Algebraically solving the inequality $\frac{1}{x} - 1 > 0$ 
$$\frac{1}{x}-1>0$$

$$\therefore \frac{1}{x} > 1$$
$$\therefore 1 > x$$
However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, but as I was unable to factories the LHS, I did so. How can I get this result algebraically?
 A: You have 
$$\frac{1}{x} - 1 > 0$$
Forming a common denominator yields
$$\frac{1 - x}{x} > 0$$
The inequality is true if the numerator and denominator have the same sign.
A: Continuing from the step:
$$\frac1x\gt1$$
Now, to multiply the inequality by any non zero number we need to know its sign. So, taking two cases,

Case 1: $x\gt0$
Multiplying  by $x$ on both sides will not affect the sign. Thus,
$$1\gt x$$
Due to the assumption,
$$1\gt x\gt0$$

Case 2: $x\lt0$
Multiplying by $x$ on both sides will reverse the sign. Thus,
$$1\lt x$$
But $x\lt 0$. Thus, no solution is there in this case.

Clearly, the case $x=0$ is not defined. The solution then is,
$$0\lt x\lt1$$
A: Your answer would be correct were we assuming $x$ is positive. You must keep in mind, though, that the inequality fails for $x<0$ (recall that the inequality sign switches when we multiply by a negative number). 
For $x>0$, we have a solution $1>x$. In other words, $0<x<1$ is a solution. 
If we have $$\frac 1x > 1$$ and $x<0$, we get solutions $x>1$. But that's an absurdity, and so there are no solutions if $x<0$. 
A: Consider the function $f(x)=1/x-1=(1-x)/x$ defined for $x\neq0$.  As $ f$ is continuous it can only change sign where it is zero or undefined, i.e., at $x=1$ or $ x=0$. Hence the sign of $f$ is constant on $]-\infty, 0[$, $]0,1[$, and $]1,\infty[$.
Finally compute the sign of say $f(-1)$, $f(1/2)$, and $f(2)$.
A: Hint: $x(1-x)>0$.Can you continue?
