Can you please explain what's wrong with the following reasoning?

Question

Suppose we want to solve the following inequality

$ \frac{1}{x} + 1 >0 $

So we proceed as follows

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

$ x < -1 $

Thus $ x \in (-\infty, -1)$

But the solution to the inequality is

$ \Re - [-1, 0) $

Answer 1

The problem is that $$1/x >−1$$ does not imply $$ x<-1$$

For example for $x=5$ we have $1/5 >-1$ but $5>-1$

One has to be careful with inequalities involving variables.

Answer 2

The problem is the sign of $x$. When $x<0$ you must flip the direction of the inequality when multiplying through.

There are two cases. Firstly for $x>0$; \begin{align*} & \frac{1}{x}>-1\\ \Rightarrow & 1>-x\\ \Rightarrow & -1<x\\ \Rightarrow & x>-1\\ \end{align*}

So $x>0$ and $x>-1$, together these conditions imply that $x>0$.

Then for $x<0$;

\begin{align*} & \frac{1}{x}>-1\\ \Rightarrow & 1<-x\\ \Rightarrow & -1>x\\ \Rightarrow & x<-1\\ \end{align*}

Putting the two cases together we have;

$$x\in(-\infty,-1)\cup(0,\infty)=\mathbb{R}\setminus[-1,0]$$

EDIT: Note that your given solution includes $x=0$, but mine doesn't. As $x\rightarrow 0$, the inequality holds, but it is undefined at zero itself which is why I have exluded this point.

Answer 3

To avoid case analysis write $$ \frac{1}{x}+1>0\iff\frac{1+x}{x}>0\iff\frac{x(1+x)}{x^2}>0. $$ From here we see that $x(1+x)>0$ i.e. $x>0$ or $x<-1$.

Answer 4

That's a tricky one.

Here is a possible solution. Clearly $x\ne -1$ follows from first inequality. Now from 2nd to 3rd inequality note that $\frac{1}{x}>-1\implies -x<1$ when $x$ is nonnegative i.e. $x$ can be zero or any positive number i.e. $x\in [0,\infty)$, again $\frac{1}{x}>-1\implies 1<-x$ when $x$ is any negative number such that $x\ne 0$ and $x\notin [-1,0]$ if so then it will contradict the last inequality. Hence, $x\in (-\infty,-1)$ . Thus in general we have: $x\in\mathbb{R}-[-1,0)$.

Hope that works.