# Question ca15a

Dec 2, 2017

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 $\le n$ , including all prime numbers. Then n!+1# will not be a product of any number $\le n$, so will be prime. We can continue the reasoning, and always get new prime numbers, so the number o prime numbers is infinite.

Dec 2, 2017

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;
$p = \left\{{p}_{1} , {p}_{2} , {p}_{3} , \ldots , {p}_{n}\right\}$
And we know ${p}_{1} {p}_{2} {p}_{3.} . . {p}_{n}$ is certainly not prime as it has factors ${p}_{1} , {p}_{2} , . . {p}_{n}$

But we can consider ${p}_{1} {p}_{2} {p}_{3.} . . {p}_{n} + 1$

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$$\le$$m$$\le$$n$
then $\frac{{p}_{1} {p}_{2} {p}_{3.} \ldots {p}_{n} + 1}{p} _ m$ is not natural as this is equal to;
$\frac{1}{p} _ m + \frac{{p}_{1} {p}_{2.} . . {p}_{n}}{p} _ m$
were we know $\frac{{p}_{1} {p}_{2.} . . {p}_{n}}{p} _ m$ is natural as $m \le n$
But $\frac{1}{p} _ m \notin \mathbb{Z}$ for $m \ge 1$
So hence $\frac{1}{p} _ m + \frac{{p}_{1} {p}_{2.} . . {p}_{n}}{p} _ m$ $\notin \mathbb{Z}$

So hence $1 + {p}_{1} {p}_{2.} . . {p}_{n}$ 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