Max is Half of Sum Plus Absolute Difference

Theorem

For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:

$\max \set {a, b} = \frac 1 2 \paren {a + b + \size {a - b} }$

$\ds \max \set {a, b}$

$=$

$\ds a + b - \min \set {a, b}$

$\ds $

$=$

$\ds a + b - \frac 1 2 \paren {a + b - \size {a - b} }$

$\ds $

$=$

$\ds \frac 1 2 \paren {2 a + 2 b - \paren {a + b - \size {a - b} } }$

$\ds $

$=$

$\ds \frac 1 2 \paren {2 a + 2 b - a - b + \size {a - b} }$

$\ds $

$=$

$\ds \frac 1 2 \paren {a + b + \size {a - b} }$

$\blacksquare$