Max is Half of Sum Plus Absolute Difference
Jump to navigation
Jump to search
Theorem
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
- $\max \set {a, b} = \frac 1 2 \paren {a + b + \size {a - b} }$
Proof
\(\ds \max \set {a, b}\) | \(=\) | \(\ds a + b - \min \set {a, b}\) | Sum Less Minimum is Maximum | |||||||||||
\(\ds \) | \(=\) | \(\ds a + b - \frac 1 2 \paren {a + b - \size {a - b} }\) | Min is Half of Sum Less Absolute Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {2 a + 2 b - \paren {a + b - \size {a - b} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {2 a + 2 b - a - b + \size {a - b} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \paren {a + b + \size {a - b} }\) |
$\blacksquare$