Proof: By Induction

(related to Proposition: Multiplication of Natural Numbers Is Associative)

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 \(n,m,p\in\mathbb N\).

Thank you to the contributors under CC BY-SA 4.0!




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