 64) = √100 = 10, not 6 + 8 = 14.

"It works for every triangle." Only for right triangles. A triangle with sides 4, 5 and 6 has 4² + 5² = 41, not 36. The next lesson shows that this test can tell you whether a triangle has a right angle at all.

What to remember

• In a right triangle with legs a and b and hypotenuse c, a² + b² = c². The hypotenuse is opposite the right angle and is the longest side.
• The rearranged-square proof and Garfield's trapezoid proof both find one area in two ways, which forces a² + b² = c² for every right triangle.
• Hypotenuse: c = √(a² + b²). Leg: b = √(c² − a²).
• Answers are often irrational, such as 90√2 ≈ 127.28 feet from home plate to second base; round sensibly for the situation.

Builders use the theorem backwards: measure the three sides of a corner, and if they fit a² + b² = c², the corner is square. That reverse statement, the converse, is the next lesson, along with right triangles hiding inside boxes.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.B.6, 8.G.B.7). thecorestandards.org
2. Lamb, E. (2012, November 19). A presidential Pythagorean proof. Scientific American (Observations blog). scientificamerican.com
3. Euclid. (c. 300 BCE/1996). Elements, Book I, Proposition 47 (D. E. Joyce, Ed.). Clark University. mathcs.clarku.edu
4. Occupational Safety and Health Administration. (n.d.). 29 CFR 1926.1053, Ladders, paragraph (b)(5)(i). U.S. Department of Labor. osha.gov
5. Mosher, R. (n.d.). Baseball diamond construction (quoting Official Baseball Rule 2.01). School of Mathematics, University of Minnesota. mosher.math.umn.edu
6. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 7: A proof of the Pythagorean theorem. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Is the corner square?

You are building a raised garden bed out of boards, and the corners have to be right angles or the bed will be lopsided. You have a tape measure but no tool for measuring angles. Mark a point 3 feet along one board from the corner, and a point 4 feet along the other. Measure straight across between the marks. If the distance is exactly 5 feet, the corner is square. If it is 5 feet and 2 inches, it is not.

Why does that work? The last lesson said that a right triangle's sides satisfy a² + b² = c². The corner check runs that statement backwards: if the sides satisfy a² + b² = c², the triangle has a right angle. That backwards statement is called the converse, and it needs its own reason. This lesson gives one, then uses the theorem twice in a row to measure across the inside of a box. It covers Common Core standards 8.G.B.6, explaining a proof of the converse of the Pythagorean theorem, and 8.G.B.7, applying the theorem in two and three dimensions.

The converse, and why it is true

Converse of the Pythagorean theorem: if a triangle has sides a, b and c with a² + b² = c², then the triangle has a right angle, opposite the side c.

A converse is not automatically true just because the original statement is. ("If it is a dog, it has four legs" is true; "if it has four legs, it is a dog" is not.) So here is the argument, which Euclid gave as Proposition 48 of Book I, right after the theorem itself.

Step 1. Start with a triangle whose sides a, b and c satisfy a² + b² = c². You do not know yet whether it has a right angle.

Step 2. Build a second triangle that definitely has a right angle, with legs a and b. Call its hypotenuse h.

Step 3. By the Pythagorean theorem, h² = a² + b². But a² + b² = c², so h² = c², and since lengths are positive, h = c.

Step 4. Now the two triangles have the same three side lengths: a, b and c. Triangles with the same three side lengths are copies of each other: one can be moved onto the other by rigid motions, as in Lesson 9. So they have the same angles, and since the second triangle has a right angle, so does the first.

In short: If the squares of the two shorter sides add to the square of the longest, the triangle is a right triangle. That gives you a test for right angles that needs only a tape measure.

Using the test

Always use the longest side as c, because the hypotenuse is the longest side of a right triangle.

The last two rows fail in different directions. When a² + b² is less than c², as with 7, 8, 11, the longest side is too long for a right angle, so the angle opposite it is wider than 90°. When a² + b² is more than c², as with 6, 7, 9, every angle is less than 90°. Either way, no right angle.

For the garden bed: 3² + 4² = 25 = 5², so a 5-foot diagonal means a right angle. A diagonal of 5 feet 2 inches is longer than 5, so the corner is opened wider than a right angle and needs pushing in.

Pythagorean triples

Three whole numbers that fit a² + b² = c² are called a Pythagorean triple: 3, 4, 5 and 5, 12, 13 are the best known. People found them very early. A clay tablet from Old Babylonian Mesopotamia, now at Columbia University and known as Plimpton 322, is a table that, according to the curators of New York University's "Before Pythagoras" exhibition, shows the Babylonians had a method for finding such triples more than a thousand years before Euclid. Scholars still debate how the table was calculated and what it was for.

You can make new triples from old ones by scaling. Multiply 3, 4, 5 by 2 to get 6, 8, 10: 36 + 64 = 100. By 4: 12, 16, 20: 144 + 256 = 400. This works because the bigger triangle is a dilation of the smaller one, and dilations keep angles, including the right angle. A builder checking a big corner can use 6, 8, 10 or 9, 12, 15 for more accuracy.

Right triangles inside a box

A shoebox measures 30 cm long, 20 cm wide and 12 cm tall. What is the longest pencil that fits inside, lying in a straight line from one bottom corner to the opposite top corner?

The pencil is the hypotenuse of a right triangle standing up inside the box. One leg is the box's height, 12 cm, going straight up the far corner. The other leg is the diagonal of the box's floor, from the starting corner to the corner directly below the pencil's tip. You need that floor diagonal first, so the theorem is used twice.

Step 1. The floor is a 30 by 20 rectangle. Its diagonal d satisfies d² = 30² + 20² = 900 + 400 = 1,300.

Step 2. The pencil, p, is the hypotenuse of a right triangle with legs d and 12, because the height is straight up from the floor: p² = d² + 12² = 1,300 + 144 = 1,444.

Step 3. p = √1,444 = 38 cm.

Notice that you never needed d itself, only d² = 1,300. Putting the two steps together: the diagonal of a box is √(length² + width² + height²).

Worked examples

Worked example 1. Is a triangle with sides 10, 24 and 26 a right triangle?

Step 1. The longest side is 26, so c = 26.

Step 2. 10² + 24² = 100 + 576 = 676, and 26² = 676. Yes: it is the 5, 12, 13 triple doubled.

Worked example 2. A carpenter checks a corner by marking 6 feet along one wall and 8 feet along the other, and measures 10 feet 3 inches between the marks. Is the corner square?

Step 1. For a square corner the diagonal should be √(36 + 64) = 10 feet exactly.

Step 2. 10 feet 3 inches is too long, so the angle is wider than 90°. Not square.

Worked example 3. A room is 12 feet long, 9 feet wide and 8 feet high. How long is the diagonal from one floor corner to the opposite ceiling corner?

