Triangle Inequality/Real Numbers/Proof 4
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Theorem
Let $x, y \in \R$ be real numbers.
Let $\size x$ denote the absolute value of $x$.
Then:
- $\size {x + y} \le \size x + \size y$
Proof
We do a case analysis.
$(1): \quad x \ge 0, y \ge 0$
| \(\ds x\) | \(\ge\) | \(\ds 0\) | ||||||||||||
| \(\ds y\) | \(\ge\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x + y}\) | \(=\) | \(\ds x + y\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size x + \size y\) |
$\Box$
$(2): \quad x \le 0, y \le 0$
| \(\ds x\) | \(\le\) | \(\ds 0\) | ||||||||||||
| \(\ds y\) | \(\le\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x + y}\) | \(=\) | \(\ds -x - y\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size x + \size y\) |
$\Box$
$(3): \quad x \ge 0, y \le 0$
We have that $\size x = x$ and $\size y = -y$.
In this case we show:
- $\size {x + y} \le \max \set {\size x, \size y}$
Let $\size x \le \size y$.
Then:
| \(\ds x\) | \(\le\) | \(\ds -y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\le\) | \(\ds y + x\) | as $x \ge 0$ | ||||||||||
| \(\ds \) | \(\le\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x + y}\) | \(=\) | \(\ds -\paren {x + y}\) | |||||||||||
| \(\ds \) | \(\le\) | \(\ds -y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \size y\) |
Let $\size x \ge \size y$.
Then:
| \(\ds x\) | \(\ge\) | \(\ds -y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\ge\) | \(\ds x + y\) | as $y \le 0$ | ||||||||||
| \(\ds \) | \(\ge\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {x + y} =\) | \(=\) | \(\ds x + y\) | |||||||||||
| \(\ds \) | \(\le\) | \(\ds x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \size x\) |
We have $\max \set {a, b} \le a + b$ for positive real numbers $a$ and $b$.
The result follows by taking $a = \size x$ and $b = \size y$.
$\Box$
$(4): \quad x \le 0, y \ge 0$
Follows by symmetry from the case $(3)$.
$\blacksquare$