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!
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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).
amWhy
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$\dfrac{x}{y}>0$ 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 $x>0$.
Maazul
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For some basic information about writing math at this site see e.g. here, here, here and here.
– Cortizol Jun 29 '13 at 20:11