i was trying to proof that $\sqrt{2}$ was irrational without looking the standard proof, and after hours of head banging i ended up with an argument for the irrationality of the square root of any prime, the argument goes like this:
[1/premise]for any prime number $P, \sqrt{P} = \frac{a}{b}$ were [2]$a,b$ are coprime
[3] every natural number can be described by a product of primes. $x = p_0^{y_0} p_1^{y_1} p_0^{y_2} ... p_n^{y_n}$
[4]the exponents of all the prime factors of any perfect square are all even
$x^2 = x*x = (p_0^{y_0} p_1^{y_1} p_0^{y_2} ... p_n^{y_n}) * (p_0^{y_0} p_1^{y_1} p_0^{y_2} ... p_n^{y_n}) = p_0^{2y_0} p_1^{2y_1} p_0^{2y_2} ... p_n^{2y_n}$
by the premise $\sqrt{P} = \frac{a}{b}$ we can square both sides and multiply by $b^2$, then we get $a^2=P*b^2$, so $a^2$ is divisible by $P$, and by [2] we know that b is not.
therefore the prime factors of $a^2 = P^1 * (p_0^{2y_0} p_1^{2y_1} p_0^{2y_2} ... p_n^{2y_n})$, the exponent of $P$ is 1, entering in contraction with [4] so the premise is false.