# Proof

Because the multiplication of integers is commutative, it is without loss of generality sufficient to show the right cancellation property, i.e. $x\cdot z=y\cdot z\Longleftrightarrow x=y,~~~~~~(x,y,z\in\mathbb Z,~z\neq 0).$

By definition of integers, the integer numbers $$x,y,z$$ are some equivalence classes of ordered pairs of natural numbers represented by some natural numbers $$a,b,c,d,e,f\in\mathbb N$$

$\begin{array}{rcl}x&:=&[a,b],\\y&:=&[c,d],\\z&:=&[e,f]\\\end{array}$

### $$(1)$$ Proof of "$$\Rightarrow$$"

Assume the equation $$x\cdot z=y\cdot z$$ holds and that $$z\neq 0$$. We have to show that $$x=y$$. By definition of multiplying integers, we have $\begin{array}{rcl} x\cdot z&=&[ae + bf,~ af + be],\\ y\cdot z&=&[ce + df,~ cf + de]. \end{array}$

By assumption, $$x\cdot z=y\cdot z$$, i.e. $\begin{array}{rcl}ae+bf&=&ce+df\quad\quad(i)\\ af+be&=&cf+de\quad\quad(ii) \end{array}$

Also by assumption, $$z\neq 0$$, i.e. $$e$$ and $$f$$ cannot both be equal $$0$$, thus $$z$$ is either a positive integer, or a negative integer.

If $$z$$ is a positive integer, then $$[e,f]=[e',0]$$ for some natural number $$e'\neq 0$$, and we have

$\begin{array}{rcrcl}(i)&\Longleftrightarrow&ae'&=&ce'\\ (ii)&\Longleftrightarrow&be'&=&de' \end{array}$ Because the multiplication of natural numbers is cancellative, it follows $$a=c$$ and $$b=d$$, thus the integer $$x$$ is equal to the integer $$y$$, if $$z$$ is positive.

Otherwise $$z$$ is a negative integer, thus $$[e,f]=[0,f']$$ for some natural number $$f'\neq 0$$, and it follows

$\begin{array}{rcrcl}(i)&\Longleftrightarrow&bf'&=&df'\\ (ii)&\Longleftrightarrow&af'&=&cf' \end{array}$ Again due to the fact that multiplying natural numbers is cancellative, it follows $$b=d$$ and $$a=c$$, thus the integer $$x$$ is equal to the integer $$y$$, if $$z$$ is negative.

Altogether, it follows from $$x\cdot z=y\cdot z$$ that $$x=y$$ for all integers $$x,y$$ and any integer $$z\neq 0$$. Thus, the multiplication of integers is cancellative.

### $$(2)$$ Proof of "$$\Leftarrow$$"

The reasoning is exactly the same with the cases $$z$$ is a positive or $$z$$ is a negative integer.

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### References

#### Bibliography

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