Grouping, identities and inverses

When you add a long bill in your head, you rarely go strictly from left to right. You pick the pairs that are easy and do them first. This les­son explains why that is allowed for adding and mul­ti­ply­ing, and why it is not allowed for tak­ing away and divid­ing. It then meets two spe­cial num­bers, $0$ and $1$, and the part­ners that undo a num­ber. You will use all of these ideas again in alge­bra, where let­ters take the place of num­bers.

You have already met swap­ping. This les­son is about group­ing. Group­ing means choos­ing which pair of num­bers you work out first.

Brack­ets tell you what to do first

Brack­ets are the curved marks around part of a sum. They say: do this bit first. In $(2 + 3) + 4$ you add $2$ and $3$ first. In $2 + (3 + 4)$ you add $3$ and $4$ first.

First pair first

$$\begin{aligned}&(2 + 3) + 4 \\ &= 5 + 4 \\ &= 9\end{aligned}$$

Sec­ond pair first

$$\begin{aligned}&2 + (3 + 4) \\ &= 2 + 7 \\ &= 9\end{aligned}$$

Both answers are $9$. The num­bers stayed in the same order. Only the group­ing changed. Addi­tion does not mind how you group.

Group­ing to make sums easy

This is more than a curios­ity. It lets you choose the easy pair. Look at $17 + 25 + 75$.

Choos­ing the friendly pair

$$\begin{aligned}&(17 + 25) + 75 = 42 + 75 = 117 \\ &17 + (25 + 75) = 17 + 100 = 117\end{aligned}$$

The sec­ond way is far eas­ier, because $25 + 75$ makes a round $100$. The asso­cia­tive prop­erty promises that both ways give the same answer, so you are free to take the easy road.

Swap­ping and group­ing are not the same

These two ideas get mixed up, so here is the dif­fer­ence.

Swap­ping is about two num­bers chang­ing places. You saw $3 + 5 = 5 + 3$. Both give $8$. That one is called the com­mu­ta­tive prop­erty.

Group­ing is about which pair you do first. You saw $(2 + 3) + 4 = 2 + (3 + 4)$. The order of the num­bers never moved. That one is called the asso­cia­tive prop­erty.

Remem­ber. Com­mu­ta­tive is about swap­ping two num­bers. Asso­cia­tive is about which pair you do first.

Mul­ti­ply­ing works the same way

First pair first

$$\begin{aligned}&(2 \times 3) \times 4 \\ &= 6 \times 4 \\ &= 24\end{aligned}$$

Sec­ond pair first

$$\begin{aligned}&2 \times (3 \times 4) \\ &= 2 \times 12 \\ &= 24\end{aligned}$$

Both answers are $24$. So mul­ti­pli­ca­tion does not mind how you group either.

The same trick helps with mul­ti­ply­ing. Look at $25 \times 4 \times 7$.

$$\begin{aligned}&(25 \times 4) \times 7 \\ &= 100 \times 7 \\ &= 700\end{aligned}$$

Grouped the other way, you would need $25 \times 28$, which is harder. Both give $700$, but one is done in a moment.

Tak­ing away does mind

Now try the same trick with sub­trac­tion. Watch what hap­pens.

First pair first

$$\begin{aligned}&(8 - 3) - 2 \\ &= 5 - 2 \\ &= 3\end{aligned}$$

Sec­ond pair first

$$\begin{aligned}&8 - (3 - 2) \\ &= 8 - 1 \\ &= 7\end{aligned}$$

You get $3$ one way and $7$ the other way. They are not equal. So sub­trac­tion is not asso­cia­tive.

Here is why. In the sec­ond sum the brack­ets turn a take away into an add. You take away $1$ in total instead of $5$, so you are left with more.

