(related to Proposition: Position of Minus Sign in Rational Numbers Representations)
We want to show the equations \[\frac{-a}b=\frac a{-b},\quad\quad\frac {-a}{-b}=\frac ab,\quad\quad\frac {0}{-b}=\frac 0b\quad\quad ( * )\]
By the definition of rational numbers, two rational numbers \(\frac ab\) and \(\frac cd\), with some integers \(a,b,c,d\in\mathbb Z\) and \(b\neq 0\), \(d\neq 0\) are equal if and only if the products of their respective nominators and denominators equal each other, formally
\[\frac ab=\frac cd\Longleftrightarrow ad=cb.\]
Thus, \(( * )\) transforms into
\[{-a}({-b})=ab,\quad\quad{-a}b=a({-b}),\quad\quad{0}b=0({-b})\quad\quad ( * * )\]
Without loss of generality, we can assume the integers \(a,b\in\mathbb Z\) to be positive. Thus, \(-a\) and \(-b\) are negative integers. By definition of integers, we can represent them by ordered pairs of the natural numbers \(0,\alpha,\beta\) in the following way:
\[\begin{array}{rcl} 0&:=&[0,0],\\ a&:=&[\alpha,0],\\ -a&:=&[0,\alpha],\\ b&:=&[\beta,0],\\ -b&:=&[0,\beta]. \end{array}\]
We will prove \(( * * )\) by verifying that the products of the respective integers are indeed equal, making use of the commutativity of adding natural numbers:
\[\begin{array}{rcll} {-a}({-b})&=&[0,\alpha]\cdot [0,\beta]&\text{by definition of integers}\\ &=&[0\cdot 0 + \alpha\cdot \beta,~ 0\cdot \beta + \alpha\cdot 0]&\text{by definition of multiplying integers}\\ &=&[\alpha\cdot \beta + 0\cdot 0,~ \alpha\cdot 0 + 0\cdot \beta]&\text{by commutativity of adding natural numbers}\\ &=&[\alpha,0]\cdot [\beta,0]&\text{by definition of multiplying integers}\\ &=&ab&\text{by definition of integers} \end{array}\]
\[\begin{array}{rcll} {-a}b&=&[0,\alpha]\cdot [\beta,0]&\text{by definition of integers}\\ &=&[0\cdot \beta + \alpha\cdot 0,~ 0\cdot 0 + \alpha\cdot \beta]&\text{by definition of multiplying integers}\\ &=&[\alpha\cdot 0 + 0\cdot \beta,~ \alpha\cdot \beta + 0\cdot 0]&\text{by commutativity of adding natural numbers}\\ &=&[\alpha,0]\cdot [0,\beta]&\text{by definition of multiplying integers}\\ &=&a(-b)&\text{by definition of integers} \end{array}\]
\[\begin{array}{rcll} {0}b&=&[0,0]\cdot [\beta,0]&\text{by definition of integers}\\ &=&[0\cdot \beta + 0\cdot 0,~ 0\cdot 0 + 0\cdot \beta]&\text{by definition of multiplying integers}\\ &=&[0\cdot 0 + 0\cdot \beta,~ 0\cdot \beta + 0\cdot 0]&\text{by commutativity of adding natural numbers}\\ &=&[0,0]\cdot [0,\beta]&\text{by definition of multiplying integers}\\ &=&0(-b)&\text{by definition of integers} \end{array}\]
