Absolute value notation in $\lvert\,x\,\rvert$

Question

I know that $\lvert x\rvert < 1$ can be rewritten as $-1 < x < 1$ What about $\lvert x\rvert > 1$? Can this also be rewritten as $-1 < x < 1$?

Thanks

Answer 1

Consider $x=0$, which has $-1 < x < 1$ but not $\lvert\,x\,\rvert > 1$.

$\lvert\,x\,\rvert > 1$ can be written as "$x < -1$ or $1 < x$".

Answer 2

1. Draw the graph of $y=|x|$.
2. Identify on the $y$-axis the region corresponding to $y>1$, i.e. $|x|>1$.
3. Project horizontally to a region on the graph (first pair of arrows).
4. Project said region vertically to the $x$-axis (second pair of arrows).
5. Identify the solutions: $x < -1$ or $x > 1$.

Answer 3

No. Recall that $|x|$ is defined piecewise by $$|x| := \begin{cases} x & \text{if $x \ge 0$, and} \\ -x & \text{if $x < 0$.} \end{cases} $$ Examining the first inequality, i.e. $|x| < 1$, with respect this definition, we have two cases to consider: either $x \ge 0$, or $x < 0$. In the first case, we conclude that $$ (|x| = x < 1 \ \text{and}\ 0 \le x) \implies 0 \le x < 1. $$ In the second case, we have $$ (|x| = -x < 1 \implies x > -1 \ \text{and}\ 0 > x) \implies -1 < x < 0.$$ Since either of these cases holds, we get that $-1 < x < 0$ or $0 \le x < 1$, which can be written more concisely as $$ -1 < x < 1. $$

In dealing with the inequality $|x| > 1$, we can't really make the same simplifications. In the case that $x \ge 0$, we have $$ |x| = x > 1, $$ and in the case that $x < 0$, we hvae $$ |x| = -x > 1 \implies x < -1.$$ Again, either case could hold, so we have $$ x < -1 \ \text{or}\ x > 1.$$ There really isn't any shorthand that makes this easier to write (at least, not without using interval notation which, honestly, is probably the right way to go).