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When you have a statement like $x/y>1$, is there anyway to simplify? All I know is that you cant multiply both sides by $y$ since you don't know the real value of $y$.

I realize the answer is that $x$ and $y$ have to have the same sign and the absolute value of $x$ has to be greater than the absolute value of $y$. But how does one arrive at this truth algebraically?

If you multiply by $y$ can you then see what happens in two scenarios, when $y$ is positive and you multiply both sides by $y$ and when $y$ is negative and you multiply both sides by a negative $y$? I.e., we have \begin{equation} x > y \end{equation} \begin{equation} x < -y \end{equation}

How does one interpret these signs? In the second equation, if $y = -3$, then $x$ can equal $2$ which is not a valid solution.

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  • $\begingroup$ You essentially have to split into 4 cases considering all possible signs of x and y. $\endgroup$ – 1110101001 Feb 2 '14 at 21:39
  • $\begingroup$ What's an organized way to do this? $\endgroup$ – Jwan622 Feb 3 '14 at 5:43
  • $\begingroup$ Actually in this problem the sign of x doesn't matter since you are only multiplying by y. So you only need to consider the case where y is positive and the case where y is negative. However, if it were a more complex inequality then you might need to consider all four cases. $\endgroup$ – 1110101001 Feb 4 '14 at 1:30
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As you say, if $y>0$ then $x>y$, and if $y<0$ then $x<y$ (note: not $x<-y$; you multiplied both sides by $y$ after all, so that's what you get on the right). Sketching that set on the real plane should tell you how you get that English "$x$ and $y$ have to have the same sign and the absolute value of $x$ has to be greater than the absolute value of $y$".

The case $(2, -3)$ that you give doesn't satisfy the above: $x>y$, true; but that only takes effect if $y>0$, and here that isn't true, so that inequality doesn't matter.

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  • $\begingroup$ I see. I made a few mistakes here. When I multiplied both sides by negative Y, i made Y negative too in addition to flipping the signs. That was unnecessary. This makes sense now. $\endgroup$ – Jwan622 Feb 3 '14 at 5:17

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