Ordering of rational numbers (Math 0)

So far while learn­ing about how to com­bine ra­tio­nal num­bers, we have seen ad­di­tion, sub­trac­tion, mul­ti­pli­ca­tion and di­vi­sion. There is one fi­nal ma­jor thing we can do to a pair of ra­tio­nal num­bers: to com­pare them. Once you know what you’re look­ing for, it is very easy to com­pare cer­tain pairs of ra­tio­nal num­bers; in this page, we’ll look at how to ex­tend that.

In­tu­itively, if I gave you an ap­ple in one hand, and four ap­ples in the other hand noteMy hands are rather large., you would be able to tell me that the four-ap­ples hand was hold­ing more ap­ples. This is the kind of com­par­i­son we are try­ing to gen­er­al­ise, and we will do it from one sim­ple ob­ser­va­tion.

The ob­ser­va­tion we will make is that it is very easy to de­ter­mine whether a ra­tio­nal num­ber is nega­tive or not. noteRe­call that “nega­tive” meant “it is ex­pressed in anti-ap­ples rather than ap­ples”. In­deed, we just need to see if we’re hold­ing an anti-thing or not. This might be hard if we have to do some calcu­la­tions first—for ex­am­ple, it’s not im­me­di­ately ob­vi­ous whether \(\frac{16}{107} - \frac{3}{20}\) is anti- or not—but we’ll as­sume that we’ve already done all the calcu­la­tions to re­duce an ex­pres­sion down to just a sin­gle ra­tio­nal num­ber. (In this ex­am­ple, we can use the sub­trac­tion tech­niques to work out that \(\frac{16}{107} - \frac{3}{20} = -\frac{1}{2140}\). That’s ob­vi­ously nega­tive, be­cause it’s got a nega­tive sign out the front.)

To sum­marise, then, what I have just as­serted is that it is easy to see whether a ra­tio­nal num­ber is nega­tive or not; we say that a num­ber which is pos­i­tive noteThat is, ex­pressed in ap­ples rather than anti-ap­ples. is “greater than \(0\)”. Re­call that \(0\) was the name we gave to the ra­tio­nal num­ber which is “no ap­ples at all”; then what this is say­ing is that if I have a pos­i­tive num­ber of ap­ples in one hand, and no ap­ples at all in the other, then ac­cord­ing to the in­tu­ition ear­lier, I have more ap­ples in the first hand than in the no-ap­ples hand. (Hope­fully you see that this is true; if not, let us know, be­cause this is one of those strange ar­eas where it’s very hard for a math­e­mat­i­cian to re­mem­ber not un­der­stand­ing it im­me­di­ately and in­tu­itively, since we’ve each been do­ing this for decades. We’re do­ing our best to re­mem­ber what parts of the maths are gen­uinely difficult and weird, but we might get it wrong.)

Similarly, we say that \(0\) is less than any pos­i­tive num­ber, and write \(0 < \frac{5}{16}\), for in­stance. The lit­tler quan­tity always goes on the lit­tler end of the ar­row. (The num­ber \(0\) it­self is nei­ther nega­tive nor pos­i­tive. It’s just \(0\). There­fore we can’t write \(0 < 0\) or \(0 > 0\); it’s ac­tu­ally the case that \(0=0\), and this ex­cludes the other two op­tions of \(<) or \(>\).)

One weird trick to com­pare any two ra­tio­nals noteMath­e­mat­i­ci­ans hate it!

Now that we can com­pare any ra­tio­nal with \(0\), we will work out how to com­pare any ra­tio­nal with any other ra­tio­nal.

The key in­sight is that adding the same num­ber of ap­ples noteOr anti-ap­ples. to each hand should not change the rel­a­tive fact of whether there are more ap­ples in one hand or the other. For a real-world ex­am­ple, on a bal­ance scale it doesn’t mat­ter whether you add five grams or even fifty kilo­grams onto each of the two pans; the re­sult of the weight com­par­i­son will be the same. (Now you should prob­a­bly for­get the scales metaphor again, be­cause the weight of an­ti­mat­ter be­haves in a way that doesn’t lend it­self nicely to what we’re try­ing to do.)


So let’s say we want to com­pare \(\frac{5}{6}\) and \(\frac{3}{4}\). Which of the two is big­ger?

Well, what we can do is add \(\frac{3}{4}\) of an anti-ap­ple to both hands. By the prin­ci­ple that “adding the same amount to each hand doesn’t change their quan­tity rel­a­tive to each other”, the re­sult of the com­par­i­son be­tween \(\frac{5}{6}\) and \(\frac{3}{4}\) is just the same as the re­sult of the com­par­i­son be­tween \(\frac{5}{6} - \frac{3}{4}\) and \(\frac{3}{4} - \frac{3}{4}\): that is, be­tween \(\frac{1}{12}\) and \(0\). That’s easy, though, since we already know how to com­pare \(0\) with any­thing!

So \(\frac{5}{6}\) is big­ger than \(\frac{3}{4}\), since \(\frac{1}{12}\) is big­ger than \(0\): we write \(\frac{5}{6} > \frac{3}{4}\).

Com­par­i­sons with anti-apples

In this sec­tion, we will just close our eyes, swal­low grimly, and hope for the best.

If we want to com­pare \(-\frac{59}{12}\) and \(\frac{4}{7}\), what should hap­pen? If you already have the right in­tu­ition built in, then this will be ob­vi­ous, but be­fore you know how to do it, it’s re­ally not clear at all. After all, \(-\frac{59}{12}\) is “a large amount of anti-ap­ple” (it’s nearly five whole anti-ap­ples!) but \(\frac{4}{7}\) is “a small amount of ap­ple” (not even one whole ap­ple).

