How do you solve $\frac{1}{x} \le 1$? I know it's probably a stupid question, but I'm confused. I have a set {$x\in\mathbb R, \frac{1}{x} \le 1$} that I want to represent as interval/s.
Thinking about it logically, I know that the set is $x\in]-\infty, 0[$U$[1, +\infty[$.
However, when trying to solve the inequality, I can't seem to get the answer. What am I doing wrong?
I take $\frac{1}x \le 1$, and I split it into 2 cases:


*

*if $x > 0$, then $x \ge 1$,

*if $x < 0$, then $x \le 1$,
which is every element of $\mathbb R$. Where am I going wrong? Thanks.

 A: In your second analysis you must intersect the conditions within each case.
In 1. you got $x>0$ and $x\geq 1$. The conjuction of these two is $x\geq 1$.
In 2. you got $x<0$ and $x\leq 1$. The conjunction of these two is $x<0$. The idea is that the solution $x\leq 1$ must be taken into account together with the assumptions that were made to reach it, $x<0$.

It is more common to forget the assumptions when applying nonequivalent transformations, like multiplying by $x$ in this inequality. Applying equivalent transformations the need to intersecting stays with you until the end.
$$\frac{1}{x}\leq1\Leftrightarrow 0\leq 1-\frac{1}{x}=\frac{x-1}{x}$$
You see that $\frac{x-1}{x}$ is non-negative, when either both factors are non-negative, or both are non-positive.
A: It is correct indeed the solutions are $x<0$ and $x\ge 1$ since


*

*for $x > 0$ we have $x \ge 1\implies x\ge1$

*for $x < 0$ we have $x \le 1\implies x<0$

A: Consider the two cases $x>0$ and $x<0$ separately.
If $x>0$ then the direction of the inequality remains if you multiply by $x$, hence
$$\frac1x \geq 1 \implies \frac1x\times x\geq 1\times x\implies 1\geq x. $$
If $x<0$, then the equality changes direction if you multiply by $x$, hence
$$\frac1x \geq 1\implies \frac 1x\times x\leq 1\times x \implies 1\leq  x, $$
which is true for all $x<0$, hence we have either $1\geq x$ or $x<0$.
A: For positive values of $x$, $$1/x \le 1 \iff x\ge 1$$
For negative values of $x$, $1/x$ is negative so it is less than $1.$
For $x=0$, $1/x$ is undefined  
Thus the answer is $$x\in [1,\infty)$$
