# Doing and undoing

An investigation looking at doing and undoing mathematical operations focusing on doubling, halving, adding and subtracting.

## Problem

Image

I wonder if you've ever used string or rope to make a knot, and then undone it afterwards. In this activity, we're going to look at undoing some maths that's been done.

To begin with, we're going to imagine taking a starting number and then doubling it. We want to undo the doubling by starting with just the answer and trying to get back to our starting number.

When I doubled my starting number, I got 6. How can I undo the doubling and get back to my starting number?

When I doubled my starting number, I got 6. How can I undo the doubling and get back to my starting number?

Suppose that I did this again with a new starting number, and when I doubled it I got 16. How can I undo the doubling?

How would you undo these three children's maths?

Danesh says:

"I took 4 away. How can I get back to my starting number?"

Meg says:

"I added 8. How can I get back to my starting number?"

Chris says:

"I halved. How can I get back to my starting number?"

If these children all ended up with the number 12, what were their starting numbers?

## Getting Started

Think carefully about what has happened to the number.

What sort of thing do you think would take it back to the number that you started with?

What sort of thing do you think would take it back to the number that you started with?

## Student Solutions

We had a few solutions sent in for this challenge. Simran from Beauchamp College England wrote;

D = Danesh

D-4=12

To 'undo' add 4

D=12+4

D=16

M = Meg

M+8=12

To 'undo' minus 8

M=12-8

M=4

C = Chris

C/2=12

To 'undo' times 2

C=12*2

C=24

Ore from Sir Joseph Williamsons Mathematical School wrote

In all these problems you must do the opposite of what has been done to the

original number to 12.

For Danesh 12+4=16

For Meg 12-8=4

And Chris 12*2=24

From Martha at Langtons Junior School

You need to do the inverse (opposite)operation. For instance Chris halved

his starting number and finished with twelve. The inverse of halving is

doubling so you double twelve so Chris's starting number is 24. it is the

same for the other sums.

Right at the end of the month we received this contribution from Daniel at Moss Hall Primary School in England in which he describes very thoroughly and with excellent mathematical skills what he did.

Let d = Danesh's starting number, m = Meg's starting number, and c = Chris' starting number. We use letters to represent their starting numbers since we do not know their values yet. These are called unknowns/variables in an equation.

For example, $x + 4 = 5$ . $x = 1$. We have 'solved' the equation and found the value of the unknown/variable.

If Danesh started with d, then took away 4 from it, and finished with $12$, we can write this as the expression:

$d - 4 = 12$ (he started with d, took away $4$, which made it equal to $12$).

This is a linear equation and can be solved by adding 4 to both sides:

$d - 4 + 4 = 12 + 4$

$d = 16$

Danesh started with $16$, subtracted $4$, and finished with $12$.

Meg started with m and added $8$ to it. She ended up with $12$. This translates to:

$m + 8 = 12$

We want to get rid of the $8$ and leave m by itself so we know it's value. To do this, we have to minus $8$ from both sides of the equation. Since one side'equals' the other side in any equation, what we do to one side has to be done to the other side so that both sides remain equal.

$m + 8 = 12$

$m + 8 - 8 = 12 - 8$

$m = 4$

Meg started with $4$, added $8$, and ended up with $12$.

Finally, to find Chris' starting number, we use the same techniques as above. You should have noticed the rough outline of what we should do by now. First, we construct an equation relating the variables and the outcome. In this case, the outcome of the mathematical operations is $12$.

Since Chris 'halved' his starting number to get $12$, we simply right $c/2 =12$. Dividing by two is the same as multiplying by $1/2$, because c, is technically $c/1$. Any integer (whole number) is in fact a fraction, but we just right it without the fraction to make life easy for our selves. For

example, $5$, is actually $5/1$. $(5/1=5)$ but we just write $5$. In this case, c is actually $c/1$, and halving is multiplying by $1/2$, so $c/1 \times 1/2 = c / 2 = 12$ (we are told they all finish with $12$). To solve this, we multiply both side by $2$. Doing this gets rid of the $2$ on the left side, because $c/2 \times 2/1 = 2c/2 = c$. And the right side ends up as $12 \times 2 = 24$. Remember, an equation needs both sides to remain equal, what we do to one side has to be

done to the other, otherwise the equations will not remain equal most of the time. (Try and find something that leaves both sides the same when it is applied only to one side! Experiment :D)

We conclude from the working out above, that the numbers the three people

started with are:

$d = 16$

$m = 4$

$c = 24$

A good idea when you have finished solving an equation is to substitute your value back into the equation, for example for Chris' equation, we got $24$, so to check if our answer is right, we put it in place of c (since $c =24$): $c/2 = 12$ substituting $24$, we get: $24/2 = 12$ , $12 = 12$

Our answer is correct.

When faced with a word problem, try to first of all convert the words into

mathematics, and then solve the problem that way.

Thank you for these and those of you who explored this challenge of inverse operations.

D = Danesh

D-4=12

To 'undo' add 4

D=12+4

D=16

M = Meg

M+8=12

To 'undo' minus 8

M=12-8

M=4

C = Chris

C/2=12

To 'undo' times 2

C=12*2

C=24

Ore from Sir Joseph Williamsons Mathematical School wrote

In all these problems you must do the opposite of what has been done to the

original number to 12.

