Subtraction of rational numbers (Math 0)

So far, we have met the idea of a ra­tio­nal num­ber, treat­ing them as chunks of ap­ples, and how to add them to­gether. Now we will dis­cover how the idea of the anti-ap­ple (by anal­ogy with the in­te­gers’ anti-cow) must work.

The anti-apple

Just as we had an anti-cow, so we can have an anti-ap­ple. If we com­bine an ap­ple with an anti-ap­ple, they both an­nihilate, leav­ing noth­ing be­hind. We write this as \(1 + (-1) = 0\).

A very use­ful thing for you to pon­der for thirty sec­onds (though I will give you the an­swer soon): given that \(\frac{a}{n}\) means “di­vide an ap­ple into \(n\) equal pieces, then take \(a\) copies of the re­sult­ing lit­tle-piece”, what would \(\frac{-1}{n}\) mean? And what would \(-\frac{1}{n}\) mean?

\(\frac{-1}n\) would mean “di­vide an ap­ple into \(n\) equal pieces, then take \(-1\) copies of the re­sult­ing lit­tle-piece”. That is, turn it into an anti-lit­tle-piece. This anti-lit­tle-piece will an­nihilate one lit­tle-piece of the same size: \(\frac{-1}{n} + \frac{1}{n} = 0\).

\(-\frac{1}{n}\), on the other hand, would mean “di­vide an anti-ap­ple into \(n\) equal pieces, then take \(1\) copy of the re­sult­ing lit­tle-anti-piece”. But this is the same as \(\frac{-1}{n}\): it doesn’t mat­ter whether we do “con­vert to anti, then di­vide up the ap­ple” or “di­vide up the ap­ple, then con­vert to anti”. That is, “lit­tle-anti-piece” is the same as “anti-lit­tle-piece”, which is very con­ve­nient. <div><div>

What about chunks of ap­ple? If we com­bine half an ap­ple with half an anti-ap­ple, they should also an­nihilate, leav­ing noth­ing be­hind. We write this as \(\frac{1}{2} + \left(-\frac{1}{2}\right) = 0\).

How about a bit more ab­stract? If we com­bine an ap­ple with half an anti-ap­ple, what should hap­pen? Well, the ap­ple can be made out of two half-chunks (that is, \(1 = \frac{1}{2} + \frac{1}{2}\)); and we’ve just seen that half an ap­ple will an­nihilate half an anti-ap­ple; so we’ll be left with just one of the two halves of the ap­ple. More for­mally, \(1 + \left(-\frac{1}{2}\right) = \frac{1}{2}\); or, writ­ing out the calcu­la­tion in full,

$$1 + \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} + \left(-\frac{1}{2}\right) = \frac{1}{2}$$

Let’s go the other way round: if we com­bine an anti-ap­ple with half an ap­ple, what hap­pens? It’s pretty much the same as the op­po­site case ex­cept flipped around: the anti-ap­ple is made of two anti-half-chunks, and the half ap­ple will an­nihilate one of those chunks, leav­ing us with half an anti-ap­ple: that is, \((-1) + \frac{1}{2} = -\frac{1}{2}\).

We call all of these things sub­trac­tion: “sub­tract­ing” a quan­tity is defined to be the same as the ad­di­tion of an anti-quan­tity.

Gen­eral procedure

Since we already know how to add, we might hope that sub­trac­tion will be eas­ier (since sub­trac­tion is just a slightly differ­ent kind of adding).

In gen­eral, we write \(-\frac{1}{n}\) for “the \(\frac{1}{n}\)-sized build­ing block, but made by di­vid­ing an anti-ap­ple (in­stead of an ap­ple) into \(n\) equal pieces”. Re­mem­ber from the pon­der­ing above that this is ac­tu­ally the same as \(\frac{-1}{n}\), where we have di­vided an ap­ple into \(n\) equal pieces but then taken \(-1\) of the pieces.

Then in gen­eral, we can just use the in­stant ad­di­tion rule that we’ve already seen. noteRe­call: this was \(\frac{a}{m} + \frac{b}{n} = \frac{a\times n + b \times m}{m \times n}\). Re­mem­ber that the or­der of op­er­a­tions in the in­te­gers is such that in the nu­mer­a­tor, we calcu­late both the prod­ucts first; then we add them to­gether. In fact,

$$\frac{a}{m} - \frac{b}{n} = \frac{a}{m} + \left(\frac{-b}{n}\right) = \frac{a \times n + (-b) \times m}{m \times n} = \frac{a \times n - b \times m}{m \times n}$$

Why are we jus­tified in just plug­ging these num­bers into the for­mula, with­out jus­tifi­ca­tion? You’re quite right if you are du­bi­ous: you should not be con­tent merely to learn this for­mu­lan­oteThis goes for all of maths! It’s not sim­ply a col­lec­tion of ar­bi­trary rules, but a proper pro­cess that we use to model our thoughts. Be­hind ev­ery pithy, un­mem­o­rable for­mula is a great ed­ifice of mo­ti­va­tion and rea­son, if you can only find it.. In the rest of this page, we’ll go through why it works, and how you might con­struct it your­self if you for­got it. I took the choice here to pre­sent the for­mula first, be­cause it’s a good ad­ver­tise­ment for why we use the \(\frac{a}{n}\) no­ta­tion rather than talk­ing about “\(\frac{1}{n}\)-chunks” ex­plic­itly: it’s a very com­pact and neat way of ex­press­ing all this talk of anti-ap­ples, in the light of what we’ve already seen about ad­di­tion.

Very well: what should \(\frac{a}{m} - \frac{b}{n}\) be? We should first find a smaller chunk out of which we can build both the \(\frac{1}{m}\) and \(\frac{1}{n}\) chunks. We’ve seen already that \(\frac{1}{m \times n}\) will work as a smaller chunk-size.

Now, what is \(\frac{a}{m}\) ex­pressed in \(\frac{1}{m \times n}\)-chunks? Each \(\frac{1}{m}\)-chunk is \(n\) of the \(\frac{1}{m \times n}\)-chunks, so \(a\) of the \(\frac{1}{m}\)-chunks is \(a\) lots of ”\(n\) lots of \(\frac{1}{m \times n}\)-chunks”: that is, \(a \times n\) of them.

Similarly, \(\frac{b}{n}\) is just \(b \times m\) lots of \(\frac{1}{m \times n}\)-chunks.

So, ex­pressed in \(\frac{1}{m \times n}\)-chunks, we have \(a \times n\) lots of pos­i­tive chunks, and \(b \times m\) lots of anti-chunks. There­fore, when we put them to­gether, we’ll get \(a \times n - b \times m\) chunks (which might be nega­tive or pos­i­tive or even zero—af­ter all the an­nihila­tion has taken place we might end up with ei­ther nor­mal or anti-chunks or maybe no chunks at all—but it’s still an in­te­ger, be­ing a num­ber of chunks).

So the to­tal amount of ap­ple we have is \(\frac{a \times n - b \times m}{m \times n}\), just like we got out of the in­stant for­mula.

the rest of the page, in­clud­ing ex­am­ples; make an ex­er­cises page, in­clud­ing sub­tract­ing nega­tives and so on