If gcd(a,b)=1 then also gcd(a^2,b^2)=1.
You got: a^2|pb^2 ==> a^2|p ==>a=1 ==>1=pb^2(contradiction!)
read here to understand the first move: Euclid's lemma - Wikipedia, the free encyclopedia
Heres a proof I am getting stuck with
Prove sqrt(p) is irrational, p is set of all primes
so heres what Ihave
prove sqrt(p) is rational
sqrt(p) = a/b a,b set of integers, b not equal to 0, and a/b is in lowest form
p = a^2/b^2
p*b^2 = a^2.
a^2/p = b^2.
a/sqrt(P) = b
Kind of at a standstill here, since b cannot be rational when dividing by the sqrt of a prime. how can I say this?
If gcd(a,b)=1 then also gcd(a^2,b^2)=1.
You got: a^2|pb^2 ==> a^2|p ==>a=1 ==>1=pb^2(contradiction!)
read here to understand the first move: Euclid's lemma - Wikipedia, the free encyclopedia