Multiplying and dividing real numbers

When­ever you work out the area of a square whose side is $\sqrt{2}$ cm, or share a length like $8$ m into $\sqrt{2}$ equal parts in a geom­e­try prob­lem, you are mul­ti­ply­ing and divid­ing real num­bers. The good news is that the rules you use for whole num­bers carry straight over. This les­son checks each rule on num­bers you know, then uses it on roots, where it really earns its keep.

Real num­bers are all the ratio­nal and irra­tional num­bers together, as you saw in The num­ber sys­tem. Whole ones, frac­tions, neg­a­tive ones, and roots like $\sqrt{2}$ all count. We write the real num­bers as $R$ for short.

You can already add real num­bers. You can already take one away from another. Now you will mul­ti­ply them and divide them.

Start with num­bers you know

Take $3 \times 5 = 15$. Now turn it round. You get $5 \times 3 = 15$.

The answer did not change. Try another pair. $2 \times 7 = 14$ and $7 \times 2 = 14$.

Here is why. Think of $3 \times 5$ as three rows of five dots. Turn the page side­ways. Now it is five rows of three dots.

The dots are the same dots. So the answer has to be the same.

Swap­ping is safe for every pair of num­bers. So we write it with let­ters. For any real num­bers $a$ and $b$, $ab = ba$.

The let­ters are not new maths. They just mean any num­ber you like.

Your chap­ter has a name for this. Mul­ti­pli­ca­tion is com­mu­ta­tive, and that word just means you may swap.

Adding is com­mu­ta­tive too. Tak­ing away and divid­ing are not. $8 \div 2 = 4$, but $2 \div 8$ is not $4$.

Brack­ets do not mat­ter either

Look at $2 \times 3 \times 5$. You can start at either end.

The first two num­bers first

$$\begin{aligned}&(2 \times 3) \times 5 \\ &= 6 \times 5 \\ &= 30\end{aligned}$$

The last two num­bers first

$$\begin{aligned}&2 \times (3 \times 5) \\ &= 2 \times 15 \\ &= 30\end{aligned}$$

Both ways give $30$. With let­ters, $(ab)c = a(bc)$.

Your chap­ter calls this asso­cia­tive. Adding is asso­cia­tive as well. Tak­ing away and divid­ing are not.

These two facts work together. A long mul­ti­pli­ca­tion can be done in any order you like.

Roots mul­ti­ply in a neat way

A square root has one job. $\sqrt{9}$ is the pos­i­tive num­ber that gives $9$ when you mul­ti­ply it by itself.

You know that num­ber. It is $3$, and $3 \times 3 = 9$.

So $\sqrt{9} \times \sqrt{9} = 9$. The root sign went away and left a plain num­ber.

Try it with $\sqrt{5}$. You do not know that num­ber as a dec­i­mal. But $\sqrt{5} \times \sqrt{5} = 5$ all the same.

Remem­ber. For any real num­ber $a$ that is not neg­a­tive, $\sqrt{a} \times \sqrt{a} = a$. A root times itself gives back the num­ber under the root.

The num­ber under the root must not be neg­a­tive. No real num­ber times itself gives a neg­a­tive answer. So $\sqrt{-4}$ is not a real num­ber at all.

This one rule makes root sums easy. Keep it beside you for the rest of the page.

A first exam­ple

Mul­ti­ply the plain num­bers first. Then mul­ti­ply the roots.

Mul­ti­ply $3\sqrt{2}$ by $4\sqrt{2}$

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

The order was changed to bring the two roots together. Swap­ping and regroup­ing are both safe, so noth­ing was bro­ken.

The two roots became a sin­gle $2$. Then $12 \times 2 = 24$.

A sec­ond exam­ple

This one works the same way.

Mul­ti­ply $5\sqrt{3}$ by $2\sqrt{3}$

$$\begin{aligned}&5\sqrt{3} \times 2\sqrt{3} \\ &= 5 \times 2 \times \sqrt{3} \times \sqrt{3} \\ &= 10 \times 3 \\ &= 30\end{aligned}$$

$\sqrt{3}$ is not a whole num­ber. You still never need its dec­i­mal. The rule does all the work.

A check you can do two ways

Some roots are whole num­bers in dis­guise. $\sqrt{4}$ is one, because $2 \times 2 = 4$.

Mul­ti­ply $5\sqrt{4}$ by $2\sqrt{4}$

$$\begin{aligned}&5\sqrt{4} \times 2\sqrt{4} \\ &= 5 \times 2 \times \sqrt{4} \times \sqrt{4} \\ &= 10 \times 4 \\ &= 40\end{aligned}$$

Now do it with­out the rule. Here $\sqrt{4} = 2$, so $5\sqrt{4} = 10$ and $2\sqrt{4} = 4$.