For Danesh 12+4=16

For Meg 12-8=4

And Chris 12*2=24

From Martha at Langtons Junior School

You need to do the inverse (opposite)operation. For instance Chris halved

his starting number and finished with twelve. The inverse of halving is

doubling so you double twelve so Chris's starting number is 24. it is the

same for the other sums.

Right at the end of the month we received this contribution from Daniel at Moss Hall Primary School in England in which he describes very thoroughly and with excellent mathematical skills what he did.

Let d = Danesh's starting number, m = Meg's starting number, and c = Chris' starting number. We use letters to represent their starting numbers since we do not know their values yet. These are called unknowns/variables in an equation.

For example, $x + 4 = 5$ . $x = 1$. We have 'solved' the equation and found the value of the unknown/variable.

If Danesh started with d, then took away 4 from it, and finished with $12$, we can write this as the expression:

$d - 4 = 12$ (he started with d, took away $4$, which made it equal to $12$).

This is a linear equation and can be solved by adding 4 to both sides:

$d - 4 + 4 = 12 + 4$

$d = 16$

Danesh started with $16$, subtracted $4$, and finished with $12$.

Meg started with m and added $8$ to it. She ended up with $12$. This translates to:

$m + 8 = 12$

We want to get rid of the $8$ and leave m by itself so we know it's value. To do this, we have to minus $8$ from both sides of the equation. Since one side'equals' the other side in any equation, what we do to one side has to be done to the other side so that both sides remain equal.

$m + 8 = 12$

$m + 8 - 8 = 12 - 8$

$m = 4$

Meg started with $4$, added $8$, and ended up with $12$.

Finally, to find Chris' starting number, we use the same techniques as above. You should have noticed the rough outline of what we should do by now. First, we construct an equation relating the variables and the outcome. In this case, the outcome of the mathematical operations is $12$.

Since Chris 'halved' his starting number to get $12$, we simply right $c/2 =12$. Dividing by two is the same as multiplying by $1/2$, because c, is technically $c/1$. Any integer (whole number) is in fact a fraction, but we just right it without the fraction to make life easy for our selves. For

example, $5$, is actually $5/1$. $(5/1=5)$ but we just write $5$. In this case, c is actually $c/1$, and halving is multiplying by $1/2$, so $c/1 \times 1/2 = c / 2 = 12$ (we are told they all finish with $12$). To solve this, we multiply both side by $2$. Doing this gets rid of the $2$ on the left side, because $c/2 \times 2/1 = 2c/2 = c$. And the right side ends up as $12 \times 2 = 24$. Remember, an equation needs both sides to remain equal, what we do to one side has to be

done to the other, otherwise the equations will not remain equal most of the time. (Try and find something that leaves both sides the same when it is applied only to one side! Experiment :D)

We conclude from the working out above, that the numbers the three people

started with are:

$d = 16$

$m = 4$

$c = 24$

A good idea when you have finished solving an equation is to substitute your value back into the equation, for example for Chris' equation, we got $24$, so to check if our answer is right, we put it in place of c (since $c =24$): $c/2 = 12$ substituting $24$, we get: $24/2 = 12$ , $12 = 12$

Our answer is correct.

When faced with a word problem, try to first of all convert the words into

mathematics, and then solve the problem that way.

Thank you for these and those of you who explored this challenge of inverse operations.

## Teachers' Resources

**Why do this problem?**

This activity possibly presents a new way for many pupils to think about the arithmetic they do. The idea of inverse operations is core mathematical concept and this activity offers opportunities to explore them in a meaningful way. The task also presents the chance to allow pupils to have the freedom to be curious and explore all kinds of
mathematics further, as well as persevere in their attempts to "undo" various calculations.

### Possible approach

One way would be to start by saying "Here's the number $4$. I'll double it", and inviting the pupils to say what the answer is and then how to get back to the $4$.

You could try some different numbers and repeat the process with doubling each time as the operation.

The operation can then be changed to an addition or subtraction one.

This can now lead to the bigger question about whether same inverse operation works for every starting number.

If your pupils are secure then use the same rule but choose a different, and this time mystery, starting number and tell them the finishing number. Invite them to think about what calculation they'd do to get back to your mystery starting number.

You could try some different numbers and repeat the process with doubling each time as the operation.

The operation can then be changed to an addition or subtraction one.

This can now lead to the bigger question about whether same inverse operation works for every starting number.

If your pupils are secure then use the same rule but choose a different, and this time mystery, starting number and tell them the finishing number. Invite them to think about what calculation they'd do to get back to your mystery starting number.

### Key questions

What number do you get?

If you 'undo' the operation what do you get?

What do you have to do to 'undo' adding $6$?

What do you have to do to 'undo' doubling?

Tell me about your ideas.

### Possible extension

For those who show fluency in the activity then a further challenge is to say that the starting number has gone through two operations. Can the children find out how to undo them each in turn? What order do they have to undo them in?

### Possible support

Children may need to stick to small numbers so that they are not overloaded by trying to remember number facts. It may help to have a 'machine box' and think about the numbers going in and the numbers coming out, and what happens if you undo the action by putting in the numbers backwards.