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While studying about inequalities, I came across the following definition ($\forall a > 0)$:

$$ \begin{alignat}{1} & |x| > a \iff \{ x \mid x < -a \text{ or } x > a \} \\ & |x| < a \iff \{ x \mid -a < x < a \} \end{alignat} $$

Naturally, as $\{ -a < x < a \}$ could be rewritten as $\{ -a < x \text{ and } x < a \}$, I wonder if is valid to rewrite $\{ x < -a \text{ or } x > a \}$ as $\{ -a > x > a \}$.

I don't know if that would be valid because, while $\{ -a < x < a \}$ represents only one interval, $\{ -a > x > a \}$ would represent two in a single expression. Is that notation valid?

Luiz Felipe
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2 Answers2

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Well, the double inequality $-a > x > a$ implies in particular $-a > a$, which is a rather strange property for a positive number, don't you think?

J.-E. Pin
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  1. On the number line, inequalities are, by construction, transitive. So, $$\forall a, b, c\in\mathbb R,\quad[(a<b\; \text{ and }\; b<c)\implies a<c].$$

    For $a>0,$ the statement $$-a > x > a$$ is therefore false, since, by transitivity, it implies that $a<-a,$ which is a contradiction.

  2. More to the point: $$-a > x > a$$ actually reads as $$x<-a\; \text{ AND }\; x>a$$ instead of as the intended $$x<-a\; \text{ OR }\; x>a.$$

ryang
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  • $-a > x > a$ is well-formed (given the usual convention that it means $-a > x \land x > a)$. It is false if $>$ is given its usual arithmetic interpretation, but that does not mean that it is logically ill-formed. Please be more careful about your use of logical terminology, see https://en.wikipedia.org/wiki/Well-formed_formula. – Rob Arthan Jul 13 '21 at 01:31
  • @RobArthan Thanks for the catch. – ryang Jul 13 '21 at 05:18