Prove: The relation $R$ on $\mathbb{N}$ given by $mRn$ iff there are natural numbers $p$, $q$ with $m^p$ = $n^q$ is reflexive, symmetric and transitive.
Proving $R$ is reflexive:
Proof. Suppose $m$ ∈ $\mathbb{N}$. Then $m^p$ = $m^p$. Therefore $m R m$. So $R$ is reflexive
Proving $R$ is symmetric:
Proof. Suppose m, n ∈ $\mathbb{N}$ and $mRn$. Let $m=2$ and $n=2$. Also let $p=1$ and $q=1$. Then $m^p$ = $n^q$ $\implies$ $2^1$ = $2^1$. And we can also say that $n^q$ = $m^p$ $\implies$ $2^1$ = $2^1$. So $nRm$. So $R$ is symmetric.
Update on symmetric proof: (since I must show the general case)
Proof. Suppose m, n ∈ $\mathbb{N}$ and $mRn$. Then $m^p$ = $n^q$. And we can also say that $n^q$ = $m^p$. So $nRm$. So $R$ is symmetric.
- Is proof for reflexive case correct?
- Is proof for symmetric case correct?
Your argument for reflexivity is correct, though you really ought either to pick a particular $p$ — $1$ would be the obvious choice — or point out that $m^p=m^p$ for all $p\in\Bbb Z^+$. Your argument for symmetry is not: to show that $R$ is symmetric, you must show for every pair of $m$ and $n$ that if $m\mathrel{R}n$, then it’s also true that $n\mathrel{R}m$. In other words, you must show that if there are positive integers $p$ and $q$ such that $m^p=n^q$, then there are also positive integers $r$ and $s$ such that $n^r=m^s$. This is pretty trivial, but you have to do it in general, for arbitrary $m$ and $n$, not for one specific pair of $m$ and $n$.