It is well known that √2 is irrational, and by modifying the proof (replacing ‘even’ with ‘divisible by 3‘), one can prove that √3 is irrational, as well. On the other hand, clearly √n2=n for any positive integer n. It seems that any positive integer has a square root that is either an integer or irrational number.

- How do we prove that if a∈N, then √a is an integer or an irrational number?
I also notice that I can modify the proof that √2 is irrational to prove that 3√2,4√2,⋯ are all irrational. This suggests we can extend the previous result to other radicals.

- Can we extend 1? That is, can we show that for any a,b∈N, a1/b is either an integer or irrational?

**Answer**

**Theorem**: If a and b are positive integers, then a1/b is either irrational or an integer.

If a1/b=x/y where y does not divide x, then a=(a1/b)b=xb/yb is not an integer (since yb does not divide xb), giving a contradiction.

I subsequently found a variant of this proof on Wikipedia, under Proof by unique factorization.

The bracketed claim is proved below.

**Lemma**: If y does not divide x, then yb does not divide xb.

Unique prime factorisation implies that there exists a prime p and positive integer t such that pt divides y while pt does not divide x. Therefore pbt divides yb while pbt does not divide xb (since otherwise pt would divide x). Hence yb does not divide xb.

[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]

**Attribution***Source : Link , Question Author : Community , Answer Author : Cloudscape*