Solving $x - \frac{1}{x} < 0.$ I need to solve for x: 
$$x - \frac{1}{x} < 0.$$
I realize that $x\neq 0$. 
The following is always true: 
$$x-\frac{1}{x} < 0 \iff x < \frac{1}{x}.$$
If $x > 0$: 
$$x < \frac{1}{x} \iff x^2 < 1 \iff x < \pm 1$$
$$0<x<1$$
If $x<0$:
$$x < \frac{1}{x} \iff x^2 > 1 \iff x > \pm 1$$
$$-1<x<0.$$
So my answer is that $-1<x<1$ with $x \neq 0$. But this is not true. The real answer is $x<1$ and $0<x<1$. Why? Where is my logic wrong?

We have:
$$x - \frac{1}{x} < 0$$
$$\frac{x^2 - 1}{x} < 0 \text{ (common denominator)}$$
$$\frac{(x-1)(x+1)}{x} < 0 \text{ (common denominator)}$$
Now you can make a chart:

And we get what we are looking for
