Adding and subtracting real numbers

What this les­son is about

You have added whole num­bers, frac­tions and dec­i­mals for years. Real num­bers include all of those, and also num­bers such as $\sqrt{2}$ whose dec­i­mals never end or repeat. You meet them when you find the diag­o­nal of a square, the side of a right-angled tri­an­gle, or a length on a geom­e­try dia­gram in Class 9 and beyond.

This les­son shows that adding and sub­tract­ing real num­bers fol­lows the same rules you already trust. The new skill is decid­ing when two roots can be com­bined and when an answer must be left as it stands.

A few words you will need

Whole num­bers are the count­ing num­bers with noth­ing after the point. They are $0, 1, 2, 3$ and so on.

The mark $\sqrt{\;}$ asks you a ques­tion. Which num­ber, mul­ti­plied by itself, gives what is inside?

So $\sqrt{9} = 3$, because $3 \times 3 = 9$. A num­ber writ­ten like this is called a root.

The chap­ter builds the real num­bers out of six sets. Here they are.

So the real num­bers are the ratio­nal and irra­tional num­bers put together. That is the chap­ter's own def­i­n­i­tion of $R$.

You write $\sqrt{2} \in R$. Read it as $\sqrt{2}$ belongs to $R$.

The num­ber $e = 2.71\ldots$ belongs to $R$ too. Its dig­its never end. It has a name because it turns up often in later work.

Adding and sub­tract­ing num­bers you know

You have been adding for years. You know that $3 + 5 = 8$.

You can swap the two num­bers over. The answer stays the same, so $5 + 3 = 8$.

Sub­tract­ing is dif­fer­ent. To sub­tract is to take away, and $9 - 4 = 5$.

Now swap those two. You get $4 - 9$, which is $-5$. That is below nought, so the order mat­ters.

Nought is gen­tle when you add. $7 + 0 = 7$, and $0 + 7 = 7$, and $7 - 0 = 7$.

But $0 - 7 = -7$. Tak­ing a num­ber away from nought turns it into its oppo­site.

Every num­ber has an oppo­site. The oppo­site of $7$ is $-7$, and $7 + (-7) = 0$.

Group­ing does not mat­ter when you add. $(2 + 3) + 4 = 9$, and $2 + (3 + 4) = 9$.

Group­ing does mat­ter when you sub­tract. $(9 - 4) - 2 = 3$, but $9 - (4 - 2) = 7$.

Remem­ber. Real num­bers fol­low the same adding rules you learnt with small num­bers. The num­bers look dif­fer­ent, but the adding does not.

The same rules, writ­ten with let­ters

A let­ter can stand for any real num­ber. Noth­ing new is hap­pen­ing below.

The chap­ter num­bers these rules. The num­ber­ing here is the chap­ter's own.

Rules 2 and 3 have one gap each for sub­trac­tion. Swap­ping gives the same answer when the two num­bers are equal, as in $6 - 6 = 6 - 6$. Regroup­ing gives the same answer when the last num­ber is nought, as in $(9 - 4) - 0 = 9 - (4 - 0)$. Oth­er­wise the answers dif­fer.

What the chap­ter says in words

It says real num­bers are closed under addi­tion, sub­trac­tion and mul­ti­pli­ca­tion. They are not closed under divi­sion.

It says they are com­mu­ta­tive under addi­tion and mul­ti­pli­ca­tion. They are not com­mu­ta­tive under sub­trac­tion or divi­sion.

It says they are asso­cia­tive under addi­tion and mul­ti­pli­ca­tion. They are not asso­cia­tive under sub­trac­tion or divi­sion.

The mul­ti­pli­ca­tion and divi­sion halves of those three sen­tences belong to the next two lessons.

Adding things of the same kind

Pic­ture $2$ apples and $5$ apples. Together you have $7$ apples.

You added the counts. The word apples did not change at all.

Now try $2$ tens and $5$ tens. That is $7$ tens, and $20 + 50 = 70$.

Both times you added the num­ber in front. The thing being counted stayed put.

Apples and pears are not the same thing. You can­not fold them into one count.

Adding and sub­tract­ing roots

A root like $\sqrt{3}$ is a real num­ber. It is not a whole num­ber, but it is still one num­ber.

So $2\sqrt{3}$ means two lots of $\sqrt{3}$. And $5\sqrt{3}$ means five lots of $\sqrt{3}$.

