Division Rules and Why You Cannot Divide by Zero

Subtraction turned out to be addition, since $a - b$ is just $a + (- b)$ , and the opposite of $b$ was the partner that made it work. That leaves the last operation, division. Here is a guess worth holding onto. If subtraction was another form of addition, division will be another form of multiplication, and it will need a partner of its own.

Problem

A blueprint app has one button that scales every length by 5, blowing a drawing up to five times its size. You tap it by accident. You need a second scaling that puts the drawing back exactly as it was, neither bigger nor smaller. Multiplying by

1

is the scaling that changes nothing, so what number must you scale by to undo the

\times 5

Show a hint

What single number multiplies with 5 to give 1?

Show the full solution

Scaling up by 5 and back down acts like multiplying by 1, so you need the number that turns 5 into 1.

5 \times \frac{1}{5} = 1

The undo scales by

\frac{1}{5}

, the reciprocal of 5.

Multiplying by 5 stretches 1 out to 5, and multiplying by

\frac{1}{5}

pulls it straight back. A number and its reciprocal undo each other, landing on 1, just as a number and its opposite land on 0.

Problem

Here is a pattern worth predicting before you compute. Last chapter, the opposite of the opposite of a number brought you right back to the number, since two flips across zero return home. Now guess the reciprocal of a reciprocal, then test the guess. What is the reciprocal of $\frac{1}{6}$ ?

Show a hint

The reciprocal of $\frac{1}{6}$ is the number that multiplies with $\frac{1}{6}$ to give 1.
Flip $\frac{1}{6}$ over.

Show the full solution

Flip

\frac{1}{6}

and it becomes 6.

\frac{1}{6} \times 6 = 1

So the reciprocal of

\frac{1}{6}

6

. Flipping twice returns the original, so the reciprocal of a reciprocal is the number itself.

Problem

One number has no reciprocal at all. Try to find the reciprocal of

0

, the partner that would multiply with 0 to give 1. Hunt for it, and say what goes wrong.

Show the full solution

The reciprocal of 0 would have to turn 0 into 1, but

0 \times (any number) = 0

, never 1. So 0 has no reciprocal, the only number without one.

Problem

Last chapter, exactly one number was its own opposite, the number 0, sitting right on its own mirror image. Now ask the matching question for reciprocals. How many numbers are their own reciprocal, equal to the number that multiplies with them to give 1? Test small cases and see what you find.

Show a hint

A number $x$ is its own reciprocal when $x \times x = 1$ .
Try 1. Then remember the sign rule and try $- 1$ .

Show the full solution

A number is its own reciprocal when it times itself gives 1.

1 \times 1 = 1

, and

(- 1) \times (- 1) = 1

, so 1 and

- 1

both qualify. Nothing else does, since any number above 1 has a reciprocal below 1. That leaves exactly

2

such numbers.

With the reciprocal in hand, we can build division out of multiplication. First we check the new rule against a division you already know the answer to, then use it on a division that is awkward to do the old way.

Problem

A relay team of

4

runners splits a

12

km course into equal legs. Plain division says each leg is

12 \div 4 = 3

km, no surprise. Now compute it the new way instead. Take

\frac{1}{4}

of 12, that is

12 \times \frac{1}{4}

. How long is each leg this way?

Show the full solution

Each runner covers one quarter of 12 km, found by multiplying.

12 \times \frac{1}{4} = \frac{12}{4} = 3

Dividing 12 by 4 and multiplying 12 by

\frac{1}{4}

are two names for one number.

Problem

Undo a

\times 6

: compute

24 \div 6.

Show a hint

Dividing by 6 undoes multiplying by 6.

Show the full solution

24 \div 6 = 4

, since

4 \times 6 = 24

Problem

A jar holds 3 cups of flour. Your scoop holds $\frac{1}{4}$ cup. How many full scoops fill the jar? In symbols, find

3 \div \frac{1}{4} .

Show a hint

Dividing by $\frac{1}{4}$ means multiplying by the reciprocal of $\frac{1}{4}$ .
The reciprocal of $\frac{1}{4}$ is 4, so you are really computing $3 \times 4$ .

Show the full solution

Turn the division into multiplication by the reciprocal of

\frac{1}{4}

, which is 4.

3 \div \frac{1}{4} = 3 \times 4 = 12

The answer grew past 3 because we divided by a number below 1, whose reciprocal is bigger than 1.

Each of the twelve cells is one

\frac{1}{4}

-cup scoop, and three cups hold all twelve. So

3 \div \frac{1}{4} = 12

, and dividing by a small piece gives a big count.

Problem

How many halves fill 6? Find

6 \div \frac{1}{2} .

Show a hint

Dividing by $\frac{1}{2}$ is multiplying by $2$ .

Show the full solution

6 \div \frac{1}{2} = 6 \times 2 = 12

, so twelve halves make six.

