Why is $|x|<2$ a conjunction ($x<2$ AND $-x<2$) while $|x|>2$ is a disjunction ($x>2$ OR $-x>2$)? My question is regarding $|x|<2$ vs $|x|>2$. When solving the first, we have that $$x<2 \quad \mathbf{and}\quad-x<2$$ while when we solve the second we have $$x>2 \quad\mathbf{or}\quad -x>2$$
My question is

Why is the first one a conjunction whereas the second is a disjunction? What is the logic behind that?

I am assuming it has something to do with the definition of the absolute value.
 A: It comes from DeMorgan's laws applied to the other inequality:
$$|x|>2 \iff \textrm{not }(|x|\leq 2)\tag{trichotomy}$$
$$\iff \textrm{not }(x\leq 2 \textrm{ and } -x \leq 2)\tag{other inequality}$$
$$\iff (\textrm{not }x\leq 2) \textrm{ or }(\textrm{not }-x \leq 2)\tag{DeMorgan}$$
$$\iff (x>2)\textrm{ or }(-x > 2)\tag{trichotomy}$$
A: Each of those cases is really two cases:
$|x|>2$ is the same as the union of $x<0\land-x>2$ and $x\geq 0\land x>2$.  This simplifies to $x<-2\cup 2<x$.
$|x|<2$ is the same as the union of $x<0\land-x<2$ and $x>0\land x<2$.  This simplifies to $-2<x<0\cup0\leq x< 2$.  However, as the two domains are consecutive, their union can be simplified into a single domain $-2<x<2$
A: HINT
Since $|x|\geq 0$, we can square both sides to obtain:
\begin{align*}
|x| < 2 & \Longleftrightarrow |x|^{2} < 4\\\\
& \Longleftrightarrow x^{2} - 4 < 0\\\\
& \Longleftrightarrow (x - 2)(x + 2) < 0\\\\
& \Longleftrightarrow x\in(-2,2).
\end{align*}
Based on such case, can you handle the other inequality?
