# How to prove: if a,b∈Na,b \in \mathbb N, then a1/ba^{1/b} is an integer or an irrational number?

It is well known that $$√2\sqrt{2}$$ is irrational, and by modifying the proof (replacing ‘even’ with ‘divisible by $$33$$‘), one can prove that $$√3\sqrt{3}$$ is irrational, as well. On the other hand, clearly $$√n2=n\sqrt{n^2} = n$$ for any positive integer $$nn$$. It seems that any positive integer has a square root that is either an integer or irrational number.

1. How do we prove that if $$a∈Na \in \mathbb N$$, then $$√a\sqrt a$$ is an integer or an irrational number?

I also notice that I can modify the proof that $$√2\sqrt{2}$$ is irrational to prove that $$3√2,4√2,⋯\sqrt[3]{2}, \sqrt[4]{2}, \cdots$$ are all irrational. This suggests we can extend the previous result to other radicals.

1. Can we extend 1? That is, can we show that for any $$a,b∈Na, b \in \mathbb{N}$$, $$a1/ba^{1/b}$$ is either an integer or irrational?

Theorem: If $a$ and $b$ are positive integers, then $a^{1/b}$ is either irrational or an integer.

If $a^{1/b}=x/y$ where $y$ does not divide $x$, then $a=(a^{1/b})^b=x^b/y^b$ is not an integer (since $y^b$ does not divide $x^b$), 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 $y^b$ does not divide $x^b$.

Unique prime factorisation implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$. Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^b$ (since otherwise $p^t$ would divide $x$). Hence $y^b$ does not divide $x^b$.

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