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Grade 3 · 3.OA.A.4 · 3.OA.B.6

Multiplication and division undo each other

A child knows that 7 \times 9 = 63. Which division facts come free with it, and is there any division that no multiplication fact can ever answer?

Where you are 16 in Multiplication and division

Why this exists

In Division facts every division was turned into a multiplication with a box: 54 \div 9 asks 9 \times \square = 54. That habit works, but it leaves two questions open.

The first is how much one fact is worth. A child who knows 7 \times 9 = 63 knows more than one fact. The same three numbers give a second multiplication and two divisions, the way Fact families gave four adding and taking-away sentences from one train of cubes. Seeing all four at once means the multiplication table and the division facts are one thing to learn, not two.

The second is why the box trick is allowed at all. It works because multiplying and dividing by the same number undo each other: multiply 8 by 6 and you get 48, divide 48 by 6 and you are back at 8. Undoing is how a division gets checked, and how a hidden starting number gets found.

And the undoing has one place where it breaks. Multiplying by 0 turns every number into 0, so nothing can undo it. That is the whole reason a number cannot be divided by zero, and it is the last part of this lesson.

The lesson takes three days. Part 1 builds multiplication and division fact families. Part 2 shows that multiplying and dividing by the same number undo each other. Part 3 explains what zero does to division.

Definition

Multiplication and division by the same number are inverse operations: each one undoes the other. Multiply a number by 6, then divide the result by 6, and you get the number you started with. Divide by 6, then multiply by 6, and again you are back where you began.

In symbols this is the statement from What division means: for whole numbers a, b and c, with b not zero,

a \div b = c \quad \text{exactly when} \quad b \times c = a

A multiplication and division fact family is the set of sentences made from one product and its two factors. For the factors 7 and 9 and the product 63, the family is

\begin{aligned} 7 \times 9 &= 63 \qquad 9 \times 7 = 63 \\ 63 \div 7 &= 9 \qquad 63 \div 9 = 7 \end{aligned}

Dividing a number by zero has no answer. In this lesson that covers both 5 \div 0 and 0 \div 0. The usual word is that division by zero is undefined: no number is the answer, so the division names nothing.

Intuition

Part 1: one array, four sentences

Ruia fills an ice cube tray. The tray has 7 rows of cups with 9 cups in each row.

7 rows 9 in each row
Figure 1One tray, 63 cups. Read by rows it is 7 \times 9; read by columns it is 9 \times 7; shared back into rows or columns it is 63 \div 7 = 9 and 63 \div 9 = 7.

The tray is an array, so it can be read in four ways without counting a single cup again.

  • Read by rows: 7 rows of 9 is 7 \times 9 = 63.
  • Turn the tray on its side and read by rows again: 9 rows of 7, so 9 \times 7 = 63. This is the Properties of multiplication.
  • Share the 63 cubes back into the 7 rows: each row gets 9, so 63 \div 7 = 9.
  • Share them into the 9 columns instead: each column gets 7, so 63 \div 9 = 7.

That is the family. Its shape matches the adding families from Fact families. There, two parts made a whole, and the whole minus one part left the other. Here, two factors make a product, and the product divided by one factor leaves the other. The product plays the part of the whole.

Two kinds of family are smaller. When both factors are the same, as in 8 \times 8 = 64, turning the tray gives the same sentence, so the family has only two: 8 \times 8 = 64 and 64 \div 8 = 8. And a family with 1 in it, such as 1, 6 and 6, still has four: 1 \times 6 = 6, 6 \times 1 = 6, 6 \div 1 = 6, and 6 \div 6 = 1.

To test whether three numbers with no zero among them form a family, multiply the two smaller ones. If the product is the biggest number, they are a family. If it is not, no sentence in the family can be true. (A family with a zero in it works differently, as Part 3 shows.)

Part 2: doing and undoing

At the robot club, Emiri builds a number robot. You give it a number and it multiplies by 6. Zodwa builds a second robot that divides by 6. Feed the first robot’s answer to the second one.

8 × 6 48 ÷ 6 8
Figure 2Multiplying by 6 and then dividing by 6 brings 8 back to 8.

Start with 8. The first robot makes 8 \times 6 = 48. The second robot makes 48 \div 6 = 8. The 8 came back.

Here is why it always comes back. Multiplying 8 by 6 makes 6 equal groups with 8 in each. Dividing 48 by 6 shares those 48 things into 6 equal groups, and the groups that come out are exactly the groups that went in: 8 in each. The second robot does not need to know where the 48 came from. It only splits back what the first robot put together.

