Sum of Maximum and Minimum

Theorem

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

$a + b = \max \set {a, b} + \min \set {a, b}$

Proof

From the definitions of max and min:

$\max \set {a, b} = \begin{cases}

b: & a \le b \\ a: & b \le a \end{cases}$ and

$\min \set {a, b} = \begin{cases}

a: & a \le b \\ b: & b \le a \end{cases}$

Let $a < b$.

Then:

$\max \set {a, b} + \min \set {a, b} = b + a$

Let $a > b$.

Then:

$\max \set {a, b} + \min \set {a, b} = a + b$

Finally, let $a = b$.

Then:

$\max \set {a, b} = \min \set {a, b} = a = b$

Hence:

$\max \set {a, b} + \min \set {a, b} = 2a = 2b = a + b$

$\blacksquare$

Note that this result does not apply to $a, b \in \C$ as there is no concept of ordering on the complex numbers $\C$.