# Proof

The (natural number) one $$1$$ exists, since it is the first ordinal succeeding of the natural number $$0$$, i.e.

$1:=0^+.$ By the definition of the multiplication of natural numbers "$$\cdot$$", the commutativity of addition of natural numbers, and because $$0$$ is neutral with respect to this addition operation, we have vor all $$n\in\mathbb N$$
$\begin{array}{rcll} n \cdot 1&=& n\cdot 0^+&1\text{ is the successor of }0\\ &=&(n\cdot 0)+n&\text{by definition of multiplying natural numbers}\\ &=&0+n&\text{by definition of multiplying natural numbers}\\ &=&n+0&\text{by commutativity of addition of natural numbers}\\ &=&n&0\text{ is neutral with respect to addition of natural numbers} \end{array}$ In other words, $$1$$ is neutral with respect to this operation (does not change the natural number $$n$$, if multiplied by it).

It remains to be shown that also the equation $$1\cdot n=n$$ holds for all $$n\in\mathbb N$$. It follows immediately from the commutativity of multiplying natural numbers.

Github: ### References

#### Bibliography

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