The robots also work in the other order. Divide 48 by 6 to get 8, then multiply 8 by 6 to get 48 again. Sharing into 6 groups and then putting the 6 groups back together leaves the total where it was.

Adding and taking away are a pair in the same way: add 5, take away 5, and you are back. That is why addition and subtraction share fact families, and why multiplication and division do.

Undoing has two everyday uses.

Checking a division. To check 42 \div 7 = 6, undo it: 6 \times 7 = 42, which is the number we started with, so the division was right. If the product had come out as anything other than 42, the quotient would be wrong.

Finding a hidden start. If a robot multiplies by 4 and prints 36, the number that went in is found by undoing: 36 \div 4 = 9. To run a robot backward, use the other operation with the same number.

Part 3: zero

Zero divided by a number is zero. Division facts showed this: sharing nothing among 5 shelves leaves every shelf empty, and 5 \times 0 = 0.

A number divided by zero has no answer. Take 5 \div 0. As a missing factor it asks

0 \times \square = 5

But zero times any number is 0. Zero groups hold nothing, however big the groups are meant to be. So no number fits the box, and 5 \div 0 has no answer. The stories say the same. Sharing 5 ice cubes into 0 cups is impossible, because there are no cups to put them in. Asking how many groups of 0 make 5 has no answer, because any number of empty groups is still empty.

Zero divided by zero has no answer either, for the opposite reason. As a missing factor, 0 \div 0 asks

0 \times \square = 0

and now every number fits: 0 \times 1 = 0, 0 \times 7 = 0, 0 \times 9 = 0. A division is supposed to name one number. Here nothing picks out one number from all the others, so 0 \div 0 is left without an answer too.

The robots show why. Build a robot that multiplies by 0.

2 5 9 × 0 0
Figure 3Every number that goes into the times-zero robot comes out as 0, so nothing can tell which number went in.

Feed it 2, 5 or 9: out comes 0 every time. Now try to build a robot that undoes it. Given the 0, it would have to say which number went in, and the 0 carries no trace of it. Multiplying by 0 cannot be undone, so there is no “divide by 0” to undo it with.

This also changes the fact family of a number and zero. For 8 and 0, the true sentences are 8 \times 0 = 0, 0 \times 8 = 0 and 0 \div 8 = 0. The fourth one would be 0 \div 0 = 8, and that is not a fact, so this family stops at three.

Results

A fact family. If a \times b = c and neither a nor b is zero, then all four sentences are true:

\begin{aligned} a \times b &= c \qquad b \times a = c \\ c \div a &= b \qquad c \div b = a \end{aligned}

When a and b are the same number, the family has two sentences, a \times a = c and c \div a = a.

Undoing. For any whole number n and any whole number b that is not zero,

(n \times b) \div b = n

and when c \div b is a whole number,

(c \div b) \times b = c

Zero. For any whole number b that is not zero, 0 \div b = 0. For any whole number a, a \div 0 has no answer.

Why it is true

The fact family. An array with a rows of b has c things. Turned on its side it is b rows of a, with the same c things, so b \times a = c. Sharing the c things into the a rows puts b in each row, so c \div a = b; sharing them into the b columns puts a in each, so c \div b = a. When a and b are equal, the turned array is the same array, so the second multiplication and the second division repeat the first ones.

Undoing. n \times b is b equal groups of n. Dividing by b splits that total into b equal groups, and the only equal split into b groups is the one we started from, with n in each. So the division gives n. In the other order, c \div b is the size of each group when c is shared into b equal groups, and b of those groups together are all of c.

Zero. 0 \div b asks b \times \square = 0. Any number above 0 in each of b groups would give more than 0, so the box is 0. For a \div 0 the question is 0 \times \square = a, and 0 \times \square is 0 whatever the box holds. If a is not 0, no number works. If a is 0, every number works. Either way there is no one number to be the answer. \blacksquare

Worked examples

Examples 1, 2 and 3 go with Part 1, Examples 4 and 5 with Part 2, and Examples 6 and 7 with Part 3.

Example 1

The ice cube tray family

Ruia’s ice cube tray has 7 rows with 9 cups in each row. Write the four sentences in the fact family this tray shows, and say what each one means for the tray.

Solution

The tray is an array of 7 rows of 9. The product 7 \times 9 is 63, so the tray holds 63 cubes when it is full.