Problem

Over

6

equal stages of a dive, a submarine’s depth changed by a total of

- 240

m (a downward move counts as negative, as it did two lessons ago). The stages were identical, so what was the depth change per stage? Compute

- 240 \div 6.

Show a hint

Division is multiplication by the reciprocal, so this is $- 240 \times \frac{1}{6}$ .
One negative factor flips the sign of the product, just as it did for multiplication.

Show the full solution

Rewrite the division as multiplication by

\frac{1}{6}

. One negative factor keeps the result negative.

- 240 \div 6 = - 240 \times \frac{1}{6} = - 40

Division uses the same sign rules as multiplication, because it is multiplication by the reciprocal.

Problem

Structure beats grinding, as always. Compute

(16 \times 21) \div (8 \times 3)

without doing the two multiplications first. Look for parts of the top that pair off neatly with parts of the bottom.

Show a hint

Dividing by $8 \times 3$ is multiplying by $\frac{1}{8} \times \frac{1}{3}$ , so pair the 16 with the 8 and the 21 with the 3.
$16 \div 8 = 2$ and $21 \div 3 = 7$ .

Show the full solution

Pair each top factor with the bottom factor it cancels.

(16 \times 21) \times \frac{1}{8} \times \frac{1}{3} = (16 \times \frac{1}{8}) \times (21 \times \frac{1}{3}) = 2 \times 7 = 14

Grinding out $336 \div 24$ gives 14 as well, but pairing first means you never build the big products.

Problem

Pair, do not grind:

(25 \times 18) \div (5 \times 9) .

Show a hint

Pair 25 with 5, and 18 with 9.

Show the full solution

Pair the factors.

25 \div 5 = 5

and

18 \div 9 = 2

, so the value is

5 \times 2 = 10

Problem

Sometimes the top and bottom share a factor so ugly you would dread computing it. Find

(637 \times 50) \div (637 \times 10) .

The 637 is bait. Do not take it.

Show a hint

637 sits on both the top and the bottom. What does $637 \times \frac{1}{637}$ equal?
Whatever 637 is, it cancels to 1, leaving only $50 \div 10$ .

Show the full solution

The 637 on top meets a $\frac{1}{637}$ on the bottom, and that pair cancels to 1.

637 \times 50 \times \frac{1}{637} \times \frac{1}{10} = 1 \times (50 \div 10) = 5

A factor sitting on both the top and the bottom always cancels to 1, however big it is, so you never multiply it out.

Problem

A signal runs through a pipeline that multiplies it by

14

, then by

27

, then by

10

. Three filters then divide it by

7

, by

9

, and by

5

. By what single factor does the signal change overall? Compute

(14 \times 27 \times 10) \div (7 \times 9 \times 5)

by pairing, not by grinding.

Show a hint

Turn every division into multiplication by a reciprocal, then hunt for pairs.
14 with 7, 27 with 9, 10 with 5. Each pair divides cleanly.

Show the full solution

Spread the three divisions out as reciprocals and pair each top factor with the bottom factor that cancels it.

(14 \div 7) \times (27 \div 9) \times (10 \div 5) = 2 \times 3 \times 2 = 12

The signal comes out 12 times bigger, and none of the six original multiplications had to be carried out.

Problem

Addition and multiplication let you reorder and regroup at will. Division is stricter, and it is worth catching in the act. First, is

12 \div 3

the same as

3 \div 12

? Then settle a grouping question. Compute

24 \div (4 \div 2)

and compare it to

(24 \div 4) \div 2

. Enter the value of

24 \div (4 \div 2) .

Show a hint

Work inside the parentheses first. $4 \div 2 = 2$ , so the left side is $24 \div 2$ .
Compare with $(24 \div 4) \div 2 = 6 \div 2$ .

Show the full solution

Inside the parentheses, $4 \div 2 = 2$ , so

24 \div (4 \div 2) = 24 \div 2 = 12

Move the parentheses and you get $(24 \div 4) \div 2 = 6 \div 2 = 3$ , a different answer from the same three numbers. Order fails too, since $12 \div 3 = 4$ while $3 \div 12 = \frac{1}{4}$ . Division flips only the number you divide by, so shuffling the numbers changes which one gets flipped.

Problem

Mind the signs.

(- 56 \times 9) \div (7 \times (- 3)) .

Show a hint

Negative over negative is positive, then pair off.

Show the full solution

Two negatives make the result positive, and then

(56 \div 7) \times (9 \div 3) = 8 \times 3 = 24

Problem

Put every tool in one place. Compute

(- 45 \times 16) \div (- 9 \times 4)

in your head. Mind the signs, convert the division, then pair off.

Show a hint

Two negative numbers are involved, so the result is positive. Now just handle the sizes.
Write it as $45 \times 16 \times \frac{1}{9} \times \frac{1}{4}$ and pair 45 with 9, then 16 with 4.

