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I tried proving the same way as the $\sqrt2$ is irrational. First I did, $2^{\frac{1}{m}}= \frac{p}{q}$ then $2= (\frac{p}{q})^m 2q^m = p^m$ assuming that $p$ is an even number, which $p=2k p^m = (2k)^m$, so $p^m = (2k)^m = 2q^m$. From there, i'm not sure what to do next. Can someone help me? Thank you!

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    See http://mathoverflow.net/a/42519/103918 :) – 2017-02-02

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It's nearly the same:

$\sqrt[m]{2}=\frac ab\\\implies2=\frac{a^m}{b^m}\\\implies2b^m=a^m\\\implies a=2k\\\implies2b^m=(2k)^m\ \leftarrow\text{ you got to this step}\\\implies2b^m=2^mk^m\\\implies b^m=2^{m-1}k^m\\\implies b=2n$

Hence, contradiction, since $a$ and $b$ must be coprime.

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    Thank you! Quick question , from the 2nd to last steps, why is the exponent 2^(m-1), and how did it go from 2^m-1 k^m to 2n? – 2017-02-02
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    @C.Chan I divided both sides by $2$.$$2b^m=2^mk^m\implies b^m=2^mk^m/2=2^{m-1}k^m$$and note that there are factors of $2$ on the right, so there must be factors of $2$ on the left. – 2017-02-02
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    oh okay! I got it! Thank you!! (: – 2017-02-02
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    No problem! :-) – 2017-02-02
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    Of course this crucial steps uses $m>1$ (perhaps too silently) – 2017-02-02
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    @HagenvonEitzen (perhaps the problem also asks specifically for $m\ge2$, perhaps also too silently?) – 2017-02-02
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Here is a mildly complicated proof.

If the equation $X^m-2=0$ has a rational root, then it should in fact be an integer root. Then it must be possible that a positive integer root to exits. The only positive integer less than 2 is 1 which is not a root of this equation. For integers $k\ge 2$ we have $k^m>2$ so they are not roots. So there are no integer roots, and hence no rational root either.

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    You are suggesting the rational roots theorem? – 2017-02-02
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    Yes. This essentially assumes your argument. – 2017-02-02
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I think there is another approach which is simpler. It is based on the idea that if integers $a, b$ have no common factor then the integers $a^{m}, b^{n} $ have no common factor for all positive integers $m, n$ (this can be established easily using unique prime factorization of integers).

Now suppose that $2^{1/m}$ is rational say $a/b$ where $a, b$ have no common factor and $b>1$. Then $a^{m}/b^{m}=2$ or $a^{m}/b^{m - 1}= 2b$. Now $a^{m}, b^{m-1}$ have no common factor and $b>1$ hence the LHS is a fraction and RHS ($2b$) is an integer. This contradiction shows that $2^{1/m}$ is irrational.

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    Isn't this essentially my answer? – 2017-02-02
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    @SimplyBeautifulArt: no, this is a fundamentally different answer and in my opinion simpler to follow. Your argument tries to obtain contradiction by showing that $a, b$ are not coprime. I try to show that a fraction is equal to an integer. – 2017-02-02
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    @SimplyBeautifulArt: the approach in my answer can be used to prove a general result that if $x$ is a real root of an equation $x^{n} +a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}=0$ where $a_{1},a_{2},\ldots,a_{n}$ are integers then either $x$ is an integer (and a factor of $a_{n}$) or $x$ is irrational. – 2017-02-02
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    Hm, ok then :-) Would you say its a mix between my answer and PVanchinathan's answer or something like that? – 2017-02-02
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    @SimplyBeautifulArt: pvanchinathan's answer uses the result I mentioned in my comments and the technique in this answer is used to prove it. – 2017-02-02