# Proof: By Induction

We shall show the associativity of "$$\cdot$$" by induction. For arbitrary $$n,m\in\mathbb N$$, it follows from the definition of multiplication that

### Base case

$n\cdot (m\cdot 0)=n\cdot 0=0=(n\cdot m)\cdot 0.$

### Induction step

Now, let assume that $$n\cdot (m\cdot p_0)=(n\cdot m)\cdot p_0$$ has been proven for all $$p_0\le p$$. Then it follows $\begin{array}{rcl} n\cdot (m\cdot p^+)&=&n((m\cdot p)+m)\quad\quad(i)\\ &=&n\cdot (m\cdot p)+ n\cdot m\quad\quad(ii)\\ &=&(n\cdot m)\cdot p + n\cdot m\quad\quad(iii)\\ &=&(n\cdot m)\cdot (p + 1)\quad\quad(iv)\\ &=&(n\cdot m)\cdot p^+. \end{array}$ Here, we have used in step $$(i)$$ the definition of multiplication, in steps $$(ii)$$ and $$iii)$$ the left-distributivity property of multiplication of natural numbs $$p^+$$ denotes the unique successor of $$p$$.

This proves the associativity of "$$\cdot$$" for all $$n,m,p\in\mathbb N$$.

Github: ### References

#### Bibliography

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