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How to prove $$\left\{\frac{p^a}{q^b}:p,q\in\mathbb{Z}\right\}=\left\{\frac{j^{\gcd(a,b)}}{k^{\gcd(a,b)}}:j,k\in\mathbb{Z}\right\}$$ for $a,b\in\mathbb{N}$?

First, is this identity correct?

For example if $a=1$

$$\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}=\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}$$

If we set $j=p$ and $k=q$

$$\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}=\left\{\frac{j(k^{b-1})}{k(k^{b-1})}\right\}=\left\{\frac{j}{k^b}:j,k\in\mathbb{Z}\right\}=\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}$$

However, this is not a formal way of proving this case for $a=1$. I have to show

$$\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}\subseteq\left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\} \land \left\{\frac{j}{k}:j,k\in\mathbb{Z}\right\}\subseteq\left\{\frac{p}{q^b}:p,q\in\mathbb{Z}\right\}$$

How do I do this for $a=1$? How do I do this for $a\in\mathbb{N}$?

Vsotvep
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Arbuja
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  • Any ideas? Is my Identity correct? – Arbuja Sep 15 '19 at 11:17
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    Perhaps using the right tags would help. This looks like number theory, and has nothing to do with set-theory (apart from borrowing notation), proof-theory, logic or real-analysis. – Vsotvep Sep 15 '19 at 12:41
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    My first approach would be to take arbitrary $a,b\in\Bbb N$ and $p,q\in\Bbb Z$, and try to find a suitable $j,k\in\Bbb Z$ such that ${p^a}\cdot k^{gcd(a,b)}={q^a}\cdot j^{gcd(a,b)}$. If that succeeds, you have half of the proof already. The other direction would be similar, taking arbitrary $a,b,j,k$ and finding suitable $p,q$. (I don't know if it will work, though) – Vsotvep Sep 15 '19 at 12:48

1 Answers1

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Yes, it is true.

The containment $\subseteq$ is easy to show.

For $\supseteq$, take any $a,b,i,j$. Let $x_1,x_2,y_1,y_2$ be such that $ax_1+by_1=ax_2+by_2=d=\operatorname{gcd}(a,b)$ and $x_1,y_2\geq 0\geq y_1,x_2$. Take $p=i^{x_1}j^{-x_2}$, $q=i^{-y_1}j^{y_2}$. Then $p^aq^{-b}=i^{ax_1+by_1}j^{-ax_2-by_2}=i^dj^{-d}$ and you are done.

tomasz
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  • The reason I asked this question was because $\left|\left{\frac{j^{\gcd(a,b)}}{k^{\gcd(a,b)}}:j,k\in\mathbb{Z},\gcd\left(j^{\gcd(a,b)},k^{\gcd(a,b)}\right)=1,k^{\left(\gcd(a,b)\right)}\le n\right}\right|$ can be converted into an asymptotic series using the answer here , I wanted to use this as to solve this question – Arbuja Sep 15 '19 at 18:58
  • I meant (https://math.stackexchange.com/questions/3357690/how-to-calculate-the-density-of-any-subset-of-mathbbq-using-a-particular-fo)[this] – Arbuja Sep 15 '19 at 20:45
  • Is my post clear? Any ideas? – Arbuja Sep 16 '19 at 20:27
  • My apologies, I had to take back the acceptnace mark. Just to check, if $i=7$, $j=9$, $d=1$, $a=1$ and $b=5$ what are all possible values of $p$ and $q$? – Arbuja Sep 22 '19 at 22:57
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    You did not ask for all possible values in your question, so I don't see why would you take back the acceptance mark... Anyway, in this case, this is easy. Since $a=1$, the possible $p$ and $q$ are given by $p=7\cdot 27\cdot c^5, q=3c$ where $c$ is a nonzero integer. In general, it is very similar. Given any $p_0,q_0$, you can generate all of them as $p(c)=\pm p_0\cdot c^b$, $q(c)=\pm q_0\cdot c^a$. – tomasz Sep 23 '19 at 11:58
  • How do we determine $p_0$ and $q_0$ for any fixed $i$, $j$, $a$ and $b$? Do we just search or can we do this mathematically. – Arbuja Sep 23 '19 at 13:58
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    There seems to be a typo. I think you meant $p=i^xj^{-y},\ q=i^{-y}j^x$. I don't see how you get $p,q\in \mathbb{Z}$ though. – saulspatz Sep 23 '19 at 15:25
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    Oh wait. I guess this was supposed to be the case $y\leq0\leq x$. – saulspatz Sep 23 '19 at 15:33
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    @saulspatz: You're right. There was a mistake: you can't just take one pair of $x,y$, you actually need two. – tomasz Sep 23 '19 at 22:04