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