Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4/Lemma 1

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Lemma for Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4

Construction

Let $\N$ denote the set of natural numbers.

Let $\beta$ be an object such that $\beta \notin \N$

Let $M = \N \cup \set \beta$.

Let us extend the operation of natural number addition from $\N$ to $M$ by defining:

\(\ds 0 + \beta\) \(=\) \(\ds \beta + 0 = \beta\)
\(\ds \beta + \beta\) \(=\) \(\ds \beta\)
\(\ds n + \beta\) \(=\) \(\ds \beta + n = n\)


The algebraic structure:

$\struct {M, +}$

is a commutative monoid such that $0$ is the identity.


Proof

Closure

It is directly apparent from the definition that $\struct {M, +}$ is closed.

$\Box$


Commutativity

We know from Natural Number Addition is Commutative that:

$\forall a, b \in \N: a + b = b + a$

It is also seen from the definition that:

$\forall a \in \N: a + \beta = \beta + a$

Thus $+$ is commutative on $M$.

$\Box$


Associativity

We know from Natural Number Addition is Associative that:

$\forall a, b, c \in \N: \paren {a + b} + c = a + \paren {b + c}$

It remains to check the other cases.

In the below, $a$ and $b$ are arbitrary non-zero elements of $\N$.


Because of commutativity it is sufficient to test the various combinations of $a$, $b$ and $0$ with $\beta$.


So:

\(\ds \paren {a + b} + \beta\) \(=\) \(\ds a + b\)
\(\ds a + \paren {b + \beta}\) \(=\) \(\ds a + b\)


\(\ds \paren {a + \beta} + \beta\) \(=\) \(\ds a + \beta\)
\(\ds \) \(=\) \(\ds a\)
\(\ds a + \paren {\beta + \beta}\) \(=\) \(\ds a + \beta\)
\(\ds \) \(=\) \(\ds a\)


\(\ds \paren {a + 0} + \beta\) \(=\) \(\ds a + \beta\)
\(\ds \) \(=\) \(\ds a\)
\(\ds a + \paren {0 + \beta}\) \(=\) \(\ds a + \beta\)
\(\ds \) \(=\) \(\ds a\)


\(\ds \paren {0 + \beta} + \beta\) \(=\) \(\ds \beta + \beta\)
\(\ds \) \(=\) \(\ds \beta\)
\(\ds 0 + \paren {\beta + \beta}\) \(=\) \(\ds 0 + \beta\)
\(\ds \) \(=\) \(\ds \beta\)


\(\ds \paren {\beta + \beta} + \beta\) \(=\) \(\ds \beta + \beta\)
\(\ds \) \(=\) \(\ds \beta\)
\(\ds \beta + \paren {\beta + \beta}\) \(=\) \(\ds \beta + \beta\)
\(\ds \) \(=\) \(\ds \beta\)

All combinations of $a$, $b$ and $0$ with $\beta$ have been verified for associativity.

Hence $+$ is associative on $M$.

$\Box$


Identity Element

We have that Identity Element of Natural Number Addition is Zero.

That is:

$\forall a \in \N: a + 0 = a = 0 + a$

We also have that:

$\beta + 0 = \beta = 0 + \beta$

and it is seen that $0$ is the identity element for $\struct {M, +}$.


Hence the result.

$\blacksquare$


Sources

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