Negative of Absolute Value/Corollary 1
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Corollary to Negative of Absolute Value
Let $x, y \in \R$ be a real numbers.
Let $\size x$ be the absolute value of $x$.
Then:
- $\size x < y \iff -y < x < y$
Proof
Necessary Condition
Let $\size x < y$.
Then from Negative of Absolute Value:
- $x \le \size x$
and:
- $\size x \ge -x$
So $x < y$ and $-x < y$, and so $x > -y$ from Ordering of Inverses in Ordered Monoid.
It follows that $-y < x < y$.
$\Box$
Sufficient Condition
Let $-y < x < y$.
Then $x < y$ and $-x < y$.
For all $x$:
- $\size x = x$
or:
- $\size x = -x$
Thus it follows that $\size x < y$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.20 \ (1)$