# Proof

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

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### References

#### Bibliography

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