Proving $\sqrt 3$ is irrational.

$3q^2 = p^2$, so $3|p$ (as in the case of $\sqrt2$).

Hence $q^2 = 3k$ for some $k$, and then $3|q$

Contradiction.

Here are some proofs I've found (link at bottom):

If $\sqrt 3 = m/n$: $$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1} = \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$ and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).

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$x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$, thus: $$x(3+x) = 2+x$$

$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$

And thus

$$x = [1,2,1,2,\dots]$$

So $\sqrt{3}$ os irrational.

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These proofs and others: How to prove that $\sqrt 3$ is an irrational number?

It works, but can be simplified: $\rm\:mod\ 2\!: p\equiv p^2 = 3q^2 \equiv q,\:$ so $\rm\:p,q,\:$ being coprime, are odd. $\rm\:mod\ 4\!:\ odd\equiv \pm 1,\:$ so $\rm\:odd^2\equiv 1,\:$ so $\rm\: 1\equiv p^2 = 3q^2\equiv 3\ \Rightarrow\ 4\:|\:3-1\:\Rightarrow\Leftarrow$