Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
$a^{1/2}$ is either an integer or an irrational number.

Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?

share|cite|improve this question

marked as duplicate by Arturo Magidin, Qiaochu Yuan, J. M., Aryabhata, Pete L. Clark Nov 26 '10 at 20:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Hint: Fermat's last theorem comes to mind when you want to prove it for n>2. Then proof it for n=2 and you got it. – Max Muller Nov 25 '10 at 20:39
@Max: pretty much the worst way to do this problem... – Qiaochu Yuan Nov 26 '10 at 1:09
@Qiaochu: I believe there was a mention of that in the "mosquito-nuking" thread in MO... – J. M. Nov 26 '10 at 1:11
I think that the most important thing is that the mosquito, bacteria or whatever is dead because of the FLT-nuke. – Max Muller Nov 26 '10 at 17:58
Look, @Bill, we're trying to eradicate an exceedingly ambiguous tag here... I can see why you'd add abstract-algebra, but your removal of algebra-precalculus sounds iffy to me. But hey, you're the "expert", good sir. – J. M. Jan 3 '12 at 2:53

Yes. In fact, for every integer $k$ and every $n\gt 1$, the $n$th root of $k$ is either an integer or irrational.

One way to prove it is to use exactly the same idea as for proving the square root of $2$ is irrational: suppose $\sqrt[n]{k} =\frac{p}{q}$, with $p$ and $q$ integers, relatively prime. Then $q^nk = p^n$. Now think about the prime factorizations: every prime that divides $q$ must divide $p$, but $p$ and $q$ are relatively prime, so $q=1$. That means that you must have $k=p^n$ with $p$ an integer. That is, the only way for the $n$th root of $k$ to be a rational is if $k$ is an $n$th power of an integer.

Or you can use the Rational Root Test: an $n$th root of $k$ is a root of the polynomial $x^n - k$. But a rational root of a polynomial with integer coefficients that is written in lowest term $\frac{p}{q}$ must have denominator $q$ that divides the leading coefficient and numerator $q$ that divides the constant coefficient. So any rational root of $x^n-k$ must be an integer.

Getting this back to your question, since $2$ is not an $n$th power of an integer for any $n\gt 1$, $\sqrt[n]{2}$ is not a rational for any integer $n\gt 1$.

share|cite|improve this answer

If $\rm\ \sqrt[n] c = a/b,\ \ gcd(a,b) = 1\ $ then $\rm\ c\ b^n = a^n\ \Rightarrow \ b\:|\: a^n\:.\: $ But $\rm\ gcd(a,b) = 1\ \Rightarrow\ gcd(a^n,b) = 1\ $ by Euclid's Lemma. Thus $\rm b = 1\ $ hence $\rm\:\ c \ =\ a^n\:.\ $ In particular $\rm\ c = 2\ \Rightarrow\ n = 1\:.$

There are many possible variations on such irrationality proofs, e.g. using the Rational Root Test or directly using Unique Factorization of integers, or using the principality of denominator ideals. Perhaps the most elegant is to employ Dedekind's notion of the conductor ideal - which yields a one-line proof that a $\rm PID$ is integrally closed, i.e. satisfies the monic case of the rational root test.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.