(related to Proposition: Multiplying Negative and Positive Integers)
According to the definition of negative and positive integers, we can represent an integer \(x\) by an ordered pair of a natural number \(a\) and \(0\) in three ways: \[x=\begin{cases}[a,0],~a\neq 0&\Longleftrightarrow \text{ if }x\text{ is a positive integer}\\ [0,0],&\Longleftrightarrow \text{ if }x\text{ equals 0}\\ [0,a],~a\neq 0&\Longleftrightarrow \text{ if }x\text{ is a negative integer}\\ \end{cases}\]
According to the definition of multiplying integers, we have for two integers \(x=[a,b]\) and \(y=[c,d]\):
\[\begin{array}{ccl} x\cdot y:=[a,b] \cdot [c,d] &:=& [a\cdot c + b\cdot d,~ a\cdot d + c\cdot b]=[ac + bd,~ ad + bc], \end{array} \]
Because the multiplication of integers is commutative, it is sufficient to prove the following four cases:
Let \(x=[a,0]\) and \(y=[c,0]\). It follows
\[[a,0] \cdot [c,0] =[a\cdot c + 0\cdot 0,~ a\cdot 0 + c\cdot 0]=[ac,0],\] which is a positive integer.
Let \(x=[0,b]\) and \(y=[c,0]\). It follows
\[[0,b] \cdot [c,0] =[0\cdot c + b\cdot 0,~ 0\cdot 0 + c\cdot b]=[0,bc],\] which is a negative integer.
Let \(x=[0,b]\) and \(y=[0,d]\). It follows
\[[a,0] \cdot [c,0] =[0\cdot 0 + b\cdot d,~ 0\cdot d + 0\cdot b]=[bd,0],\]
which is a positive integer.
Let \(x=[0,0]\) and \(y=[c,d]\). It follows
\[[0,0] \cdot [c,d] =[0\cdot c + 0\cdot d,~ 0\cdot d + c\cdot 0]=[0,0],\]
which is zero.
