Inequalities $x/y > 0$ I am sure I am just being stupid but I can't get the illogic in this problem:
$$
\begin{align}
\frac{x}{y} &> 0\\
\frac{x}{y}\cdot y &> 0\cdot y\\
x &> 0   
\end{align}
$$
However:  when $x= -1,\; y= -1$   ;  $-1/-1 > 0$  is true  but  $x = -1 > 0$ is not!   
 A: We address for which values of $x, y$ the following inequality holds $$xy > 0$$
Note that when you multiply by $y$, when $y<0$ then $y\cdot \dfrac xy> y\cdot 0 \implies x < 0$. That is, multiplying each side of an inequality by a negative number reverses the direction of the inequality.
So, the inequality $\;\dfrac xy > 0\;$ holds if and only if


*

*both $ x > 0$ and $ y>0$, 


OR 


*

*both $x\lt 0$ and $y < 0$.


Indeed, the inequality holds for all pairs, and only those ordered pairs $(x, y)$ that are located strictly within the first and third quadrants of the Cartesian Plane (excluding, of course, the $x, y$ axes).
A: $\dfrac{x}{y}>0$ is not true for all $x,y$. It is NOT true when 
i) $y=0$
ii) $x=0$
iii) $x<0$ and $y>0$
iv) $x>0$ and $y<0$
So you can only multiply $\dfrac{x}{y}>0$ by $y$ on both sides of the inequality if the above conditions are excluded. In general $\left(\dfrac{x}{y}\times y\right) > \left(0 \times y\right)$ is not true for all $y$. So you cannot do this to conclude $x>0$.
