Reverse Triangle Inequality/Real and Complex Fields/Proof 2
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Theorem
Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.
Then:
- $\cmod {x - y} \ge \size {\cmod x - \cmod y}$
Proof
From proof $2$ of corollary $1$ to this result, which is derived independently:
- $\size {x - y} \ge \size x - \size y$
There are two cases:
$(1): \quad \size x \ge \size y$
We have :
- $\size {\size x - \size y} = \size x - \size y$
and the proof is finished.
$\Box$
$(2): \quad \size y \ge \size x$
We have:
- $\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$
But:
- $\size {y - x} = \size {x - y}$
and:
- $\size {\size y - \size x} = \size {\size x - \size y}$
From this we have:
- $-\size {\size x - \size y} \ge -\size {x - y}$
Since, by Negative of Absolute Value, we have that:
- $\size x - \size y \ge -\size {\size x - \size y}$
it follows that:
- $-\size {x - y} \le \size x - \size y \le \size {x - y}$
The result follows.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.20 \ (2)$