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. Why is that allowed for adding and mul­ti­ply­ing, but not for tak­ing away and divid­ing? We'll find out, and then meet two spe­cial num­bers, $0$ and $1$, along with the part­ners that undo a num­ber. You'll use every one of these ideas again in alge­bra, where let­ters take the place of num­bers.

You've already met swap­ping. Now we turn to group­ing, which sim­ply 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, and their mes­sage is sim­ple: 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 and only the group­ing changed, so addi­tion clearly does­n't mind how you group.

Group­ing to make sums easy#

This is more than a curios­ity; it lets you pick 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#

Stu­dents often mix these two ideas up, so let's be clear about the dif­fer­ence.

Swap­ping is about two num­bers chang­ing places. You saw $3 + 5 = 5 + 3$, where both give $8$. That is the com­mu­ta­tive prop­erty.

Group­ing is about which pair you do first. You saw $(2 + 3) + 4 = 2 + (3 + 4)$, where the order of the num­bers never moved. That is 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 does­n't mind how you group either.

The same trick makes mul­ti­ply­ing eas­ier. 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'd need $25 \times 28$, which is a lot harder. Both give $700$, but one of them is done in a blink.

Tak­ing away does mind#

Now try the same trick with sub­trac­tion, and watch care­fully 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. They are not equal, so sub­trac­tion is not asso­cia­tive.

Why the dif­fer­ence? 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.

Let's test the idea on one more pair.

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

Once again the two answers dif­fer. In fact, a sin­gle exam­ple where the answers dif­fer is enough to show a rule does not hold, and 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, so divi­sion is not asso­cia­tive.

The rea­son: in the sec­ond sum the brack­ets shrink the num­ber you divide by. You divide $12$ by $3$ instead of by $6$, so you end up with 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 let us state the rule for all of them at once. Pick any three real num­bers and 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, like a guest who vis­its and changes noth­ing. Look at these four sums.

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

Adding $0$ left the num­ber alone, and it did­n't mat­ter which side the $0$ sat on. So $0$ is called the addi­tive iden­tity.

Mul­ti­ply­ing by $1$ also left the num­ber alone, whichever side the $1$ sat on. So $1$ is called the mul­ti­plica­tive iden­tity.

Here are both rules writ­ten 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 $7$, just on the other side. So if you add $7$ and $-7$, you land right 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 kind of 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, because you can­not share some­thing into zero groups.

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

How­ever big the other num­ber is, the answer stays zero, so no num­ber times zero can ever give $1$. Zero sim­ply 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#

Here are three last facts. They look easy, but they're 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 dif­fer­ence is the whole 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#

Show 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. $\displaystyle 5 \times \frac{1}{5} = \frac{5}{5} = 1$