(related to Proposition: Multiplying Negative and Positive Rational Numbers)
According to the definition of negative and positive rational numbers, we can represent a rational number \(x\) by an ordered pair of 0 and a positive integer \(b > 0\) or two positive integers \(a > 0,~b > 0\) in three ways: \[x=\begin{cases}\frac ab,~b\neq 0&\Longleftrightarrow \text{ if }x\text{ is a positive rational number}\\ \frac 0b,~b\neq 0&\Longleftrightarrow \text{ if }x\text{ equals 0}\\ \frac {-a}b,~b\neq 0&\Longleftrightarrow \text{ if }x\text{ is a negative rational number}\\ \end{cases}\]
Due to the definition of multiplying rational numbers, we have for two rational numbers \(x=\frac ab\) and \(y=\frac cd\), \(b\neq 0\), \(d\neq 0\):
\[\begin{array}{ccl} x\cdot y:=\frac {ac}{bd}.\quad\quad ( * ) \end{array} \]
Because the multiplication of rational numbers is commutative, it is sufficient to prove the following four cases, applying the rules of multiplying negative and positive integers:
Let \(x=\frac ab\) and \(y=\frac cd\) with \(a > 0,~b > 0,~c > 0,~d > 0\). It follows in \( ( * ) \) that \(ac > 0\) and \(bd > 0\), thus the product is a positive rational number.
Let \(x=\frac ab\) and \(y=\frac cd\) with \(a < 0,~b > 0,~c > 0,~d > 0\). It follows in \( ( * ) \) that \(ac < 0\) and \(bd > 0\), thus the product is a negative rational number.
Let \(x=\frac ab\) and \(y=\frac cd\) with \(a < 0,~b > 0,~c < 0,~d > 0\). It follows in \( ( * ) \) that \(ac > 0\) and \(bd > 0\), thus the product is a positive rational number.
Let \(x=\frac ab\) and \(y=\frac cd\) with \(a = 0,~b > 0,~d > 0\) and an arbitrary \(c\in\mathbb Z\). It follows in \( ( * ) \) that \(ac = 0\) and \(bd > 0\), thus the product is zero.
