Most readers of this blog probably know what a rational number is: it’s a number that can be represented as a ratio (hence rational) of two integers. In other words, a fraction. Examples are 3/4, 99/2, and -20837/231, and even 4 (it can be represented as the ratio 4/1). The only restriction is that you can’t have zero in the denominator (the bottom of the ratio) since division by zero, of course, is undefined.
You’ve probably also heard of irrational numbers, which, of course, are numbers that aren’t rational. But are you sure that irrational numbers exist? Why? Just because some teacher told you, or you’ve heard of them before? The ancient Greeks thought irrational numbers didn’t exist—that every quantity could be described by a ratio—and they were in for a rude surprise when they themselves proved this to be false! I won’t go into that history in more detail here; it’s actually quite confused and unclear as to who did what when and so on. Maybe I’ll try writing about that in a future post. For now, I’ll just offer some incontrovertible proof that at least one number exists which isn’t rational. Probably many of you have seen it already, but it’s a good place to start, as I intend to write more posts on this topic in the near future.
Ready? OK, I’m thinking of a number which we’ll call s. In particular, s is the length of one side of a square whose area is 2. This is how the Greeks would have thought of it, and certainly no one can argue that it isn’t a number! There obviously must be a square with an area of 2, and the sides of that square obviously must have a length, and that length is a number. (Astute readers will note that I’m playing rather fast and loose with the term “number”. Which is true. So sue me.) In more modern terms, we would say that s is the positive solution to the equation , or just .
Now, suppose that s is a rational number, so that s can be represented by a ratio . Remember, we want to show that s isn’t rational, so I’ll show that making this assumption leads to an absurdity, from which we can conclude that it was an incorrect assumption in the first place. This is called a proof by contradiction—assume the logical opposite of what you want to prove, and show that this leads to an absurdity, like , or something being both true and not true at the same time.
We’ll assume that p/q is in lowest terms, that is, that p and q have no common factors. It’s ok to assume this, since any rational number can always be turned into an equivalent one in lowest terms.
We know that squaring s yields 2, so substituting p/q for s gives us the equation . Rearranging, we find that , so is even. But if is even, so is p (you should convince yourself of this), so we can write for some other integer r. Substituting, we find that , and therefore . But this means that is even, and therefore so is q—but this is absurd, because we assumed that p and q have no common factor, so they can’t both be even! Therefore, our original assumption—that s is rational—must be false.
In upcoming posts I plan to explore this and other related topics, possibly including: the connection between rational numbers and decimal expansions, the period of repeating decimals, repunits, set cardinality and orders of infinity, some famously irrational numbers (like and e), and maybe even some continued fractions. There are many fascinating topics connected in one way or another to this distinction between rational and irrational numbers, so it should be interesting—I hope you’ll come along for the ride!
As a further challenge, can you extend the above proof to show that the square root of any prime number is irrational? What about the square root of any positive integer which is not a perfect square? (This second one is a bit trickier, because if n is not prime, you cannot assume that if n divides then it must divide p as well—for example, 12 divides , but it does not divide 6.)