How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well. On the other hand, clearly $\sqrt{n^2} = n$ 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 $a \in \mathbb N$, then $\sqrt a$ is an integer or an irrational number?
I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ 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 $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
Solution 1:
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)]
Solution 2:
These (standard) results are discussed in detail in
http://math.uga.edu/~pete/4400irrationals.pdf
This is the second handout for a first course in number theory at the advanced undergraduate level. Three different proofs are discussed:
1) A generalization of the proof of irrationality of $\sqrt{2}$, using the decomposition of any positive integer into a perfect $k$th power times a $k$th power-free integer, followed by Euclid's Lemma. (For some reason, I don't give all the details of this proof. Maybe I should...)
2) A proof using the functions $\operatorname{ord}_p$, very much along the lines of the one Carl Mummert mentions in his answer.
3) A proof by establishing that the ring of integers is integrally closed. This is done directly from unique factorization, but afterwards I mention that it is a special case of the Rational Roots Theorem.
Let me also remark that every proof I have ever seen of this fact uses the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) in some form. [Edit: I have now seen Robin Chapman's answer to the question, so this is no longer quite true.] However, if you want to prove any particular case of the result, you can use a brute force case-by-case analysis that avoids FTA.