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 $\alpha \beta = \beta \alpha$.

Case where $\alpha = 0^*$
Since real multiplication is left distributive over addition, we have that:
 * $\beta 0^* = \beta \left({0^* + 0^*}\right) = \beta 0^* + \beta 0^*$

Hence:
 * $0^* \beta = 0^* = \beta 0^*$

Case where $\alpha > 0^*$, $\beta \ge 0^*$
If $\beta = 0^*$, this has already been proven.

Assume that $\beta > 0^*$.

Then we can choose $p_0 \in \alpha$, $q_0 \in \beta$ such that $p_0 > 0$, $q_0 > 0$.

By the definition of real multiplication, $q_0 p_0 \in \beta \alpha$.

Suppose that $q \in \alpha \beta$.

If $q \le p_0 q_0$, then by the definition of a Dedekind cut, we have that $q \in \beta \alpha$.

Otherwise, $q > 0$, and there exist $r \in \alpha$, $r > 0$ and $s \in \beta$ such that $q = rs$.

We have that $s > 0$.

Since rational multiplication is commutative, we have that $q = sr \in \beta \alpha$.

Since $q$ was arbitrary, it follows that $\alpha \beta \subseteq \beta \alpha$.

By symmetry of the variables, we can also conclude that $\beta \alpha \subseteq \alpha \beta$.

Hence, by the definition of set equality, we have that $\alpha \beta = \beta \alpha$.

Case where $\alpha > 0^*$, $\beta < 0^*$

 * $\alpha \beta = -\left({\alpha \left({-\beta}\right)}\right) = -\left({\left({-\beta}\right) \alpha}\right) = \beta \alpha$

Case where $\alpha < 0^*$

 * $\alpha \beta = -\left({\left({-\alpha}\right) \beta}\right) = -\left({\beta \left({-\alpha}\right)}\right) = \beta \alpha$