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I have seen induction and the fundamental theorem of arithmetic used for this proof, but is it wrong to use this much simpler idea? $$a^n \mid b^n \: \: \therefore \:\: k = \bigg(\frac{b}{a} \bigg)^n \in \mathbb{Z} \implies \sqrt[n]{k} \in \mathbb{Z}$$ since $x^n \in \mathbb{Z} \iff x \in \mathbb{Z}$

Is there anything wrong with this idea? Is it perhaps recursive and thus invalid?

Airdish
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  • The last thing you wrote is absolutly wrong. take $x=\sqrt{2}$ and $n=2$ – 35T41 Aug 03 '16 at 14:11
  • @35T41 Sorry I should have added some more conditions, what I mean is that if $x^n \in \mathbb{Z} \implies x \in \mathbb{Z} \iff x \in \mathbb{Q} : :$. Here $\frac {b}{a}$ is clearly in $\mathbb{Q}$ – Airdish Aug 03 '16 at 14:14
  • It's true that if $x$ is rational then $x^n\in\Bbb Z$ if and only if $x\in\Bbb Z$. But that's exactly equivalent to what you're trying to prove - the proof of this fact you want to use is going to use FTA or one of the other things you say you want to avoid. – David C. Ullrich Aug 03 '16 at 14:15
  • @DavidC.Ullrich Isn't there a separate proof of that fact? I assumed that this problem was a corollary. – Airdish Aug 03 '16 at 14:16
  • I guess I should clarify. When I said the two were exactly equivalent what I meant was this: The two are exactly equivalent. You see how one of them implies the other; the converse is just as easy. Think about it. – David C. Ullrich Aug 03 '16 at 14:18
  • @DavidC.Ullrich I understand what you mean. What I meant was that I thought there was already an age-old proposition of the fact based on dividing rational numbers into an integer part and a decimal part. – Airdish Aug 03 '16 at 14:20
  • I'm clearly wrong. Forgive me, I'm not very well-versed in these things. – Airdish Aug 03 '16 at 14:21
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    You might be interested to know that in algebraic number theory there are examples of number rings in which the implication $a^n | b^n \implies a | b$ fails. Of course such number rings do not satisfy the fundamental theorem of arithmetic. So, something deeper is going on. – Lee Mosher Aug 03 '16 at 14:22
  • @LeeMosher That certainly is intriguing. I've not yet delved into algebraic number theory, but I'd love to once I have more experience. – Airdish Aug 03 '16 at 14:25

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You have reformulated the divisibility statement into an equivalent fractional form. Although the fractional form of the statement may seem to be intuitively clearer (even obvious), it does require rigorous proof, since it is not true in all rings. For example, in the ring $\, R = \Bbb Z[\sqrt{12}] = \{j + k\sqrt{12}: j,k\in\Bbb Z\}\,$ we have $\,x^2 = (\sqrt{12}/2)^2 = 3\in R\,$ but $\,x\not\in R.\ $

One simple way to proceed is to note that $\,x\in\Bbb Q,\,\ x^n\in\Bbb Z\,\Rightarrow\, x\in\Bbb Z\,$ is an immediate corollary of the Rational Root Test, so you can use that if you have it available. Alternatively you can prove it directly using the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations, i.e. $\,\Bbb Z\,$ is a UFD), or various closely related properties such as Euclid's Lemma, the Bezout identity for the gcd, various gcd laws, etc.

Community
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Bill Dubuque
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  • Ah, so I'm guessing historically and from the ground-up my idea is circular reasoning. Thank you for your answer. – Airdish Aug 03 '16 at 17:33
  • @Airdish It's not circular as long as you do prove the fractional equivalent, e.g. using the rational root theorem. It is often useful to transform problems into equivalent forms that are easier to prove. – Bill Dubuque Aug 03 '16 at 17:57