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.

Mathematics Asked by Simplex1 on October 15, 2020

$$a,bin mathbb{Z}$$

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

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.

Correct answer by user2661923 on October 15, 2020

If $$frac{a_ia_j}{b_kb_l}$$ were reducible then there would exist a prime number $$pinmathbb{N}$$ such that

$$a_ia_j=pcdotalpha;$$,

$$b_kb_l=pcdotbeta;$$.

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.

Answered by Angelo on October 15, 2020

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.

Answered by Gribouillis on October 15, 2020

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