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!

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 $p$ $\square$