(related to Proposition: Multiplication of Integers)
Let \(x\) and \(y\) be integers. By definition, it means that they are equivalence classes represented by some natural numbers \(x=\lbrack a,b\rbrack \), \(y=\lbrack c,d\rbrack \), \(a,b,c,d\in \mathbb N\).
Note that \(ac\), \(bd\), \(ad\) and \(cb\) are all natural numbers, since they are the products of the respective natural numbers \(a,b,c,d\in \mathbb N\). Also, \(ac + bd\) and \(ad + cb\) are natural numbers, since they are the sums of the respective natural numbers \(ac ,bd\) and \(ad ,cb\). Therefore, the product $$\begin{array}{ccl} x\cdot y:=\lbrack ac + bd,~ ad + bc\rbrack \end{array} $$ exists, because it denotes some new integer, as it is represented by the natural numbers \(ac + bd\) and \(ad + cb\).
It remains to be shown that the multiplication of integers does not depend on the specific representatives of the numbers \(x\) and \(y\). Suppose, we have different representatives $$\begin{array}{rcl} x=\lbrack a_1,b_1\rbrack =\lbrack a_2,b_2\rbrack ,~y=\lbrack c_1,d_1\rbrack =\lbrack c_2,d_2\rbrack .&&(*) \end{array}$$ Without loss of generality, we can assume \(a_1\ge a_2\) and \(c_1\ge c_2\). It follows from the definition of integers that there exist some natural numbers \(i,j\) with $$\begin{array}{rl} a_1=a_2+i,&c_1=c_2+j,\\ b_1=b_2+i,&d_1=d_2+j.\\ \end{array}\quad (*)$$
We have to show that $$x\cdot y=\lbrack a_1c_1 + b_1d_1,~ a_1d_1 + b_1c_1\rbrack =\lbrack a_2c_2 + b_2d_2,~ a_2d_2 + b_2c_2\rbrack .$$
In the following, we will use the following mathematical definitions and concepts: * definition of integers, * definition of adding integers (hypothesis), * associativity law of adding natural numbers, * distributivity law for adding natural numbers, * commutativity law of adding natural numbers, and * cancellation law of adding natural numbers:
$$\begin{array}{rcll} x\cdot y&=&\lbrack a_1,b_1\rbrack \cdot \lbrack c_1,d_1\rbrack &\text{by definition of integers}\\ &=&\lbrack a_1c_1 + b_1d_1,~ a_1d_1 + b_1c_1\rbrack &\text{by hypothesis}\\ &=&\lbrack (a_2+i)\cdot (c_2+j)+(b_2+i)\cdot (d_2+j),\\ &&~(a_2+i)\cdot (d_2+j) + (b_2+i)\cdot (c_2+j)\rbrack &\text{according to }(*)\\ &=&\lbrack a_2c_2+a_2j+ic_2+ij+b_2d_2+b_2j+id_2+ij,\\ &&~a_2d_2+a_2j+id_2+ij+b_2c_2+b_2j+ic_2+ij\rbrack &\text{by associativity and distributivity law for adding natural numbers}\\ &=&\lbrack a_2c_2+b_2d_2+a_2j+ic_2+ij+b_2j+id_2+ij,\\ &&~a_2d_2+b_2c_2+a_2j+ic_2+ij+b_2j+id_2+ij\rbrack &\text{by commutativity law for adding natural numbers}\\ &=&\lbrack a_2c_2+b_2d_2+(a_2j+ic_2+ij+b_2j+id_2+ij),\\ &&~a_2d_2+b_2c_2+(a_2j+ic_2+ij+b_2j+id_2+ij)\rbrack &\text{by associativity law for adding natural numbers}\\ &=&\lbrack a_2c_2+b_2d_2+\cancel{(a_2j+ic_2+ij+b_2j+id_2+ij)},\\ &&~a_2d_2+b_2c_2+\cancel{(a_2j+ic_2+ij+b_2j+id_2+ij)}\rbrack &\text{by cancellation law for adding natural numbers}\\ &=&\lbrack a_2c_2+b_2d_2,a_2d_2+b_2c_2\rbrack &\text{by definition of integers}\\ &=&\lbrack a_2,b_2\rbrack \cdot \lbrack c_2,d_2\rbrack &\text{by hypothesis}\\ \end{array}$$
