(related to Proposition: Multiplication Of Rational Numbers)
Note that since the products of integers \(ac\) and \(bd\) are also integers, it follows from the definition of rational numbers that for the rational numbers \(x,y \in \mathbb Q\) with \(x:=\frac ab\), \(y:=\frac cd\) for some integers \(a,b,c,d\in\mathbb Z\) with \(b\neq 0\) and \(d\neq 0\), also the number \[x\cdot y=\frac{ac}{bd} \] is a rational number. It remains to be shown that the definition of multiplication of rational numbers is well-defined, i.e. it does not depend on the specific representatives of \(x\) and \(y\). Assume we have different representatives \[x=\frac{a_1}{b_1}=\frac{a_2}{b_2},~y=\frac{c_1}{d_1}=\frac{c_2}{d_2}.\quad\quad( * )\] It follows from the definition of rational numbers that \(a_1=\frac{a_2b_1}{b_2}\) and \(c_1=\frac{c_2d_1}{d_2}.\) Therefore, we have by virtue of commutativity of multiplying integers and because integer one is neutral with respect to the multiplication of integers that \[\begin{array}{rcll} x\cdot y&=&\frac{a_1}{b_1}\cdot\frac{c_1}{d_1}&\text{by definition of rational numbers}\\ &=&\frac{\frac{a_2b_1}{b_2}}{b_1}\cdot\frac{\frac{c_2d_1}{d_2}}{d_1}&\text{according to }(*)\\ &=&\frac{\frac{a_2b_1}{b_2}}{1}\cdot \frac 1{b_1}\cdot\frac{\frac{c_2d_1}{d_2}}{1}\cdot \frac 1{d_2}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2b_1\cdot 1}{1\cdot b_2\cdot b_1}\cdot\frac{c_2d_1\cdot 1}{1\cdot d_2\cdot d_1}&\text{by definition of multiplying rational numbers}\\ &=&\frac{a_2\cdot 1}{b_2}\cdot\frac{b_1}{b_1}\cdot\frac{c_2\cdot 1}{d_2}\cdot\frac{d_1}{d_1}&\text{by commutativity of multiplying integers}\\ &=&\frac{a_2}{b_2}\cdot\frac{b_1}{b_1}\cdot\frac{c_2}{d_2}\cdot\frac{d_1}{d_1}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ &=&\frac{a_2\cdot 1}{b_2\cdot 1}\cdot\frac{c_2\cdot 1}{d_2\cdot 1}&\text{by definition of rational numbers}\\ &=&\frac{a_2}{b_2}\cdot\frac{c_2}{d_2}&\text{because }1\text{ is neutral with respect to multiplication of integers}\\ \end{array}\]
