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.

The good news is that adding and sub­tract­ing real num­bers fol­lows the same rules you already trust. The only new skill is decid­ing when two roots can be com­bined, and when an answer must sim­ply 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, shown here.

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

You write $\sqrt{2} \in R$ and read it as "$\sqrt{2}$ belongs to $R$".

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

Adding and sub­tract­ing num­bers you know#

Let's warm up with what you already know. You've been adding for years, and you know that $3 + 5 = 8$. Swap the two num­bers over and the answer stays the same, so $5 + 3 = 8$.

Sub­tract­ing is a dif­fer­ent story. To sub­tract is to take away, and $9 - 4 = 5$. But swap those two and you get $4 - 9$, which is $-5$. That's below nought, so here the order mat­ters.

Nought is gen­tle when you add: $7 + 0 = 7$, and $0 + 7 = 7$, and $7 - 0 = 7$. But $0 - 7 = -7$, because 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 does­n't mat­ter when you add: $(2 + 3) + 4 = 9$, and $2 + (3 + 4) = 9$. It 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, so don't be put off: noth­ing new is hap­pen­ing below. The rules are num­bered just as the chap­ter num­bers them.

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#

The chap­ter puts it this way. Real num­bers are closed under addi­tion, sub­trac­tion and mul­ti­pli­ca­tion, but not under divi­sion.

They are com­mu­ta­tive under addi­tion and mul­ti­pli­ca­tion, but not under sub­trac­tion or divi­sion.

And they are asso­cia­tive under addi­tion and mul­ti­pli­ca­tion, but not under sub­trac­tion or divi­sion.

The mul­ti­pli­ca­tion and divi­sion halves of those sen­tences are wait­ing for you in the next two lessons.

Adding things of the same kind#

Pic­ture a fruit seller with $2$ apples in one bas­ket and $5$ apples in another. Together he has $7$ apples. You added the counts, but the word apples did­n't 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, and the thing being counted stayed put. But apples and pears are not the same thing, and you can't fold them into one count.

Adding and sub­tract­ing roots#

A root like $\sqrt{3}$ is a real num­ber. It isn't a whole num­ber, but it is still one num­ber, and you can count lots of it.

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 $1\sqrt{3}$, with a count of $1$.

Both $2\sqrt{3}$ and $5\sqrt{3}$ are count­ing $\sqrt{3}$s, so you add the counts, exactly as you did with 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 $\sqrt{3}$ did­n't change; only the count in front of it did.

Sub­tract­ing works the same way. This time 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

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

And some­times the counts can­cel each other out com­pletely.

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. With no root left to write, you sim­ply 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, but the $\sqrt{5}$ is not.

Since $2 \times 2 = 4$ and $3 \times 3 = 9$, and five sits between four and nine, $\sqrt{5}$ must sit between $2$ and $3$.

In fact $\sqrt{5}$ is irra­tional: its dig­its run on for ever and never set­tle into a pat­tern.

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

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

$\sqrt{2}$ 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.

It's nat­ural to want a sum to end in one tidy num­ber. Here, though, that would mean round­ing, and round­ing throws a lit­tle of the truth away. So resist the urge!

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, but 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, and both groups came to nought. A long ques­tion with a very short answer!

Try these#

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

Show 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 $7 + \sqrt{10}$.