Here is one more pair to test the idea.

$$\begin{aligned}(20 - 10) - 5 &= 10 - 5 = 5 \\ 20 - (10 - 5) &= 20 - 5 = 15\end{aligned}$$

Once again the two answers dif­fer. One exam­ple where the answers dif­fer is enough to show a rule does not hold. Math­e­mati­cians call such an exam­ple a coun­terex­am­ple.

Divid­ing minds too

First pair first

$$\begin{aligned}&(12 \div 6) \div 2 \\ &= 2 \div 2 \\ &= 1\end{aligned}$$

Sec­ond pair first

$$\begin{aligned}&12 \div (6 \div 2) \\ &= 12 \div 3 \\ &= 4\end{aligned}$$

You get $1$ one way and $4$ the other way. So divi­sion is not asso­cia­tive.

Here is why. In the sec­ond sum the brack­ets shrink the num­ber you divide by. You divide $12$ by $3$ instead of by $6$, so you get more.

Writ­ing it with let­ters

So far you have used ordi­nary num­bers you can count and mea­sure with. All the ratio­nal and irra­tional num­bers together are called the real num­bers, as you saw in The num­ber sys­tem. That takes in whole num­bers, neg­a­tive num­bers, frac­tions, and num­bers like $\pi$.

Let­ters can now say the rule for all of them at once. Pick any three real num­bers. Call them $a$, $b$ and $c$.

When two let­ters sit side by side, it means mul­ti­ply them. So $ab$ is a short way of writ­ing $a \times b$. You can­not do that with dig­its, because $23$ would look like twenty-three.

Remem­ber. Real num­bers are asso­cia­tive under addi­tion and mul­ti­pli­ca­tion. They are not asso­cia­tive under sub­trac­tion or divi­sion.

The num­bers that change noth­ing

Some num­bers leave oth­ers exactly as they were. Look at these four sums.

The word iden­tity means same­ness. The num­ber keeps being itself.

Adding $0$ left the num­ber alone. It did not mat­ter which side the $0$ sat on. So $0$ is called the addi­tive iden­tity.

Mul­ti­ply­ing by $1$ also left the num­ber alone. Again it did not mat­ter which side the $1$ sat on. So $1$ is called the mul­ti­plica­tive iden­tity.

Here are both rules in let­ters.

The num­bers that undo

An inverse is a part­ner that undoes a num­ber.

$-7$ means seven below zero. It sits the same dis­tance from zero as $7$, but on the other side.

Add $7$ and $-7$ and you land back on $0$.

Adding the part­ner

$$\begin{aligned}&7 + (-7) \\ &= 0\end{aligned}$$

So $-7$ is the addi­tive inverse of $7$. In let­ters, $a + (-a) = 0$. The part­ner takes you back to $0$, the addi­tive iden­tity.

Mul­ti­ply­ing has its own part­ner. Mul­ti­ply $2$ by $\displaystyle \frac{1}{2}$ and you land back on $1$.

Mul­ti­ply­ing by the part­ner

$$\displaystyle \begin{aligned}&2 \times \frac{1}{2} \\ &= \frac{2}{2} \\ &= 1\end{aligned}$$

So $\displaystyle \frac{1}{2}$ is the mul­ti­plica­tive inverse of $2$. In let­ters, $\displaystyle a \times \frac{1}{a} = 1$, as long as $a$ is not zero.

More part­ners

Every real num­ber has an addi­tive inverse, and frac­tions are no excep­tion.

$$\displaystyle \begin{aligned}&-\frac{2}{5} + \frac{2}{5} \\ &= 0\end{aligned}$$

To find a mul­ti­plica­tive inverse of a frac­tion, turn it upside down. The frac­tion $\displaystyle \frac{3}{4}$ has the part­ner $\displaystyle \frac{4}{3}$.

$$\displaystyle \begin{aligned}&\frac{3}{4} \times \frac{4}{3} \\ &= \frac{12}{12} \\ &= 1\end{aligned}$$

Why $a$ must not be zero

There are two rea­sons, and both are easy to see.