Step 1. Floor diagonal: 12² + 9² = 144 + 81 = 225, so 15 feet.

Step 2. Room diagonal: 15² + 8² = 225 + 64 = 289, so 17 feet.

Worked example 4. A cube has edges of 1 m. How long is its diagonal?

√(1 + 1 + 1) = √3 ≈ 1.73 m, another irrational length.

Worked example 5. Find a Pythagorean triple whose shortest side is 15.

Scale 3, 4, 5 by 5: 15, 20, 25. Check: 225 + 400 = 625 = 25². Scaling 5, 12, 13 by 3 gives another: 15, 36, 39.

Common misconceptions

"Test a² + b² = c² with any side as c." c must be the longest side. With 5, 12, 13, trying 13² + 5² = 12² fails even though the triangle is a right triangle.

"If the sides do not fit a² + b² = c², they cannot make a triangle." They can make a triangle; it just has no right angle. 7, 8, 11 is a perfectly good triangle with one wide angle.

"A converse is always true when the original statement is." Some are not; this one is, because of the argument with the second triangle.

"The diagonal of a box is √(length² + width²)." That is only the diagonal of the floor. The height has to come in too: √(length² + width² + height²).

"Scaling a Pythagorean triple spoils it." Multiplying all three sides by the same number gives a similar triangle with the same right angle, so 6, 8, 10 is a triple just like 3, 4, 5.

Pulling it together

• Converse: if a² + b² = c², with c the longest side, the triangle has a right angle opposite c.
• Why: a right triangle built with legs a and b has hypotenuse √(a² + b²) = c, so the two triangles share all three sides and are copies.
• The converse checks corners with a tape measure: 3, 4, 5 or any scaled triple.
• Pythagorean triples are whole-number solutions such as 3, 4, 5 and 5, 12, 13; any multiple of a triple is a triple.
• In three dimensions, use the theorem twice: the diagonal of a box is √(l² + w² + h²).

On a coordinate grid, every slanted segment is the hypotenuse of a right triangle whose legs run along the grid lines. The next lesson uses that to measure the straight-line distance between any two points, and compares it with the distance along the streets of a city grid.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.B.6, 8.G.B.7). thecorestandards.org
2. Euclid. (c. 300 BCE/1996). Elements, Book I, Proposition 48 (D. E. Joyce, Ed.). Clark University. mathcs.clarku.edu
3. Institute for the Study of the Ancient World. (2010). Plimpton 322. In Before Pythagoras: The culture of Old Babylonian mathematics [Exhibition]. New York University. isaw.nyu.edu
4. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 9: The converse. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 10: Applications of the Pythagorean theorem. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

Fourteen blocks or ten

On a map of a town laid out in square blocks, your home is at (1, 2) and the skate park is at (7, 10), with one unit for each block. Walking there along the streets, you must go 6 blocks east and 8 blocks north, in some order: 14 blocks however you zigzag. A drone delivering your forgotten helmet flies straight. How far does it fly?

The drone's path is the hypotenuse of a right triangle whose legs are the walk's two parts: 6 across and 8 up. So its length is √(6² + 8²) = √(36 + 64) = √100 = 10 blocks. The drone saves 4 blocks.

Two kinds of distance came up there, and this lesson compares them. It covers Common Core standard 8.G.B.8: applying the Pythagorean theorem to find the distance between two points in a coordinate system.

Two ways to measure distance

The distance along the grid lines has a name of its own. MathWorld lists it as the taxicab metric, also called the city block distance or Manhattan distance: the distance a taxi would drive on a grid of streets, which is the horizontal change plus the vertical change. The straight-line distance is the one used in geometry, and when people say "the distance between two points" they mean this one.

Read the last two columns down. The straight-line distance is never longer than the grid distance, and it is only equal when the two points are on the same grid line, as with (2, 3) and (2, 9). Every other time, the straight line cuts the corner. That is the Pythagorean theorem at work: the hypotenuse is shorter than the two legs added together.

What matters here: The straight-line distance between two points is the hypotenuse of the right triangle formed by the horizontal and vertical changes between them.

The distance formula

The horizontal leg of the triangle is the change in x, and the vertical leg is the change in y, exactly the run and the rise from Lesson 14. For points (x₁, y₁) and (x₂, y₂):

distance = √((x₂ − x₁)² + (y₂ − y₁)²).

A change can be negative, as when you go left or down, but squaring makes it positive, so the order of the points does not matter. From (−4, 3) to (5, −9) the changes are 9 and −12; squared, 81 and 144; total 225; distance 15. From (5, −9) back to (−4, 3) the changes are −9 and 12, which square to the same numbers.

You do not have to memorise the formula. Drawing the right triangle on the grid and using the Pythagorean theorem gives the same answer every time.

Worked examples

Worked example 1. Find the distance between (−3, 4) and (5, −2).

Step 1. Across: 5 − (−3) = 8. Up or down: −2 − 4 = −6.

Step 2. Distance = √(8² + (−6)²) = √(64 + 36) = √100 = 10.

Worked example 2. Find the distance between (−2, −1) and (3, 4).

Step 1. Across: 5. Up: 5.

Step 2. Distance = √(25 + 25) = √50 ≈ 7.07. The exact answer is irrational, as with the diagonal of a square, because these two points are opposite corners of a 5-by-5 square.

Worked example 3. In a game, your character is at (−4, 3) and a power-up is at (5, −9). Each unit is 1 metre. The power-up can be collected from up to 15 m away. Can you collect it without moving?

Step 1. Across 9, down 12: distance √(81 + 144) = √225 = 15 m.

Step 2. Exactly 15 m, so yes, just.

Worked example 4. A triangle has corners A(1, 1), B(4, 5) and C(8, 2). Find its side lengths and decide whether it has a right angle.

Step 1. AB: across 3, up 4: √25 = 5.

Step 2. BC: across 4, down 3: √25 = 5.

Step 3. AC: across 7, up 1: √50 ≈ 7.07.

Step 4. Converse test with the longest side: AB² + BC² = 25 + 25 = 50 = AC². So there is a right angle at B, the corner opposite AC. The triangle is also isosceles, with two sides of 5. Its perimeter is 5 + 5 + √50 ≈ 17.07.

Worked example 5. Is the point (6, 8) exactly 10 units from the origin? Is (7, 7)?

Step 1. (6, 8): √(36 + 64) = √100 = 10. Yes.

Step 2. (7, 7): √(49 + 49) = √98 ≈ 9.9. No, just inside. All the points exactly 10 units from the origin form a circle; (6, 8) is on it and (7, 7) is not.

Common misconceptions

"The distance is the change in x plus the change in y." That is the grid distance, the walk along the streets. The straight-line distance is √(change in x² + change in y²), which is shorter unless the points share a grid line.

"√(36 + 64) = 6 + 8." Add the squares first, then take one square root: √100 = 10, not 14.

