# Proof: By Induction

The proof is in two steps:

### Step $$(1)$$ Showing by induction of $$n$$ that $$n\cdot 0=0\cdot n$$ for all $$n\in\mathbb N$$.

It follows from the definition of multiplication that $0\cdot 0=0\cdot 0\text{ (Base case)}.$ Now, let assume that $$n_0\cdot 0=0\cdot n_0$$ has been proven for all $$n_0\le n$$, where we use "$$\le$$" as the order relation of natural numbers. Then it follows again from the definition of multiplication that $n^+ \cdot 0=0.\quad\quad ( * )$ Here, $$n^+$$ denotes the unique successor of $$n$$. On the other side, we have by definition of multiplication and by assumption $0\cdot n^+=(0\cdot n)+0=(n\cdot 0)+0=0+0=0.\quad\quad ( * * )$ In the last step ($$0 + 0=0$$) we have used the definition of addition of natural numbers. The comparison of $$( * )$$ and $$( * * )$$ proves the commutativity law $$n\cdot 0=0\cdot n$$ for all $$n\in\mathbb N$$.

### Step $$(2)$$ Showing by induction of $$m$$ that $$n\cdot m=m\cdot n$$ for all $$m\in\mathbb N$$.

It follows from the step $$(1)$$ that $n\cdot 0=0\cdot n\text{ (Base case)}.$ Now, let assume that $$n\cdot m_0=m_0\cdot n$$ has been proven for all $$m_0\le m$$. Then it follows from the definition of multiplication and by assumption that $n\cdot m^+=(n\cdot m) + n=(m\cdot n) + n.\quad\quad ( \ast )$ From the definition of multiplication and the definition of addition, we have $$n\cdot 1=(n\cdot 0)+n=0+n=n$$. Using that result, applying $$\ast$$ and also the right-distributivity law for natural numbers we get $(m\cdot n)+n=(m\cdot n)+(n\cdot 1)=(m+1)\cdot n=m^+\cdot n\quad\quad ( \ast \ast )$ The comparison of $$( \ast )$$ and $$( \ast\ast )$$ proves the commutativity law $$n\cdot m=m\cdot n$$ for all $$n,m\in\mathbb N$$.

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### References

#### Bibliography

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