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So I have a small question about the following proof,

Prove that $\sqrt [m]{2}$ irrational for every integer $m\ge2$

So far this is what I have,

Assume towards contradiction, $\sqrt [m]{2}=p/q$ for $m≥2$. Then $2=\frac{p^m}{q^m}$ so $2q^m=p^m$. Then $p^m$ is even so $p$ is even such that $p=2k$ for some integer, $k$. Then $2q^m=p^m=(2k)^m$. So $q^m=2^{m-1}(k)^m$.

For the next step I was wondering if I needed a lemma that proves that $2^{m-1}$ is a multiple of $2$ before I can state that $q^m$ is even?

I would appreciate it if anyone could clarify this for me. Thank you!

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    If $m\in\mathbb N_{\ge2}$, then by the definition of exponentiation, I think one would have $2^{m-1}$ is a multiple of $2$. – 2017-02-03
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    Okay! I see. Thank you! – 2017-02-03

3 Answers 3

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$2^{m-1}$ is a multiple of $2$ for $m\geq 2$. You don't need to prove this, it follows directly from the rules of exponentiation.

$2^{m-1}=2 \times 2^{m-2}$ and since $m \geq 2$ we know that $m-2 \geq 0$. For $m=2$ the term $2^{2-2}=1$ but $2 \times 1$ is still a multiple of $2$. The result for $m>2$ follows even easier than this.

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    The last line: "The result for $m>2$ follows ***even*** easier than this." :D – 2017-02-03
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Assume there is a rational number, a/b, such that (a/b)^m=2. This can be rewritten as (a^m)/(b^m)=2. Multiplying both sides by b^m gives us a^m=2(b^m), or, a^m=b^m+b^m, which is actually Fermat's Last Theorem which has been proven to have no integer solutions. So there are no integers a and b such that (a/b)^m=2, meaning that a/b is irrational.

This proof is taken from a Numberphile video

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For a prime $\ell$ and an equation $\ell q^m=p^m$ with $m\ge2$, there’s an immediate contradiction to the uniqueness part of the Fundamental Theorem of Arithmetic.

Because: count how many appearances of the prime $\ell$ there are on the two sides of the equation: divisible by $m$ on the right, indivisible by $m$ on the left. So, a contradiction.