Proving Is mth root of 2 an irrational number for every integer $m\ge 2$?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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.