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Let $a_1...a_k$ be positive integers that are pairwise relatively prime. Assume that $\sqrt[m]{a_1...a_k}\in\mathbb{N}$ for some $m\in\mathbb{N}$. Show that $\sqrt[m]{a_i}\in\mathbb{N}$ for each $i$.

I want to know if this proof is correct. I have trouble with constructing rigorous proofs.

So I know that for root m to be an element of $\mathbb{N}$, $a_1...a_k$ must have m copies of any natural number, say $s\in\mathbb{N}$. However, since each $a_i$ must be pairwise relatively prime, each $a_i$ cannot share any common factors, this includes s. But if any $a_i$ does not have a factor of s, then it cannot be taken out of the root which would produce a product $a_1...a_k\notin\mathbb{N}$. So each $a_i$ must be reducible by root m.

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I am new to such proofs as well, and though I see the logic that you are employing I think most such proofs are usually done by induction when trying to be fully rigorous. just my 2 cents. –  MSEoris Dec 16 '12 at 16:20
    
My teacher has actually told me to try and not use induction in number theory questions. However, maybe this is a special case where induction works? So proposition: $\sqrt[m]{a_i}\in\mathbb{N}$ for each $i$. And then just perform induction? –  Gbean Dec 16 '12 at 16:55

1 Answer 1

up vote 1 down vote accepted

Your proof is correct. Here is another way to write things.

Claim: For each $j$, if $p|a_j$ is a prime, then $p^{lm}$ completely divides $a_j$ for some integer $l$.

Proof: Suppose that $p|a_j$. Let $l$ be the largest integer such that $p^{lm}$ divides $a_1...a_k$. Since $\sqrt[m]{a_1...a_k}\in\mathbb{N}$, it follows that $p^{ml+1}$ does not divide the product of the $a_i$. However, since the $a_i$ are pairwise relatively prime, we must have $p^{ml}|a_j$, proving the claim.

From the claim, it follows for each $j$, each prime in the factorization of $a_j$ must appear to a power which is a multiple of $m$. Hence $\sqrt[m]{a_j}\in\mathbb{N}$.

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