User:Abcxyz/Sandbox/Real Numbers/Real Addition is Commutative

Theorem
Let $\R$ denote the set of real numbers.

Let $+$ denote addition on $\R$.

Then $+$ is commutative on $\R$.

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the real numbers, as axiomatically defined as a Dedekind complete totally ordered field.

By the field axioms, $+$ is commutative on $\R$.

Proof 2
Let $\R$ denote the set of real numbers, as constructed from Cauchy sequences.

Let $+$ denote addition on $\R$.

From Rational Addition is Commutative, it directly follows that $+$ is commutative on $\R$.

Proof 3
Let $\R$ denote the set of real numbers, as constructed from Dedekind cuts.

Let $+$ denote addition on $\R$.

From Rational Addition is Commutative, it directly follows that $+$ is commutative on $\R$.

Proof 4
Let $\left({\R, \le}\right)$ denote the ordered set of real numbers, as defined as the Dedekind completion of the rational numbers.

Let $+$ denote addition on $\R$.

Let $\left({\left({\R, \le}\right), \phi}\right)$ be the Dedekind completion of the ordered set $\left({\Q, \le}\right)$ of rational numbers.

Then, for all $x, y \in \R$:

The result follows from Rational Addition is Commutative.