And $10 \times 4 = 40$ once more. The two paths agree.

This sum was picked because you can do it both ways. Most roots, like $\sqrt{3}$, are not whole num­bers in dis­guise. Then the rule is the only way through.

A harder exam­ple: a bracket with a root inside

Share the out­side num­ber out, then use the root rule on any root times itself.

Mul­ti­ply out $\sqrt{2}(3 + \sqrt{2})$

$$\begin{aligned}&\sqrt{2}(3 + \sqrt{2}) \\ &= \sqrt{2} \times 3 + \sqrt{2} \times \sqrt{2} \\ &= 3\sqrt{2} + 2\end{aligned}$$

The $3\sqrt{2}$ and the $2$ can­not be joined into one num­ber. One has a root and the other does not, just as $3$ apples and $2$ oranges stay as they are.

Three roots in a row

Mul­ti­ply $2\sqrt{5} \times 3\sqrt{5} \times \sqrt{5}$

$$\begin{aligned}&2\sqrt{5} \times 3\sqrt{5} \times \sqrt{5} \\ &= 2 \times 3 \times (\sqrt{5} \times \sqrt{5}) \times \sqrt{5} \\ &= 6 \times 5 \times \sqrt{5} \\ &= 30\sqrt{5}\end{aligned}$$

Only two of the three roots could pair up. The third has no part­ner, so it stays in the answer.

A num­ber out­side a bracket

Some­times a num­ber sits out­side a bracket. It must mul­ti­ply every part inside.

Look at $6 \times (2 + 5)$. You can do the bracket first.

The bracket first

$$\begin{aligned}&6 \times (2 + 5) \\ &= 6 \times 7 \\ &= 42\end{aligned}$$

You can also share the $6$ out. Give it to the $2$ and to the $5$.

Shar­ing the 6 out

$$\begin{aligned}&6 \times 2 + 6 \times 5 \\ &= 12 + 30 \\ &= 42\end{aligned}$$

Both ways give $42$. That is why shar­ing out is safe.

With let­ters we write $a(b + c) = ab + ac$.

Your chap­ter calls this dis­trib­u­tive. Mul­ti­pli­ca­tion is dis­trib­u­tive over addi­tion.

It works when the bracket has a take away in it. $6 \times (5 - 2) = 30 - 12 = 18$.

With let­ters, $a(b - c) = ab - ac$.

Shar­ing out works with roots as well. So $3(\sqrt{2} + 4) = 3\sqrt{2} + 12$.

The $3$ reached both parts. Noth­ing inside the bracket was left out.

Divid­ing real num­bers

Divid­ing undoes mul­ti­ply­ing. $12 \div 4 = 3$ because $3 \times 4 = 12$.

Roots divide in a tidy way as well. Any num­ber divided by itself is $1$, and a root is just a num­ber.

So $\sqrt{6}$ over $\sqrt{6}$ is $1$. The num­ber under the root has to be big­ger than zero here. You will see about zero soon.

Divide $8\sqrt{6}$ by $2\sqrt{6}$

$$\displaystyle \begin{aligned}&\frac{8\sqrt{6}}{2\sqrt{6}} \\ &= \frac{8}{2} \times \frac{\sqrt{6}}{\sqrt{6}} \\ &= 4 \times 1 \\ &= 4\end{aligned}$$

The roots can­celled each other. Only $8 \div 2 = 4$ was left to do.

When the root is on the bot­tom

A root on the bot­tom of a frac­tion is hard to read. You can move it to the top.

Mul­ti­ply the top and the bot­tom by the same root.

Doing the same thing to the top and the bot­tom is really mul­ti­ply­ing by $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}$. That is $1$, and mul­ti­ply­ing by $1$ changes noth­ing.

It is like cut­ting a cake into more slices. You still hold the same amount of cake.

Divide $8$ by $\sqrt{2}$

$$\displaystyle \begin{aligned}&\frac{8}{\sqrt{2}} \\ &= \frac{8 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\ &= \frac{8\sqrt{2}}{2} \\ &= 4\sqrt{2}\end{aligned}$$

The bot­tom turned into $2$ because $\sqrt{2} \times \sqrt{2} = 2$. The root now sits on top.

A num­ber like $\sqrt{2}$ can­not be writ­ten as one inte­ger over another. Num­bers that can are called ratio­nal.

