How to approve that relation is equivalence relation or partial order relation? Given set ${A = N \setminus \{0\}}$ and defined relation R, so that for each ${xRy, x, y \in A}$ just if exist natural number $i \gt 0$ such as ${\frac{y}{x} = 2^i}$ 
The task is: Prove that relation - equivalence relation or partial order relation.

As far as I understood we can't prove any of this because of 


*

*To prove that relation is an equivalence relation we need to prove that the relation is reflexive + symmetric + transitive

*To prove that relation is a partial order relation we need to prove that relation is reflexive + antisymmetric + transitive
According to the given conditions let's try to prove that relation is a reflexsive. Reflection means that $x$ should be equal with $y$, but if this condition is satisfied then it means ${\to \frac{y}{x} = 2^{i=0}}$,but according to the condition of the problem $i \gt 0$. 
So, it means that there is no pair where $x = y$ and $i \gt 0$, so we can't prove that it is reflexive, so we can't prove that it is equivalence relation nor partial order relation...
What am I doing wrong?
 A: No, "reflexive" doesn't mean that "$x$ should be equal with $y$". Reflexive means that for each $x$, $x$ is related to itself. So you want to show that there exists a $j$ such that $\frac{x}{x}=2^j$. Your wording is incorrect, because it makes "$x$ equal to $y$" look like a conclusion. It is not; it is a premise for the property you are trying to prove.
As you note, no such $j$ exists if we restrict $j$ to positive integers, so $R$ is not a reflexive relation. 
For transitivity, suppose that $x$ is related to $y$ and $y$ is related to $z$. That means that there exist $j$ and $k$, both positive integers, with
$$ \frac{x}{y}= 2^j\qquad\text{and}\qquad \frac{y}{z}=2^k.$$
We want to see whether $\frac{x}{z}$ is a power of $2$. Try multiplying $\frac{x}{y}$ by $\frac{y}{z}$ to verify that both relations are transitive.
Is $R$ anti-symmetric? Yes! Because if $x$ and $y$ are related, and we also have $y$ and $x$ related, that means that there exist $i\gt 0$ and $j\gt 0$ such that $\frac{x}{y}=2^i$ and $\frac{y}{x}=2^j$; but this requires $i=-j$, which cannot occur if $i$ and $j$ are both positive. That means that the premise of antisymmetry can never be satisfied... which means that the relation is anti-symmetric, by vacuity.
So $R$ is transitive and anti-symmetric, but not reflexive.
The conditions "reflexive, anti-symmetric, and transitive" model what is called a partial order, with the prototypical one being $\subseteq$ among sets. There is a closely related concept, that of strict partial order, which is modeled by $\subsetneq$. Such a relation is a binary relation that satisfies


*

*Anti-reflexivity: For each $x$, $x$ is not related to $x$.

*Transitivity: If $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.


(You could also require antisymmetry, but a transitive antireflexive [also known as arreflexive] relation must be asymmetric: either $x$ is not related to $y$, or $y$ is not related to $x$). 
One can pass from a strict partial order $\prec$ to a partial order $\preceq$ and vice-versa as follows:


*

*If $\prec$ is a strict partial order, then defining "$x\preceq y$ if and only if either $x=y$ or $x\prec y$" yields a partial order $\preceq$.

*If $\preceq$ is a partial order, then defining "$x\prec y$ if and only if $x\preceq y$ and $x\neq y$" yields a strict partial order $\prec$.
Your relation $R$ is a strict partial order. 