"Subtract the coordinates of the same point." Subtract x from x and y from y: x₂ − x₁ and y₂ − y₁. Working out x₁ − y₁ mixes up across and up.

"A negative change gives a negative distance." Each change is squared before adding, so the sign disappears. Distances are never negative.

"Points on the same vertical line need the formula." They can use it, but there is no triangle: the distance is just the difference in y, as from (2, 3) to (2, 9), which is 6.

The takeaway

• The straight-line distance between two points is the hypotenuse of a right triangle whose legs are the horizontal and vertical changes between them.
• distance = √((x₂ − x₁)² + (y₂ − y₁)²); the order of the points does not matter.
• Grid distance, also called taxicab or city block distance, adds the two changes. It equals the straight-line distance only when the points share a grid line.
• Side lengths from coordinates let you find perimeters and use the converse to spot right angles.

From flat figures, the next module moves to solids: how much a can, a cone of popcorn or a basketball holds, and the result Archimedes asked to have carved on his tomb.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.B.8). thecorestandards.org
2. Barile, M. (n.d.). Taxicab metric. MathWorld, a Wolfram Web Resource. mathworld.wolfram.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 8, Lesson 11: Finding distances in the coordinate plane. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., and Honeycutt Mathis, A. (2020). Section 11.1, Distance and midpoint formulas; circles. Intermediate Algebra 2e. OpenStax. openstax.org

How much does the can hold?

A can is 6.6 cm across and 12.2 cm tall. How much does it hold? The can is a cylinder: two equal circles, one on top of the other, joined by a curved side. A cylinder is built like a stack of identical coins, and that picture gives its volume.

In seventh grade you found the volume of a prism, such as a box, as the area of its base times its height: each centimetre of height adds one more layer the size of the base. A cylinder works the same way. Its base is a circle, so each centimetre of height adds a layer of area πr², and the volume is

V = πr²h,

where r is the radius of the base and h is the height. For the can, the radius is half of 6.6, so r = 3.3 cm.

Step 1. Base area: π × 3.3² = π × 10.89 ≈ 34.21 cm².

Step 2. Volume: 34.21 × 12.2 ≈ 417.4 cm³.

A cubic centimetre is a millilitre, so the can holds about 417 mL when filled to the very top. This lesson covers Common Core standard 8.G.C.9: knowing the formulas for the volumes of cones, cylinders and spheres, and using them to solve real-world and mathematical problems. Cylinders and cones come first; spheres are the next lesson.

The method for a cylinder

Worked example 1. A cylinder has radius 5 cm and height 10 cm. Find its volume.

Step 1. Base area: π × 5² = 25π cm².

Step 2. Volume: 25π × 10 = 250π cm³.

Step 3. As a decimal: 250 × 3.1416 ≈ 785.4 cm³.

Writing the answer as 250π keeps it exact; the decimal is for measuring out real water. Using 3.14 for π gives 785 cm³, close enough for most purposes.

Worked example 2: a diameter instead of a radius. A cylindrical pencil cup is 8 cm across and 15 cm tall. Find its volume.

Step 1. The radius is half the diameter: r = 4 cm.

Step 2. V = π × 4² × 15 = π × 16 × 15 = 240π ≈ 754 cm³.

Had you used 8 as the radius, you would have got π × 64 × 15 = 960π, four times too much. Squaring a doubled radius quadruples the answer.

The cone: a third of the cylinder

A cone has a circular base and a curved side that narrows to a single point, the apex. Its height is the straight-up distance from the base to the apex, not the slanted distance along its side.

Take a cone and a cylinder with the same base and the same height, fill the cone with rice or water, and pour it into the cylinder. Illustrative Mathematics' lesson on cones describes exactly this experiment. It takes three cones to fill the cylinder. So a cone's volume is one third of its cylinder's:

V = (1/3)πr²h.

The same one third turns up for a pyramid and the prism with the same base and height, and a proof that it must be exactly one third, not roughly, comes in later courses.

Worked example 3. A snow cone cup is a cone with radius 3 cm and height 4 cm. How much does it hold, level with the top?

Step 1. The matching cylinder: π × 3² × 4 = 36π cm³.

Step 2. The cone holds a third: 36π ÷ 3 = 12π ≈ 37.7 cm³.

The upshot: Cylinder: base area times height. Cone with the same base and height: one third of that. Both start from the area of the circular base, πr².

Working backwards

Worked example 4: a missing height. A cylinder has volume 200π cm³ and radius 5 cm. How tall is it?

Step 1. V = πr²h, so 200π = π × 25 × h.

Step 2. Divide both sides by 25π: h = 200 ÷ 25 = 8 cm.

Check: π × 25 × 8 = 200π.

Worked example 5: a missing radius. A cylinder holds 360π cm³ and is 10 cm tall. What is its radius?

Step 1. 360π = π × r² × 10, so r² = 360 ÷ 10 = 36.

Step 2. r = √36 = 6 cm, the positive square root, since a radius is a length. Lesson 5's square roots are at work here.

Worked example 6: a cone's missing height. A cone has volume 50π cm³ and radius 5 cm. Find its height.

Step 1. 50π = (1/3)π × 25 × h.

Step 2. Multiply both sides by 3: 150π = 25πh, so h = 6 cm.

Comparing containers

Worked example 7. A cinema sells popcorn in a cone 10 cm across and 12 cm tall, or in a cylindrical cup 8 cm across and 10 cm tall, for the same price. Which holds more?

Step 1. Cone: r = 5, so V = (1/3)π × 25 × 12 = 100π ≈ 314 cm³.

Step 2. Cup: r = 4, so V = π × 16 × 10 = 160π ≈ 503 cm³.

Step 3. The cup holds 60π ≈ 188 cm³ more, even though the cone is wider and taller. The cone's narrowing costs it two thirds of its cylinder.

The last two rows explain a common surprise: doubling the height doubles the volume, but doubling the radius multiplies it by four, because the radius is squared.

Common misconceptions

"Use the diameter in the formula." The formula needs the radius, half the diameter. Using the diameter makes the answer four times too big.

"V = 2πrh." 2πr is the circumference of the base, a length. The volume needs the base's area, πr².

"A cone is half its cylinder." It is one third. The rice experiment takes three cones, not two.

"A cone's height is the slanted side." The height is the perpendicular distance from the base to the apex. The slanted side is longer, as the Pythagorean theorem would show.

"Volumes are in square units." Volume counts cubes, so its units are cubic: cm³, m³, or millilitres and litres.

Summing up

• A cylinder's volume is its base area times its height: V = πr²h.
• A cone holds a third of the cylinder with the same base and height: V = (1/3)πr²h.
• Use the radius, not the diameter, and the perpendicular height, not the slant. Answers in cubic units, exact in terms of π or rounded.
• To find a missing height, divide the volume by the base area (and multiply by 3 for a cone). To find a missing radius, solve for r² and take the positive square root.
• Doubling the height doubles the volume; doubling the radius multiplies it by four.