This move leaves a plain whole num­ber on the bot­tom. So it is called ratio­nal­is­ing.

Two more divi­sions

Divide $15$ by $5\sqrt{3}$

Divide the plain num­bers first, then move the root to the top.

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

Divide $12$ by $\sqrt{6}$

$$\displaystyle \begin{aligned}&\frac{12}{\sqrt{6}} \\ &= \frac{12\sqrt{6}}{\sqrt{6} \times \sqrt{6}} \\ &= \frac{12\sqrt{6}}{6} \\ &= 2\sqrt{6}\end{aligned}$$

Check by mul­ti­ply­ing back: $2\sqrt{6} \times \sqrt{6} = 2 \times 6 = 12$. Divid­ing undoes mul­ti­ply­ing, so the check works.

Why divi­sion is the awk­ward one

There is a word for the next idea. Closed means the answer is always a real num­ber again.

Adding, tak­ing away and mul­ti­ply­ing are all closed. $2 + 3 = 5$. $2 - 5 = -3$. $2 \times 3 = 6$.

Every one of those answers is still a real num­ber. You can­not land on some­thing that is not a real num­ber.

Your chap­ter says it like this. Real num­bers are closed under addi­tion, sub­trac­tion and mul­ti­pli­ca­tion.

Divi­sion is the odd one out. Real num­bers are not closed under divi­sion.

That is not because the answer escapes. It is because one divi­sion has no answer at all.

Think about what divi­sion asks. $8 \div 2$ asks how many $2$s fit inside $8$.

Now ask how many zeros fit inside $8$. Zeros add noth­ing at all.

You could take a mil­lion of them and still have noth­ing. You never reach $8$.

So $8 \div 0$ has no answer. Divid­ing by zero is not allowed.

Remem­ber. You may divide by any real num­ber except zero. Never put zero on the bot­tom of a frac­tion.

The rules on one page

Prac­tice

Try these on paper. The answers come after them.

1. Work out $\sqrt{7} \times \sqrt{7}$.
2. Mul­ti­ply $2\sqrt{3}$ by $3\sqrt{3}$.
3. Work out $5(3 + 4)$.
4. Divide $10\sqrt{5}$ by $2\sqrt{5}$.
5. Mul­ti­ply out $4(\sqrt{2} + 3)$.
6. Write $\displaystyle \frac{6}{\sqrt{3}}$ with no root on the bot­tom.

Com­mon mis­takes

• Writ­ing $\sqrt{5} \times \sqrt{5} = \sqrt{25}$ and stop­ping. It is right but unfin­ished: $\sqrt{25} = 5$.
• Shar­ing the out­side num­ber with only the first part of a bracket. $3(\sqrt{2} + 4)$ is $3\sqrt{2} + 12$, not $3\sqrt{2} + 4$.
• Swap­ping the order in a divi­sion. $8 \div 2 = 4$, but $2 \div 8$ is a quar­ter.
• Mul­ti­ply­ing only the bot­tom by the root when ratio­nal­is­ing. Top and bot­tom must both be mul­ti­plied, or the value changes.
• Adding a root to a plain num­ber, as in $3\sqrt{2} + 2 = 5\sqrt{2}$. They are dif­fer­ent kinds of num­ber and stay apart.
• Divid­ing by zero. It has no answer at all.

Key terms

Real num­bers ($R$)

All the ratio­nal and irra­tional num­bers together.

Com­mu­ta­tive

The order can be swapped with­out chang­ing the answer: $ab = ba$.

Asso­cia­tive

The group­ing can be changed with­out chang­ing the answer: $(ab)c = a(bc)$.

Dis­trib­u­tive

A num­ber out­side a bracket mul­ti­plies every part inside: $a(b + c) = ab + ac$.

Square root

$\sqrt{a}$ is the num­ber that is not neg­a­tive and gives $a$ when mul­ti­plied by itself.

Ratio­nal­is­ing

Mul­ti­ply­ing top and bot­tom by a root so that no root is left on the bot­tom.

Closed

An oper­a­tion is closed when its answer is always a num­ber of the same set.

Answers

1. $\sqrt{7} \times \sqrt{7} = 7$
2. $2\sqrt{3} \times 3\sqrt{3} = 6 \times 3 = 18$
3. $5(3 + 4) = 5 \times 7 = 35$
4. $\displaystyle \dfrac{10\sqrt{5}}{2\sqrt{5}} = 5 \times 1 = 5$
5. $4(\sqrt{2} + 3) = 4\sqrt{2} + 12$
6. $\displaystyle \dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$