Euclid's Theorem/Corollary 2/Proof 1

Proof
Let $\mathbb P$ be the set of all prime numbers.

there exists a largest prime number $p_m$.

Then:
 * $\mathbb P \subseteq \left[{1 \,.\,.\, p_m}\right] = \left\{{1, 2, \ldots, p_m}\right\}$

and so $\mathbb P$ is a finite set.

By Euclid's Theorem, there exists a prime number $q$ such that $q \notin \mathbb P$.

But that means $q \notin \left[{1 \,.\,.\, p_m}\right]$.

That is, $q > p_m$.

So $p_m$ is not the largest prime number after all.

Hence the result, by Proof by Contradiction.