(related to Proposition: Multiplication of Integers Is Associative)
By the definition of integers, the integers \(x,y,z \in \mathbb Z\) are identified by pairs \(x:=[a,b]\), \(y:=[c,d]\) and \(z:=[e,f]\) of natural numbers \(a,b,c,d,e,f\in\mathbb N\).
We have to show that \((x\cdot y)\cdot z=x\cdot (y\cdot z)\) is valid for all \(x,y,z\in\mathbb Z\). Multiplying the integers \(x\cdot y\), respectively \(y\cdot z\) means by definition. \[\begin{array}{ccl} x\cdot y:=[a,b] \cdot [c,d] &:=& [a\cdot c + b\cdot d,~ a\cdot d + c\cdot b]=[ac + bd,~ ad + bc],\\ y\cdot z:=[c,d] \cdot [e,f] &:=& [c\cdot e + d\cdot f,~ c\cdot f + d\cdot e]=[ce+df,~cf+de], \end{array} \] with some new integers \([ac + bd,~ ad + bc]\) and \([ce+df,~cf+de]\). Because adding natural numbers is commutative and associative, and due to the distributivity law for natural numbers, we get: \[\begin{array}{cclcl} (x\cdot y)\cdot z&=&([a,b]\cdot [c,d])\cdot [e,f]&&\text{by definition of integers}\\ &=&([ac+bd,~ad+bc])\cdot [e,f]&&\text{by definition of multiplication of integers}\\ &=&[(ac+bd) e+(ad+bc) f,~(ac+bd) f+(ad+bc) e]&&\text{by definition of multiplication of integers}\\ &=&[(ace+bde)+(adf+bcf),~(acf+bdf)+(ade+bce)]&&\text{by distributivity law for natural numbers}\\ &=&[ace+bde+adf+bcf,~acf+bdf+ade+bce]&&\text{by associativity of adding natural numbers}\\ &=&[(ace+adf)+(bcf+bde),~(acf+ade)+(bce+bdf)]&&\text{by commutativity of adding natural numbers}\\ &=&[a(ce+df)+b(cf+de),~a(cf+de)+b(cd+df)]&&\text{by distributivity law for natural numbers}\\ &=&[a,b]\cdot[ce+df,~cf+de]&&\text{by definition of multiplication of integers}\\ &=&[a,b]\cdot([c,d)\cdot[e,f])&&\text{by definition of multiplication of integers}\\ &=&x\cdot (y\cdot z)&&\text{by definition of integers} \end{array} \]
