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

$\blacksquare$


Also see