# Proof

(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 $$af=eb\Rightarrow\frac ab\sim\frac ef$$.

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

Github:

### References

#### Bibliography

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