Prove that there is an infinite number of primes.

The problem was originally called Euclid’s Theorem, named after Euclid, who proved that there were an infinite amount of primes in his book Elements (Book: IX, Proposition: 20) using contradiction. The following is an adaptation of his proof:

Assuming that there is a finite list of primes, let $m$ denote the product of all such primes and be one more than it.

$m=2\times 3\times 5\times 7\times...\times p+1$

It is then shown that $m$ must either be a prime or have prime factors larger than $p$.

Case $m$ is prime:

If $m$ is prime, and is the product of all primes in a finite list plus 1, then it must be larger than the largest prime $p$, meaning that there is a larger prime bigger than the one defined as the largest, and therefore, there cannot be a finite number of prime numbers.

Case $m$ is composite:

If $m$ is composite, it follows that it cannot have a prime factor smaller than $p$:

$m$ cannot be divided by 2 as it is $2(3\times 5\times 7\times...\times p)+1$, or one more than a multiple of 2.

$m$ cannot be divided by 3 as it is $3(2\times 5\times 7\times...\times p)+1$, or one more than a multiple of 3.

$m$ cannot be divided by 5 as it is $5(2\times 3\times 7\times...\times p)+1$, or one more than a multiple of 5.

It follows that $m$ cannot be divided by the assumed largest prime $p$ (in a finite list to comply with contradiction assumption that there is a finite number of primes) as it is $p(2\times 3\times 7\times...)+1$, or one more than a multiple of $p$.

Therefore, $m$ can only be the product of two or more primes larger than $p$, and therefore contradicts the assumption that $p$ is the largest prime in a finite list.

Therefore, there is an infinite number of primes.

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## Comments

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TopNewestNice! I think you meant to say "let $m$ denote the

productof all such primes". :)Log in to reply

Yeah, that was a mistake. Thanks for pointing it out! :)

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No problem!

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