(related to Proposition: Multiplication of Integers Is Cancellative)
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 \(a,b,c,d,e,f\in\mathbb N\)
\[\begin{array}{rcl}x&:=&[a,b],\\y&:=&[c,d],\\z&:=&[e,f]\\\end{array}\]
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.
The reasoning is exactly the same with the cases \(z\) is a positive or \(z\) is a negative integer.
