Non-Zero Natural Numbers under Addition do not form Monoid

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Theorem

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

The structure $\struct {\N_{>0}, +}$ does not form a monoid.


Proof

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

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

From Identity Element of Natural Number Addition is Zero, $0$ is the identity of $\struct {\N, +}$

From Natural Number Addition is Cancellable, all elements of $\struct {\N, +}$ are cancellable.


Aiming for a contradiction, suppose $\struct {\N_{>0}, +}$ is a monoid.

Then by definition $\struct {\N_{>0}, +}$ has an identity $e$.

From Identity of Cancellable Monoid is Identity of Submonoid, $e$ the same as the identity of $\struct {\N, +}$.

That is:

$e = 0$

But $0 \notin \N_{>0}$.

Thus it follows that $\struct {\N_{>0}, +}$ has no identity.

Hence by Proof by Contradiction it follows that $\struct {\N_{>0}, +}$ is not a monoid.

$\blacksquare$


Sources