Non-Zero Natural Numbers under Multiplication form Commutative Semigroup

Theorem

Let $\N_{>0}$ be the set of natural numbers without zero, that is, $\N_{>0} = \N \setminus \set 0$.

The structure $\struct {\N_{>0}, \times}$ forms an infinite commutative semigroup.

Proof

Semigroup Axiom $\text S 0$: Closure

From Natural Numbers have No Proper Zero Divisors:

$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$

It follows that:

$\forall m, n \in \N_{>0}: m \times n \ne 0$

and so:

$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$

That is, $\struct {\N_{>0}, \times}$ is closed.

$\Box$

Semigroup Axiom $\text S 1$: Associativity

Natural Number Multiplication is Associative.

$\Box$

Commutativity

Natural Number Multiplication is Commutative.

$\Box$

Infinite

We have that the Natural Numbers are Infinite.

Then we have that Infinite if Injection from Natural Numbers.

The mapping $s: \N \to \N_{>0}: \map s n = n + 1$ is such an injection.

Hence $\N_{>0}$ is infinite.

$\blacksquare$

Sources

• 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $2$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$