Why does $\lvert x^2 \rvert < 16$ imply $\lvert x \rvert < 4$?
Ordering axioms:
- $O1$ (reflexivity): $x \le x$
- $O2$ (anti-symmetry): $x \le y$ and $y \le x$ imply $x = y$
- $O3$ (transitivity): $x \le y$ and $y \le z$ imply $x \le z$
- $O4$ (totality): $x \le y$ or $y \le x$
- $O5$ (additive compatibility): $a \le b$ implies $a+c \le b+c$
- $O6$ (multiplicative compatibility): $0 \le a$ and $0 \le b$ imply $0 \le ab$
$1.1$ Lemma: $x^2 \ge 0$ for all $x$.
Proof: by totality, either $x \le 0$ or $x \ge 0$ (or both).
If $x \le 0$, then $0 \le -x$ by additive compatibility, so $0 \le (-x)(-x)$ by multiplicative compability, i.e. $0 \le x^2$.
If $x \ge 0$, then $0 \le (x)(x) = x^2$ also by multiplicative compability.
$1.2$ Definition: $|x|$ is defined as $x$ when $x \ge 0$ and $-x$ otherwise.
For example, $|3| = 3$ and $|-3| = -(-3) = 3$.
$1.3$ Lemma: $|x^2| = x^2$ for all $x$.
Proof: $x^2 \ge 0$ by $1.1$, so $|x^2| = x^2$ by definition $1.2$.
$1.4$ Lemma: $x \le 0$ and $y \le 0$ imply $0 \le xy$.
Proof: By additive compatibility, $0 \le -x$ and $0 \le -y$, so by multiplicative compatibility we have $0 \le (-x)(-y) = xy$.
$1.5$ Lemma: If $x \le -4$, then $x^2 \ge 16$.
Proof: We have $x+4 \le 0$ by additive compatibility. Also, $-4 \le 4$ so $-4+x \le 4+x$, whence by transitivity we have $-4+x \le 0$, i.e. $x-4 \le 0$.
Then, by $1.4$ we obtain $0 \le (x+4)(x-4)$, i.e. $0 \le x^2-16$, i.e. $x^2 \ge 16$.
$1.6$ Lemma: If $x \ge 4$, then $x^2 \ge 16$.
Proof: $x-4 \ge 0$ by additive compatibility, and then $-4 \le 4$ so $x-4 \le x+4$, so $x+4 \ge x-4 \ge 0$ by transitivity. Then, by multiplicative compatibility we have $(x+4)(x-4) \ge 0$, i.e. $x^2-16 \ge 0$, i.e. $x^2 \ge 16$.
$1.7$ Definition: $a < b$ is defined to be true if and only if $a \le b$ and $a \ne b$.
$1.8$ Lemma (additive compatibility): $a < b$ implies $a+c < b+c$.
Proof: $a<b$ implies $a \le b$ and $a \ne b$, whence $a+c \le b+c$ and $a+c \ne b+c$, whence $a+c < b+c$.
$1.9$ Lemma (multiplicative compatibility): $0 < a$ and $0 < b$ imply $0 < ab$.
Proof: $0<a$ and $0<b$ mean that $0\le a$ and $0\le b$ and $0\ne a$ and $0\ne b$, whence $0 \le ab$ and $0 \ne ab$, whence $0 < ab$.
$1.10$ Lemma: If $-4 < x < 4$, then $x^2 < 16$.
Proof: $-4 < x$ means that $0 < x+4$ by additive compatibility, and $x < 4$ means that $0 < 4-x$ by additive compatibility, whence by multiplicative compatibility we have $0 < (x+4)(4-x) = 16-x^2$, and then by additive compatibility we have $x^2 < 16$.
$1.11$ Lemma: either $x \le y$ or $y < x$, and both cannot simultaneously hold.
Proof: if $x=y$ then $x \le y$ holds, and if $x \ne y$ then either $x \le y$ holds or $y \le x$ holds by totality, but the latter case becomes $y < x$ because $y \ne x$.
If both hold, then we have $x \le y$ and $y \le x$, whence by anti-symmetry we have $x=y$, contradicting the fact that $x \ne y$ as deduced from $y < x$.
$1.12$ Lemma: If $x^2 < 16$, then $-4 < x < 4$.
Proof: Otherwise, if $x \le -4$ or $x \ge 4$, then by $1.5$ and $1.6$ we obtain $x^2 \ge 16$, contradicting $1.11$.
