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Is my proof beneath perfect and complete?

I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: $$\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.$$

I start by assuming that $m^{\frac 1n}$ is rational and non-integer. So there exist co-prime integers $a,b$ so that $$\sqrt[n]{m}=\frac{a}{b}$$ $$\implies m=\frac{a^n}{b^n}\in\mathbb{N}.$$ But since $a$ and $b$ have no common factor, $a^n$ and $b^n$ also have no common factor. So: $$\frac{a^n}{b^n}\notin\mathbb{N},$$ a contradiction.

76david76
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  • Yes, that is correct. You should just have started with "assume that it is rational and non-integer", so $b\neq 1$. Or you can prove directly that if a $n$-th root of an integer is rational, then it is an integer. – Luiz Cordeiro Sep 19 '16 at 09:53
  • I edited like you said, thanks. – 76david76 Sep 19 '16 at 09:57

2 Answers2

3

Your proof is fine. You can use essentially the same idea to prove the following more general statement:

Theorem. If $ P(X) \in \mathbf Z[X] $ is a monic polynomial, then any rational roots of $ P $ are integers. In other words, $ \mathbf Z $ is integrally closed.

Proof. Assume that $ q = a/b $ is a rational root with $ a, b $ coprime, and let $ P(X) = X^n + c_{n-1} X^{n-1} + \ldots + c_0 $. We have $ P(q) = 0 $, which gives

$$ a^n + c_{n-1} a^{n-1} b + \ldots + c_0 b^n = 0 $$

In other words, $ a^n $ is divisible by $ b $. This is a contradiction unless $ b = \pm 1 $, since then any prime dividing $ b $ also divides $ a $, contradicting coprimality. Hence, $ b = \pm 1 $ and $ q \in \mathbf Z $.

Ege Erdil
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1

It is basically right. However, in the beginning, $\sqrt[n]{m}$ shoulg make sense, i.e., it should be in $R$. For example, $\sqrt[2]{-2} \notin R$, then you proof is not proper.

Paul
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