(related to Proposition: Addition 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 adding rational Cauchy sequences, the sum \((x_n+y_n)_{n\in\mathbb N}\) exists and is a rational Cauchy sequence. Therefore, \((x_n+y_n)_{n\in\mathbb N}+I\) exists and is a real number. It remains to be shown that the addition of real numbers \[\begin{array}{rcccl} x+y&:=&(x_n+y_n)_{n\in\mathbb N}+ I \end{array}\] 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 addition of real numbers does not depend on the representatives, we have to prove that \[x+y=(x_n^\prime+y_n^\prime)_{n\in\mathbb N}+ I=(x_n+y_n)_{n\in\mathbb N}+ I.\]
We first observe that \((M, + )\) is a commutative group, which has been shown already. Moreover, we have seen that \((I, + )\) is its subgroup. Note that \((x_n)_{n\in\mathbb N} + I\) 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 \((y_n^\prime)_{n\in\mathbb N}=(y_n + b_n)_{n\in\mathbb N}\).
Using the definition of adding rational Cauchy sequences, the associativity law for adding rational Cauchy sequences, and the associativity law for adding rational numbers we get: \[\begin{array}{rcll} x+y=(x_n^\prime+y_n^\prime)_{n\in\mathbb N}+ I&=&(x_n+a_n)_{n\in\mathbb N}+(y_n+b_n)_{n\in\mathbb N}+ I&\text{because }x\text{ and }y\text{ are left cosets of }I\\ &=&((x_n+a_n)+(y_n+b_n))_{n\in\mathbb N}+ I&\text{by definition of adding rational Cauchy sequences}\\ &=&((x_n+a_n+y_n)+b_n)_{n\in\mathbb N}+ I&\text{by associativity of adding rational numbers}\\ &=&(x_n+a_n+y_n)_{n\in\mathbb N}+(b_n)_{n\in\mathbb N}+ I&\text{by definition of adding rational Cauchy sequences}\\ &=&(x_n+a_n+y_n)_{n\in\mathbb N}+I&\text{because }(b_n)_{n\in\mathbb N}\in I\\ &=&(x_n+(a_n+y_n))_{n\in\mathbb N}+I&\text{by associativity law for adding rational numbers}\\ \end{array}\] Note that, since \((M, + )\) is commutative, so is \((I, + )\). Thus, \(I\) is a normal subgroup of \(M\). Therefore, there for the rational sequence \((a_n)_{n\in\mathbb N}\in I\) convergent to \(0\) it holds that \((a_n+y_n)_{n\in\mathbb N}=(y_n+a_n)_{n\in\mathbb N}\). We can now continue our calculation by replacing \[\begin{array}{rcll} (x_n+(a_n+y_n))_{n\in\mathbb N}+I&=&(x_n+(y_n+a_n))_{n\in\mathbb N}+I&\text{because }I\text{ is a normal subgroup of }M\\ &=&((x_n+y_n)+a_n)_{n\in\mathbb N}+I&\text{by associativity law for adding rational numbers}\\ &=&(x_n+y_n)_{n\in\mathbb N}+(a_n)_{n\in\mathbb N}+I&\text{by definition of adding rational Cauchy sequences}\\ &=&(x_n+y_n)_{n\in\mathbb N}+I&\text{because }(a_n)_{n\in\mathbb N}\in I\\ \end{array}\]
This demonstrates that the addition of real numbers "\(+\)" is well-defined, since it 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 \((x_n^\prime)_{n\in\mathbb N},(y_n^\prime)_{n\in\mathbb N}\in M\).