  • 7 \times 9 = 63: 7 rows of 9 cubes make 63 cubes.
  • 9 \times 7 = 63: read the tray by columns, and it is 9 columns of 7. The same 63 cubes, by the commutative property.
  • 63 \div 7 = 9: share the 63 cubes into the 7 rows, and each row gets 9.
  • 63 \div 9 = 7: share them into the 9 columns, and each column gets 7.

Every sentence is about the same 63 cubes. Nothing new was counted.

Example 2

A whole family from one division

Zodwa knows that 56 \div 8 = 7. Write the other three sentences in its family. Then write the family of 9, 9 and 81.

Solution

The division says 56 shared into 8 equal groups gives 7 in each. Undo it: 8 groups of 7 is 8 \times 7 = 56. So 8 and 7 are the factors and 56 is the product. The family is

\begin{aligned} 8 \times 7 &= 56 \qquad 7 \times 8 = 56 \\ 56 \div 8 &= 7 \qquad 56 \div 7 = 8 \end{aligned}

For 9, 9 and 81, the factors are both 9, since 9 \times 9 = 81. Turning the factors around gives the same sentence, and so does sharing by the other factor. The family has only two sentences:

9 \times 9 = 81 \qquad 81 \div 9 = 9
Example 3

Which trios make a family?

Do these three numbers form a multiplication and division fact family? If they do, write the family. (a) 4, 8, 32. (b) 3, 6, 24. (c) 1, 5, 5.

Solution

In a family the biggest number is the product, so multiply the two smaller numbers and compare.

(a) 4 \times 8 = 32, which is the biggest number. It is a family: 4 \times 8 = 32, 8 \times 4 = 32, 32 \div 4 = 8, 32 \div 8 = 4.

(b) 3 \times 6 = 18, not 24. It is not a family. Any sentence built from these numbers would be false: 24 \div 3 is 8, not 6, because 3 \times 8 = 24.

(c) 1 \times 5 = 5, which is the biggest number (the other 5). It is a family of four: 1 \times 5 = 5, 5 \times 1 = 5, 5 \div 1 = 5, and 5 \div 5 = 1. The two divisions are different sentences, because one shares 5 into 1 group and the other shares 5 into 5 groups.

Example 4

Multiply by six, then divide by six

Emiri’s robot multiplies by 6 and Zodwa’s robot divides by 6. Emiri puts 7 into her robot and gives its answer to Zodwa’s robot. What does each robot print? Explain why the last number had to be 7.

Solution

Emiri’s robot prints 7 \times 6 = 42. Zodwa’s robot gets 42 and prints 42 \div 6. To find it, ask 6 \times \square = 42; since 6 \times 7 = 42, the box is 7. So Zodwa’s robot prints 7.

It had to be 7. The 42 was made as 6 groups of 7. Dividing by 6 shares 42 into 6 equal groups, and the equal groups are the ones Emiri’s robot made, with 7 in each. Multiplying by 6 and dividing by 6 undo each other, so the number that comes out is the number that went in.

Example 5

Running a robot backward

(a) A robot multiplies by 4. It printed 36. What number went in? (b) Another robot divides by 7. It printed 5. What number went in?

Solution

To run a robot backward, use the operation that undoes it, with the same number.

(a) The robot multiplied by 4, so undo it by dividing by 4: 36 \div 4 = 9. The number that went in was 9. Check by running the robot forward: 9 \times 4 = 36.

(b) The robot divided by 7, so undo it by multiplying by 7: 5 \times 7 = 35. The number that went in was 35. Check: 35 \div 7 = 5, because 7 \times 5 = 35.

Example 6

Zero cubes, and five cubes into no cups

(a) Ruia shares 0 ice cubes equally among 9 cups. How many go in each cup? (b) Explain why 5 \div 0 has no answer.

Solution

(a) This is 0 \div 9, which asks 9 \times \square = 0. If each cup got even 1 cube, the 9 cups would hold 9 cubes, not 0. Only 0 works: 9 \times 0 = 0. Each cup gets 0 cubes, and 0 \div 9 = 0.

(b) 5 \div 0 asks 0 \times \square = 5. Try numbers: 0 \times 1 = 0, 0 \times 5 = 0, 0 \times 9 = 0. Zero groups always hold 0, so the product is 0 whatever goes in the box, and it is never 5. No number answers 5 \div 0. In the story, 5 cubes cannot be shared into 0 cups, because there are no cups for them.

Example 7

Zero divided by zero, and the times-zero robot

(a) What numbers fit the box in 0 \times \square = 0, and what does that say about 0 \div 0? (b) A robot multiplies by 0. It printed 0. What number went in?