The third shape in the standard is the sphere, the shape of a basketball or a planet. Its formula was found by Archimedes, who was so proud of it that he asked for it to be marked on his tomb. That is the next lesson.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.C.9). thecorestandards.org
2. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 13: The volume of a cylinder. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
3. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 15: The volume of a cone. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Marecek, L., Anthony-Smith, M., and Honeycutt Mathis, A. (2020). Section 9.6, Solve geometry applications: Volume and surface area. Prealgebra 2e. OpenStax. openstax.org

A column in the brambles

In 75 BCE, the Roman orator Cicero was serving as a quaestor, a financial official, in Sicily. He went looking for the grave of Archimedes, the great mathematician of Syracuse, who had died more than a century before. Cicero later wrote, in his Tusculan Disputations, that the people of Syracuse knew nothing about it and even denied that it existed. But he remembered some lines of verse said to be inscribed on the tomb, describing a sphere and a cylinder set on top of it. Searching among the many tombs near one of the city gates, he spotted a small column just showing above the bushes, with the shape of a sphere and a cylinder on it. Men with sickles cleared a path, and the verses were still there, half worn away.

Why a sphere and a cylinder? According to Plutarch, Archimedes had asked for exactly that monument, marking the discovery he was proudest of. In his book On the Sphere and Cylinder he proved that a sphere fills exactly two thirds of the cylinder that just contains it. This lesson uses his result to find the volume of any sphere. It finishes Common Core standard 8.G.C.9: knowing the formulas for the volumes of cones, cylinders and spheres, and using them to solve problems.

From Archimedes' cylinder to the formula

Picture a ball of radius r that fits snugly inside a can: touching the sides all the way round, the bottom and the lid. The can is a cylinder with radius r, and its height is the ball's full height, its diameter, 2r.

Step 1. The cylinder's volume: base area times height, πr² × 2r = 2πr³.

Step 2. Archimedes' result: the sphere is two thirds of that. (2/3) × 2πr³ = (4/3)πr³.

So the volume of a sphere of radius r is

V = (4/3)πr³.

Notice the cube. A cylinder's or a cone's volume multiplies a radius by a radius by a height; a sphere's multiplies three lengths that are all r. Doubling a sphere's radius multiplies its volume by 2 × 2 × 2 = 8.

Worth holding on to: A sphere fills two thirds of the cylinder that wraps it exactly, so its volume is (2/3) × 2πr³ = (4/3)πr³. Archimedes thought this result was worth a tombstone.

A family of three

Cut the snug can and the ball in half across the middle. Now you have a hemisphere, half a sphere, of radius r, inside a cylinder of radius r and height r. Put a cone with the same base and height beside them. Their volumes:

Cone, hemisphere and cylinder of the same radius and height hold amounts in the ratio 1 : 2 : 3. A cone and a hemisphere together exactly fill the cylinder. Illustrative Mathematics' lessons on estimating a hemisphere and on the volume of a sphere build the same picture.

Worked examples

Worked example 1. Find the volume of a sphere with radius 3 cm.

Step 1. r³ = 3 × 3 × 3 = 27.

Step 2. V = (4/3)π × 27 = 36π ≈ 113.1 cm³.

Check with the cylinder: the snug can holds 2π × 27 = 54π, and two thirds of 54π is 36π.

Worked example 2. A ball is 24 cm across. About how much air is inside it?

Step 1. The radius is 12 cm, and 12³ = 1,728.

Step 2. V = (4/3)π × 1,728 = 2,304π ≈ 7,238 cm³, about 7.2 litres.

Worked example 3: a bowl. A hemispherical bowl has an inside radius of 6 cm. How much soup does it hold when full?

Step 1. A whole sphere of radius 6 would hold (4/3)π × 216 = 288π.

Step 2. The bowl is half: 144π ≈ 452.4 cm³, about 450 mL.

Worked example 4: the Earth. NASA's fact sheet gives the Earth's mean radius as 6,371 km. Treating the Earth as a sphere, find its volume.

Step 1. 6,371³ ≈ 2.586 × 10¹¹ km³, in the scientific notation of Lesson 3.

Step 2. V ≈ (4/3) × 3.1416 × 2.586 × 10¹¹ ≈ 1.083 × 10¹² km³.

Check: the same fact sheet lists the volume as 108.321 × 10¹⁰ km³, which is 1.08321 × 10¹² km³. The formula agrees with NASA to four figures.

Worked example 5: working backwards. A sphere has volume 288π cm³. What is its radius?

Step 1. (4/3)πr³ = 288π. Divide both sides by π, then multiply by 3/4: r³ = 288 × 3/4 = 216.

Step 2. r = ³√216 = 6 cm, using the cube roots of Lesson 5.

Worked example 6: an ice cream cone. A cone of radius 3 cm and height 10 cm is filled level with ice cream and topped with a hemisphere of ice cream of radius 3 cm. If it all melts, will it overflow the cone?

Step 1. The cone holds (1/3)π × 9 × 10 = 30π cm³.

Step 2. The ice cream is 30π + (2/3)π × 27 = 30π + 18π = 48π cm³.

Step 3. 48π is more than 30π, so yes: about 18π ≈ 57 cm³ drips over the edge.

Worked example 7: balls in a can. Three balls of radius r fit exactly in a can of radius r and height 6r. What fraction of the can do they fill?

Step 1. Balls: 3 × (4/3)πr³ = 4πr³. Can: πr² × 6r = 6πr³.

Step 2. 4πr³ ÷ 6πr³ = 2/3. Archimedes' fraction again, because the can is three snug cans stacked up.

Common misconceptions

"V = (4/3)πr²." Volume needs three lengths multiplied, so the radius is cubed. r² would give square units, not cubic ones.

"Use the diameter as r." The formula uses the radius. A ball 24 cm across has r = 12; using 24 would give 8 times the true volume.

"It is (3/4)πr³." The fraction is four thirds, which comes from two thirds of 2πr³. Three quarters is a common slip.

"Double the radius, double the volume." The radius is cubed, so doubling it multiplies the volume by 8.

"A sphere holds as much as the can it fits in." It fills only two thirds; the gaps around the curve make up the other third.

What you now know

• Archimedes proved that a sphere fills two thirds of the cylinder that just contains it, and Cicero found the sphere and cylinder marking his tomb in 75 BCE.
• Sphere: V = (4/3)πr³. Hemisphere: half of that, (2/3)πr³.
• For the same radius r and height r, a cone, a hemisphere and a cylinder hold amounts in the ratio 1 : 2 : 3.
• To find a radius from a volume, solve for r³ and take the cube root.
• Doubling a sphere's radius multiplies its volume by 8.

