# Proof

The proof will consist of four steps:

### Step 1: Demonstrate that the multiplication of natural numbers "$$\cdot$$" is associative.

This follows from the corresponding proposition.

### Step 2: Demonstrate that the multiplication of natural numbers "$$\cdot$$" is commutative.

This follows from the corresponding proposition.

### Step 3: Demonstrate that $$( \mathbb N\setminus\{0\}, \cdot )$$ forms a commutative monoid.

• Step 1 shows that "$$\cdot$$" is associative, so $$( \mathbb N\setminus\{0\}, \cdot )$$ forms a semigroup.
• Following the definition of multiplication, $$n \cdot 1=1\cdot n=n$$.
• Thus, the element $$1\in\mathbb N\setminus\{0\}$$ is the identity element of this semigroup.
• Together with step 2, this shows that $$( \mathbb N\setminus\{0\}, \cdot )$$ forms a commutative monoid.

### Step 4: Show that $$( \mathbb N\setminus\{0\}, \cdot )$$ is cancellative.

This follows from the corresponding proposition.

These four steps show altogether that the set $$( \mathbb N\setminus\{0\}, \cdot )$$, i.e. the set of natural numbers without the element $$0$$, together with the multiplication "$$\cdot$$" as a binary operation, forms a cancellative commutative monoid.

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013