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Suppose I have something like $\lvert x^2 \rvert < 16$. The properties of absolute value state that $\lvert x^2 \rvert = \lvert x \rvert^2$. While this makes sense, I'm having trouble understanding the rather basic fact that this would imply that $\lvert x \rvert < 4$. Does this follow from simply taking the principal root of both sides, if that's even a valid step? I guess I'm not seeing how we wouldn't end up with something like $\pm \lvert x \rvert < \pm 4$. In other words, what it is that allows us to drop the $\pm$ symbol?

My apologies if this is really basic. (I think the tag I've used here is appropriate for this reason.) I would appreciate any help on this.

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  • $\begingroup$ $|x|$ can't be negative. That's why But note: $|x| < 4$ is equivalent to $-4 < x < 4$. $\endgroup$ – quasi Dec 31 '17 at 2:03
  • $\begingroup$ But isn't it possible that $- \lvert x \rvert \cdot - \lvert x \rvert = \lvert x^2 \rvert$? $\endgroup$ – Matt.P Dec 31 '17 at 2:05
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    $\begingroup$ The inequality that $|x^2|<16$ is equivalent to $|x|^2 < 4^2$, and since $|x|$ is nonnegative, it follows that $|x| < 4$. $\endgroup$ – quasi Dec 31 '17 at 2:09
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    $\begingroup$ The function $f(t) = t^2$ is strictly increasing for $t \ge 0$. Thus, letting $t=|x|$, $$f(|x|) < f(4) \iff |x| < 4$$ $\endgroup$ – quasi Dec 31 '17 at 2:11
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    $\begingroup$ Or, the long, complicated, un-intuitive way: $$\displaystyle |x|^2 \lt 16 \iff |x|^2-16 \lt 0 \iff \big(|x|-4\big)\big(|x|+4\big) \lt 0 $$ But $\,|x| \ge 0\,$, so $\,|x|+4 \gt 0\,$, then divide by the strictly positive $\,|x|+4\,$ and get $\,|x|-4 \lt 0\,$. $\endgroup$ – dxiv Dec 31 '17 at 2:18
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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.

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  • $\begingroup$ And here I though my answer was overkill! (+1) $\endgroup$ – Xander Henderson Dec 31 '17 at 3:10
  • $\begingroup$ @XanderHenderson I removed the remaining "trivial", and now everything is proved. $\endgroup$ – Kenny Lau Dec 31 '17 at 3:13
  • $\begingroup$ This is excellent! Thank you very much. $\endgroup$ – Matt.P Dec 31 '17 at 3:52
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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.

enter image description here

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 $$ |x| = \begin{cases} x & \text{if $x\ge 0$, and}\\ -x & \text{if $x < 0$.} \\ \end{cases} $$

  1. 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 $0 \le x < 4$.
  2. 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 $$ |x^2| < 16 \iff -4 < x < 4. $$

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  • $\begingroup$ This answer was absolutely phenomenal. Thank you very much for this. $\endgroup$ – Matt.P Dec 31 '17 at 2:56
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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: $$-4<x<4.$$

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Actually, $|x|^2 < 16 \implies ||x|| < 4 \implies |x| < 4$

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There are three things you are overlooking.

1) $x^2=|x^2|=|x|^2$.

2) $x^2 \ge 0$. Always. $x^2 <0$ is not an option.

3) $|x|<4$ if and only if $-4 <x <4$. $x$ being less than $-4$ or greater than $4$ is simply not an option.

so there are four possibilities:

a) $x\le -4$. Then $x <0$ so $|x|=-x>0$ and $|x|\ge 4$ and $x^2=|x|^2\ge 16$.

That is not an option if $x^2 <16$.

b) $-4 <x<0$. Then, as above, $x <0$ and so $|x|=-x>0$ and $0 <|x|<4$ and $x^2=|x|^2 <16$.

This is possible.

c) $0\le x <4$. Then $x \le 0$ so $x=|x|$ and $0 \le x=|x|< 4$ and $x^2 <16$

This is possible.

d) $x\ge 4$. The $x\ge 0$ and $|x|=x\ge 4 $. And $x^2 \ge 16$.

This is not an option if $x^2 <16$.

Of the four options $x^2 <16$ is only compatible with $-4 <x <4$ which means exactly $|x|<4$.

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Trickery:

$|x^2| =|x|^2$ using

$|x^2| = |xx| = |x||x| = |x|^2 $.

Recall: $|ab| =|a||b|$, $a,b$ real.

Rewriting:

$|x|^2 -16\lt 0$, or

$(|x| -4)(|x|+4)\lt 0.$

The above product of $2$ factors is

less than $0$ $\iff$

$1$ of the factors is less than $0$, and the other greater than $0$.

Since $|x| + 4 \gt 0$, we have:

$|x| -4 \lt 0$, or $|x| \lt 4$.

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Use the double sign $\pm$ carefully: $$x^2=16 \Rightarrow x_{1,2}=\pm 4;\\ x^2<16 \not\Rightarrow \pm x_{1,2}<\pm 4 \ \ \text{or} \ \ x_{1,2}<\pm 4 \ \ \text{but:}\\ \pm x_{1,2}<4 \Rightarrow -x<4 \ \text{and} \ x<4 \Rightarrow x>-4 \ \text{and} \ x<4 \Rightarrow -4<x<4.$$ Hence: $$|x^2|<16 \Rightarrow \\ |x|^2<16 \Rightarrow \\ \pm|x|<4 \Rightarrow \\ -|x|<4 \ \text{or} \ |x|<4 \Rightarrow \\ \underbrace{|x|>-4}_{\text{always true}} \ \text{or} \ |x|<4 \Rightarrow \\ |x|<4 \Rightarrow \\ -4<x<4.$$ Alternatively, you can label $y=x^2$ right away: $$|x^2|<16 \Rightarrow |y|<16 \Rightarrow -16<y<16 \Rightarrow \underbrace{-16<x^2}_{\text{always true}}<16 \Rightarrow \\ x^2<16 \Rightarrow (x-4)(x+4)<0 \Rightarrow -4<x<4.$$

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