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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.

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    $\begingroup$ Hint: $|x|$ is always positive. $\endgroup$
    – Micah
    Feb 20, 2013 at 18:40
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    $\begingroup$ @Micah non-negative! $\endgroup$ Feb 20, 2013 at 18:40
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    $\begingroup$ Is x defined over integers, real numbers, complex, or some other number system? $\endgroup$
    – JB King
    Feb 20, 2013 at 19:41
  • $\begingroup$ @JBKing $x \in \mathbb{R}$, if that's what you were asking. I've added the replacement set restriction to the original post. $\endgroup$
    – Calculemus
    Feb 22, 2013 at 18:10

4 Answers 4

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$$-|x|<2\stackrel{\text{multiplication by}\,\,(-1)}\Longleftrightarrow |x|>-2$$

So: for what values of (real) $\,x\,$ it is true that $\,|x|>-2\,$ ? Hint: this is a rather huge subset of the reals...

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    $\begingroup$ I wanted to post my answer below as a comment in reply to yours, but I exceeded the character limit. $\endgroup$
    – Calculemus
    Feb 22, 2013 at 19:01
  • $\begingroup$ One more question regarding your hint: since the set of real numbers is dense, is the claim that the set of all real numbers is "huge", or the implication that $\mathbb{R} \gt \mathbb{R}_{\ge 0}$ (both of which are subsets of the set of real numbers) correct? $\endgroup$
    – Calculemus
    Feb 22, 2013 at 20:34
  • $\begingroup$ "The real numbers dense"...where? Anyway, with that "huge" thing I was hinting that the inquality is true for all the real numbers...;) $\endgroup$
    – DonAntonio
    Feb 23, 2013 at 0:01
  • $\begingroup$ I have learned earlier from the same book that the set of rational numbers is dense (meaning there is an infinity of rational numbers between any two rational numbers), and that the rational numbers are a subset of the real numbers, so I assumed that the set of real numbers are dense also. I think the set of all real numbers and the set of positive real numbers are both "huge" (infinite) so I misunderstood your hint, because I thought my original (but incorrect) answer was "huge" (infinite) as well. Is it wrong to say the real numbers are dense? $\endgroup$
    – Calculemus
    Feb 23, 2013 at 0:21
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    $\begingroup$ No, it's not wrong, but without referring to what it is meaningless: a subset $\,A\,$ of a topological space $\,X\,$ is dense if $\,\forall\,U\subset X\,\,,\,\,\emptyset\neq U\,\,\text{open}\,\,,\,\,A\cap U\neq\emptyset\,$ . Pretty boringly, $\,\Bbb R\,$ is dense in itself... $\endgroup$
    – DonAntonio
    Feb 23, 2013 at 0:24
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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?

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    $\begingroup$ Yes, though I think in my answer things look clearer as $\,|x|>t\,$ , for any $\,t\in\Bbb R_{\,-}\,$ is a true statement always $\endgroup$
    – DonAntonio
    Feb 22, 2013 at 19:14
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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.

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  • $\begingroup$ I will try the mechanical approach at some point but I think I need to review the exposition on logic more thoroughly. $\endgroup$
    – Calculemus
    Feb 22, 2013 at 19:00
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$ \{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}\}$

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    $\begingroup$ Somewhat tangential questions: (1) Is there a reason why you used the sign for equality in one case and the sign for equivalence in the other? (2) Does the colon denote the same thing as the vertical line that I used in my original notation ("such that")? $\endgroup$
    – Calculemus
    Feb 22, 2013 at 18:58

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