First, $\displaystyle \frac{1}{0}$ has no mean­ing. You can­not share some­thing into zero groups.

Sec­ond, any­thing times zero is zero. $5 \times 0 = 0$. $100 \times 0 = 0$.

How­ever big the other num­ber is, the answer is still zero. So no num­ber times zero can ever give $1$. Zero has no mul­ti­plica­tive inverse.

Remem­ber. An inverse takes you back to an iden­tity. $-a$ takes you back to $0$. $\displaystyle \frac{1}{a}$ takes you back to $1$, and here $a$ can­not be zero.

Three small facts

Three last facts. They look easy, but they are worth say­ing out loud.

1. Take noth­ing away and noth­ing changes. $9 - 0 = 9$, so $a - 0 = a$.
2. Take a num­ber away from noth­ing and you are left with its neg­a­tive. $0 - 4 = -4$, so $0 - a = -a$.
3. One times a num­ber leaves it alone. $1 \cdot 5 = 5$, so $1 \cdot a = a$.

The small dot in $1 \cdot a$ is one more way to write mul­ti­ply.

Your turn

Work each one out. Then check your­self below.

The first two pairs give the same answer. The next two pairs do not, and that is the point of this les­son.

The last two ques­tions show a num­ber meet­ing its part­ner. $-15$ is the addi­tive inverse of $15$. $\displaystyle \frac{1}{5}$ is the mul­ti­plica­tive inverse of $5$.

Com­mon mis­takes

• Mix­ing up the two prop­er­ties. Mov­ing num­bers to new places is com­mu­ta­tive. Mov­ing only the brack­ets is asso­cia­tive.
• Regroup­ing a sub­trac­tion or a divi­sion. $(8 - 3) - 2$ and $8 - (3 - 2)$ are dif­fer­ent.
• Think­ing $0$ is the iden­tity for mul­ti­ply­ing. $6 \times 0 = 0$, not $6$. The iden­tity for mul­ti­ply­ing is $1$.
• Giv­ing the wrong inverse. The addi­tive inverse of $5$ is $-5$; the mul­ti­plica­tive inverse is $\displaystyle \frac{1}{5}$.
• Look­ing for a mul­ti­plica­tive inverse of $0$. There is none.

Key terms

Asso­cia­tive prop­erty

Chang­ing the group­ing does not change the answer: $(a + b) + c = a + (b + c)$ and $(ab)c = a(bc)$.

Com­mu­ta­tive prop­erty

Swap­ping two num­bers does not change the answer, as in $3 + 5 = 5 + 3$.

Addi­tive iden­tity

The num­ber $0$, because $a + 0 = a$.

Mul­ti­plica­tive iden­tity

The num­ber $1$, because $a \times 1 = a$.

Addi­tive inverse

The part­ner $-a$ that gives $a + (-a) = 0$.

Mul­ti­plica­tive inverse

The part­ner $\displaystyle \frac{1}{a}$ that gives $\displaystyle a \times \frac{1}{a} = 1$, for $a$ not zero.

Coun­terex­am­ple

One exam­ple that shows a rule does not always hold.

Answers

1. $(4 + 6) + 5 = 10 + 5 = 15$
2. $4 + (6 + 5) = 4 + 11 = 15$
3. $(5 \times 2) \times 3 = 10 \times 3 = 30$
4. $5 \times (2 \times 3) = 5 \times 6 = 30$
5. $(9 - 4) - 3 = 5 - 3 = 2$
6. $9 - (4 - 3) = 9 - 1 = 8$
7. $(16 \div 4) \div 2 = 4 \div 2 = 2$
8. $16 \div (4 \div 2) = 16 \div 2 = 8$
9. $12 + 0 = 12$
10. $8 \times 1 = 8$
11. $0 - 7 = -7$
12. $15 + (-15) = 0$
13. $\displaystyle 5 \times \frac{1}{5} = \frac{5}{5} = 1$