With volume done, the course turns to its last topic, data. When will Old Faithful erupt next? Rangers in Yellowstone answer that with a pattern in two columns of numbers, and the next lesson plots them.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Geometry (8.G.C.9). thecorestandards.org
2. Rorres, C. (n.d.). Tomb of Archimedes: Sources (Cicero, Tusculan Disputations V.64-66; Plutarch, Marcellus). Archimedes home page, Courant Institute, New York University. math.nyu.edu
3. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 20: The volume of a sphere. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
4. Illustrative Mathematics. (2019). Grade 8, Unit 5, Lesson 19: Estimating a hemisphere. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Williams, D. R. (2024, November 15). Earth fact sheet (mean radius and volume). NASA Goddard Space Flight Center. nssdc.gsfc.nasa.gov

The geyser that keeps its own timetable

Old Faithful, in Yellowstone National Park, is one of only a handful of geysers that park rangers predict. The National Park Service says it does not erupt on an exact schedule, but the rangers have a method, and their classroom activity for students spells it out: time how long an eruption lasts. If it lasts less than 3 minutes, expect the next one in about 68 minutes; if it lasts more than 3 minutes, expect it in about 94 minutes.

Where does a rule like that come from? From data: pairs of numbers, one eruption's length and the wait that followed it, recorded again and again. The statistics software R comes with a well-known set of 272 such pairs for Old Faithful. Here are the first sixteen, with eruption lengths rounded to the nearest tenth of a minute:

Data like these, two measurements on each of many things, are called bivariate data. This lesson shows how to picture them and what patterns to look for. It covers Common Core standard 8.SP.A.1: constructing and interpreting scatter plots for bivariate measurement data and describing patterns such as clustering, outliers, positive or negative association, and linear or nonlinear association.

Building the scatter plot

A scatter plot shows each pair as one point. Put the variable you know first, or the one you use to predict, across the horizontal axis; here that is eruption length, since the rangers time the eruption and then predict the wait. Put the other variable up the vertical axis.

Step 1. Choose scales that fit the data: lengths run from 1.8 to 4.7 minutes, so an axis from 1.5 to 5 works; waits run from 47 to 88, so an axis from 40 to 90 works.

Step 2. Plot each pair. Eruption 1 becomes the point (3.6, 79): across to 3.6, up to 79.

Step 3. Repeat for all sixteen, then step back and look at the shape of the cloud.

The axes do not have to start at 0. What matters is that each axis has an even scale, so equal distances mean equal differences.

What the cloud shows

Plot the sixteen points and two things jump out.

Clusters. The points fall into two groups with a gap between them. Seven short eruptions, from 1.8 to 2.9 minutes, sit low on the graph, with waits from 47 to 62 minutes; their average wait is 375 ÷ 7 ≈ 54 minutes. Nine long eruptions, from 3.3 to 4.7 minutes, sit high, with waits from 74 to 88 minutes; their average is 741 ÷ 9 ≈ 82 minutes. Nothing falls between 2.9 and 3.3 minutes. That gap is why the rangers' rule splits at 3 minutes.

Positive association. Across the whole plot, points further right tend to be higher: longer eruptions tend to be followed by longer waits. When larger values of one variable tend to go with larger values of the other, the variables have a positive association. When larger values of one tend to go with smaller values of the other, the association is negative. When there is no trend either way, there is no association.

Linear or not. If the cloud is spread around a straight line, the association is linear; if it bends around a curve, it is nonlinear. The Old Faithful points rise roughly along a straight line, so the association is positive and roughly linear. The next lesson draws that line.

Outliers. A point far away from the overall pattern is an outlier. None of these sixteen is far from its cluster. But suppose a seventeenth record said a 2.0-minute eruption was followed by a 95-minute wait. It would sit alone at the top left, far from the short-eruption cluster, and you would want to check it: a timing mistake, or a genuinely odd eruption?

Bottom line: To describe a scatter plot, say whether it shows clusters and outliers, whether the association is positive, negative or absent, and whether it is linear or nonlinear, and say what each feature means in the situation.

A pattern is not a cause, though here there is one

An association says the two variables tend to move together. On its own it does not say that one causes the other. For Old Faithful, the Park Service gives a physical reason for the pattern: a short eruption discharges less water and heat, so the geyser recharges sooner, while a long eruption empties more and needs more time to refill. That explanation comes from geology, not from the scatter plot. With other data, two things can rise together because a third thing drives both, or by chance.

The data are also not the last word on the geyser. The Park Service notes that earthquakes have changed Old Faithful's intervals over the decades, and gives a median interval of 102 minutes as of January 2025, longer than most of the waits in this older data set. A pattern describes the data it came from.

Worked examples

Worked example 1. A class records hours of screen time after 9 p.m. and hours of sleep for ten students (invented data for practice): (0, 9.5), (0.5, 9), (1, 8.5), (1, 9), (1.5, 8), (2, 7.5), (2, 8), (2.5, 7), (3, 6.5), (3.5, 6.5). Describe the association.

Step 1. As screen time goes up, sleep tends to go down: a negative association.

Step 2. The points fall close to a straight line, roughly half an hour less sleep for each extra hour of screens: negative and linear, with no clusters or outliers.

Worked example 2. Ten students' shoe sizes and their scores on a maths quiz (invented) are plotted, and the points are scattered with no upward or downward trend. Describe it.

There is no association: knowing a student's shoe size tells you nothing about their score.

Worked example 3. A ball is dropped and its height is recorded every quarter second: (0, 20), (0.25, 19.7), (0.5, 18.8), (0.75, 17.2), (1, 15.1), (1.25, 12.3), (1.5, 9.0), (1.75, 5.0). Describe the association.

Negative, because height falls as time passes, and nonlinear, because the drops get bigger each quarter second, so the points bend along a curve.

Worked example 4. In the Old Faithful table, which variable should go on the horizontal axis if you want to predict the wait?

Eruption length: it is known first and used to predict the wait, so it goes across and the wait goes up.

Common misconceptions

"A scatter plot should join the dots." Each point is a separate eruption. Joining them in table order draws a meaningless zigzag. Look at the overall cloud.

"Positive association means every point is higher than the one before." It means higher points tend to go with points further right. Eruption 8 lasted 3.6 minutes and eruption 5 lasted 4.5 minutes, yet both were followed by 85-minute waits: individual points vary around the trend.

"Association proves causation." A pattern only shows that two things tend to go together. A cause has to be shown another way, as the Park Service does with the geology of the geyser.

"An outlier is a mistake and should be deleted." It might be a mistake, or it might be the most interesting point in the data. Check it before deciding.

"No association means the points make a flat line." A flat line of points would mean the second variable stays the same. No association looks like a shapeless scatter.

Looking back

• Bivariate data pair two measurements on each item. A scatter plot shows each pair as a point, with the predicting variable usually across.
• Describe the pattern: clusters and outliers; positive, negative or no association; linear or nonlinear.
• Old Faithful's eruptions form two clusters, short with short waits and long with long waits, with a positive, roughly linear association, which is why the rangers' rule splits at 3 minutes.
• Association is not causation. An explanation has to come from knowledge of the situation.