A root on its own is one lot of it. So $\sqrt{3}$ means the same as $1\sqrt{3}$, and its count is $1$.

Both $2\sqrt{3}$ and $5\sqrt{3}$ are count­ing $\sqrt{3}$s. So you add the counts, just like the apples.

Adding counts of the same root

$$\begin{aligned}&2\sqrt{3} + 5\sqrt{3} \\ &= (2 + 5)\sqrt{3} \\ &= 7\sqrt{3}\end{aligned}$$

The $\sqrt{3}$ did not change. Only the count in front of it changed.

Sub­tract­ing works the same way. Here you take $4\sqrt{5}$ from $9\sqrt{5}$.

Sub­tract­ing counts of the same root

$$\begin{aligned}&9\sqrt{5} - 4\sqrt{5} \\ &= (9 - 4)\sqrt{5} \\ &= 5\sqrt{5}\end{aligned}$$

You worked out $9 - 4 = 5$ and kept the $\sqrt{5}$ as it was.

Bare roots have a count of $1$, so they add in the same way.

Adding two bare roots

$$\begin{aligned}&\sqrt{2} + \sqrt{2} \\ &= 1\sqrt{2} + 1\sqrt{2} \\ &= (1 + 1)\sqrt{2} \\ &= 2\sqrt{2}\end{aligned}$$

Some­times the counts can­cel each other out.

When the counts can­cel

$$\begin{aligned}&3\sqrt{2} - 3\sqrt{2} \\ &= (3 - 3)\sqrt{2} \\ &= 0\sqrt{2} \\ &= 0\end{aligned}$$

Nought lots of any­thing is nought. So no root is left to write, and you write $0$.

Remem­ber. You can only add or sub­tract counts of the same root. $\sqrt{3}$ and $\sqrt{5}$ are dif­fer­ent things, in the way apples and pears are.

When the answer stays as it is

Now look at $4 + \sqrt{5}$. The $4$ is a whole num­ber. The $\sqrt{5}$ is not.

$2 \times 2 = 4$. $3 \times 3 = 9$. Five sits between four and nine, so $\sqrt{5}$ sits between $2$ and $3$.

$\sqrt{5}$ is irra­tional. Its dig­its run on for ever, and they never set­tle into a pat­tern.

So the two parts are not the same kind of thing. It is like adding $4$ apples and one pear.

You have both, so you write both down. The answer is $4 + \sqrt{5}$.

$\sqrt{2}$ and $\sqrt{3}$ are two dif­fer­ent roots. So $\sqrt{2} + \sqrt{3}$ stays as it is too.

Order does not mat­ter here, because rule 2 lets you swap. So $4 + \sqrt{5}$ and $\sqrt{5} + 4$ are the same answer.

Remem­ber. Leav­ing $4 + \sqrt{5}$ alone is the answer. It is exact, and it is fin­ished.

By rule 1 that answer is still a real num­ber. Adding two real num­bers always gives a real num­ber.

Most peo­ple want a sum to end in one tidy num­ber. Here that would mean round­ing, and round­ing throws a lit­tle of the truth away.

How to decide

1. Look at the two parts. Ask what each one is count­ing.
2. If they count the same root, add or sub­tract the num­bers in front. A root with no num­ber in front counts as one.
3. Write that root down again, unchanged.
4. If the counts come to noth­ing, the answer is just $0$. No root is left to write.
5. If the parts are not the same kind, leave the sum as it stands.

More worked exam­ples

These go one step fur­ther. Each one uses only the steps from the list above.

Col­lect­ing two kinds of root

Sim­plify $6\sqrt{2} + 3\sqrt{5} - 2\sqrt{2} + \sqrt{5}$. There are two kinds here, so gather each kind on its own, the way you would sort apples from pears.

$$\begin{aligned}&6\sqrt{2} + 3\sqrt{5} - 2\sqrt{2} + \sqrt{5} \\ &= (6 - 2)\sqrt{2} + (3 + 1)\sqrt{5} \\ &= 4\sqrt{2} + 4\sqrt{5}\end{aligned}$$

The two parts still count dif­fer­ent roots, so the answer stays in two parts. Rule 2 let you move the terms into their groups, and rule 3 let you bracket them.

