Describe $x/(x+1) < 0$ with an interval

Question

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?.

Answer 1

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 $x\in (-1,0)\subset (-2,0)$.

Answer 2

$$\frac{1}{|1+x|}>1$$

Does not imply:

$$\frac{1}{1+x}>1$$

Answer 3

HINT: $$\frac{x}{x+1}<0$$ is equivalent to $$x>0$$ and $$x+1>0$$ or $$x<0$$ and $$x+1>0$$

Answer 4

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 $\frac{1}{x+1} = -2$.