# Addition of rational numbers (Math 0)

We’ve already met the idea that ra­tio­nal num­bers are what we can make by putting to­gether the build­ing blocks which we get by di­vid­ing a sin­gle ap­ple into some num­ber of equally-sized pieces; and re­mem­ber that the build­ing blocks have a spe­cial no­ta­tion, as $$\frac{1}{\text{number}}$$. Re­call also that if we take $$5$$ of a cer­tain build­ing block, for in­stance, we write $$\frac{5}{\text{number}}$$.

It’s clear that if you take one ap­ple and an­other ap­ple, and put them to­gether, you’ll get two ap­ples. So we should hope that that’s the same if we di­vided up the ap­ples into pieces first (with­out re­mov­ing any of the pieces).

We write $$a+b$$ for “take the ra­tio­nal num­ber $$a$$ and put it next to the ra­tio­nal num­ber $$b$$, and count up what you’ve got”. Be­cause I’m bad at draw­ing, we’ll pre­tend ap­ples are just perfect cir­cles. noteThis is ac­tu­ally a de­sign de­ci­sion: even­tu­ally we’ll want to get away from con­sid­er­ing ap­ples, and this more ab­stract rep­re­sen­ta­tion will be use­ful.

For ex­am­ple, we should hope that $$\frac{2}{2} + \frac{3}{3} = 2$$.

What about in cases which aren’t of the form $$\frac{n}{n}$$ for some in­te­ger $$n$$? Well, how about $$\frac{5}{3} + \frac{8}{3}$$.

If we had five $$\frac{1}{3}$$-blocks, and we put them to­gether with an­other eight $$\frac{1}{3}$$-blocks, we would hope to have $$5+8=13$$ blocks. So we should hope that $$\frac{5}{3} + \frac{8}{3} = \frac{13}{3}$$.

And in gen­eral, if both our quan­tities are made up from the same size of build­ing-block (in the above case, the blocks are $$\frac{1}{3}$$-sized), we should just be able to take the two nu­mer­a­tors noteRe­mem­ber, that was the word math­e­mat­i­ci­ans use for the num­ber of blocks we have. and add them to­gether.

Very well. But what if the two quan­tities are not made up from the same size of build­ing-block? That is, the de­nom­i­na­tors are differ­ent? For ex­am­ple, $$\frac{5}{3} + \frac{5}{4}$$?

Now it’s not so clear. How might we ap­proach this? You should muse on this for thirty sec­onds be­fore read­ing on; it will be good for your soul.

The way we do it is to find some smaller block-size, out of which we can make both build­ing-blocks.

# Example

Our ex­am­ple is $$\frac{5}{3} + \frac{5}{4}$$. So we want to make the $$\frac{1}{3}$$ block and the $$\frac{1}{4}$$ block out of some smaller block.

Now, I won’t tell you yet how I got this, but you can check for your­self that $$\frac{1}{12}$$-blocks will make both—$$\frac{1}{3}$$and $$\frac{1}{4}$$-blocks: be­cause three $$\frac{1}{12}$$-blocks to­gether make some­thing the same size as a $$\frac{1}{4}$$-block, and four $$\frac{1}{12}$$-blocks to­gether make some­thing the same size as a $$\frac{1}{3}$$-block.

There­fore $$\frac{1}{3} = \frac{4}{12}$$:

And $$\frac{1}{4} = \frac{3}{12}$$:

Now, if $$\frac{1}{3} = \frac{4}{12}$$ - that is, if one $$\frac{1}{3}$$-piece is the same size as four $$\frac{1}{12}$$-pieces—then it should be the case that five $$\frac{1}{3}$$-pieces come to five lots of four $$\frac{1}{12}$$-pieces. That is, to twenty $$\frac{1}{12}$$-pieces: $$\frac{5}{3} = \frac{20}{12}$$. pic­ture which I haven’t got round to de­tailing yet

Similarly, $$\frac{5}{4} = \frac{15}{12}$$, be­cause five $$\frac{1}{4}$$-pieces is the same as five lots of three sets of $$\frac{1}{12}$$-pieces, and $$5 \times 3 = 15$$.

