Division of rational numbers (Math 0)

So far in our study of the ar­ith­metic of ra­tio­nal num­bers, we’ve had ad­di­tion (“putting ap­ples and chunks of ap­ples side by side and count­ing what you’ve got”), sub­trac­tion (“the same, but you’re al­lowed anti-ap­ples too”), and mul­ti­pli­ca­tion (“make a ra­tio­nal num­ber, but in­stead of start­ing from \(1\) ap­ple, start from some other num­ber”).

Divi­sion is what re­ally sets the ra­tio­nal num­bers apart from the in­te­gers, and it is the math­e­mat­i­cian’s an­swer to the ques­tion “if I have some ap­ples, how do I share them among my friends?”.

What’s wrong with the in­te­gers?

If you have an in­te­ger num­ber of ap­ples (that is, some num­ber of ap­ples and anti-ap­ples—no chunks al­lowed, just whole ap­ples and anti-ap­ples), and you want to share them with friends, some­times you’ll get lucky. If you have four ap­ples, for in­stance, then you can share them out be­tween your­self and one friend, giv­ing each per­son two ap­ples.

But some­times (of­ten, in fact) you’ll get un­lucky. If you want to share four ap­ples be­tween your­self and two oth­ers, then you can give each per­son one ap­ple, but there’s this pesky sin­gle ap­ple left over which you just can’t share.

What the ra­tio­nals do for us

The trick, ob­vi­ous to any­one who has ever eaten a cake, is to cut the lef­tover ap­ple into three equally-sized pieces and give each per­son a piece. Now we have shared out all four ap­ples equally.

But in or­der to do so, we’ve left the world of the in­te­gers, and in get­ting out our knife, we started work­ing in the ra­tio­nals. How much ap­ple has ev­ery­one re­ceived, when we shared four ap­ples among three peo­ple (that is, my­self and two friends as re­cip­i­ents of ap­ple)?

Every­one got \(\frac{4}{3}\).

In­deed, ev­ery­one got one whole ap­ple; and then we chopped the re­main­ing ap­ple into three \(\frac{1}{3}\)-chunks and gave ev­ery­one one chunk. So ev­ery­one ended up with one ap­ple and one \(\frac{1}{3}\)-chunk.

By our in­stant ad­di­tion rule <div><div>note:If you’ve for­got­ten it, check out the ad­di­tion page again; it came from work­ing out a chunk size out of which we can make both the \(\frac{1}{3}\)-chunk and the \(1\)-chunk.%%, \($1 + \frac{1}{3} = \frac{1}{1} + \frac{1}{3} = \frac{3 \times 1 + 1 \times 1}{3 \times 1} = \frac{3+1}{3} = \frac{4}{3}\)$

There’s an­other way to see this, if (laud­ably!) you don’t like just ap­ply­ing rules. We could cut ev­ery ap­ple into three pieces at the be­gin­ning, so we’re left with four col­lec­tions of three \(\frac{1}{3}\)-sized chunks. But now it’s easy to share this among three peo­ple: just give ev­ery­one one of the \(\frac{1}{3}\)-chunks from each ap­ple. We gave ev­ery­one four chunks in to­tal, so this is \(\frac{4}{3}\). %%

The ra­tio­nals provide the nat­u­ral an­swer to all “shar­ing” ques­tions about ap­ples.

We write “ra­tio­nal num­ber \(x\) di­vided by ra­tio­nal num­ber \(y\)” as \(\frac{x}{y}\): that is, \(x\) ap­ples di­vided amongst \(y\) peo­ple. (We’ll soon get to what it means to di­vide by a non-in­te­ger num­ber of peo­ple; just roll with it for now.) If space is a prob­lem, we can write \(a/n\) in­stead. No­tice that our fa­mil­iar no­ta­tion of “$\frac{1}{m}$-sized chunks” is ac­tu­ally just \(1\) ap­ple di­vided amongst \(m\) peo­ple: it’s the re­sult of di­vid­ing \(1\) into \(m\) equal chunks. So the no­ta­tion does make sense, and it’s just an ex­ten­sion of the no­ta­tion we’ve been us­ing already.

Divi­sion by a nat­u­ral number

In gen­eral, \(\frac{a}{m}\) ap­ples, di­vided amongst \(n\) peo­ple, is ob­tained by the “other way” above. Cut the \(\frac{a}{m}\) into \(n\) pieces, and then give ev­ery­one an equal num­ber of pieces.

