# Proof: By Induction

• For $$n,m,p\in\mathbb N$$ we will first prove the "right-distributivity law" $$(n+m)\cdot p=(n\cdot p)+(m\cdot p)$$. The validity of the "left-distributivity law" $$p\cdot(n+m)=(p\cdot n)+(p\cdot m)$$ then follows from the commutativity of the multiplication of natural numbers.
• Let $$n,m,p\in\mathbb N$$ be arbitrary natural numbers.
• Considering the introduced binary operations on natural numbers addition "$$+$$" and multiplication "$$\cdot$$", we will prove the right-distributivity law $(n+m)\cdot p=(n\cdot p)+(m\cdot p)$ by induction over $$p\in\mathbb N$$.

#### Base Case $$p=0$$.

• For $p=0$, $(n+m)\cdot 0=0=0+0=(n\cdot 0 + m\cdot 0).$

#### Induction step $$p\mapsto p^+:= p + 1$$.

• Let the rule $$(n+m)\cdot p=(n\cdot p + m\cdot p)$$ be proven for all $$p\ge 0$$.
• According to the definition of multiplication of natural numbers we have $$(n+m)\cdot p^+:=(n+m)\cdot p + (n+m).\quad\quad( * )$$
• On the other hand, it follows from the commutativity of the addition of natural numbers that $$\begin{array}{ccl} (n\cdot p^+ + m\cdot p^+)&=&n\cdot p + n + m\cdot p + m\\ &=&n\cdot p + m\cdot p + n + m,\end{array}$$ which equals $$( * )$$, by hypothesis.
• It follows that $(n+m)\cdot p^+=(n\cdot p^+ + m\cdot p^+).$

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### References

#### Bibliography

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