# Proof

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:

$\begin{array}{rcll} {-a}({-b})&=&[0,\alpha]\cdot [0,\beta]&\text{by definition of integers}\\ &=&[0\cdot 0 + \alpha\cdot \beta,~ 0\cdot \beta + \alpha\cdot 0]&\text{by definition of multiplying integers}\\ &=&[\alpha\cdot \beta + 0\cdot 0,~ \alpha\cdot 0 + 0\cdot \beta]&\text{by commutativity of adding natural numbers}\\ &=&[\alpha,0]\cdot [\beta,0]&\text{by definition of multiplying integers}\\ &=&ab&\text{by definition of integers} \end{array}$

$\begin{array}{rcll} {-a}b&=&[0,\alpha]\cdot [\beta,0]&\text{by definition of integers}\\ &=&[0\cdot \beta + \alpha\cdot 0,~ 0\cdot 0 + \alpha\cdot \beta]&\text{by definition of multiplying integers}\\ &=&[\alpha\cdot 0 + 0\cdot \beta,~ \alpha\cdot \beta + 0\cdot 0]&\text{by commutativity of adding natural numbers}\\ &=&[\alpha,0]\cdot [0,\beta]&\text{by definition of multiplying integers}\\ &=&a(-b)&\text{by definition of integers} \end{array}$

$\begin{array}{rcll} {0}b&=&[0,0]\cdot [\beta,0]&\text{by definition of integers}\\ &=&[0\cdot \beta + 0\cdot 0,~ 0\cdot 0 + 0\cdot \beta]&\text{by definition of multiplying integers}\\ &=&[0\cdot 0 + 0\cdot \beta,~ 0\cdot \beta + 0\cdot 0]&\text{by commutativity of adding natural numbers}\\ &=&[0,0]\cdot [0,\beta]&\text{by definition of multiplying integers}\\ &=&0(-b)&\text{by definition of integers} \end{array}$

Github: ### References

#### Bibliography

1. Piotrowski, Andreas: Own Research, 2014