# Proof

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: $\begin{array}{cclcl} (x\cdot y)\cdot z&=&([a,b]\cdot [c,d])\cdot [e,f]&&\text{by definition of integers}\\ &=&([ac+bd,~ad+bc])\cdot [e,f]&&\text{by definition of multiplication of integers}\\ &=&[(ac+bd) e+(ad+bc) f,~(ac+bd) f+(ad+bc) e]&&\text{by definition of multiplication of integers}\\ &=&[(ace+bde)+(adf+bcf),~(acf+bdf)+(ade+bce)]&&\text{by distributivity law for natural numbers}\\ &=&[ace+bde+adf+bcf,~acf+bdf+ade+bce]&&\text{by associativity of adding natural numbers}\\ &=&[(ace+adf)+(bcf+bde),~(acf+ade)+(bce+bdf)]&&\text{by commutativity of adding natural numbers}\\ &=&[a(ce+df)+b(cf+de),~a(cf+de)+b(cd+df)]&&\text{by distributivity law for natural numbers}\\ &=&[a,b]\cdot[ce+df,~cf+de]&&\text{by definition of multiplication of integers}\\ &=&[a,b]\cdot([c,d)\cdot[e,f])&&\text{by definition of multiplication of integers}\\ &=&x\cdot (y\cdot z)&&\text{by definition of integers} \end{array}$

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