(for fun) Any non-perfect square has an irrational 2nd root

[1MileCrash]

Homework Statement

I'm trying to see if I can prove that any non-square number's square root is irrational. I'm using only what I already know how to do ( I like trying to prove things myself before looking up the best proof), so it's going to be round-about.

Attempt#1 Eventually required me to proving something that was really the equivalent of the theorem itself, so I deleted it.

Attempt #2:

Let K be a number that is not a perfect square. Assume that sqrt(K) is rational. Then there exists integers n,l such that sqrt(K) = n/l

So
K = n^2/l^2
Kl^2 = n^2

However, since K is not a perfect square, its prime factorization must have at least one prime that occurs an odd number of times (otherwise, you could separate its prime factors into two identical groups, making it a perfect square). Since l^2 is a perfect square, it is guaranteed not to have any primes that occur an odd number of times it its prime factorization. Therefore, the prime factorization of kl^2 still has at least one prime that occurs an odd number of times, namely, the same one(s) that occur in K, since an odd plus an even is always odd.

However, we have noted that it is equal to the perfect square n^2. But since n^2 is a perfect square, it cannot have any factors in its prime factorization that occur an odd number of times, yet we have shown that must have at least one factor in its prime factorization that occurs an odd number of times. This is a contradiction, so our assumption that sqrt(K) is rational cannot be true.

Therefore, sqrt(K) is irrational where K is a non-perfect square.

QED

Valid?

 

[micromass]

Yes, it's valid. I would like to emphasize that it is crucial here that the prime factorization is unique.