Prove transitivity or not of some relation

Question

I'm trying to prove if this equation is an equivalence relation or not.

$R =\{(x,y) \in N\times N: \mbox{There exist }m,n \in N\mbox{ such that } x^m = y^n\}$

It's relatively easy to prove both reflexivity and symmetry of this relation. But I am unsure how to prove transitivity. I was thinking that $y^n = x^ac, a,c\in N$ but am unsure how to translate this into a solution.

Answer 1

$xRy \iff x^n = y^m$ and $yRz \iff y^s = z^t$ for natural $s,t,n,m$. Thus take $$ (y^s)^m = (y^m)^s = (x^n)^s = x^{ns}\: \text{and} \: (y^s)^m = (z^t)^m = z^{tm} $$ which gives $x^a = z^b$ for natural $a=ns, b = tm$, so $xRz$.

Answer 2

If $(x,y),(y,z)\in R$ implies: $$x^n=y^m\quad and \quad y^\hat{n}=z^\hat{m}$$ This will lead to: $$x^{n*\hat{n}} = y^{m * \hat{n}}=x^{m* \hat{m}}$$ Just a sketch, but I bet you can proof everything neccessary.

Answer 3

Suppose $x \sim y $ such that $x^a = y^b$ and $y \sim z$ such that $y^c = z^d$.

Then $(y^c)^b = y^{cb} = (z^d)^b = z^{db} \rightarrow (x^a)^c = x^{ac} = z^{db}$. This proves that $x \sim z$.

Edit: This relation can be seen after doing a few examples like on $3 \sim 9$ and $9 \sim 81$:

$3^2 = 9^1$ and $9^2 = 81^1$ such that $3^4 = (3^2)^2 = 9^2 = 81^2$ which shows $3 \sim 81$.