Let q≥2 be a positive integer. If for all integers a and b, whenever q|ab, q|a or q|b, then √q is irrational.
I know I should do a proof by contradiction so i should assume √q is rational, and that for all integers a and b if q|ab then q|a or q|b.
So since √q is rational it can be written as m/n where m and n are integer and are relatively prime.
So √q=m/n. Squaring both sides q=m^2/n^2. Then q(n^2)=m^2.
I got stuck at this point. I think im suppose to show q|m and q|n to show a contradiction but im not sure how to get there.
Anything helps!! thanks!
Proof by contradiction is only needed partly.
If the premise holds, i.e., if for all a and b: q | ab => q|a or q|b, then q is a prime. For suppose that q is not prime; then you can take a as one of the prime factors of q, and b = q/a.
From that, it's trivial to see √q is irrational - the proof for any prime q is the same as the classical proof that √2 is irrational.