Prove that $\sqrt p$ is irrational for all prime $p$

Woah! What the hell happened to me, I’m all serif and academic ! … huh.

Well anyway, today I thought it’d be fun to share this little proof, plus try to pull off the TeX look just for the hell of it! :p

I’m going to slow this off using proof by contradiction, by the way.

Proof

Assume that $\sqrt p$ is rational - that is $\sqrt p = \frac a b$, where $a$ and $b$ are coprime integers.

If $a$ divides by $p$:

$a$ is of the form $kp, k\in\mathbb Z$

For this to be true, $b^2$ must divide by $p$, $\therefore b$ must divide by $p$.

This would mean that $a$, $b$ share factor $p$, so are not coprime! ↯

If $a$ does not divide by $p$:

$a$ is of the form $kp+n, k\in\mathbb Z, n$ does not divide by $p$.

$n$ does not divide by $p \therefore n^2$ does not divide by $p$.

$\therefore \frac{n^2} p$ is not an integer, $\therefore b^2$ is not an integer, $\therefore b$ is not an integer! ↯

Conclusion

$a$ cannot divide by $p$, but also cannot not divide by $p$.

This is impossible, so $\sqrt p$ cannot be rational!

$\therefore \sqrt p$ is irrational for all prime $p$ $\square$