Thread: proving √q is irrational by contradiction

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!

Originally Posted by junebug5389

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.

That's right, just do it: your equation shows that divides , hence divides . Write to conclude that it divides as well.

maybe im missing a simple definition but i was not sure how to show since q|m^2 that q|m. or is it just implied that since q|m^2 that q|m and thats all i have to say. sorry this might be a stupid question...

Originally Posted by junebug5389

maybe im missing a simple definition but i was not sure how to show since q|m^2 that q|m.

This is because of your assumption on (with , that's why I wrote ).

By the way, this assumption is equivalent to saying that is prime.

Originally Posted by junebug5389

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.

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.