Describe $x/(x+1) < 0$ with an interval I am doing this exercise in a Calculus textbook. Describe all solutions $\frac{x}{x+1} < 0$ as an interval. I think I did successfully and got the solution $x \in (-1,0)$ by considering signs of $x$ and $x+1$. But then I tried this alternative approach.
$$\frac{x}{x+1} < 0$$
Doing long division gives
$$1-\frac{1}{x+1}< 0$$
And so
$$\frac{1}{x+1} > 1$$
Since LHS is greater than $1$, I take absolute value.
$$|\frac{1}{x+1}| > 1$$
Now I take the reciprocal
$$|x+1| < 1 $$
And so $x \in (-2,0)$
So, where is the mistake?.
 A: $$\frac{1}{|1+x|}>1$$
Does not imply:
$$\frac{1}{1+x}>1$$
A: If you want to arrive to the exact solution set at the end, you want a chain of equivalences such as
$$\frac{x}{x+1}<0 \iff 1- \frac1{x+1}<0 \iff \cdots\iff x\in (-1,0).$$
In that case you can be sure that the values of $x$ that verify the original inequality are those in $(-1,0)$ and not any other.
So you should check that you can revert every step. You would then find out that while
$$\frac1{x+1}>1 \implies \left|\frac1{x+1} \right|>1$$
(because $1>0$), it is not true that 
$$\color{red}{\left|\frac1{x+1} \right|>1 \implies \frac1{x+1}>1} $$
(for instance $-\tfrac32$ satisfies the left inequality but not the right inequality.
So, since you have several equivalences but one step only works to the right you have only proved that
$$\frac{x}{x+1}<0 \implies x\in (-2,0),$$
which is true, since you actually now that the solution is $x\in (-1,0)\subset (-2,0)$.
A: Note as an example $x > 1 \implies |x| > 1$ but $|x| > 1$ does not imply $x > 1$! We need biconditionals here for moving from one step to other. Here your proof in wrong because 
$$\frac{1}{x+1} > 1 \iff \frac{1}{|x+1|} > 1$$
is false, because first part of biconditional is satisfied: $\frac{1}{x+1} > 1 \implies \frac{1}{|x+1|} > 1$ but the other part of biconditional $\frac{1}{|x+1|} > 1 \implies\frac{1}{x+1} > 1 $ is not. 
Take for example, $x = -1.5$. Then $\frac{1}{|x+1|} = 2$ but $\frac{1}{x+1} = -2$.
A: HINT: $$\frac{x}{x+1}<0$$ is equivalent to $$x>0$$ and $$x+1>0$$ or
$$x<0$$ and $$x+1>0$$
