Every integer $n>1$ is either a prime number or a product of primes.
Proof. Sieve of Eratosthenes algorithm gives an impression of this result. However here is a simple proof by Euclid: By induction on n, for $n=2$ this is true. Let's suppose it's true for every integer $<n$. If $n$ is not prime then, there exists an integer $d \ne n$ such that $d|n$, hence $n=kd$ where $k,d < n$ and $k,d>1$. By induction hypothesis $k,d$ are products of primes, so is $n$.
There are infinitely many prime numbers.
Proof. Euclid. let $N=1+\prod_{i=0}^{n}p_i$. Hence $N > 1$, so it's either prime or a product of primes. $N$ is neither prime nor a product of primes which contradicts previous result.
If a prime doesn't not divide a, then $(p,a)=1$.
Proof. Let $(p,a) =d$, then $d \ne p$ and $d | p$, hence $d=1$.
If a prime p divides ab, then $p|a$ or $p|b$.
Proof. Suppose $p \not | a$, hence $(p,a)=1$, then according to Euclid's lemma, $p|b$.
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