A root that hides a like root

Some­times a root can be rewrit­ten so that it matches another. Since $12 = 4 \times 3$ and $\sqrt{4} = 2$, the root $\sqrt{12}$ is the same as $2\sqrt{3}$.

$$\begin{aligned}&\sqrt{12} + \sqrt{3} - 3\sqrt{3} \\ &= 2\sqrt{3} + 1\sqrt{3} - 3\sqrt{3} \\ &= (2 + 1 - 3)\sqrt{3} \\ &= 0\end{aligned}$$

The counts can­cel, so noth­ing is left to write except $0$.

Whole num­bers and roots together

Sim­plify $(5 + 2\sqrt{3}) + (1 - 7\sqrt{3}) - (6 - 5\sqrt{3})$. Open the brack­ets first. The minus in front of the last bracket changes the sign of both parts inside it.

$$\begin{aligned}&(5 + 2\sqrt{3}) + (1 - 7\sqrt{3}) - (6 - 5\sqrt{3}) \\ &= 5 + 2\sqrt{3} + 1 - 7\sqrt{3} - 6 + 5\sqrt{3} \\ &= (5 + 1 - 6) + (2 - 7 + 5)\sqrt{3} \\ &= 0 + 0\sqrt{3} \\ &= 0\end{aligned}$$

The whole num­bers were col­lected with each other and the roots with each other. Both groups came to nought.

Try these

Work each one out. Some of them stay as they are, and that is fine.

1. $3\sqrt{7} + 4\sqrt{7}$
2. $8\sqrt{2} - 5\sqrt{2}$
3. $6 + \sqrt{3}$
4. $\sqrt{5} + \sqrt{5}$
5. $11\sqrt{6} - 11\sqrt{6}$
6. $\sqrt{10} + 7$

Remem­ber. Count what is alike. Keep the root the same, unless the counts can­cel. Leave the rest as it stands.

Com­mon mis­takes

• Adding the num­bers under the root: $\sqrt{2} + \sqrt{3}$ is not $\sqrt{5}$. Check with a cal­cu­la­tor: about $1.414 + 1.732 = 3.146$, but $\sqrt{5}$ is about $2.236$.
• Adding a whole num­ber to the count: $4 + \sqrt{5}$ is not $5\sqrt{5}$. The $4$ is not count­ing roots.
• For­get­ting that a bare root has a count of $1$, so $\sqrt{5} + \sqrt{5}$ is writ­ten as $\sqrt{10}$ instead of $2\sqrt{5}$.
• Writ­ing $0\sqrt{2}$ as the final answer instead of $0$.
• Swap­ping or regroup­ing in a sub­trac­tion as if it were an addi­tion.
• Round­ing a root to a dec­i­mal when the ques­tion wants the exact answer.

Key terms

Real num­bers

All the ratio­nal and irra­tional num­bers together, writ­ten $R$.

Irra­tional num­ber

A num­ber that can­not be writ­ten as $\displaystyle \frac{p}{q}$; its dec­i­mals never end and never repeat, like $\sqrt{2}$.

Root

A num­ber writ­ten with the sign $\sqrt{\;}$; it is the num­ber which, mul­ti­plied by itself, gives what is inside.

Like roots

Roots with the same num­ber inside, such as $2\sqrt{3}$ and $5\sqrt{3}$. Only these can be com­bined.

Clo­sure

Adding or sub­tract­ing two real num­bers always gives a real num­ber.

Addi­tive iden­tity

Nought, because adding it leaves any num­ber unchanged.

Addi­tive inverse

The oppo­site of a num­ber; the two add to nought, as $7 + (-7) = 0$.

Answers

1. $3\sqrt{7} + 4\sqrt{7} = (3 + 4)\sqrt{7} = 7\sqrt{7}$
2. $8\sqrt{2} - 5\sqrt{2} = (8 - 5)\sqrt{2} = 3\sqrt{2}$
3. $6 + \sqrt{3}$ stays as it is: one part is a whole num­ber and one is a root.
4. $\sqrt{5} + \sqrt{5} = (1 + 1)\sqrt{5} = 2\sqrt{5}$
5. $11\sqrt{6} - 11\sqrt{6} = 0\sqrt{6} = 0$
6. $\sqrt{10} + 7$ stays as it is. By rule 2 it may also be writ­ten $7 + \sqrt{10}$.