# Irrationality of $\sqrt{3}$ [duplicate]

No doubt an easy question: I'm trying to follow Wikipedia's (second) proof of the irrationality of $\sqrt{3}$ and it relies on the notion that since $3n^2 = m^2$ is divisible by 3 then so is $m$. Why is this so and can you help me prove it?

A prime $p$ has property $p|ab\Rightarrow p|a\text{ or }p|b$. This for integers $a$ and $b$. So $3|m^2$ gives $3|m$

• Is that Euclid's lemma? Which is not entirely obvious or trivial? Then I don't feel quite so stupid. – Tom Aug 28 '14 at 20:26
• Is there a more elementary proof for $p=3$ specifically, along the lines of the corresponding one used in the proof that $\sqrt{2}$ is irrational? – Tom Aug 28 '14 at 20:42
• @Tom $\,\ 3\mid x^2\,\Rightarrow\,3\mid x\,$ can be proved by direct case-analyis as follows $\tag*{}$ ${\rm mod}\ 3\!:\,\ x^2 \equiv 0\, \Rightarrow\, x\equiv 0,\,\ {\rm by}\,\ x\not\equiv 0\, \Rightarrow\, x \equiv \pm 1\, \Rightarrow\, x^2 \equiv 1\ \$ – Bill Dubuque Aug 28 '14 at 20:53
• Is there anything wrong with saying that any $x$ must be one of $3k$, $3k+1$ and $3k+2$, but only $(3k)^2$ is divisible by 3 because $(3k+1)^2 = 3(3k^2+2k) + 1$ and $(3k+2)^2 = 3(3k^2+4k+1)+1$? I guess that's basically the same as modular arithmetic (which I'm less familiar with)? – Tom Aug 29 '14 at 12:47
• There is nothing wrong with that and in essence it is the same solution mentioned by @Bill. Now you have a solid proof of yourself :). – drhab Aug 29 '14 at 12:50

$3$ divides the multiples of $3$ $\Rightarrow 3 \mid 3 n^2 \Rightarrow 3 \mid m^2$

Knowing that:

$\text{ When } p \text{ is a prime and } p \mid a^n \Rightarrow p \mid a$

Since $3$ ia a prime, $3 \mid m$.