Sum of Maximum and Minimum

Theorem

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

$a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$

Proof

From the definitions of max and min:

$\max \left({a, b}\right) = \begin{cases} b: & a \le b \\ a: & b \le a \end{cases}$

and

$\min \left({a, b}\right) = \begin{cases} a: & a \le b \\ b: & b \le a \end{cases}$

• Let $a < b$.

Then:

$\max \left({a, b}\right) + \min\left({a, b}\right) = b + a$

• Let $a > b$.

Then:

$\max \left({a, b}\right) + \min \left({a, b}\right) = a + b$

• Finally, let $a = b$.

Then:

$\max \left({a, b}\right) = \min \left({a, b}\right) = a = b$

Hence:

$\max \left({a, b}\right) + \min \left({a, b}\right) = 2a = 2b = a + b$

$\blacksquare$

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