Solution

(a) Every number fits. 0 \times 3 = 0, 0 \times 8 = 0, and the same for any number in the box. A division names one number, and here nothing picks one out, so 0 \div 0 has no answer. It fails for the opposite reason from 5 \div 0: there, no number works; here, every number does.

(b) There is no way to tell. Feeding the robot 2, 5 or 9 prints 0 every time, so the 0 could have come from any number at all. Running the robot backward would need a “divide by 0” that names one number, and part (a) shows there is none. Multiplying by 0 is the one multiplication that cannot be undone.

Boundaries

  • Never divide by zero. Dividing by a number undoes multiplying by it for every number except 0. A number divided by 0 has no answer, and 0 \div 0 has no answer either.
  • Zero divided by a number is fine. 0 \div 9 = 0. The order in a division matters: 0 \div 9 is 0, while 9 \div 0 has no answer.
  • Divide-then-multiply needs an even split, for now. With whole numbers, (c \div b) \times b = c only when b fits into c exactly. 25 \div 4 is not a whole number, since 4 \times 6 = 24 and 4 \times 7 = 28. That case belongs to Remainders and what they mean.
  • Not every family has four sentences. Equal factors give two (8 \times 8 = 64, 64 \div 8 = 8). A number and zero give three, because the fourth would divide by zero.
  • Three numbers are a family only if they multiply. When none of them is zero, the two smaller numbers must have the biggest as their product. 3, 6 and 24 are not a family.

Where this goes next

The undoing habit is used in every division that follows. Equal group problems turns stories into a multiplication with a box and finds the box by dividing. Dividing by a single digit checks each longer division by multiplying the quotient back by the divisor.

Undoing grows up into algebra. To find the number that, times 4, makes 36, we divided by 4. In Inverse operations and the properties of equality the same move is written as dividing both sides of 4x = 36 by 4. Later, Multiplying and dividing integers gets its rules for dividing negative numbers straight from the multiplying rules, because each division is still a multiplication undone.

The ban on dividing by zero never goes away. It is the reason Simplifying a rational expression and where it is undefined keeps out every number that makes a bottom zero.

Exercises

Exercises 1 to 7 go with Part 1, 8 to 15 with Part 2, and 16 to 21 with Part 3. The reasoning group draws on all three.

Part 1: fact families

  1. An ice cube tray has 5 rows with 8 cups in each row. Write the four sentences in its fact family.
  2. Write the fact family for 3, 9 and 27.
  3. Zodwa knows that 72 \div 8 = 9. Write the other three sentences in its family.
  4. Write the fact family for 7, 7 and 49. How many sentences does it have, and why?
  5. Write the fact family for 1, 8 and 8.
  6. Which of these are fact families? (a) 6, 7, 42. (b) 4, 9, 32. (c) 5, 5, 25.
  7. A fact family has the numbers 3 and 24, and 24 is the product. What is the third number? Write the family.

Part 2: doing and undoing

  1. A robot multiplies by 8. Emiri puts in 3. What does it print? What robot would turn the answer back into 3?
  2. Start at 9. Multiply by 5, then divide by 5. Give the number after each step.
  3. Start at 16. Divide by 2, then multiply by 2. Give the number after each step.
  4. A robot multiplies by 7. It printed 21. What number went in?
  5. A robot divides by 9. It printed 4. What number went in?
  6. Use multiplication to check each division. Is it right? (a) 28 \div 4 = 7. (b) 45 \div 5 = 8.
  7. Ruia’s robot multiplies by 3, then Emiri’s multiplies by 2. Ruia puts in 4. What comes out of each robot? What one division brings the last number back to 4?
  8. A number times 6 is 30. Use division to find the number, then check it.

Part 3: zero

  1. Find 0 \div 3 and check it by multiplying.
  2. Explain why 9 \div 0 has no answer.
  3. Which of these have an answer? Give it when there is one. (a) 0 \div 6. (b) 6 \div 0. (c) 0 \div 1. (d) 1 \div 0.
  4. (a) Which numbers fit the box in \square \times 0 = 4? (b) Which numbers fit the box in \square \times 0 = 0? Give three.
  5. Zodwa shares 0 ice cubes equally among 4 glasses. How many cubes go in each glass?
  6. Write every true sentence in the fact family of 8 and 0, with 0 as the product. Why does the family have only three?