Here’s where the “close our eyes” hap­pens. Let’s just go by the prin­ci­ple that adding the same amount of ap­ple to both hands shouldn’t change any­thing, and we’ll add \(\frac{59}{12}\) ap­ples to both sides.

Then the \(-\frac{59}{12}\) be­comes \(0\), and the \(\frac{4}{7}\) be­comes the rather grue­some \(\frac{461}{84}\). noteThis is all good prac­tice for you to get fluent with adding and sub­tract­ing. But we already know how to com­pare \(0\) with things, so we can see that \(\frac{461}{84}\) is big­ger than \(0\).

There­fore we must have \(\frac{4}{7}\) be­ing big­ger than \(-\frac{59}{12}\).

By the same to­ken, any amount of anti-ap­ple is always less than any amount of ap­ple, and in­deed any amount of anti-ap­ple is always less than \(0\).

Another example

How about com­par­ing \(\frac{-3}{5}\) and \(\frac{9}{-11}\)? The first thing to do is to re­mem­ber that we can take the minus signs out­side the frac­tions, be­cause an anti-chunk of ap­ple is the same as a chunk of anti-ap­ple.

That is, we are try­ing to com­pare \(-\frac{3}{5}\) and \(-\frac{9}{11}\).

Add \(\frac{3}{5}\) to both, to com­pare \(0\) and \(-\frac{9}{11} + \frac{3}{5} = -\frac{12}{55}\).

Add \(\frac{12}{55}\) to both again, to com­pare \(\frac{12}{55}\) and \(0\).

Clearly the \(\frac{12}{55}\) is big­ger, so it must be that \(\frac{-3}{5}\) is big­ger than \(\frac{9}{-11}\).

In­stant rule

Just as we had an in­stant rule for ad­di­tion, so we can make an in­stant rule for com­par­i­son.

If we want to see which of \(\frac{a}{b}\) and \(\frac{c}{d}\) is big­ger, it is enough to see which of \(0\) and \(\frac{c}{d} - \frac{a}{b}\) is big­ger.

But \($\frac{c}{d} - \frac{a}{b} = \frac{c \times b - a \times d}{b \times d}\)$

Sadly from this point there are ac­tu­ally two cases to con­sider, be­cause we might have pro­duced some­thing that looks like any of the fol­low­ing:

  • \(\frac{5}{6}\)

  • \(\frac{-4}{7}\)

  • \(\frac{3}{-8}\)

  • \(\frac{-2}{-9}\)

(That is, there could be minus-signs scat­tered all over the place.)

How­ever, there is a way to get around this, and it hinges on the fact from the Divi­sion page that \(\frac{-1}{-1} = 1\).

If \(b\) is nega­tive, then we can just write \(\frac{a}{b} = \frac{-a}{-b}\), and now \(-b\) is pos­i­tive! For ex­am­ple, \(\frac{5}{-6}\) has \(b=-6\); then that is the same as \(\frac{-5}{6}\). Similarly, \(\frac{-7}{-8}\) is the same as \(\frac{7}{8}\).

Like­wise we can always write \(\frac{c}{d}\) so that the nu­mer­a­tor is pos­i­tive: as \(\frac{-c}{-d}\) if nec­es­sary.

So, we have four cases:

  • If \(b, d\) are both pos­i­tive, then \(\frac{c \times b - a \times d}{b \times d}\) is pos­i­tive pre­cisely when \(c \times b - a \times d\) is pos­i­tive as an in­te­ger; i.e. when \(cb > ad\).

  • If \(b\) is pos­i­tive and \(d\) is nega­tive, then \($\frac{c}{d} - \frac{a}{b} = \frac{-c}{-d} - \frac{a}{b} = \frac{(-c) \times b - a \times (-d)}{b \times (-d)}\)$ where the de­nom­i­na­tor noteRe­mem­ber, that’s the thing on the bot­tom of the frac­tion: in this case, \(b \times (-d)\). is pos­i­tive. the rest of this sec­tion and bul­let point

Why did I say “Math­e­mat­i­ci­ans hate it”?noteAside from par­o­dy­ing In­ter­net ban­ner ads, that is. A di­ver­sion on pedagogy

The way I’ve done this is all com­pletely cor­rect, but it’s slightly back­wards from the way a math­e­mat­i­cian would usu­ally pre­sent it. (Not suffi­ciently back­wards that math­e­mat­i­ci­ans should hate it, but I couldn’t re­sist.)

Usu­ally, when find­ing a way of com­par­ing ob­jects, math­e­mat­i­ci­ans would prob­a­bly do what we’ve done above: find an easy way of com­par­ing some of the ob­jects, and then try to ex­tend it to cover all the ob­jects.

But if a math­e­mat­i­cian already knows a way of com­par­ing ob­jects and is just writ­ing it down for the benefit of other math­e­mat­i­ci­ans, they would usu­ally write down the com­plete “cover all the ob­jects” method right at the start, and would then go on to show that it does in­deed cover all the ob­jects and has all the right prop­er­ties.

This has the benefit of pro­duc­ing very terse de­scrip­tions with the min­i­mum nec­es­sary amount of writ­ing; but it’s very bad at helping other peo­ple un­der­stand where it came from. If you know where some­thing came from, you stand a bet­ter chance of be­ing able to recre­ate it your­self if you for­get the bot­tom line, and you might well re­mem­ber the bot­tom line bet­ter, too. This is why we’ve done it slightly back­wards here.


  • Mathematics

    Math­e­mat­ics is the study of num­bers and other ideal ob­jects that can be de­scribed by ax­ioms.