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.

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

We wish to show that $\left({\alpha \beta}\right) \gamma = \alpha \left({\beta \gamma}\right)$.

Case where $\alpha = 0^*$
The result follows because $0^*$ is the identity of $\left({\R, +}\right)$.

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

which implies that:
 * $\alpha \left({0^* \gamma}\right) = \alpha 0^* = 0^* = 0^* \gamma = \left({\alpha 0^*}\right) \gamma$

Case where $\alpha > 0^*$, $\beta > 0^*$, $\gamma \ge 0^*$
Suppose $q \in \left({\alpha \beta}\right) \gamma$.

We have that $\alpha \beta > 0^*$.

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

By definition, there exist $u \in \alpha$, $u > 0$, and $v \in \beta$ such that $r = uv$.

Since $r > 0$, it follows that $v > 0$.

Hence, $vs \in \beta \gamma$.

Since rational multiplication is associative, we have that $q = \left({uv}\right) s = u \left({vs}\right) \in \alpha \left({\beta \gamma}\right)$.

Since $q$ was arbitrary, we have that $\left({\alpha \beta}\right) \gamma \subseteq \alpha \left({\beta \gamma}\right)$.

Now, suppose $q' \in \alpha \left({\beta \gamma}\right)$.

By definition, there exist $r' \in \alpha$, $r' > 0$, and $s' \in \beta \gamma$ such that $q' = r's'$.

By definition, there exist $u' \in \beta$, $u' > 0$, and $v' \in \gamma$ such that $s' = u'v'$.

We have that $r'u' > 0$ and $r'u' \in \alpha \beta$.

Since rational multiplication is associative, we have that $q' = r' \left({u'v'}\right) = \left({r'u'}\right) v' \in \left({\alpha \beta}\right) \gamma$.

Since $q'$ was arbitrary, we have that we have that $\alpha \left({\beta \gamma}\right) \subseteq \left({\alpha \beta}\right) \gamma$.

By the definition of set equality, it follows that $\left({\alpha \beta}\right) \gamma = \alpha \left({\beta \gamma}\right)$.

Case where $\alpha > 0^*$, $\beta < 0^*$
We have that $-\beta > 0^*$, $\alpha \left({-\beta}\right) > 0^*$.

Hence:

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