Show the full solution

One negative on top and one on the bottom make the result positive, so drop both signs and pair the factors.

45 \times 16 \times \frac{1}{9} \times \frac{1}{4} = (45 \div 9) \times (16 \div 4) = 5 \times 4 = 20

Signs first, then division turned into multiplication, then pairs that cancel. Three lessons in one short computation.

Practice these ideas

Practice

Compute

(9 + 18 + 27 + 36) \div (36 + 27 + 18 + 9) .

Show the solution

The numbers above the line and below it are identical, only listed in reverse, so the two sums are equal. Any nonzero amount divided by itself is

1

, you never have to add a single thing.

Practice

Compute

5, 600, 000 \div 70, 000

by cancelling, not by long division.

Show the solution

Strike four matching zeros from the top and the bottom, and

5, 600, 000 \div 70, 000

becomes

560 \div 7 = 80

. Cancelling equal factors of 10 from top and bottom never changes a quotient.

Practice

Compute

(317 \times 317) \div 317

without ever multiplying 317 by itself.

Show the solution

Dividing a product by one of its own factors simply hands back the other factor, so the answer is

317

. Multiplying by a number and then dividing by it is a round trip straight back to the start.

Practice

A ribbon is

6

metres long. How many

\frac{1}{5}

-metre bookmarks can you cut from it? Compute

6 \div \frac{1}{5} .

Show the solution

Five fifths fit in every metre, so six metres give

6 \times 5 = 30

bookmarks. Dividing by

\frac{1}{5}

means multiplying by its reciprocal, 5, and the count grows past 6 because the piece you cut is tiny.

Practice

Compute

20 \div \frac{2}{5} .

Show the solution

Flip the divisor and multiply.

20 \div \frac{2}{5} = 20 \times \frac{5}{2} = \frac{100}{2} = 50

Practice

What number is 4 less than the value of

48 \div \frac{3}{4}

Show the solution

48 \div \frac{3}{4} = 48 \times \frac{4}{3} = 64

, and

64 - 4 = 60

. Dividing by a number below 1 makes the result bigger, which catches almost everyone off guard the first time.

Practice

Compute

(13 \times 40 \times 9) \div (9 \times 13)

by cancelling first.

Show the solution

The 9 and the 13 on top each meet a copy on the bottom and cancel to 1, leaving the lone

40

. Neither big product was ever worth forming.

Practice

Compute

(20 \times 50 \times 90) \div (2 \times 5 \times 9) .

Show the solution

Pair and divide.

20 \div 2 = 10

50 \div 5 = 10

, and

90 \div 9 = 10

, so the whole expression is

10 \times 10 \times 10 = 1000

. Scaling every factor up by ten multiplies the quotient by ten, once for each factor.

Practice

Compute

(4, 000, 000 + 36) \div 4

by sharing the division across the sum.

Show the solution

Divide each piece on its own.

4, 000, 000 \div 4 + 36 \div 4 = 1, 000, 000 + 9 = 1, 000, 009

. Splitting the dividend into friendly chunks beats one monstrous long division.

Practice

Compute

(72 + 56 + 40 + 24) \div (9 + 7 + 5 + 3)

by comparing the two sums.

Show the solution

Every top term is eight times its partner (

72 = 8 \times 9

56 = 8 \times 7

40 = 8 \times 5

24 = 8 \times 3

), so the whole top is eight times the whole bottom. A quantity divided by one-eighth of itself is

8

, and not a single sum had to be worked out.

Practice

Compute

96 \div \frac{1}{2} \div \frac{1}{2} \div \frac{1}{3},

working left to right.

Show the solution

Working left to right,

96 \div \frac{1}{2} = 192

, then

192 \div \frac{1}{2} = 384

, then

384 \div \frac{1}{3} = 1152

. Two doublings and a tripling multiply 96 by

2 \times 2 \times 3 = 12

Practice

Compute

(45 \times 28) \div (35 \times 18)

without multiplying either product out.

Show the solution

Cancel in stages. 45 and 35 share 5, leaving 9 and 7, and 28 and 18 share 2, leaving 14 and 9. Now

(9 \times 14) \div (7 \times 9)

, where the 9 cancels and

14 \div 7 = 2

, so the answer is

2

. Clearing shared factors first turns four awkward numbers into a single digit.

Practice

Watch what division does as the divisor shrinks.

12 \div 1 = 12

, then

12 \div \frac{1}{2} = 24

, then

12 \div \frac{1}{4} = 48

, then

12 \div \frac{1}{100} = 1200

. The closer the divisor gets to

0

, the larger the quotient grows, with no ceiling anywhere. That is the real reason you cannot divide by

0

, since there is no single number the quotient would settle on, only values that grow without end.