Natural Numbers Bounded Below under Addition form Commutative Semigroup

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m \in \N$ where $\N$ is the set of natural numbers.

Let $M \subseteq \N$ be defined as:

$M := \left\{{x \in \N: x \ge m}\right\}$

That is, $M$ is the set of all natural numbers greater than or equal to $m$.

Then the algebraic structure $\left({M, +}\right)$ is a commutative semigroup.


Proof

We have that:

Natural Number Addition is Associative
Natural Number Addition is Commutative

From Restriction of Associative Operation is Associative, $+$ is associative on $\left({M, +}\right)$.

From Restriction of Commutative Operation is Commutative, $+$ is commutative on $\left({M, +}\right)$.

It remains to be shown that $+$ is closed on $\left({M, +}\right)$.


Let $a, b \in M$.

Then $\exists r, s \in \N: a = m + r, b = m + s$.

Thus:

\(\displaystyle a + b\) \(=\) \(\displaystyle \left({m + r}\right) + \left({m + s}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle m + \left({m + r + s}\right)\)
\(\displaystyle \) \(\ge\) \(\displaystyle m\)

So $a + b \in M$ and so $+$ is closed on $\left({M, +}\right)$.

$\blacksquare$