Proving $\sqrt 3$ is irrational. [duplicate]

Solution 1:

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

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

Contradiction.

Solution 2:

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., $\sqrt 3 < 2$).

$x = \sqrt{3} - 1$ 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.

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