How can I show that there are an infinite number of primes by using the Fundamental Theorem of Arithmetic?
Proof of there are an infinite number of primes by using the Fundamental Theorem of Arithmetic
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You start by assuming the opposite. Let's say there are a finite amount of prime numbers, in fact, let's write them in a list.
$P_1$, $P_2$, $P_3$, ... $P_n$
Note, this is a complete list.
Now let's form a new number $a$, by multiplying all of our prime numbers and adding $1$. According to the Fundamental Theorem of Arithmetic, every integer greater than $1$ is either prime or a unique factorization of primes. So let's try both of these possibilities.
Possibility 1: $a$ is prime. However, we previously wrote all the primes on our complete list, so this is a contradiction.
Possibility 2: $a$ is composite. However, if it is, it needs to be a unique factorization of the primes on our list. But, it won't divide $P_1$ exactly, $P_2$, $P_3$, or any $P_n$ for that matter. So it is a violation of the Fundamental Theorem of Arithmetic, and therefore a contradiction.
Because we get a contradiction when we assume there are finitely many primes, it must be the opposite, or there must be infinitely many primes.
Q.E.D
The standard way is by assuming that there are only a finite number of primes and deducing that if all terms of the form $\prod_{p \in P} p^{a(p)}$ are counted, there are not enough of them.
I don't remember the details, but it might go something like this:
The number of integers of the form $\prod_{p \in P} p^{a(p)}$ which are less that $x$ are of the form $\prod_{p \in P} p^{a(p)} \le x$ or $\sum_{p \in P} a(p)\ln p \le \ln x$.
We certainly have $a(p) \le \ln x/\ln p$.
The total number of possibilities is certainly less than, where $N$ is the number of primes, $\prod_{p \in P}(\ln x)/(\ln p) =(\ln x)^{N}\prod_{p \in P}1/(\ln p) $.
However, for every finite $N$, $(\ln x)^{N}$ will be less than $x$ for large enough $x$.
So $N$ must be infinite.
I just did this off the top of my head, so I'm not completely sure that it is correct, but it seems OK to me.
Anyway, there it is.