solve the inequality $\frac{1}{x}<1$ My trial :
For $x>0$, $\frac{1}{x}<1\Longrightarrow \frac{1}{x}x<1x\Longrightarrow 1<x$
For $x<0$, $\frac{1}{x}<1\Longrightarrow \frac{1}{x}x>1x\Longrightarrow 1>x$
Hence :$x<1\vee x>1$
 A: The last step is incorrect. The first line is correct. The second is correct too, but when you get to $1>x$ remember you were assuming $x<0$. So, $1>x$ same as $x<1$ is the interval $(−\infty,1)$ and $x<0$ is the interval $(−\infty,0)$, their intersection, i.e. when $x<0$ and $1>x$ hold simultaneously is $(−∞,0)$. In this case, since $x<0$ is a stronger restriction than $1>x$, the $1>x$ becomes redundant, but you need to keep the $x<0$ that you started with. 
So the second line results in $x<0$. 
In the first line you are assuming $x>0$ and then obtaining $x>1$. This time $x>1$ is the stronger condition, and $x>0$ happens to be redundant. So from the first line you get $x>1$ as a valid solution. 
Thus the answer would be $x<0\vee x>1$. 
In general (not directly related to this problem) when two inequalities hold together simultaneously, then you should intersect the respective intervals. 
For example, $x>6$ and $11>x$ are satisfied simultaneously by those $x$ that are in $(6,\infty)\cap(-\infty,11)=(6,11)$ also written as $6<x<11$. 
Edit prompted by the comments.
Put $A=x>0$, $B=x<0$, $C=x>1$, $D=x<1$. Then consider cases as follows:
$(A∧(A→C))∨(B∧(B→D))$. This reduces to $(A∧C)∨(B∧D)$.
As it happens for the above example, $A∧C⟺C$, and $B∧D⟺B$, so we end up with $C∨B$, that is $x>1 \lor x<0$. This is what I mean when I say don't forget that $x<0$. We should not only be looking at $x<0→x<1$ but at $(x<0)∧((x<0)→(x<1))$ which reduces to $x<0$. 
A: You've done nearly everything correctly and in a nice careful way. However, be careful when you combine them. Your first line proves:

If $x>0$ then $\frac{1}x<1$ if and only if $1<x$.

Your second line proves

If $x<0$ then $\frac{1}x<1$ if and only if $x<1$.

The proper way to combine these is into the statement:

$\frac{1}x<1$ if and only if either:

*

*$x>0$ and $1<x$

*$x<0$ and $x<1$

where we incorporate the conditions on either of our results into our lines. In particular, we may simplify this to

$\frac{1}x < 1$ if and only if either $1<x$ or $x<0$.

Which is different from what you got, since you forgot to combine the condition $x<0$ into the second line's result.
