Maximum Function in terms of Absolute Value
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Theorem
Let $x$ and $y$ be real numbers.
Then:
- $\ds \max \set {x, y} = \frac 1 2 \paren {\paren {x + y} + \size {x - y} }$
Proof
We aim to show that:
- $\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} } = \begin{cases}y & x \le y \\ x & x > y\end{cases}$
Let $x \le y$.
Then:
- $x - y \le 0$
Then, from the definition of the absolute value, we have:
- $\size {x - y} = y - x$
So:
| \(\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} }\) | \(=\) | \(\ds \frac 1 2 \paren {\paren {x + y} + \paren {y - x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \times 2 y\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds y\) |
Let $x > y$.
Then:
- $x - y > 0$
Then, from the definition of the absolute value, we have:
- $\size {x - y} = x - y$
So:
| \(\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} }\) | \(=\) | \(\ds \frac 1 2 \paren {\paren {x + y} + \paren {x - y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 2 \times 2 x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
So:
- $\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} } = \begin{cases}y & x \le y \\ x & x > y\end{cases}$
as required.
$\blacksquare$