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

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