If $\frac{a}{b}$ is irreducible, then the quotient of the product of any $2$ factors of $a$ and any $2$ factors of $b$ are irreducible.

Question

$a,b\in \mathbb{Z}$

Factors of $a$: $a_1,a_2,...,a_n$.

Factors of $b$: $b_1,b_2,...,b_m$

Prove that if $\frac{a}{b}$ is irreducible, then $\frac{a_ia_j}{b_kb_l}$ is irreducible for all $i,j,k,l$.

I proved the case where $i=j, k=l$ using the fact that the square root of any non-perfect square is irrational.

I can't prove the more general case though.

EDIT: not just prime factors, just all the integer factors. E.g. 12: 1,2,3,4,6,12

Answer 1

Alternative approach : proof by contradiction.

Suppose that $\frac{a_i a_j}{b_k b_l}$ is not irreducible.
Then $\exists \;$ prime $\;p \;\ni p|(a_i \times a_j)$ and $p|(b_k \times b_l) \Rightarrow$
$p|a$ and $p|b \Rightarrow \frac{a}{b}$ is not irreducible.

Answer 2

If a prime number $p$ divides both $a_i a_j$ and $b_k b_l$, then it divides $a_i$ or $a_j$ hence $a$, and it also divides $b_k$ or $b_l$ hence $b$. This is impossible hence there is no such prime number and $\frac{a_i a_j}{b_k b_l}$ is irreductible.

Answer 3

If $\frac{a_ia_j}{b_kb_l}$ were reducible then there would exist a prime number $p\in\mathbb{N}$ such that

$a_ia_j=p\cdot\alpha\;$,

$b_kb_l=p\cdot\beta\;$.

So the prime number $p$ would be a factor of $a_i$ or $a_j$ and it would also be a factor of $b_k$ or $b_l$.

Consequently $p$ would be a factor of $a$ and a factor of $b$ too.

For this reason the fraction $\frac{a}{b}$ would not be irreducible and it is a contradiction.

So it is impossible that $\frac{a_ia_j}{b_kb_l}$ is reducible and it means that $\frac{a_ia_j}{b_kb_l}$ is irreducible.