Basic algebra inequality mistake (I probably didn't break algebra) I'm not sure what I have done here. I'm guessing it's something to do with the absolute values but I really have no idea what it is.


$$\begin{array}{ccc}
    y>\frac1y &  & y>\frac1y \\
    y^2>1 &  & 1>\frac1{y^2}  \\
    |y|>1 &  & 1>\frac1{|y|}  \\    
    \frac1{|y|}>1 &  & \downarrow \\        
    \searrow &  & \swarrow  \\            
    & 1 < \left|\dfrac1y\right| < 1 & 
  \end{array}
$$

 A: The error is on the left.
You went from
$$|y|>1$$
to
$$\left|\frac{1}{y}\right|>1$$
This is incorrect. You should have divided both sides by $|y|$, leaving
$$1>\frac{1}{|y|}$$
This agrees with the right-hand result.
However, something happened with the absolute value. Try the case where $y=-2$. In this case, the final equation holds, but the original equation does not. So for $y<0$, we instead have
$$1>\frac{1}{|y|}$$
Why did this happen? It was the first step in each one, multiplying (or dividing) by $y$ to get $y^2$. This loses the distinction between the signs.
A: Check out this graph here:
http://m.wolframalpha.com/input/?i=1%2Fx+%2C+x&x=10&y=14
The values you are looking for are the ones where the hyperbola is below the straight line. 
Values between -1 and zero (not inclusive)
 And values between one and infinity (also not inclusive)
You have to be carefull with inequalities, you can't use them in the same way you treat equality.
For example $y^2 > 1$ is not correct. Since these include values between -1 and infinity. Which is not what was shown on the graph above. 
Best to draw the graph before drawing conclusions. 
