Closure and the commutative property

Every time you work out a sum, you are qui­etly trust­ing two facts. The first is that the answer will be a num­ber you already know how to han­dle. The sec­ond is that, for some oper­a­tions, you may write the two num­bers in either order. Here we give those two facts their proper names: clo­sure and the com­mu­ta­tive prop­erty. You will meet both again and again, in men­tal arith­metic, when you rearrange a bill at the shop, and later when you sim­plify alge­bra such as $3x + 5 = 5 + 3x$.

Real num­bers in one place#

All the ratio­nal and irra­tional num­bers together form the real num­bers. We write the whole set as $R$.

A ratio­nal num­ber can be writ­ten as one inte­ger over another, like $\displaystyle \frac{3}{4}$ or $\displaystyle \frac{-7}{2}$. An inte­ger is a count­ing num­ber, or zero, or one with a minus sign. An irra­tional num­ber can­not be writ­ten that way.

The dec­i­mal of an irra­tional num­ber never stops and never repeats; two famous exam­ples are $\pi$ and $e = 2.71\ldots$

So whole num­bers, frac­tions and dec­i­mals are all real, and so are num­bers below zero, like $-3$.

We are going to ask two sim­ple-sound­ing ques­tions about real num­bers.

1. If you work with two real num­bers, is the answer real too?
2. May you swap the two num­bers round and still get the same answer?

Ques­tion one: does the answer stay in $R$?#

Pick any two real num­bers, say 7 and 4. Now add them, take one from the other, and mul­ti­ply them.

Work­ing with 7 and 4

$$\begin{aligned}7 + 4 &= 11 \\ 7 - 4 &= 3 \\ 7 \times 4 &= 28\end{aligned}$$

Every answer is a real num­ber. You never fell out of $R$.

Big­ger num­bers behave the same way: $10(4) = 40$, and $40$ is real. (Here $10(4)$ means $10 \times 4$; a num­ber right beside a bracket means mul­ti­ply.)

Clo­sure does­n't care whether the num­bers are whole, either. Try two frac­tions, one of them neg­a­tive.

Work­ing with $\displaystyle -\frac{3}{4}$ and $\displaystyle \frac{5}{2}$

$$\displaystyle \begin{aligned}-\frac{3}{4} + \frac{5}{2} &= -\frac{3}{4} + \frac{10}{4} = \frac{7}{4} \\ -\frac{3}{4} \times \frac{5}{2} &= -\frac{15}{8}\end{aligned}$$

Both answers are frac­tions, and every frac­tion is real. Even an irra­tional num­ber can­not push you out: $\pi + 1$ and $2\pi$ are still real num­bers, because they still sit some­where on the num­ber line.

That is what clo­sure means. A set is closed under an oper­a­tion when the answer always stays inside the set.

Remem­ber. Closed means you can­not escape. You start in $R$ and you land in $R$.

Divi­sion is the odd one#

Divi­sion is a lit­tle dif­fer­ent. It works every sin­gle time but one: you can­not divide by zero.

Divid­ing real num­bers

$$\begin{aligned}10 \div 2 &= 5 \\ 7 \div 4 &= 1.75\end{aligned}$$

Both answers are real. $1.75$ isn't a whole num­ber, but it is still real.

The next bit is extra, but worth know­ing: why does divid­ing by zero fail?

Think about what $6 \div 0$ is ask­ing: which num­ber times zero gives 6? No num­ber can do that, because zero times any­thing is zero.

So $6 \div 0$ has no answer at all. There's noth­ing to write down, and so no real num­ber to land on.

Zero is the only trou­ble­maker. Every other divi­sion of two real num­bers gives a real num­ber.

Remem­ber. Real num­bers are closed under addi­tion, sub­trac­tion and mul­ti­pli­ca­tion. They are not closed under divi­sion.

The same idea in let­ters#

Let $a$ and $b$ stand for any two real num­bers. Then:

Clo­sure

$$\begin{aligned}&a + b \in R \\ &a - b \in R \\ &a \times b \in R \\ &a \div b \ \text{may or may not be in } R \\ \text{and } a \div b \in R \ \text{when } b &\neq 0\end{aligned}$$

Read $a + b \in R$ as: $a$ plus $b$ belongs to $R$.

Read the last line as: $a$ divided by $b$ belongs to $R$ as long as $b$ is not zero.

The fourth line says "may" for a rea­son: there is no answer at all when $b$ is zero, and that is the only thing that can go wrong.

Ques­tion two: may you swap them round?#

Add 3 and 5, and then add the same two num­bers the other way round.

Adding both ways

$$\begin{aligned}3 + 5 &= 8 \\ 5 + 3 &= 8\end{aligned}$$

The same answer twice, so the order made no dif­fer­ence. Mul­ti­ply­ing behaves just the same way.

Mul­ti­ply­ing both ways

$$\begin{aligned}3 \times 5 &= 15 \\ 5 \times 3 &= 15\end{aligned}$$

There's a word for this: com­mu­ta­tive. It sim­ply means you may swap the two num­bers round. Think of adding up a shop­ping bill; whether you add the rice first or the dal first, the total is the same.

Remem­ber. Com­mu­ta­tive means the order of the two num­bers does not change the answer.