Reasoning

  1. Explain, using an array, why knowing one multiplication fact such as 6 \times 4 = 24 gives three more facts without any new counting.
  2. Explain why multiplying any number by 5 and then dividing by 5 always gives back the number you started with.
  3. A robot multiplies by 0. Can any robot undo it? Explain.
  4. Is it always true that dividing by a number and then multiplying by the same number gives back the start? Give an example where it works, and say when it does not.
Answers
  1. 5 \times 8 = 40, 8 \times 5 = 40, 40 \div 5 = 8, 40 \div 8 = 5. The tray holds 5 \times 8 = 40 cubes.
  2. 3 \times 9 = 27, 9 \times 3 = 27, 27 \div 3 = 9, 27 \div 9 = 3.
  3. 8 \times 9 = 72, 9 \times 8 = 72, 72 \div 9 = 8. Undoing 72 \div 8 = 9 gives 9 \times 8 = 72.
  4. 7 \times 7 = 49 and 49 \div 7 = 7. Two sentences, because both factors are 7: turning them around, or dividing by the other factor, repeats the same sentence.
  5. 1 \times 8 = 8, 8 \times 1 = 8, 8 \div 1 = 8, 8 \div 8 = 1.
  6. (a) A family, since 6 \times 7 = 42. (b) Not a family, since 4 \times 9 = 36, not 32. (c) A family of two, since 5 \times 5 = 25: 5 \times 5 = 25 and 25 \div 5 = 5.
  7. 8, since 3 \times 8 = 24 (equally, 24 \div 3 = 8). The family is 3 \times 8 = 24, 8 \times 3 = 24, 24 \div 3 = 8, 24 \div 8 = 3.
  8. It prints 3 \times 8 = 24. A robot that divides by 8 turns it back: 24 \div 8 = 3.
  9. 9 \times 5 = 45, then 45 \div 5 = 9. Back at 9.
  10. 16 \div 2 = 8, then 8 \times 2 = 16. Back at 16.
  11. 3. Undo multiplying by 7 by dividing by 7: 21 \div 7 = 3. Check: 3 \times 7 = 21.
  12. 36. Undo dividing by 9 by multiplying by 9: 4 \times 9 = 36. Check: 36 \div 9 = 4.
  13. (a) Right, since 7 \times 4 = 28. (b) Not right, since 8 \times 5 = 40, not 45. The quotient is 9, because 9 \times 5 = 45.
  14. Ruia’s robot prints 4 \times 3 = 12; Emiri’s prints 12 \times 2 = 24. Dividing by 6 brings it back: 24 \div 6 = 4. (Dividing by 2 and then by 3 also works: 24 \div 2 = 12, 12 \div 3 = 4.)
  15. 30 \div 6 = 5. Check: 5 \times 6 = 30.
  16. 0 \div 3 = 0. Check: 3 \times 0 = 0.
  17. 9 \div 0 asks 0 \times \square = 9. Zero times any number is 0, never 9, so no number fits and there is no answer.
  18. (a) 0 \div 6 = 0. (b) No answer. (c) 0 \div 1 = 0. (d) No answer. A zero on the left of \div gives 0; a zero on the right gives no answer.
  19. (a) No number fits, because any number times 0 is 0, not 4. (b) Every number fits, for example 1 \times 0 = 0, 6 \times 0 = 0 and 10 \times 0 = 0.
  20. 0 cubes, since 0 \div 4 = 0 and 4 \times 0 = 0.
  21. 8 \times 0 = 0, 0 \times 8 = 0, 0 \div 8 = 0. The fourth would be 0 \div 0 = 8, and 0 \div 0 has no answer, because every number times 0 is 0.
  22. Draw 6 rows of 4. Read by rows it is 6 \times 4 = 24; turned on its side it is 4 rows of 6, so 4 \times 6 = 24. Sharing the 24 things back into the 6 rows gives 4 each, so 24 \div 6 = 4, and into the 4 columns gives 6 each, so 24 \div 4 = 6. Every sentence is about the same 24 things.
  23. Multiplying a number by 5 makes 5 equal groups of that number. Dividing by 5 shares the total into 5 equal groups, and those are the groups that were made, so each holds the starting number.
  24. No. It prints 0 for every number that goes in, so the 0 does not say which number it came from. Undoing would need 0 \div 0 to name one number, and every number fits 0 \times \square = 0.
  25. Not always. It works when the divisor is not zero and the division comes out even: 18 \div 3 = 6 and 6 \times 3 = 18. It fails when dividing by 0, which has no answer, and with whole numbers it fails when the split is not even, as in 25 \div 4.
kg1128 · updated 2026-10-21 · CC-BY-SA-4.0