Positive Real Numbers under Max Operation form Monoid

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Theorem

Let $\R_{\ge 0}$ be the set of positive (that is, non-negative) real numbers.

Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the max operation on $\R_{\ge 0}$.


Then $\left({\R_{\ge 0}, \max}\right)$ is a monoid whose identity is $0$.


Proof

From Real Numbers are Totally Ordered, $\R$ is a totally ordered set.

From Max Operation on Toset is Semigroup, $\left({\R_{\ge 0}, \max}\right)$ is a semigroup.

By definition of $\R_{\ge 0}$:

$\forall x \in \R_{\ge 0}: 0 \le x$

Thus by definition of the max operation:

$\forall x \in \R_{\ge 0}: \max \left({0, x}\right) = x = \max \left({x, 0}\right)$

So $0$ is the identity of $\left({\R_{\ge 0}, \max}\right)$ by definition.

The result follows by definition of monoid.

$\blacksquare$


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