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

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

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

Then $\times$ is associative 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 associative 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 Associative, it directly follows that $\times$ is associative on $\R$.

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

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

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

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