[summary: "Ordering" is the idea that some quantities of apple are "bigger" than others.]

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 %%note:My 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. %%note:Recall 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* %%note:That 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 %%note:Mathematicians 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 %%note:Or 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.)

## Example

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}$~$. %%note:This 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 %%note:Remember, that's the thing on the bottom of the fraction: in this case, $~$b \times (-d)$~$.%% is positive. [todo: the rest of this section and bullet point]

## Why did I say "Mathematicians hate it"?%%note:Aside 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.