Re­mem­ber, \(\frac{a}{m}\) is made of \(a\) copies of pieces of size \(\frac{1}{m}\); so what we do is cut all of the \(\frac{1}{m}\)-chunks in­di­vi­d­u­ally into \(n\) pieces, and then give ev­ery­one \(a\) of the lit­tle pieces we’ve made. But “cut a \(\frac{1}{m}\)-chunk into \(n\) pieces” is just “cut an ap­ple into \(n\) pieces, but in­stead of do­ing it to one ap­ple, do it to a \(\frac{1}{m}\)-chunk”: that is, it is \(\frac{1}{m} \times \frac{1}{n}\), or \(\frac{1}{m \times n}\).

So the an­swer is just \($\frac{a}{m} / n = \frac{a}{m \times n}\)$


pic­to­rial example

Divi­sion by a nega­tive integer

What would it even mean to di­vide an ap­ple be­tween four anti-peo­ple? How about a sim­pler ques­tion: di­vid­ing an ap­ple be­tween one anti-per­son­noteRe­mem­ber­ing that di­vid­ing \(x\) ap­ples be­tween one per­son just gives \(x\), since there’s not even any cut­ting of the ap­ples nec­es­sary.? (The an­swer to this would be \(\frac{1}{-1}\).) It’s not ob­vi­ous!

Well, the thing to think about here is that if I take an ap­ple, and share it among one per­son noteMy­self, prob­a­bly. I’m very self­ish., then I’ve got just the same ap­ple as be­fore (cut into no chunks): that is, \(\frac{1}{1} = 1\). Also, if I take an ap­ple and share it among one per­son who is not my­self (and I don’t give any ap­ple to my­self), then we’ve also just got the same un­sliced ap­ple as be­fore: \(\frac{1}{1} = 1\) again.

But if I take an ap­ple, and give it to an anti-per­son­noteBe­ing very care­ful not to touch them, be­cause I would an­nihilate an anti-per­son!, then from their per­spec­tive I’m the anti-per­son, and I’ve just given them an anti-ap­ple. There’s a law of sym­me­try built into re­al­ity: the laws of physics are in­var­i­ant if we re­flect “thing” and “anti-thing” through­out the uni­verse.noteTech­ni­cally this is not quite true: the ac­tual sym­me­try is on re­flect­ing charge, par­ity and time all to­gether, rather than just par­ity alone. But for the pur­poses of this dis­cus­sion, let’s pre­tend that the uni­verse is par­ity-sym­met­ric. The anti-per­son sees the uni­verse in a way that’s the same as my way, but where the “anti” sta­tus of ev­ery­thing (and ev­eryan­tithing) is flipped.

write “the uni­verse is un­changed on swap­ping anti-sta­tus” in the Nega­tive In­te­gers chunk too, around where we work out what −5+3 is, and say “re­mem­ber this!”

Put an­other way, “anti-ness” is not an ab­solute no­tion but a rel­a­tive one. I can only de­ter­mine whether some­thing is the same anti-ness or the op­po­site anti-ness to my­self.

Sear this into your mind: the laws of ra­tio­nal-num­ber “physics” are the same no mat­ter who is ob­serv­ing. If I ob­serve a trans­ac­tion, like “I, a per­son, give some­one an ap­ple”, then an ex­ter­nal per­son will ob­serve “The au­thor, a per­son, gave some­one an ap­ple”, while an ex­ter­nal anti-per­son will ob­serve “The au­thor, an anti-per­son, gave an anti-per­son an anti-ap­ple”.

From the ex­ter­nal per­son’s per­spec­tive, they saw “some­one (of the same anti-ness as me) gave some­one else (of the same anti-ness as me) an ap­ple (of the same anti-ness as me)”. From the anti-per­son’s per­spec­tive, ev­ery­thing is rel­a­tively the same: “some­one (of the op­po­site anti-ness to me) gave some­one else (of the op­po­site anti-ness to me) an ap­ple (of the op­po­site anti-ness to me)”.

So \(\frac{-1}{-1}\), be­ing “one anti-ap­ple shared among one anti-per­son”, can be viewed in­stead from the per­spec­tive of an anti-per­son; they see one ap­ple be­ing given to one per­son: that is, \(\frac{-1}{-1}\) is equal to \(1\).

Armed with the fact that \(\frac{-1}{-1} = 1\), we can just ap­ply our usual mul­ti­pli­ca­tion rule that \(\frac{a}{m} \times \frac{b}{n} = \frac{a \times b}{m \times n}\), to de­duce that \($\frac{1}{-m} = \frac{1}{-m} \times 1 = \frac{1}{-m} \times \frac{-1}{-1} = \frac{-1 \times 1}{-m \times -1} = \frac{-1}{m}\)$

The law of sym­me­try-of-the-uni­verse ba­si­cally says that \(\frac{a}{-b} = \frac{-a}{b}\).

di­vi­sion by ra­tio­nal num­ber di­vi­sion as “in­verse of multiplication

di­vi­sion by zero