$1.13$ Lemma: for $b > 0$, $|x| < b$ if and only if $-b < x < b$.
Proof: Assume $|x| < b$.
From $1.11$, either $x \ge 0$ or $x < 0$. If $x \ge 0$, then $|x| = x$ by definition of absolute, so $x < b$; $-b < x$ because $-b < -x$ and $-x < x$, which is because $0 < x+x$, which is because $0<x$ and $x<x+x$, which is because $0<x$. If $x < 0$, then $|x| = -x$, so we have $-x < b$, i.e. $-b < x$. We also have $x < b$ because $x < 0$ and $0 < b$.
Assume $-b < x < b$.
From $1.11$, either $x \ge 0$ or $x < 0$. If $x \ge 0$, then $|x| = x$, so $|x| < b$ because $x < b$. If $x < 0$, then $|x| = -x$, so $|x| < b$ because $-x < b$ because $-b < x$.
$1.14$ Theorem: $|x^2| < 16 \iff |x| < 4$.
From $1.10$ and $1.12$ we have $x^2 < 16 \iff -4 < x < 4$.
From $1.3$ we can rewrite it as $|x^2| < 16 \iff -4 < x < 4$ since $|x^2| = x^2$.
From $1.13$ we have $-4 < x < 4 \iff |x| < 4$, so we are done.
If we want to use the simplest tools possible (rather than the simplest argument possible), then we can go directly to the definition of the absolute value: $$ |x^2| = \begin{cases} x^2 & \text{if $x^2 \ge 0$, and} \\ -x^2 & \text{if $x^2 < 0$. } \end{cases} $$ But $x^2 \ge 0$ for all real numbers, which means that the second case is irrelevant. That is, $$ |x^2| = x^2. $$ Then we have $$ |x^2| < 16 \implies x^2 - 16 < 0. $$ The graph of the function $p(x) = x^2 - 16$ a parabola which opens upward (see the picture below), which means that it is negative between its zeros.
Since the zeros of $p$ are given by $x = \pm 4$, it follows that $|x^2| < 16$ if and only if $x$ is between $-4$ and $4$, exclusive. That is, $$ |x^2| < 16 \iff -4 < x < 4. $$
An alternative approach, based on your attempt, is to note that $|x^2| = |x|^2 < 16$ implies that $|x| < 4$. We extract only the positive square root, since we know that $|x|$ must be nonnegative. But now we need to apply the definition of the absolute value again, which implies that $$ |x| = \begin{cases} x & \text{if $x\ge 0$, and}\\ -x & \text{if $x < 0$.} \\ \end{cases} $$
- In the first case, we have that $x \ge 0$, from which it follows that $$ |x| = x < 4. $$ But then, combining the two inequalities, we have $0 \le x < 4$.
- In the second case, we have that $x < 0$, from which it follows that $$ |x| = -x < 4 \implies x > -4 $$ (note that we multiply each side by $-1$, which changes the direction of the inequality). But then $-4 < x < 0$.
Since either case (1) or case (2) might hold, we take the union of the two sets represented, which implies that $$ |x^2| < 16 \iff -4 < x < 4, $$ which is exactly the result we got above.
Finally, since I think that dxiv's approach is more intuitive than they give it credit for, I'll expand on their approach. The "difficulty" in this problem is the absolute value function, so let's sweep that under the rug for a minute and hide it in a new variable so that we don't have to think about it: let $y = |x|$ so that $$ |x|^2 = y^2 < 16 \iff y^2 - 16 < 0 \iff (y-4)(y+4) < 0. $$ This is possible if and only if $y-4$ and $y+4$ have different signs. But $y$ is the absolute value of $x$, which implies that $y > 0$. Hence $y+4$ must be positive. Therefore we must have $$ y - 4 < 0 \iff |x| - 4 < 0 \iff |x| < 4. $$ But then $-4 < x < 4$ (using the argument above). Therefore $$ |x^2| < 16 \iff -4 < x < 4. $$
First you can drop the absolute value in $|x^2|$ since the square of a real number is always positive or $0$.
Second, you must remember the square root is an increasing function on $\mathbf R^+$. So from $x^2\le 16$, you deduce that $$\sqrt{x^2\mathstrut}=|x|<\sqrt{16\mathstrut}=4,$$ which in turn is equivalent to: $$-4<x<4.$$