Prove that if p is not divisible by 5 then p2 is not divisib

Prove that if p is not divisible by 5 then p^2 is not divisible by 5. Prove that Squareroot 5 epsilon R Q (i.e. is irrational )

Let, p=r_1^a1..r_k^ak

r_1,..,r_k are the prime factors of p

Hence, p^2=r_1^2a1..r_k^2ak

So, if 5|p^2 then one of r1,...,rk must be equal to 5

Let, r1=5 without loss of generality

2a1>=2

Hence, a1>=1

Hence, 5|p

We proved here the contrapositive which equivalent to the given statement

Let, sqrt{5} belong to Q

So there exist , p,q in Q so that gcd(p,q)=1

sqrt{5}=p/q

5q^2=p^2

So, 5|p^2, hence, 5|p from part a.

So, p=5m for some integer m

So,

5q^2=5^2m^2

q^2=5m^2

Hence, 5|q^2 and hence, 5|q

But, gcd(q,p)=1

So we have a contraditction

Hence, sqrt{5} belongs to R-Q