Minimum Function in terms of Absolute Value

Theorem

Let $x$ and $y$ be real numbers.

Then:

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

We aim to show that:

$\ds \frac 1 2 \paren {\paren {x + y} - \size {x - y} } = \begin{cases}x & x \le y \\ y & x > y\end{cases}$

Let $x \le y$, then:

$x - y \le 0$

so, by the definition of absolute value, we have:

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

So, for $x \le y$, we have:

$\ds \frac 1 2 \paren {\paren {x + y} - \size {x - y} }$

$=$

$\ds \frac 1 2 \paren {\paren {x + y} - \paren {y - x} }$

$\ds $

$=$

$\ds \frac {2 x} 2$

$\ds $

$=$

$\ds x$

Now let $x > y$, then:

$x - y > 0$

so, by the definition of absolute value, we have:

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

So, for $x > y$ we have:

$\ds \frac 1 2 \paren {\paren {x + y} - \size {x - y} }$

$=$

$\ds \frac 1 2 \paren {\paren {x + y} - \paren {x - y} }$

$\ds $

$=$

$\ds \frac {2 y} 2$

$\ds $

$=$

$\ds y$

so:

$\ds \frac 1 2 \paren {\paren {x + y} - \size {x - y} } = \begin{cases}x & x \le y \\ y & x > y\end{cases}$

as required.

$\blacksquare$