# Proof

(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 $$(x_n^\prime)_{n\in\mathbb N},(y_n^\prime)_{n\in\mathbb N}\in M$$.

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013