# Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$

Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}|q|=1$.

I don't have a lot of experience working with infinite products, but I read a couple of theorems that say that absolute convergence of infinite products requires that $\prod1+|a_n|$ converges, and that $\prod1+|a_n|$ converges iff $\sum a_n$ converges.

Now $\sum_{q\in \mathbb{Q}^{\times}}|q|$ certainly does not converge, implying that my original product is not absolutely convergent. Which leaves me with the problem of being unable to rearrange its terms. But since I was never given an enumeration of my rationals to begin with, I'm a bit vexed as to how I should proceed.

Here is the link to the problem: homework 2 (problem 2). I'm not in the class, just doing the homeworks. I'm doing it for the $\mid \cdot \mid_{\infty}$ absolute value. Which is supposed to be just the normal absolute value (according to homework 1).

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Append \displaystyle to the back of the infinite product to make it look bigger and better. – Git Gud Feb 22 '13 at 7:02
@git gud thanks good tip – Thoth Feb 22 '13 at 7:03
No problem. It works with sums (which you used) and limits too. – Git Gud Feb 22 '13 at 7:04

You've misstated the problem. For all $x\in\mathbb{Q}^\times$, $$\prod_{p\text{ prime or }\infty}|x|_p$$ is an infinite product all but finitely many terms of which are 1, so it certainly converges.

Hint: Note that, if $x=p_1^{a_1}\cdots p_n^{a_n}$ where the $p_i$ are primes and $a_i\in\mathbb{Z}$, then $$|x|_{p_i}=\frac{1}{p_i^{a_i}}$$ and that $|x|_p=1$ for all other primes.

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Ohh ok! I get it, so it's for a fixed rational, and it's the absolute value that is running through all primes, is that correct? – Thoth Feb 22 '13 at 7:13
Yup, that's exactly right (as long as you mean a fixed nonzero rational)! – Stahl Feb 22 '13 at 7:13
I do! =] ok thanks everyone for the quick responses. – Thoth Feb 22 '13 at 7:14
@Electro: That's correct, it is a product over the different $p$-adic absolute values together with the usual one. – Zev Chonoles Feb 22 '13 at 7:14

It looks like the problem is that you're misinterpreting the problem. You don't want to show that the product over all elements of $\mathbb{Q}$ is $1$, you want to show that when you take a given element $x\in\mathbb{Q}^{\times}$ and multiply it's absolute value for all possible absolute values on $\mathbb{Q}$, you get $1$. For a hint, look at Zev's answer (considering the prime factorization of $x$).

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