# Prove that the square of an irreducible fraction is irreducible

• Sep 10th 2010, 07:31 AM
uberbandgeek6
Prove that the square of an irreducible fraction is irreducible
a) Let r and s be integers >= 1, and suppose r/s is in lowest terms. Prove that (r^2)/(s^2) is in lowest terms.
b) Prove that an inter N cannot equal (r/s)^2 unless s = 1.

I understand that (a) should be true, but I'm having trouble trying to explain why. Knowing that r/s is in lowest terms means that gcd(r,s) = 1, and the fraction can be written as (r*r)/(s*s). I'm not sure that I can say if s could divide r*r, than s must divide one of the r's because I am not told that s is a prime number. I was thinking maybe I would have to break r and s up into prime factors and show that they have none in common, but I'm not sure how that would work.

Also in (b), would it be similar to proving that sqrt(2) cannot equal a rational number a/b? I'm not sure if it would work the same way, since N is a nonspecific integer.
• Sep 10th 2010, 08:01 AM
undefined
Quote:

Originally Posted by uberbandgeek6
I was thinking maybe I would have to break r and s up into prime factors and show that they have none in common, but I'm not sure how that would work.

That's what I would do.

1) there does not exist a prime p such that p | r and p | s

2) for any prime p and integer n, p | n if and only if p | n^2

3) therefore....

Quote:

Originally Posted by uberbandgeek6
Also in (b), would it be similar to proving that sqrt(2) cannot equal a rational number a/b? I'm not sure if it would work the same way, since N is a nonspecific integer.

I don't think so. It follows from part (a) in the sense that a rational m/n in lowest terms is integer iff |n| = 1.
• Sep 10th 2010, 08:05 AM
Plato
One quick comment: the prime factors of \$\displaystyle r\$ are exactly the same prime factors of \$\displaystyle r^2\$.