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Proving The Irrationality of 3\sqrt{3} Via Modding By 4

Lucy and I found this proof last night. I'm sure it is not new, but hadn't seen it before.

We aim to prove that the square root of 33 is irrational, i.e. that there is no rational pp such that p2=3p^2 = 3.

Proof: By contradiction. Suppose that there was such a pp. Then, as a rational number, it could be written in lowest terms: i.e. as mn\frac{m}{n} for m,nZ+m, n \in \mathbb{Z}^+ such that mm and nn are relatively prime.

m2n2=3\frac{m^2}{n^2} = 3

Consider the parities of mm and nn. There are four options. Note that the parities of mm and nn determine the parities of m2m^2 and n2n^2.

mm even, nn even. This cannot happen since then mm and nn would share the factor 22, violating our assumption that they were relatively prime.

mm even, nn odd. This cannot happen since an even number divided by an odd number is always even, but 33 is odd.

mm odd, nn even. This cannot happen since an odd number divided by an even number is not an integer, but 33 is an integer.

mm odd, nn odd. This possibility is the only one that is not ruled out immediately. But it too cannot happen. Write m=2j+1m = 2j + 1 and n=2k+1n = 2k + 1, where j,kZ+j, k \in \mathbb{Z}^+. Then:

(2j+1)2(2k+1)2=3(2j+1)2=3(2k+1)24j2+4j+1=3(4k2+4k+1) \begin{matrix} \frac{(2j + 1)^2}{(2k + 1)^2} & = & 3 \\ (2j + 1)^2 & = & 3(2k + 1)^2 \\ 4j^2 + 4j + 1 & = & 3(4k^2 + 4k + 1) \\ \end{matrix}

Now consider the equation mod 44. The left side is equal to 11, but the right side is equal to 1-1, so we have a contradiction.

Notes: Consider using this proof to prove that the square root of 55 is irrational. We also conclude that both mm and nn must be odd, but since 51(mod4)5 \equiv 1 \pmod{4} the last bit does not work. And indeed it better not, otherwise we could prove that the square root of 99 was irrational.

This inelegant, bedraggled proof applies unevenly and will not win any beauty contests.