Max Operation is Associative
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Theorem
The Max operation is associative:
- $\map \max {\map \max {x, y}, z} = \map \max {x, \max \paren{y, z}}$
Thus we are justified in writing $\map \max {x, y, z}$.
Proof
To simplify our notation:
- Let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$
There are the following cases to consider:
- $(1): \quad x \le y \le z$
- $(2): \quad x \le z \le y$
- $(3): \quad y \le x \le z$
- $(4): \quad y \le z \le x$
- $(5): \quad z \le x \le y$
- $(6): \quad z \le y \le x$
Taking each one in turn:
- $(1): \quad$ Let $x \le y \le z$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = z\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds y \overline \wedge z\) | \(\ds = z\) |
- $(2): \quad$ Let $x \le z \le y$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge y\) | \(\ds = y\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds y \overline \wedge z\) | \(\ds = y\) |
- $(3): \quad$ Let $y \le x \le z$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = z\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = z\) |
- $(4): \quad$ Let $y \le z \le x$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = x\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = x\) |
- $(5): \quad$ Let $z \le x \le y$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge y\) | \(\ds = y\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds y \overline \wedge z\) | \(\ds = y\) |
- $(6): \quad$ Let $z \le y \le x$. Then:
\(\ds x \overline \wedge \paren{y \overline \wedge z}\) | \(=\) | \(\ds x \overline \wedge y\) | \(\ds = x\) | |||||||||||
\(\ds \paren{x \overline \wedge y} \overline \wedge z\) | \(=\) | \(\ds x \overline \wedge z\) | \(\ds = x\) |
Thus in all cases it can be seen that the result holds.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.12 \ \text{(a)}$