# Euclid's Theorem/Corollary 2

## Corollary to Euclid's Theorem

There is no largest prime number.

## Proof 1

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

Aiming for a contradiction, suppose there exists a largest prime number $p_m$.

Then:

- $\mathbb P \subseteq \closedint 1 {p_m} = \set {1, 2, \ldots, p_m}$

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 $q \notin \closedint 1 {p_m}$.

That is, $q > p_m$.

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

Hence the result, by Proof by Contradiction.

$\blacksquare$

## Proof 2

Aiming for a contradiction, suppose there exists a largest prime number $p$.

Let $b = p! + 1$.

Let $q$ be a prime number that divides $b$.

Since $p$ is the largest prime number, $q \le p$.

However, no positive integer $d \le p$ is a divisor of $b$.

Hence $q \not \le p$.

Hence the result, by Proof by Contradiction.

$\blacksquare$

## Source of Name

This entry was named for Euclid.