Euclid's Theorem

Theorem
For any finite set of prime numbers, there exists a prime number not in that set.

Proof
Let $\mathbb P$ be a finite set of prime numbers.

Consider the number: $\displaystyle n_p = \left({\prod_{p \in \mathbb P} p}\right) + 1$.

Take any $p_j \in \mathbb P$.

We have that $\displaystyle p_j \backslash \prod_{p \in \mathbb P} p$.

Hence $\displaystyle \exists q \in \Z: \prod_{p \in \mathbb P} p = q p_j$.

So:

So $p_j \nmid n_p$.

There are two possibilities:


 * $n_p$ is prime, which is not in $\mathbb P$.


 * $n_p$ is composite. But from Positive Integer Greater than 1 has a Prime Divisor‎, it must be divisible by some prime.

That means it is divisible by a prime which is not in $\mathbb P$.

So in either case there exists at least one prime which is not in the original set we created.

Hence the result, by Proof by Contradiction.

Also see

 * Furstenberg's Proof of the Infinitude of Primes