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.

Example of a sum of two rational numbers

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

Two halves and three thirds

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

Five thirds and eight thirds

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 \(\frac{5}{3} + \frac{8}{3} = \frac{13}{3}\).

Thirds, separated out

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, \(\frac{5}{3} + \frac{5}{4}\)?

Five thirds and five quarters

Five thirds and five quarters, rearranged

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.


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.

Thirds, divided into twelfths

Quarters, divided into twelfths

There­fore \(\frac{1}{3} = \frac{4}{12}\):

One third, divided into twelfths

And \(\frac{1}{4} = \frac{3}{12}\):

One quarter, divided into twelfths

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?

Square divided into tenths

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!

Square divided into tenths, one half indicated

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

Square divided into tenths, one fifth indicated

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.