How to prove that $\frac{x}{1-x}\geq0$ implies $0\leq x<1$? If $$\frac{x}{1-x}\geq{0}$$
How do we prove that $$0\leq{x}<1$$ using the laws of linear inequalities ?
Confusion:
From $\frac{x}{1-x}\geq{0}$$\implies$$x\geq{0}$ and $x\neq1$
and 
From $\frac{x}{x-1}\leq{0}$$\implies$$x\leq{0}$ and $x\neq1$
 A: Hint:
\begin{align}
\frac{x}{1-x} = \frac{x-1+1}{1-x} =-1+\frac{1}{1-x}\ge 0
\end{align}
which means
\begin{align}
\frac{1}{1-x} \geq 1.
\end{align}
Consider two cases. $1-x>0$ and $1-x<0$. 
A: If $x\lt0$, $-x\gt0$ and hence $1-x>0$. Thus, $\frac{x}{1-x}\lt 0$.
If $x\gt1$, $-x\gt-1$ and hence $1-x<0$. Thus, $\frac{x}{1-x}\lt 0$.
If $0\le x\lt1$, $-1\lt-x\le0$ and hence $0\lt1-x\le1$. Thus, $\frac{x}{1-x}\ge0$.
The solution thus is $$0\le x\lt1$$
A: Hint: $\frac{x}{1-x} \ge 0 \iff x(1-x) \ge 0\,$ and $\,x \ne 1$.
A: If $\frac ab \ge 0$ then either:
1)$a\ge 0$ and $b >0$
Or
2)$a <0$ and $b <0$
If $a=x $ and $b=1-x $ then 2)is impossible, as $x <0 \implies 1-x >1>0$.
So 1) must be true.
So $x\ge 0$ and $1-x>0$ so $1>x\ge 0 $.
OR
To address your confusion:
$\frac {x}{1-x} \ge 1$.  
If $1-x <0\implies x >1$ then $\frac {x}{1-x}*(1-x)\le 0*(1-x) $ (remember, the inequality sign flips when multiplying both sides by negative) and $x \le 0$.  This contradicts $x>1$.
if, however, $1-x\ge 0\implies x \le 1$ we have $\frac {x}{1-x}*(1-x)\ge 0*(1-z) $ and $x \ge 0$.  We can't have $x=1$ so $x <1$ and $x \ge 0$ so $0\le x < 1$.
