Euclid's Theorem

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


 * Prime numbers are more than any assigned multitude of prime numbers.

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

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

Take any $p_j \in \mathbb P$.

We have that:
 * $\displaystyle p_j \mathop \backslash \prod_{p \mathop \in \mathbb P} p$

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

So:

So $p_j \nmid n_p$.

There are two possibilities:


 * $(1): \quad n_p$ is prime, which is not in $\mathbb P$.


 * $(2): \quad n_p$ is composite.

But from Positive Integer Greater than 1 has 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 $\mathbb P$ we created.

Hence the result, by Proof by Contradiction.

Fallacy
There is a danger in this proof.

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.

That is, 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.

Also see

 * Furstenberg's Proof of Infinitude of Primes


 * Definition:Euclid Number