Prove $\sim$ is an equivalence relation: $x \sim y$ if and only if $y = 3^kx$, where $k$ is a real number The relation $\sim$ is defined on $\mathbb{Z}^+$ (all positive integers). We say $x\sim y$ if and only if $y=3^kx$ for some real number $k$.
I need to prove that $\sim$ is an equivalence relation.
To prove an equivalence relation we must certify that:


*

*It is reflexive

*It is symmetric

*It is transitive


I am not sure how to start this off since if I want to prove for:
Reflexive: I would replace $y$ with $x$, so that $x = (3^k)*x$ which is a positive integer when $k > 0$. Thus $x \sim x$ and $\sim$ is reflexive (?).
Symmetric: Suppose $x \sim y$ so that $y = (3^k)*x$ then $x = (3^k)*y$ is also an element of $\mathbb{Z}^+$. Thus $y \sim x$ and $\sim$ is symmetric (?).
Transitive: I'm not too sure about this last one. I think my entire proof is wrong anyways.
Can anyone help me on this?

Put $\rm\: G = \{ 3^r\: :\: r\in \mathbb R\}\:.\:$ It's reflexive by $\rm\: x/x = 1\in G\:.\:$ It's symmetric by $\rm\:G\:$ is closed under inverses $\rm\ x/y \in G\ \Rightarrow\ y/x\in G\:.\ $ It's transitive by $\rm\:G\:$ is closed under products $\rm\ x/y,\ y/z \in G\ \Rightarrow\ x/z \in G\:.\:$  
Therefore the proof requires only that $\rm\:G\:$ contains $1\:$ and is closed under products and inverses or, equivalently, that $\rm\:G\:$ is nonempty and $\rm\:G\:$ is closed under quotients, i.e. $\rm\:G\:$ is a commutative group. This equivalence relation (congruence) will come to the fore when you study quotient groups.
