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

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

Let $\times$ denote multiplication on $\R$.

Then $\times$ 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, $\times$ is commutative on $\R$.

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

Let $\times$ denote multiplication on $\R$.

From Rational Multiplication is Commutative, it directly follows that $\times$ is commutative on $\R$.

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

Let $\alpha, \beta \in \R$.

We wish to show that $\alpha \beta = \beta \alpha$.

If $\alpha = 0^*$ or $\beta = 0^*$, then the result is clear.

If $\alpha > 0^*$ and $\beta > 0^*$, then the result follows directly from the commutativity of rational multiplication.

If $\alpha > 0^*$ and $\beta < 0^*$, then:
 * $\alpha \beta = -\left({\alpha \left({-\beta}\right)}\right) = -\left({\left({-\beta}\right) \alpha}\right) = \beta \alpha$

If $\alpha < 0^*$, then:
 * $\alpha \beta = -\left({\left({-\alpha}\right) \beta}\right) = -\left({\beta \left({-\alpha}\right)}\right) = \beta \alpha$