# [SOLVED] real analysis - absolute value

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• Sep 13th 2009, 06:20 PM
TreyD0g
[SOLVED] real analysis - absolute value
If x, y, and are elements of R and x <=z, show that x <= y <= z if and only if |x - y| + |y - z| = |x - z|.
• Aug 20th 2011, 07:10 AM
dgomes
Re: [SOLVED] real analysis - absolute value
$\displaystyle \textbf{Proof.}$

$\displaystyle (\Rightarrow) \text{We have that } x \leq y \leq z \text{.}$

$\displaystyle \text{i) Supose } \left | x - y \right | + \left | y - z \right | < \left | x - z \right | \text{.}$

$\displaystyle \Rightarrow y - x + z - y < z - x$

$\displaystyle \Rightarrow z - x < z - x$

$\displaystyle \Rightarrow \text{Contradiction!}$

$\displaystyle \text{ii) Supose } \left | x - y \right | + \left | y - z \right | > \left | x - z \right | \text{.}$

$\displaystyle \Rightarrow y - x + z - y > z - x$

$\displaystyle \Rightarrow z - x > z - x$

$\displaystyle \Rightarrow \text{Contradiction!}$

$\displaystyle \text{Hence, } \left | x - y \right | + \left | y - z \right | = \left | x - z \right | \text{.}$

$\displaystyle (\Leftarrow) \text{We have that } \left | x - y \right | + \left | y - z \right | = \left | x - z \right | \text{.}$

$\displaystyle \text{i) Supose } y < x \leq z \text{.}$

$\displaystyle \Rightarrow x - y + z - y = z - x$

$\displaystyle \Rightarrow 2x = 2y$

$\displaystyle \Rightarrow x = y$

$\displaystyle \Rightarrow \text{Contradiction!}$

$\displaystyle \text{ii) Supose } x \leq z < y \text{.}$

$\displaystyle \Rightarrow y - x + y - z = z - x$

$\displaystyle \Rightarrow 2y = 2z$

$\displaystyle \Rightarrow y = z$

$\displaystyle \Rightarrow \text{Contradiction!}$

$\displaystyle \text{Hence, } x \leq y \leq z \text{.}$
$\displaystyle \text{Q.E.D.}$