There­fore $$\frac{5}{3} + \frac{5}{4}$$ should be just the same as $$\frac{20}{12} + \frac{15}{12}$$, which we know how to calcu­late! It is $$\frac{35}{12}$$.

# Gen­eral procedure

The pre­ced­ing ex­am­ple might sug­gest a gen­eral way for adding two frac­tions to­gether. (In the pro­cess, this should put to rest the ex­is­ten­tial dread men­tioned in the in­tro to ra­tio­nal num­bers, about whether it makes sense to add two ra­tio­nal num­bers at all.)

The gen­eral pro­ce­dure is as fol­lows:

1. Find a build­ing-block size such that the build­ing-blocks of both frac­tions can be made out of that build­ing-block

2. Ex­press each frac­tion in­di­vi­d­u­ally in terms of that build­ing-block

3. Add the two to­gether (which we can now do, be­cause they are both made of the same size of build­ing-block).

The in­ter­est­ing part is re­ally the first step.

## Find­ing a build­ing-block that will neatly make two other differ­ently-sized blocks

Let’s be con­crete here: just as above, we showed that $$\frac{1}{12}$$ works for blocks of size $$\frac{1}{3}$$ and $$\frac{1}{4}$$, so we’ll now con­sider $$\frac{1}{2}$$ and $$\frac{1}{5}$$.

We’re try­ing to di­vide up $$\frac{1}{2}$$ into copies of a smaller build­ing-block, such that with copies of that same build­ing block, we can also make $$\frac{1}{5}$$. Here is the key in­sight that tells us how to do it: let us pre­tend that we are di­vid­ing up the same ob­ject, this time a square in­stead of a cir­cle, into two pieces along one edge and into five pieces along the other edge. How many pieces have we made?

There are $$2 \times 5 = 10$$ pieces, each of them the same size (namely the $$\frac{1}{10}$$-block size), and note that as if by magic we’ve just pro­duced a way to sub­di­vide $$\frac{1}{2}$$ into pieces made up of the $$\frac{1}{10}$$-block!

And we also have the $$\frac{1}{5}$$ sub­di­vided in the same way!

This might sug­gest a gen­eral way to do this; you should cer­tainly pon­der for thirty sec­onds be­fore con­tin­u­ing.

find a way to do this with­out alge­bra Let’s say we are try­ing to make both $$\frac{1}{m}$$ and $$\frac{1}{n}$$-sized blocks out of smaller blocks. (So above, $$m$$ is stand­ing for $$2$$ and $$n$$ is stand­ing for $$5$$.)

Then we can do this by split­ting both up into $$\frac{1}{m \times n}$$-blocks.

In­deed, $$\frac{1}{n} = \frac{m}{m \times n}$$ (i.e. we can make a $$\frac{1}{n}$$-block out of $$m$$ of the $$\frac{1}{m \times n}$$ tiny-blocks), and $$\frac{1}{m} = \frac{n}{m \times n}$$ (i.e. we can make a $$\frac{1}{m}$$-block out of $$n$$ of the $$\frac{1}{m \times n}$$ tiny-blocks). <div><div>

## Aside: ar­ith­meti­cal rules

The de­scrip­tion of the “gen­eral way” above has be­come pretty full of $$\frac{1}{\text{thing}}$$, and refers less and less to the build­ing-blocks we origi­nally con­sid­ered. The rea­son for this is that $$\frac{1}{\text{thing}}$$ is an ex­tremely con­ve­nient short­hand, and the “gen­eral way” can be writ­ten as a very sim­ple rule for ma­nipu­lat­ing that short­hand. Once we have such a con­ve­nient rule, we can com­pute with frac­tions with­out hav­ing to go through the bother of work­ing out what each frac­tion re­ally and truly means (in terms of the build­ing-blocks); this saves a lot of time in prac­tice. As an anal­ogy, it’s eas­ier for you to drive a car if, ev­ery time you change gears, you don’t have to work out what’s hap­pen­ing in­side the gear­box.

