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?
 A: Isn't this relation true for all $x,y$ with the same sign? Since $3^k$, $k$ in reals, can take any positive real number?
A: 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.
A: My short answer:
Reflexive: $k = 0$.
Symmetric: negate $k$ - $y = 3^k x \Longleftrightarrow x = 3^{-k} y$.
Transitive:  $y = 3^k x$ and $z = 3^h y \Longrightarrow z = 3^{k+h} x$.
I don't see why there is anything more to it than this.
A: You are misunderstanding what needs to be shown in order to show that something is an equivalence relation. None of the arguments you present is a valid argument for the proposition you are trying to establish.
Your equivalence relation is defined on the set of positive integers. So if you write $x\sim y$, you are already assuming that $x$ and $y$ are positive integers, there is no need to prove that they are.
The relation is defined as follows: if $x$ and $y$ are positive integers, then
$$x\sim y\Longleftrightarrow \text{there exists a real number }k\text{ such that }y = 3^kx.$$
That means that in order to show that two given numbers $x$ and $y$ are related, then you need to produce a real number $k$ that witnesses the identity $y=3^kx$. For example, to show that $y=9$ and $x=3$ are related, I just need to say: "take $k=1$. Then $9 = 3^1\cdot 3$; that is, $y = 3^1x$, so $x\sim y$ holds." To show that $18\sim 6$, I say "take $k=-1$; then $6 = 3^{-1}(18)$ holds." Etc.
And if you know that $x\sim y$, then you know that there exists a real number $k$ such that $y=3^kx$.
Thus, to show that $\sim$ is reflexive, you need to show that given any positive integer $x$, you can find a real number $k$ (which may depend on $x$) such that $x = 3^k x$. You need to say who $k$ is. Your argument about "being positive when $k\gt 0$" doesn't get you there in any way.  
To  show that $\sim$ is symmetric, you have to show that if you already know that $x\sim y$, so that you know there is a real number $k$ such that $y=3^k x$, then you can find some real number $\ell$ such that $x=3^{\ell}y$. This will witness the fact that $y\sim x$ holds, showing symmetry. You already know that $x$ and $y$ are positive integers. You know that simply because you know that $x\sim y$ holds, and that means that $x$ and $y$ have to be positive integers in the first place.
To show that $\sim$ is transitive, you have to show that if you already know $x\sim y$ and $y\sim z$, then you can exhibit a real number $r$ such that $z=3^rx$; this will witness $x\sim z$. You know there is a real number $k$ such that $y=3^kx$ (because $x\sim y$); and you know there is a real number $\ell$ such that $z=3^{\ell}y$ (because $y\sim z$); now you need to produce that number $r$ somehow. Again, you already know that $x$, $y$, and $z$ are positive integers, because you already know that $x\sim y$ and $y\sim z$ are true, which means, inter alia, that $x$, $y$, and $z$ are all positive integers.
