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Published December 21, 2020

I have been reading a Calculus book recently and came across formulas for min and max:

\begin{align} min(x, y) = \frac{x + y - |y - x|}{2} \\ max(x, y) = \frac{x + y + |y - x|}{2} \end{align}

Here $$\frac{x + y}{2}$$ is the average of the two numbers or the midpoint between the two on the number line.

Also $$\frac{|y - x|}{2}$$ is half the distance between y and x. Adding it to the midpoint gives the maximum (rightmost) value and subtracting it from the midpoint gives the minimum (leftmost) value.

This is particularly interesting coming from a programming background where:

min(x, y) := x <= y ? x : y
max(x, y) := x >= y ? x : y


I never thought of coming up with a non-procedural formula for either of these.

We can prove that $$min(x, y) = \frac{x + y - |y - x|}{2}$$ and $$max(x, y) = \frac{x + y + |y - x|}{2}$$ as follows:

If $$x \leq y$$, then $$|y - x| = y - x$$. So $$x + y - |y - x| = x + y - y + x = 2x = 2 \cdot min(x, y)$$, and $$x + y + |y - x| = x + y + y - x = 2y = 2 \cdot max(x, y)$$.

In both cases, swap $$x$$ and $$y$$ to prove the formulas for $$x \geq y$$.