Truth set of $-|x| \lt 2$? An exercise in my Algebra I book (Pearson and Allen, 1970, p. 261) asks for the graph of the truth set for $-\left|x\right| \lt 2; x \in \mathbb{R}$.
I've re-stated the inequality in the equivalent form of $\left|x\right| \gt -2$. I know that the truth set of $\left|x\right| = -2$ is $\emptyset$, but I'm not certain how to handle the inequality in conjunction with the absolute value.
I suspect the truth set is $\{x \mid x \ge 0\}$, but I am not certain whether this is correct, or how to prove it using the algebraic concepts I've learned thus far. (I suspect this is a flaw with the book as this is not the first time it assumes knowledge that hasn't yet been presented.)
Is the truth set I arrived at correct? Is there a simple proof of the solution using Algebra I concepts (i.e. the field axioms, basic order properties, etc.)?
Bibliography
Pearson, H. R and Allen, F. B., 1970. Modern Algebra - A Logical Approach (Book I). Boston: Ginn and Company.
 A: Thanks to the helpful hints and comments (@Micah, @HagenvonEitzen, @RossMillikan) and a review of the earlier sections of the book, including the definition of absolute value, I've come to the conclusion that my tentative guess above was incorrect. 
Substituting a negative number such as $-1$ for $x$ in the original inequality $-\left|x\right| \lt 2$, where $x \in \mathbb{R}$, makes it a true statement, namely $-1 \lt 2$. However, $-1 \notin \left\{x \mid x \geq 0\right\}$, which was my previous guess as the truth set.
Given $\left|x\right| \gt -2$, the definition of absolute value stating that $\left|x\right| \ge 0$ for all $\mathbb{R}$, and the fact that $0 \gt -2$, the truth set should be $\left\{x \mid x \in \mathbb{R}\right\}$.
Is this correct so far?
A: Hint:  whatever $x$ is, its absolute value is at least zero.
The mechanical approach is to separate it by cases:  you want the union of $-x \lt 2 \text { and } x \ge 0$ with $x \lt 2 \text { and } x \lt 0$ because the comparison with zero tells you what to do with the absolute value sign.
A: $ \{x : -|x|< 2 $ and $ x\in\mathbb{R}\} = \{x : |x|>-2 $ and $ x\in\mathbb{R}\}$
But 
$\{x : |x|>-2 $ and $ x\in\mathbb{R}\} \equiv \{x : |x|\ge0 $ and $ x\in\mathbb{R}\}$  
A: $$-|x|<2\stackrel{\text{multiplication by}\,\,(-1)}\Longleftrightarrow |x|>-2$$
So: for what values of (real) $\,x\,$ it is true that $\,|x|>-2\,$ ? Hint: this is a rather huge subset of the reals...
