Non-Zero Natural Numbers under Addition do not form Monoid

Theorem

Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \left\{{0}\right\}$.

The structure $\left({\N_{>0}, +}\right)$ does not form a monoid.

Proof

From Natural Numbers under Addition form Commutative Monoid, $\left({\N, +}\right)$ forms a commutative monoid.

From Natural Numbers Bounded Below under Addition form Commutative Semigroup, $\left({\N_{>0}, +}\right)$ forms a commutative semigroup.

From Identity Element of Natural Number Addition is Zero, $0$ is the identity of $\left({\N, +}\right)$

From Natural Number Addition is Cancellable, all elements of $\left({\N, +}\right)$ are cancellable.

Suppose $\left({\N_{>0}, +}\right)$ is a monoid.

From Identity of Cancellable Monoid is Identity of Submonoid, if $\left({\N_{>0}, +}\right)$ has an identity then it is the same as the identity of $\left({\N, +}\right)$.

But as $0 \notin \N_{>0}$, it follows that $\left({\N_{>0}, +}\right)$ has no identity.

Hence $\left({\N_{>0}, +}\right)$ is not a monoid.

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids