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How would you prove that $\sqrt[n]{2}$ is irrational?, where $n \in \{2, 3, 4, \ldots\}$.

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marked as duplicate by Daniel Fischer Oct 18 '15 at 15:45

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up vote 4 down vote accepted

You have probably already seen the proof for $n=2$. So, let's assume that $n \geq 3$.

Seeking a contradiction, suppose that $\sqrt[n]{2}$ is rational. This, together with the positivity of $\sqrt[n]{2}$ imply that there exist $p,q \in \mathbb{N}$ such that $\sqrt[n]{2}=p/q$. Raising both sides to the $n^{th}$ power, we see that $$2=\frac{p^n}{q^n}.$$ Multiplying through by $q^n$ and using $2q^n = q^n + q^n$, we have $$q^n + q^n = p^n. $$ This violates Fermat's Last Theorem, giving a contradiction. Thus $\sqrt[n]{2}$ is irrational.

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I wanted to post this as well, saw it last night on 4chan, haha. – Ruben May 6 '14 at 6:23
This is very neat. – zscoder May 6 '14 at 8:22

By the rational root theorem, all rational roots of $x^n - 2$ are one of $\{ \pm 1 , \pm 2 \}$. It's easy to verify that none of these work if $n \ge 2$.

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It is very neat, but it is not clear that the OP has the rational root theorem at the time this question is asked. – Ross Millikan May 6 '14 at 4:07

If you’re willing to use the Fundamental Theorem of Arithmetic, which says that there’s essentially only one way to write an integer as a product of primes, then the proof drops right out.

For, if $2^{1/n}=p/q$ for any integers $p$ and $q$, take the $n$-th power and clear of fractions to get $p^n=2q^n$. How many $2$’s do you see on the left? A number that’s divisible by $n$. How many on the right? A number that’s $1$ more than a multiple of $n$. And there’s your contradiction.

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Exactly the same way that you prove $\sqrt{2}$ is irrational -- the standard proof generalizes perfectly.

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I would argue that the polynomial $$ p(x)=x^{n}-2 $$

is irreducible over $\mathbb{Q}$ using Eisenstein criterion and in particular it does not have a root in $\mathbb{Q}$

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Can the downvoter please say what is wrong with this answer ? – Belgi May 6 '14 at 20:14
For some reason I had a moment where I thought irreducible $\implies$ no rational root was false, and was too lazy to comment. Sorry. If you'd like to edit your answer I'd be glad to undo the downvote. – Jack M May 6 '14 at 23:02

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