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Proof

(related to Proposition: Algebraic Structure Of Natural Numbers Together With Multiplication)

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.

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References

Bibliography

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