Tak­ing away says no#

A num­ber line shows why the order mat­ters. Start­ing point and size of jump both change when you swap.

Tak­ing away both ways

$$\begin{aligned}5 - 3 &= 2 \\ 3 - 5 &= -2\end{aligned}$$

Those two answers are not the same: one is $2$ and the other is $-2$.

$-2$ is a real num­ber too, so sub­trac­tion has­n't escaped $R$. But the order has changed the answer, so sub­trac­tion is not com­mu­ta­tive. Here the order really mat­ters.

Divid­ing says no as well#

Divid­ing both ways

$$\begin{aligned}10 \div 2 &= 5 \\ 2 \div 10 &= 0.2\end{aligned}$$

Once again the two answers dif­fer, so divi­sion is not com­mu­ta­tive either.

A few more pairs to test#

Here are some harder pairs. Try each one both ways your­self before you peek at the answer.

• $2.5 \times 4 = 10$ and $4 \times 2.5 = 10$. The same, as mul­ti­pli­ca­tion is com­mu­ta­tive even with dec­i­mals.
• $(-6) \times 3 = -18$ and $3 \times (-6) = -18$. The same again. A minus sign does not stop you swap­ping.
• $\displaystyle \frac{1}{2} - \frac{3}{4} = -\frac{1}{4}$ but $\displaystyle \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$. Dif­fer­ent. The two answers are oppo­sites of each other.
• $8 \div 2 = 4$ but $\displaystyle 2 \div 8 = \frac{1}{4}$. Dif­fer­ent. Swap­ping a divi­sion turns the answer upside down.

Notice the pat­tern in the last two. Swap­ping a sub­trac­tion gives the neg­a­tive of the answer. Swap­ping a divi­sion gives the rec­i­p­ro­cal. So the swapped answer is related to the first one, but it is not the same.

The same idea in let­ters#

Two let­ters side by side mean mul­ti­ply. So $ab$ is short for $a \times b$.

For any two real num­bers $a$ and $b$, these two always hold:

You may swap

$$\begin{aligned}a + b &= b + a \\ ab &= ba\end{aligned}$$

And here you may not swap:

You may not swap

$$\begin{aligned}&a - b \ \text{is usually not} \ b - a \\ &a \div b \ \text{is usually not} \ b \div a\end{aligned}$$

Now and then these two can be equal by acci­dent: $5 - 5$ is $0$ either way round. But you can't count on luck, so you may not swap.

Both answers on one page#

Remem­ber. Addi­tion and mul­ti­pli­ca­tion of real num­bers are com­mu­ta­tive. Sub­trac­tion and divi­sion are not. Swap the two num­bers there and the answer usu­ally changes.

Prac­tice#

1. Work out $9 + 6$ and then $6 + 9$.
2. Work out $8 \times 5$ and then $5 \times 8$.
3. Work out $12 - 7$ and then $7 - 12$.
4. Work out $20 \div 4$ and then $4 \div 20$.
5. Is the answer to $3 + 11$ a real num­ber?
6. One of these has no answer. Is it $0 \div 6$ or $6 \div 0$?

Com­mon mis­takes#

• Think­ing a neg­a­tive answer means you have left the real num­bers. $-2$ is as real as $2$; sub­trac­tion is still closed.
• Say­ing divi­sion is closed. One case, divid­ing by zero, has no answer at all, so the set is not closed under divi­sion.
• Mix­ing up $0 \div 6$ and $6 \div 0$. The first is $0$; the sec­ond has no answer.
• Decid­ing an oper­a­tion is com­mu­ta­tive from one lucky exam­ple, such as $5 - 5$. One exam­ple where the order fails is enough to show it is not com­mu­ta­tive.
• Treat­ing $ab$ as two sep­a­rate num­bers. Let­ters writ­ten side by side mean mul­ti­ply.

Key terms#

Real num­bers ($R$)

All the ratio­nal and irra­tional num­bers together.

Ratio­nal num­ber

A num­ber that can be writ­ten as one inte­ger over another, with the bot­tom not zero.

Irra­tional num­ber

A num­ber whose dec­i­mal never stops and never repeats, such as $\pi$.

Clo­sure

A set is closed under an oper­a­tion when the answer always stays inside the set.

Com­mu­ta­tive

The order of the two num­bers does not change the answer.

Belongs to ($\in$)

The sym­bol that says a num­ber is a mem­ber of a set.

Answers#

Answers to the Prac­tice ques­tions.

1. $9 + 6 = 15$ and $6 + 9 = 15$. The same, so addi­tion is com­mu­ta­tive.
2. $8 \times 5 = 40$ and $5 \times 8 = 40$. The same, so mul­ti­pli­ca­tion is com­mu­ta­tive.
3. $12 - 7 = 5$ and $7 - 12 = -5$. Dif­fer­ent, so the order mat­ters.
4. $20 \div 4 = 5$ and $4 \div 20 = 0.2$. Dif­fer­ent, so the order mat­ters.
5. Yes. $3 + 11 = 14$, which is real, because $R$ is closed under addi­tion.
6. $6 \div 0$ has no answer. $0 \div 6 = 0$.