[summary: Subtraction is "adding anti-apples", or equivalently removing some apples you already have.]

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?

%%hidden(Show possible solution): $~$\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. %%

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. %%note:Recall: 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 formula%%note:This 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.

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