A sledgehammer proof of irrationality

Mathematical proofs are a bit like people. You can choose which ones to love and which ones to spit at. The loveable proofs are undoubtably the most important, especially as a pick-me-up on those cold lonely days. I’d like to share with you all a little proof that the $n$ th root of 2 is irrational for $n \geq 3$ . Of course, most people see the proof that the square root of 2 is irrational sometime during their study. If you want some background on irrational numbers, you can check out my earlier blog post.

First, let’s take a detour into the well known equation:

$x^2 + y^2 = z^2$

This equation has infinitely many integer solutions (we call these Pythagorean triples). Here are some examples:

$3^2 + 4^2 = 5^2$

$5^2+12^2=13^2$

What about the equation $x^3+y^3 = z^3$ or even $x^4+y^4 = z^4$ ? Well, it turns out that these have no integer solutions (except for silly ones involving 0’s)! Furthermore, for any integer $n \geq 3$ , we have that the equation

$x^n + y^n = z^n$

has no non-silly solutions in the integers. This result is known as Fermat’s Last Theorem, and was proven to be true by Andrew Wiles around 1995. Now that we know this, we can proceed with our cute little proof that $\sqrt[n]{2}$ is irrational.

We will assume to the contrary. Let $n \geq 3$ and assume that $\sqrt[n]{2}$ is rational, so there must be two positive integers $a$ and $b$ such that

$\sqrt[n]{2} = \frac{a}{b}.$

We put both sides to the power of $n$ and rearrange to get

$2 b^n = a^n.$

But this is really $b^n + b^n = a^n$ and by Fermat’s Last Theorem there are no positive integer solutions to this equation! So we have found a contradiction to FLT and so $\sqrt[n]{2}$ must be irrational for all $n \geq 3$ . What a cutie-proof!

One thought on “A sledgehammer proof of irrationality”

Melissa Lee says:

July 20, 2013 at 3:35 pm

This is the most adorable proof I have seen in ages! 😀

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