# Natural Numbers Bounded Below under Addition form Commutative Semigroup

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## 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:

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 $\left({M, +}\right)$.

$\blacksquare$