Euclid's Theorem

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

Corollary
There are infinitely many prime numbers.

Proof by Contradiction
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.

Proof of Corollary
Assume that there are only finitely many prime numbers, and that there is a grand total of $n$ primes.

Then it is possible to define the set of all primes: $\mathbb P = \left\{{p_1, p_2, \ldots, p_n}\right\}$.

From the main proof, however, we can always create a prime which is not in $\mathbb P$.

So we can never create a finite list of all the primes, because we can guarantee to construct a number which has prime factors that are not in this list.

Thus, there are infinitely many prime numbers.

Euclid Number
During the course of the proof of the corollary, we assume that all the primes (of our supposedly finite set) are multiplied together, and $1$ added. The resulting number is then either a prime or contains a prime factor not on that original list.

Such a number is known as a Euclid number.

Although this is not quite the way Euclid originally stated his original theorem, it is such a well-known and accessible proof that it is considered to have entered the mainstream of "general knowledge".

Fallacy
There is a danger in the proof of the corollary.

It is often seen to be stated that: the number made by multiplying all the primes together and adding $1$ is not divisible by any members of that set, so it is not divisible by any primes and "is therefore itself prime".

Sometimes readers think that if $P$ is the product of the first $n$ primes then $P + 1$ is itself prime.

This is not the case. For example:
 * $\left({2 \times 3 \times 5 \times 7 \times 11 \times 13}\right) + 1 = 30\ 031 = 59 \times 509$

both of which are prime, but, take note, not in that list of six primes that were multiplied together to get $30\ 030$ in the first place.