(related to Proposition: Multiplication of Real Numbers)
Let \(x\) and \(y\) be real numbers, which by definition means that they are the equivalence classes \[\begin{array}{rcl}x&:=&(x_n)_{n\in\mathbb N} + I,\\ y&:=&(y_n)_{n\in\mathbb N} + I.\end{array}\] In the above definition, \((x_n)_{n\in\mathbb N}\) and \((y_n)_{n\in\mathbb N}\) denote elements of the set \(M\) of all rational Cauchy sequences, which represent the real numbers \(x\) and \(y\), while \(I\) denotes the set of all rational sequences, which converge to \(0\).
Note that according definition of multiplying rational Cauchy sequences, the product \((x_n\cdot y_n)_{n\in\mathbb N}\) exists and is a rational Cauchy sequence. Therefore, the product \((x_n\cdot y_n)_{n\in\mathbb N}+I\) exists and is a real number. It remains to be shown that the multiplication of real numbers \[\begin{array}{rcccl} x\cdot y&:=&(x_n\cdot y_n)_{n\in\mathbb N}+ I \end{array}\] is well-defined, i.e. it does not depend on the particular choice of the representatives \((x_n)_{n\in\mathbb N}\) and \((y_n)_{n\in\mathbb N}\).
Let \((x_n)_{n\in\mathbb N},(x_n^\prime)_{n\in\mathbb N}\) as two different representatives of the real number \(x\) and let \((y_n)_{n\in\mathbb N},(y_n^\prime)_{n\in\mathbb N}\) be two different representatives of the real number \(y\). In order to show that the multiplication of real numbers does not depend on the representatives, we have to prove that \[x\cdot y=(x_n^\prime\cdot y_n^\prime)_{n\in\mathbb N}+ I=(x_n\cdot y_n)_{n\in\mathbb N}+ I.\]
We have seen that \((I, + )\) is the subgroup of the group \((M, + )\). Therefore \((x_n)_{n\in\mathbb N} + I\) and \((y_n)_{n\in\mathbb N} + I\) are left cosets of \(I\), which means that the real numbers \(x\) and \(y\) are in fact sets with different elements (albeit we usually interpret them as "numbers").
From \((x_n^\prime)_{n\in\mathbb N}\in (x_n)_{n\in\mathbb N} + I\) it follows that there exists \((a_n)_{n\in\mathbb N}\in I\) (i.e. a rational sequence convergent to \(0\)) with \((x_n^\prime)_{n\in\mathbb N}=(x_n + a_n)_{n\in\mathbb N}\). Analogously, from \((y_n^\prime)_{n\in\mathbb N}\in (y_n)_{n\in\mathbb N} + I\) it follows that there exists a \((b_n)_{n\in\mathbb N}\in I\) (i.e. another rational sequence convergent to \(0\)) with \((y_n^\prime)_{n\in\mathbb N}=(y_n + b_n)_{n\in\mathbb N}\).
In the following, we will use the following mathematical definitions and concepts: * definition of real numbers, * definition of multiplying real numbers (hypothesis), * \(I\) is an ideal of the commutative unit ring \((M, \cdot, + )\), * distributivity law for rational Cauchy sequences, and * associativity law for adding rational Cauchy sequences:
We first observe that which has been shown already. In particular, w
\[\begin{array}{rcll} x\cdot y&=&((x_n^\prime)_{n\in\mathbb N} + I)\cdot ((y_n^\prime)_{n\in\mathbb N} + I)&\text{by definition of real numbers}\\ &=&(x_n^\prime\cdot y_n^\prime)_{n\in\mathbb N}) + I&\text{by hypothesis}\\ &=&((x_n + a_n)_{n\in\mathbb N}\cdot(y_n + b_n)_{n\in\mathbb N})+ I&\text{according to }(*)\\ &=&((x_n\cdot y_n + x_n\cdot b_n+ a_n\cdot y_n+a_n\cdot b_n)_{n\in\mathbb N}+I&\text{by distributivity law for rational Cauchy sequences}\\ &=&((x_n\cdot y_n + (x_n\cdot b_n+ a_n\cdot y_n+a_n\cdot b_n))_{n\in\mathbb N}+I&\text{by associativity of adding rational Cauchy sequences}\\ &=&((x_n\cdot y_n)_{n\in\mathbb N}+I&\text{because }I\text{ is an ideal of }M\text{, i.e. }(x_n\cdot b_n+ a_n\cdot y_n+a_n\cdot b_n)\in I\\ &=&((x_n)_{n\in\mathbb N} + I)\cdot ((y_n)_{n\in\mathbb N} + I)&\text{by hypothesis}\\ \end{array}\]
This demonstrates that the multiplication of real numbers "\(\cdot\)" does not depend on the particular choice of their representatives \((x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}\in M\) or \((x_n^\prime)_{n\in\mathbb N},(y_n^\prime)_{n\in\mathbb N}\in M\).
