Min is Half of Sum Less Absolute Difference
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Theorem
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
- $\min \set {a, b} = \dfrac 1 2 \paren {a + b - \size {a - b} }$
Proof
From the definition of min:
- $\map \min {a, b} = \begin{cases} a: & a \le b \\ b: & b \le a \end{cases}$
Let $a < b$.
Then:
\(\ds \dfrac 1 2 \paren {a + b - \size {a - b} }\) | \(=\) | \(\ds \dfrac 1 2 \paren {a + b - \paren {b - a} }\) | Definition of Absolute Value | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {a + b - b + a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {2 a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {a, b}\) | Definition of Min Operation |
$\Box$
Let $a \ge b$.
Then:
\(\ds \dfrac 1 2 \paren {a + b - \size {a - b} }\) | \(=\) | \(\ds \dfrac 1 2 \paren {a + b - \paren {a - b} }\) | Definition of Absolute Value | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {a + b - a + b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {2 b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \min \set {a, b}\) | Definition of Min Operation |
$\blacksquare$