The rangers' two-number rule is crude: every eruption over 3 minutes gets the same 94-minute prediction. A straight line through the cloud can do better, giving a different prediction for every eruption length. That is the next lesson.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Statistics and Probability (8.SP.A.1). thecorestandards.org
2. National Park Service. (n.d.). Predict Old Faithful. Teachers, U.S. Department of the Interior. nps.gov
3. National Park Service. (n.d.). Old Faithful (frequently asked questions on intervals and prediction). Yellowstone National Park. nps.gov
4. R Core Team. (n.d.). Old Faithful geyser data (faithful), R datasets package documentation; data from Härdle (1991), with Azzalini and Bowman (1990). stat.ethz.ch
5. Illustrative Mathematics. (2019). Grade 8, Unit 6, Lesson 5: Describing trends in scatter plots. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com

A different prediction for every eruption

The rangers' rule from the last lesson gives only two answers: about 68 minutes after a short eruption, about 94 after a long one. But the scatter plot rose steadily, so a 4.7-minute eruption should be followed by a longer wait than a 3.3-minute one. A straight line through the cloud of points can give a different prediction for every eruption length.

The Common Core standard puts the idea plainly: straight lines are widely used to model relationships between two quantitative variables. This lesson fits one, judges it, and reads it. It covers 8.SP.A.2, informally fitting a straight line to a scatter plot and judging how closely the points follow it, and 8.SP.A.3, using the equation of a linear model to solve problems and interpreting its slope and intercept.

Fitting a line by eye

Here again are the sixteen Old Faithful eruptions, eruption length e in minutes and the wait w until the next eruption, in minutes:

(3.6, 79), (1.8, 54), (3.3, 74), (2.3, 62), (4.5, 85), (2.9, 55), (4.7, 88), (3.6, 85), (2.0, 51), (4.4, 85), (1.8, 54), (3.9, 84), (4.2, 78), (1.8, 47), (4.7, 83), (2.2, 52).

Lay a ruler across the scatter plot and move it until it follows the trend of the cloud, with roughly as many points above it as below, and the points as close to it as you can get them overall. A line of fit does not have to pass through any data point. One good choice runs through (2, 53) and (4.5, 85).

Step 1. Slope: (85 − 53) ÷ (4.5 − 2) = 32 ÷ 2.5 = 12.8.

Step 2. Intercept: 53 = 12.8 × 2 + b = 25.6 + b, so b = 27.4.

Step 3. The line of fit is w = 12.8e + 27.4.

For comparison, the method a computer uses, called least squares, gives w ≈ 12.9e + 27.9 for these sixteen points, very close to the line drawn by eye. Using all 272 eruptions in the data set, the same method gives w ≈ 10.7e + 33.5: still a rising line, a little less steep, because more data changes the details of a fit.

How well does it fit?

For each point, compare the actual wait with the line's prediction. The difference, actual minus predicted, says how far the point is above the line (positive) or below it (negative).

Across all sixteen eruptions, 9 points lie above the line, 6 lie below it and 1 sits exactly on it, and the line misses by about 4.3 minutes on average, never by more than 11.5. That is a reasonable fit: the points cluster around the line without any part of the plot sitting entirely above or below it. If most points on the left sat above the line and most on the right below, the line would be tilted wrong and should be redrawn.

Reading the slope and intercept

Slope. The slope, 12.8, has units of minutes of wait per minute of eruption. It means that each extra minute of eruption is associated with about 12.8 more minutes of waiting, on average. The standard gives its own example of this kind of reading: in a model of plant growth, a slope of 1.5 cm per hour means each extra hour of sunlight a day is associated with 1.5 cm more height.

Intercept. The intercept, 27.4, is the predicted wait after an eruption lasting 0 minutes. That makes no sense: an eruption of 0 minutes is no eruption at all, and the data only run from 1.8 to 4.7 minutes. Here the intercept is just a number that positions the line, not a fact about the geyser.

So what?: A line of fit turns a cloud of points into a rule you can use, with a slope that says how much the output tends to change per unit of input. Use it inside the range of the data, and read its numbers in the units of the situation.

Predicting, and knowing when to stop

Worked example 1. An eruption lasts 3.0 minutes. Predict the wait.

w = 12.8 × 3.0 + 27.4 = 38.4 + 27.4 = 65.8, so about 66 minutes.

Worked example 2. An eruption lasts 4.0 minutes. Predict the wait, and compare with a 3.0-minute eruption.

w = 51.2 + 27.4 = 78.6, about 79 minutes. One extra minute of eruption added 12.8 minutes to the prediction: that is the slope.

Worked example 3. Use the line to predict the wait after a 10-minute eruption. Should you trust it?

The line gives 12.8 × 10 + 27.4 = 155.4 minutes. But the National Park Service says Old Faithful's eruptions normally last between 1.5 and 5 minutes. A 10-minute eruption is far outside the data, so the line has no evidence behind it there. Predicting outside the range of the data is called extrapolation, and it is risky.

Worked example 4. A model for plants grown under lights is h = 1.5s + 10, where s is hours of light a day and h is height in centimetres after a month (invented data). Interpret the numbers and predict for 8 hours.

The slope says each extra hour of light a day goes with 1.5 cm more growth. The intercept says a plant with no light would be 10 cm tall, which may or may not make sense, depending on whether s = 0 was inside the experiment. For 8 hours: 1.5 × 8 + 10 = 22 cm.

Worked example 5. A basketball coach records daily shooting practice, in minutes, and free throws made out of 20 (invented data): (10, 9), (20, 11), (30, 12), (40, 14), (50, 15), (60, 17). Fit a line through (10, 9) and (60, 17) and interpret it.

Step 1. Slope: (17 − 9) ÷ (60 − 10) = 8 ÷ 50 = 0.16 free throws per minute of practice.

Step 2. Intercept: 9 = 0.16 × 10 + b, so b = 7.4. The line is f = 0.16p + 7.4.

Step 3. Each extra 10 minutes of practice goes with about 1.6 more free throws out of 20. The line cannot be trusted far beyond 60 minutes, since no one can make more than 20 out of 20.

Common misconceptions

"A line of fit must pass through as many points as possible." It should follow the overall trend, through the middle of the cloud. Passing exactly through a few points can leave the rest badly balanced.

"A line of fit must start at (0, 0)." The Old Faithful line crosses the vertical axis at 27.4. A line of fit goes wherever the data put it.

"Predictions from the line are exact." The line describes a tendency. Real eruptions of 3.6 minutes were followed by waits of 79 and 85 minutes, while the line predicts 73.5.

"The line works for any input." Outside the range of the data, extrapolation can give nonsense, such as a 155-minute wait after an impossibly long eruption, or more than 20 free throws out of 20.

