# Proof: By Induction

Because of the counter-example $2\cdot 0=1\cdot 0\not\Rightarrow 2=1,$ the multiplication of natural numbers is obviously not cancellative for the natural number $$0$$. However, we will prove it for all natural numbers $$z\in\mathbb N,~z\neq 0$$. Because the multiplication of natural numbers "$$\cdot$$" is commutative, it is without loss of generality sufficient to show the right-cancellation property, i.e. $x\cdot z=y\cdot z\Rightarrow x=y,~~~~~~(x,y,z\in\mathbb N,~z\neq 0).$ The proposition can be proven by induction of $$z\ge 1$$.

### Base case $$z=1$$.

For arbitrary $$x,y\in\mathbb N$$, it follows from the definition of multiplication and the definition of addition that $\begin{array}{rcl} x\cdot 1 &=&y\cdot 1\\ (x\cdot 0) + x&=&(y\cdot 0)+y\\ x&=&y \end{array}$

### Induction step $$z\to z^+:=z+1$$

Now, let assume that the inclusion $$x\cdot z_0=y\cdot z_0\Rightarrow x=y$$ has been proven for all $$1\le z_0\le z$$, where we use "$$\le$$" as the order relation of natural numbers. Then it follows again from the definition of multiplication of natural numbers that $\begin{array}{rcl} x\cdot z^+ &=&(x\cdot z)+x,\\ y\cdot z^+ &=&(y\cdot z)+y. \end{array}$

By assumption, if $$x\cdot z=y\cdot z$$, then $$x=y$$. The right sides of both equations equal each other and it follows $x\cdot z^+=y\cdot z^+\Rightarrow x=y,$ which completes the proof that the multiplication of natural numbers is cancellative.

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### References

#### Bibliography

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