Min Operation Representation on Real Numbers

Theorem

Let $x, y \in \R$.

Then:

$\min \set{x, y} = \dfrac 1 2 \paren {x + y - \size {x - y} }$

where $\min$ denotes the min operation.

Proof

From the Trichotomy Law for Real Numbers exactly one of the following holds:

$x < y$ and so $\min \set {x, y} = x$

$x = y$ and so $\min \set {x, y} = x = y$

$y < x$ and so $\min \set {x, y} = y$

By the definition of the absolute value function for each case respectively we have:

$\size {x - y} = y - x$

$\size {x - y} = 0$

$\size {x - y} = x - y$

Thus the equation holds by $+$ being commutative and associative as for each case:

$\dfrac 1 2 \paren {x + y - \paren {y - x} } = x$

$\dfrac 1 2 \paren {x + y + 0} = x = y$

$\dfrac 1 2 \paren {x + y - \paren {x - y} } = y$

$\blacksquare$

Also see

• Max Operation Representation on Real Numbers

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 18 \ \text {(a)}$