"The slope is the predicted value." The slope is a rate: how much the prediction changes for each extra unit of input. 12.8 is minutes of wait per minute of eruption, not a wait.

What to carry forward

• When a scatter plot looks linear, fit a line by eye that follows the trend with points balanced above and below, then write its equation from two points on it.
• Judge the fit by the distances of the points from the line: small and balanced is good; a pattern of misses means the line needs redrawing.
• The slope is the predicted change in the output for each extra unit of input; the intercept is the prediction at input 0, which may lie outside the data and mean nothing.
• Predict inside the range of the data. Extrapolating far outside it is risky.
• For the sixteen Old Faithful eruptions, w = 12.8e + 27.4: each extra minute of eruption goes with about 13 more minutes of waiting.

Not every variable is a measurement. "Has a curfew" and "does chores" are categories, yes or no. The last lesson asks whether two such categories go together, and finds that counts alone can mislead.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Statistics and Probability (8.SP.A.2, 8.SP.A.3). thecorestandards.org
2. R Core Team. (n.d.). Old Faithful geyser data (faithful), R datasets package documentation. stat.ethz.ch
3. National Park Service. (n.d.). Old Faithful (eruption lengths and intervals). Yellowstone National Park. nps.gov
4. Illustrative Mathematics. (2019). Grade 8, Unit 6, Lesson 4: Fitting a line to data. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illustrative Mathematics. (2019). Grade 8, Unit 6, Lesson 6: The slope of a fitted line. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
6. Illowsky, B., and Dean, S. (2023). Section 12.3, The regression equation. Introductory Statistics 2e. OpenStax. openstax.org

Twenty-four against twelve

Sixty eighth-graders answer two questions on a survey. Do you have a curfew on school nights? Do you have assigned chores at home? Among the students with a curfew, 24 say they have chores. Among the students without one, only 12 do. (The class and its answers are invented, but the two questions come straight from the Common Core standard for this lesson.)

Maya looks at those two numbers and calls the question settled: 24 is twice 12, so students with a curfew tend to have chores. Leo looks at the same survey and says it shows no connection at all. They cannot both be right. By the end of this lesson you will be able to say who is, and why, and you will use the same idea on a real table from 1912: who survived the sinking of the Titanic.

This lesson covers Common Core standard 8.SP.A.4: seeing patterns of association in data about categories by showing frequencies and relative frequencies in a two-way table; building and reading a two-way table for two categorical variables collected from the same people; and using relative frequencies for rows or columns to describe a possible association. The standard's own example is Maya and Leo's question: is there evidence that students who have a curfew also tend to have chores?

Measurements and categories

Everything else in this module has been a measurement: minutes of eruption, minutes of waiting, minutes of practice. Each point on a scatter plot was a pair of numbers. A curfew is different. The answer is yes or no, a category, not a number. A variable whose values are categories is called a categorical variable. Class of ticket, favourite sport and eye colour are categorical too.

You cannot plot yes against no on a scatter plot. Instead you count how many people fall into each pair of categories and set the counts out in a grid called a two-way table. One variable runs down the side, the other runs across the top, and totals run along the bottom and down the right-hand edge. Statistics books also call it a contingency table.

The standard says the two variables must be collected from the same subjects. Every student answers both questions, so every student lands in exactly one cell of the grid.

Worked example 1. Ten students give these answers, written as curfew/chores: yes/yes, yes/no, no/no, yes/yes, no/yes, yes/yes, no/no, yes/no, no/no, yes/yes. Build the two-way table.

Step 1. Put curfew down the side and chores across the top, with a Total row and a Total column.

Step 2. Count each pair. yes/yes appears 4 times, yes/no 2 times, no/yes once and no/no 3 times. Check: 4 + 2 + 1 + 3 = 10 students.

Step 3. Add across each row and down each column.

Check: the row totals, 6 + 4, and the column totals, 5 + 5, both make 10, the number of students. If they did not, someone would have been counted twice or missed.

Worked example 2. A survey of 80 students asked whether each has a library card and whether each reads for fun. The table arrived with gaps. Fill them in.

Step 1. Card holders who do not read for fun: 45 − 30 = 15.

Step 2. Students without a card: 80 − 45 = 35.

Step 3. Students without a card who read for fun: 42 − 30 = 12.

Step 4. Students without a card who do not read for fun: 35 − 12 = 23.

Step 5. Total who do not read for fun: 15 + 23 = 38. Check another way: 80 − 42 = 38. Every row and every column now adds up.

Maya's case: count the people

Here is the full survey of sixty.

Maya's evidence is real. Twenty-four students with a curfew have chores, and only twelve students without a curfew do: twice as many. She can add another true fact. Of the 36 students with chores, 24 have a curfew, which is two out of every three. If curfews had nothing to do with chores, she asks, why would most of the students with chores have one?

Leo's case: compare like with like

Leo points at the Total column. There are 40 students with a curfew and only 20 without. The curfew group is twice the size of the other one, so it would have twice as many of almost anything: twice as many students with chores, twice as many with brown eyes, twice as many who like pizza. To compare the groups fairly, he asks a different question: what fraction of each group has chores?

Curfew: 24 out of 40 = 24 ÷ 40 = 0.6 = 60%.

No curfew: 12 out of 20 = 12 ÷ 20 = 0.6 = 60%.

The two groups have chores at exactly the same rate, so knowing whether a student has a curfew tells you nothing about whether they have chores. And Maya's two out of three? Two out of every three students in the whole class have a curfew, 40 out of 60, so of course two out of three of the students with chores do. Among the 24 students with no chores, 16 have a curfew: two out of three again.

What settles it

A count depends on two things at once: how big a group is, and how common something is inside it. When the groups are different sizes, the counts mix the two up. A relative frequency, a count divided by a total, takes the size of the group out and leaves only how often something happens. You met relative frequency in seventh grade as the number of times something happened divided by the number of trials. Here the trials are the people in one group.

So Leo wins this argument. In this survey there is no association between curfews and chores: 60% of each group have chores.

Now look at a second survey, of 50 students (also invented).

Curfew: 18 ÷ 25 = 0.72 = 72% have chores. No curfew: 9 ÷ 25 = 0.36 = 36%. Here students with a curfew are twice as likely to have chores, so this survey does show an association: students who have a curfew tend to have chores too.

The core of it: Counts say how many; relative frequencies say how often. To decide whether two categories go together, compare how often something happens inside each group, not how many people it happens to.

Rows, columns and the whole table

You can divide a cell by three different totals, and each one answers a different question. Take the cell 18 in the second survey: the students with a curfew and chores.

Worked example 3. Find three relative frequencies for the cell 18, and say which question each one answers.

Step 1. Divide by the row total: 18 ÷ 25 = 0.72 = 72%. Of the students with a curfew, 72% have chores. This is a row relative frequency.

