(related to Proposition: Definition of Rational Numbers)

Let \((a,b),(c,d)\) be ordered pairs of integers, such that \(b\) and \(d\) do not equal the integer number zero \(0\). We will prove that the relation defined using integer products. \[\frac ab \sim \frac cd\quad:=\quad(a,b)\sim(c,d)\quad\Longleftrightarrow\quad ad = bc\] is an equivalence relation. In other words, the rational numbers as equivalence classes \(x:=\{(c,d),~(c,d)\sim (a,b)\}\) are well-defined.

\(( i )\) Reflexivity \(\frac ab\sim \frac ab\)

Clearly, \(ab=ab\).

\(( ii )\) Symmetry \(\frac ab\sim \frac cd\Leftrightarrow\frac cd\sim\frac ab\)

Since \(ad=cb\Leftrightarrow cb=ad\).

\(( iii )\) Transitivity: If \(\frac ab\sim \frac cd\) and \(\frac cd\sim \frac ef\), then \(\frac ab\sim \frac ef\)

By hypothesis, \(ad=cb\) and \(cf=ed\). Multiplying both sides of the two equations together gives us \((ad)(cf)=(cb)(ed)\), which because of the commutativity and associativity of the integer multiplication is equivalent to \((af)(cd)=(eb)(cd)\). If \(cd\neq 0\), we can conclude immediately \(af=eb\Rightarrow\frac ab\sim\frac ef\) because the cancellation property of integer multiplication. Therefore, assume \(cd=0\). Because integers form an integral domain, a product of integers can be only \(0\), if at least one factor equals \(0\). Because \(d\neq 0\) by hypothesis, we must have \(c=0\). In this case we can conclude from \(ab=cd\) that \(a=0\), because of the same argument \(b\neq 0\). Similarly, we can conclude from \(cd=ed\) that \(e=0\), because of the same argument \(f\neq 0\). Therefore, \((af)(cd)=(eb)(cd)\) is the trivial equation \(0\cdot 0=0\cdot 0\) and again \(af=eb\Rightarrow\frac ab\sim\frac ef\).

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