The sim­ple rule we have just seen is $$\frac{1}{m} + \frac{1}{n} = \frac{n}{m \times n} + \frac{m}{m \times n}$$$And we’ve already had the rule that $$\frac{a}{m} + \frac{b}{m} = \frac{a+b}{m}$$$ (Re­mem­ber, if these are not ob­vi­ous to you, you should think back to what they re­ally mean, in terms of the build­ing-blocks. The first rule is to do with di­vid­ing up a big square into lit­tle squares, from the pre­vi­ous sub­sec­tion; the sec­ond rule is just tak­ing two col­lec­tions of build­ing-blocks of the same size and putting them to­gether.) first rule has pic­ture: square di­vided into n-by-m lit­tle squares, with the left­most ver­ti­cal column high­lighted some­how, in­di­cat­ing that 1/​n = m/​(mn); sec­ond rule has pic­ture: $$a$$ things of size 1/​m, next to $$b$$ things of size 1/​m

This is al­most ev­ery rule we need to add any pair of frac­tions! We can now add sin­gle build­ing blocks of any size to­gether.

To add any pair of ra­tio­nal num­bers to­gether—say of build­ing-block size $$\frac{1}{m}$$ and $$\frac{1}{n}$$ re­spec­tively—we just need to ex­press them each in terms of the smaller build­ing-block $$\frac{1}{m \times n}$$, and add them to­gether as nor­mal.

Ear­lier, we con­sid­ered $$\frac{5}{4} + \frac{5}{3}$$.

A build­ing-block which can make both the $$\frac{1}{4}$$ and $$\frac{1}{3}$$-sized blocks is the $$\frac{1}{12}$$-block, be­cause $$3 \times 4 = 12$$.

There­fore we need to ex­press both our frac­tions as be­ing made from $$\frac{1}{12}$$-sized blocks.

$$\frac{5}{4}$$ is $$\frac{15}{12}$$, be­cause in each of the five $$\frac{1}{4}$$-blocks, there are three $$\frac{1}{12}$$-blocks, so we have a to­tal of $$5 \times 3$$ blocks of size $$\frac{1}{12}$$.

$$\frac{5}{3}$$ is $$\frac{20}{12}$$, be­cause in each of the five $$\frac{1}{3}$$-blocks, there are four $$\frac{1}{12}$$-blocks, so we have a to­tal of $$5 \times 4 = 20$$ blocks of size $$\frac{1}{12}$$.

There­fore our fi­nal an­swer is $$\frac{15}{12} + \frac{20}{12} = \frac{35}{12}$$. <div><div>

To make this into a gen­eral rule: $$\frac{a}{m} + \frac{b}{n} = \frac{a \times n}{m \times n} + \frac{b \times m}{m \times n} = \frac{a \times n + b \times m}{m \times n}$$\$

(Re­call the or­der of op­er­a­tions in the in­te­gers: the no­ta­tion $$a \times n + b \times m$$ means “do $$a \times n$$; then do $$b \times m$$; then fi­nally add them to­gether”. Mul­ti­pli­ca­tion comes be­fore ad­di­tion.)

And that’s it! That’s how we add ra­tio­nal num­bers to­gether, and it works even when any or all of $$a, b, m, n$$ are nega­tive. (Re­mem­ber, though, that $$m$$ and $$n$$ can’t be zero, be­cause it makes no sense to di­vide some­thing up into no pieces.)

It would be wise now to try the ex­er­cises. One learns math­e­mat­ics through do­ing, much more than through sim­ply read­ing; your un­der­stand­ing will be ce­mented by go­ing through some con­crete ex­am­ples.

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