Natural Numbers form Commutative Semiring

Theorem
The algebraic structure $\left({\N, +, \times}\right)$ consisting of the set of natural numbers $\N$ under addition $+$ and multiplication $\times$ forms a commutative semiring.

Proof
From Natural Numbers form Naturally Ordered Semigroup, the algebraic structure $\left ({\N, +}\right)$ is a naturally ordered semigroup.

The algebraic structure $\left({\N, +}\right)$ is a commutative monoid from Natural Numbers under Addition form Commutative Monoid.

Let $\times$ be the multiplication operation on $\left({S, \circ, \preceq}\right)$, recursively defined on $S$ by means of the mapping $g_m: S \to T$:


 * $\forall m, n \in S: n \times m = g_m \left({n}\right)$

where $g_m: S \to S$ is the unique mapping that satisfies:


 * $\forall m \in S: g_m \left({n}\right) = \begin{cases}

0 & : n = 0 \\ g_m \left({r}\right) \circ m & : n = r \circ 1 \end{cases}$

Then we have:


 * $(1): \quad$ From Multiplication in Naturally Ordered Semigroup is Closed, $\times$ is closed on $S$


 * $(2): \quad$ From Multiplication in Naturally Ordered Semigroup is Associative, $\times$ is associative on $S$


 * $(3): \quad$ From Multiplication in Naturally Ordered Semigroup is Commutative, $\times$ is commutative on $S$


 * $(4): \quad$ From Multiplication in Naturally Ordered Semigroup is Distributive, $\times$ is distributive over $\circ$.

Thus $\left({\N, \times}\right)$ forms an algebraic structure which is closed such that $\times$ is associative and commutative.

So by definition, $\left({\N, \times}\right)$ is a commutative semigroup.

The result follows from definition of commutative semiring and the distributivity of $\times$ over $+$.