(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.