# Subtraction of rational numbers (Math 0)

https://arbital.com/p/subtraction_of_rational_numbers_math_0

by Patrick Stevens Jul 7 2016 updated Aug 1 2016

In which we meet anti-apples.

[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]