Commutativity of reals using Cauchy seq. I am trying to prove that real numbers are commutative using the definition of real numbers as the equivalence class of Cauchy sequences of rational numbers.
The following is what I know and what I plan.

RTP If $x,y \in \Bbb R$, then $x+y = y+x$ and $xy = yx$.
My understanding is that if $x \in \Bbb R$ it means that $x$ is the equivalence class of the rational sequence $\{x_j\}$.
Since rationals are commutative, it's clear that $\{x_j+y_j\} \equiv \{y_j +x_j\}$ so I am planning to use this idea with multiplication as well.

Now this is my question.
Do I have to show not only the sequence which the terms are switched, but also all the equivalent sequences, as in

RTP If $\{x_j\} \equiv \{x'_j\}$ and $\{y_j\} \equiv \{y'_j\}$ then $\{x_j+y_j\} \equiv \{y'_j+x'_j\}$ and something analogous to this with multiplication ?

I know how to prove it, I just need advice if this part is actually necessary or if I am missing more.
 A: You don't need the argument in the second box if you've already shown that addition and multiplication of Cauchy sequences is well-defined on equivalence classes. Once you've shown that, it suffices to work with any arbitrary choice of representatives. (Because the operations being well-defined on equivalences classes means precisely that you can pick any pair representatives, add them, and all the resulting sequences will be in the same equivalence class.)
I assume you showed addition and multiplication were well-defined when they were first introduced, so you probably don't need the second argument. 
A: It is true that the real numbers can be shown to be equal as the limit of Cauchy sequences of rationals under the equivalence class, but have you ever thought about the fact that you need the real numbers to construct metric spaces, hence to define Cauchy sequences themselves?
It can be done, but one must take care not to use the general theory of metric spaces (which builds itself on the existence of $\mathbb R$) and do all the work by hand assuming all distances worked with are rational distances. 
As long as you show that with the definition $\{x_i\} + \{y_i\} \overset{def}= \{ x_i + y_i \}$, you have $\{x_i\} + \{ y_i \} = \{ y_i \} + \{ x_i \}$, you are okay. Both expressions on each side do not depend on the choice of representatives of their respective equivalence classes, so showing it only for one choice of representatives is enough. Same comments for multiplication. The issue you are worried with is that addition is well-defined, and it is a different concern.
Hope that helps,