Step 2. Divide by the column total: 18 ÷ 27 = 0.667 = 66.7%. Of the students with chores, 66.7% have a curfew. This is a column relative frequency.

Step 3. Divide by the grand total: 18 ÷ 50 = 0.36 = 36%. Of all 50 students, 36% have both a curfew and chores.

Three true numbers, three different questions. Read the question first, then decide which total to divide by.

Doing step 1 for every cell gives a table of row relative frequencies, and each row adds up to 100%.

Doing step 2 for every cell gives column relative frequencies instead, and then each column adds up to 100%. Of the 27 students with chores, 66.7% have a curfew and 33.3% do not. Of the 23 with no chores, 7 ÷ 23 = 30.4% have a curfew and 16 ÷ 23 = 69.6% do not. The columns tell the same story from the other side: 66.7% against 30.4%. Rows and columns always agree about whether there is an association. If the row percentages match, the column percentages match too, and if one set differs, so does the other. They just answer different questions.

Worked example 4. Go back to the library-card survey from worked example 2. Is there an association between having a library card and reading for fun?

Step 1. Decide which way to divide. The question compares students with a card against students without one, so divide by the row totals.

Step 2. Card: 30 ÷ 45 = 0.667 = 66.7% read for fun. No card: 12 ÷ 35 = 0.343 = 34.3%.

Step 3. 66.7% is nearly twice 34.3%, so yes: in this survey, students with a library card tend to read for fun more than students without one.

Notice what the table does not say: why. A library card might lead to more reading, or keen readers might be the ones who sign up for cards. As with scatter plots, an association is not a cause.

A real table: who survived the Titanic

The ocean liner Titanic sank on 15 April 1912 after hitting an iceberg. The official British inquiry that year counted the people on board by class of travel and recorded who survived. In 1995 Robert Dawson of Saint Mary's University set out the inquiry's figures for statistics teachers, and they are now built into the statistics software R as a data set of 2,201 people. Sources disagree a little about the exact numbers on board and lost, and Dawson concludes that the exact numbers may never be known, so treat these as the official figures rather than the final word.

Look only at the Survived column and the crew seem to have done best: 212 crew members survived, more than any other group. That is Maya's argument again, and it fails for the same reason. There were 885 crew, far more than the 325 people in first class.

Worked example 5. Find the survival rate for each class, and for everyone on board.

Step 1. First class: 203 ÷ 325 = 0.625 = 62.5%.

Step 2. Second class: 118 ÷ 285 = 0.414 = 41.4%.

Step 3. Third class: 178 ÷ 706 = 0.252 = 25.2%.

Step 4. Crew: 212 ÷ 885 = 0.240 = 24.0%.

Step 5. Everyone: 711 ÷ 2,201 = 0.323 = 32.3%. Check: the overall rate should lie between the lowest and highest group rates, 24.0% and 62.5%, and it does.

The crew had the most survivors but the lowest survival rate. A first-class passenger was more than twice as likely to survive as a third-class passenger. There is a strong association between class and survival.

Worked example 6. A friend says: "The crew were 29.8% of the survivors, the biggest share. So it was safest to be one of the crew." Check the number, then judge the argument.

Step 1. 212 ÷ 711 = 0.298 = 29.8%. The number is right. It is a column relative frequency: of the survivors, what fraction were crew?

Step 2. The friend's question, whether it was safer to be crew, is about each group's chance, so it needs row relative frequencies, and the crew's was the lowest of all, 24.0%. The crew made up a big share of the survivors only because they made up a big share of everyone: 885 ÷ 2,201 = 40.2% of the people on board.

Worked example 7. Split by sex, the same figures give these counts. Female: 344 survived out of 470. Male: 367 survived out of 1,731. More males survived than females. Did males have the better chance?

Step 1. Female: 344 ÷ 470 = 0.732 = 73.2%.

Step 2. Male: 367 ÷ 1,731 = 0.212 = 21.2%.

Step 3. No. Nearly three in four females survived, but only about one in five males. The count favours males only because there were nearly four times as many males on board. Both Dawson and the R documentation connect this pattern with the rule of "women and children first".

The table shows that class and survival go together. It cannot say why. Dawson reports one explanation that was offered: the third-class cabins were low in the hull of the ship. Testing an explanation like that needs evidence from outside the table.

Common misconceptions

"The biggest count shows which group did best." Not when the groups are different sizes. The crew had the most Titanic survivors, 212, and the lowest survival rate, 24.0%. Compare relative frequencies within each group.

"Row and column percentages are the same thing." In the second curfew survey, 72% of the students with a curfew have chores, but 66.7% of the students with chores have a curfew. Same cell, different totals, different questions.

"Any difference in the percentages proves an association." In a small survey, 75% against 71% could easily come from chance. A big, clear difference, like 72% against 36%, is much stronger evidence. In eighth grade you judge this by eye; later statistics courses measure it.

"An association means one thing causes the other." A curfew does not make chores appear. Families with more rules might simply set both. A two-way table shows that two categories go together, not why.

"Every column of a relative frequency table adds up to 100%." That depends on which totals you divided by. Divide by the row totals and each row adds up to 100%; divide by the column totals and each column does.

Putting it together

• Categorical data are categories, not measurements. A two-way table counts the same people by two categorical variables, and its row and column totals must agree.
• Counts mix up the size of a group with how common something is inside it. Relative frequencies, counts divided by a total, separate the two.
• To look for an association, compare the row relative frequencies, or the column ones, from group to group. A clear difference suggests an association; nearly equal values suggest none.
• Divide by the row total, the column total or the grand total, depending on the question you are asking.
• An association is not a cause. On the Titanic, 62.5% of first-class passengers survived against 25.2% in third class; the table shows the pattern, not its reasons.

That is the last lesson of the course. You began with a view counter that ran out of numbers, and you finish by reading a table from 1912. In between you worked with exponents and roots, solved equations and systems, built and compared functions, proved the Pythagorean theorem and found volumes. That is the toolkit Algebra I expects you to bring, and it is yours now.

Sources

1. National Governors Association Center for Best Practices, and Council of Chief State School Officers. (2010). Common Core State Standards for Mathematics: Grade 8, Statistics and Probability (8.SP.A.4). thecorestandards.org
2. Dawson, R. J. M. (1995). The "unusual episode" data revisited. Journal of Statistics Education, 3(3). jse.amstat.org
3. R Core Team. (n.d.). Survival of passengers on the Titanic (Titanic), R datasets package documentation. stat.ethz.ch
4. Illustrative Mathematics. (2019). Grade 8, Unit 6, Lesson 9: Looking for associations. IM 6-8 Math. Kendall Hunt. im.kendallhunt.com
5. Illowsky, B., and Dean, S. (2023). Section 3.4, Contingency tables. Introductory Statistics 2e. OpenStax. openstax.org