# Euclid's Theorem

## Theorem

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

In the words of Euclid:

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

### Corollary 1

There are infinitely many prime numbers.

### Corollary 2

There is no largest prime number.

## Proof

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

Consider the number:

$\ds n_p = \paren {\prod_{p \mathop \in \mathbb P} p} + 1$

Take any $p_j \in \mathbb P$.

We have that:

$\ds p_j \divides \prod_{p \mathop \in \mathbb P} p$

Hence:

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

So:

 $\ds n_p$ $=$ $\ds q p_j + 1$ Division Theorem $\ds \leadsto \ \$ $\ds n_p$ $\perp$ $\ds p_j$ Definition of Coprime Integers

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.

$\blacksquare$

## Historical Note

This proof is Proposition $20$ of Book $\text{IX}$ of Euclid's The Elements.

## Fallacy

There is a fallacy associated with Euclid's Theorem.

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.

## Source of Name

This entry was named for Euclid.