Question

Question #ca15a

Answer 1

There are many ways. I'll just show you one.

Explanation:

Imagine that there is a finite number of primes `n`.
The number `n!` will be the product of every number `<=n` , including all prime numbers. Then `n!+1` will not be a product of any number `<=n`, so will be prime. We can continue the reasoning, and always get new prime numbers, so the number o prime numbers is infinite.

Answer 2

Prove by contradiction:
Considering a finite number of primes

Explanation:

We well solve this via the use of contradiction, we we can let there be a finite number of prime `p` where we can let the be primes;
`p = {p_1,p_2,p_3,...,p_n}`
And we know `p_1p_2p_3...p_n` is certainly not prime as it has factors `p_1 ,p_2,..p_n`

But we can consider `p_1p_2p_3...p_n + 1`

This on the other hand is prime, as if it were not prime than we could devide it by a smaller prime and yield a natural number, we can prove this by;

let `p_m` be the `m`th prime, for `1` `<=` `m` `<=` `n`
then `(p_1p_2p_3....p_n + 1)/p_m` is not natural as this is equal to;
`1/p_m + (p_1p_2...p_n)/p_m`
were we know `(p_1p_2...p_n)/p_m` is natural as `m<=n`
But `1/p_m notin ZZ` for `m >=1`
So hence `1/p_m + (p_1p_2...p_n)/p_m` `notin ZZ`

So hence `1 + p_1p_2...p_n` has no prime factors and hence is prime

Hence for any finite set of primes we can make a new prime, hence there is not finitely many primes

Hence there can't be a finite number of primes, hence proven via contridiction