Definition:Zero (Number)

Theorem
Let $$\left({S, \circ; \preceq}\right)$$ be a naturally ordered semigroup.

Then $$\left({S, \circ; \preceq}\right)$$ has a minimal element.

This minimal element of $$\left({S, \circ; \preceq}\right)$$ is called zero and has the symbol $$0$$.

That is: $$\forall n \in S: 0 \preceq n$$.

This element $$0$$ is the identity for $$\circ$$. That is:


 * $$\forall n \in S: n \circ 0 = n = 0 \circ n$$

Proof
This follows immediately from naturally ordered semigroup: NO 1.


 * $$\left({S, \circ; \preceq}\right)$$ is well-ordered, so has a minimal element.

Now by the definition of the zero:
 * $$0 \preceq 0$$

as zero precedes everything.

Thus from Naturally Ordered Semigroup: NO 3:
 * $$0 \preceq 0 \implies \exists p \in S: 0 \circ p = 0$$

By the definition of the zero:
 * $$0 \preceq 0 \circ 0$$ and $$0 \preceq p$$

as zero precedes everything.

Thus from naturally ordered semigroup: NO 2:
 * $$0 \circ 0 \preceq 0 \circ p = 0$$

Thus:
 * $$0 \circ 0 \preceq 0 \and 0 \preceq 0 \circ 0$$

and by the antisymmetry of ordering, it follows that $$0 \circ 0 = 0$$.

Because $$\left({S, \circ; \preceq}\right)$$ is a semigroup, $$\circ$$ is associative.

So:
 * $$\forall n \in S: \left({n \circ 0}\right) \circ 0 = n \circ \left({0 \circ 0}\right) = n \circ 0$$

Thus from naturally ordered semigroup: NO 2:
 * $$\forall n \in S: n \circ 0 = n$$

Finally:
 * $$\forall n \in S: 0 \circ n = n$$

because a naturally ordered semigroup is commutative.

Thus:
 * $$\forall n \in S: n \circ 0 = n = 0 \circ n$$

and so $$0$$ is the identity for $$\circ$$.