An integer is either a perfect square or its square root is irrational. Said a different way, when you compute the square root of an integer, there are either no figures to the right of the decimal or there are an infinite number of figures to right of the decimal and they don’t repeat. There’s no middle ground. You can’t hope, for example, that the decimal expansion might stop or repeat after a hundred or so terms.
This theorem came up recently when I was talking to one of my kids about her math class, so I decided to look up the proof. It’s easier than I expected, not much harder than the familiar proof that the square root of 2 is irrational. Here goes.
Suppose a/b is a fraction in lowest terms, i.e. a and b are relatively prime, and a/b a solution to xn = c where n > 0 is an integer and c is an integer. Then
(a/b)n = an / bn = c
and so
an / b = c bn-1.
Now the right side of the equation above is an integer, so the left side must be an integer as well. But b is relatively prime to a, and so b is relatively prime to an. The only way an / b could be an integer is for b to equal 1 or -1. And so a/b must be an integer.
This proof shows that what we said about square roots extends to cube roots and in fact to all integer roots. For example, the fifth root of an integer is either an integer or an irrational number.
Update: Note that there’s another proof in the comments, one that I believe is easier to follow.
{ 6 comments… read them below or add one }
I. J. Kennedy 12.21.09 at 10:57
Another way to get the same result is to assume a/b is fraction in lowest terms and is not an integer (i.e. b ≠ 1), and consider powers (a/b)n. Clearly an and bn are relatively prime, and the denominator bn ≠ 1, so an/bn is not an integer. In other words, no (non-integer) fraction, when raised to a power, can produce an integer.
John 12.21.09 at 11:02
I think your proof is easier to follow than mine.
John Armstrong 12.21.09 at 15:09
How do you know that an and bn are relatively prime? The work amounts to the messy parts in the post. This version is only simpler because it hides the hard bits in a lemma.
I. J. Kennedy 12.21.09 at 15:43
Perhaps you’re right, but I’m not sure. In any case we know a^n and b^n are relatively prime by the Fundamental Theorem of Arithmetic. If a factors to certain primes, then multiplying a times itself isn’t going to yield any new primes. So the factorizations of a^n and b^n have no primes in common, because a and b don’t.
Denise Gaskins 01.05.10 at 08:06
I remember being amazed when I first saw this proof, at how easy it was. I wondered then (and still do) why we never had it in school. We never did any proofs except in geometry and trig.
John 01.05.10 at 10:32
Denise, I agree that is seems odd that proofs in high school are limited to geometry. (And from what I understand, proofs have largely been removed from geometry.) Number theory seems like a reasonable place to learn proofs.