'Infinite Prime Theorem' – Euclid's Theorem

Problem:

Prove that there is an infinite number of primes.

Solution:

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 mm denote the product of all such primes and be one more than it.

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

It is then shown that mm must either be a prime or have prime factors larger than pp.

Case mm is prime:

If mm is prime, and is the product of all primes in a finite list plus 1, then it must be larger than the largest prime pp, 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 mm is composite:

If mm is composite, it follows that it cannot have a prime factor smaller than pp:

mm cannot be divided by 2 as it is 2(3×5×7×...×p)+12(3\times 5\times 7\times...\times p)+1, or one more than a multiple of 2.

mm cannot be divided by 3 as it is 3(2×5×7×...×p)+13(2\times 5\times 7\times...\times p)+1, or one more than a multiple of 3.

mm cannot be divided by 5 as it is 5(2×3×7×...×p)+15(2\times 3\times 7\times...\times p)+1, or one more than a multiple of 5.

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

Note by Just C
1 week, 4 days ago

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1 vote

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Nice! I think you meant to say "let mm denote the product of all such primes". :)

David Stiff - 1 week, 4 days ago

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

Just C - 1 week, 4 days ago

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No problem!

David Stiff - 1 week, 4 days ago

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