Show equivalence on $m^q =n$ for $\Bbb Z$

Question

Gentlemen,

May I please ask for your help.

I have this question but no idea how to solve this.

The relation $\sim$ on $\mathbb{Z}^{+}$ is given by m $\sim$ n if and only if there exists $q \in$ $\mathbb{Q}^{+}$ such that $m^{q} = n$.

(i) Show that $\sim$ is an equivalence relation on $\mathbb{Z}^{+}$

(ii) List $5$ smallest elements of the equivalence class $[9]$.

I thank you very much for your time and patience.

Answer 1

Well, its an equivalence relation:

1. Reflexivity: $m\sim m$, since for $q=1$, $m^q=m$.

2. Symmetry: If $m\sim n$, i.e., $m^q=n$ for some rational number $q>0$, then $n^{1/q}=m$ with $1/q>0$ a rational number and so $n\sim m$.

3. Transitivity: if $m\sim n$ and $n\sim k$, i.e., $m^q=n$ and $n^p=k$ for some rational numbers $p,q>0$, then $k=(m^q)^p = m^{pq}$ with $pq>0$ a rational number and so with symmetry $m\sim k$.

Answer 2

(i) Showing the three properties for an equivalence relation :

1) Reflexivity: Let $m \in \mathbb{Z}^+$, then $ m \sim m$ as $m = m^q$ with $q=1$.

2) Symmetry: Let $m,n \in \mathbb{Z}$ then if $m \sim n \Rightarrow \ \exists\ q \in \mathbb{Q}^+ $ such that $ m^q = n$, then $ n^{\frac{1}{q}} = m$ and $\frac{1}{q} \in \mathbb{Q}^+$. Hence $n \sim m$.

3) Transitivity: Let $l,m,n \in \mathbb{Z}^+$ with $l \sim m$ and $m \sim n$, then $\exists$ $p,q \in \mathbb{Q}^+$ such that $ l^p = m, m^q =n \Rightarrow (l^p)^q = l^{pq} = n$, since $pq \in \mathbb{Q}^+$, hence $l \sim n$.

(ii) The class $[9]$ will have numbers $x$ in $\mathbb{Z}^+$ such that $x \sim 9$. The smallest such numbers will be $3,9,27,81,243.$ There can be only those numbers which are some power of $3$, there should be no factor of $2$ since $2^q \ne 9 = 3^2$ for any $q \in \mathbb{Q}^+$.

Hope this helps.