A simple inequality...or not? I came upon this question,

$\frac1x<\frac12$

thinking that the answer was $x>2$. But the answer turned out to be (-∞, 0) ∪ (2, +∞). Why is the answer in this form, and how do you get it?
 A: Because $x$ can be
negative as well as positive.
If $x > 0$,
then you get your
$x > 2$.
If $x < 0$,
then
$\frac1{x} < 0 < \frac12$
is true
for all negative $x$.
A: One is bound to have to split into cases for this. If we are working only over positive $x$, then it is correct to take
$$\frac{1}x<\frac{1}2$$
and take the reciprocal of both sides to get
$$x>2$$
since the reciprocal reverses the order on positive x (or more generally, when both sides are of the same sign). This is probably what you did. If you'd noted the restriction to positive $x$, you could check out the case for negative $x$ and note that $\frac{1}x<0<2$ for any $x<0$, thus any $x<0$ suffices. This gives the answer.
Another way to go about it is to consider multiplying
$$\frac{1}x<\frac{1}2$$
by $2x$. If $x$ is positive, the order is preserved giving $2<x$. If $x$ is negative, the order is reversed, giving $2>x$, which is is satisfied for all negative $x$, thus we know that a positive $x$ greater than $2$ satisfies the equation, as well as all negative $x$. This is, again, the given answer.
