User:Abcxyz/Sandbox/Real Numbers/Real Multiplication Distributes over Addition

Theorem
The operation of multiplication on the set of real numbers $\R$ is distributive over the operation of addition:
 * $\forall x, y, z \in \R:$
 * $x \times \left({y + z}\right) = x \times y + x \times z$
 * $\left({y + z}\right) \times x = y \times x + z \times x$

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 distributive over $+$ on $\R$.

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

Let $\times$ and $+$ respectively denote multiplication and addition on $\R$.

From Rational Multiplication Distributes over Addition, it directly follows that $\times$ is distributive over $+$ on $\R$.

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

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

$\times$ is Left Distributive over $+$
We wish to show that:
 * $\alpha \left({\beta + \gamma}\right) = \alpha \beta + \alpha \gamma$

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

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

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

By the definition of real addition, there exist $u \in \beta$ and $v \in \gamma$ such that $s = u + v$.

We have that $ru \in \alpha \beta$ and $rv \in \alpha \gamma$.

Since rational multiplication distributes over addition, we have that $q = ru + rv \in \alpha \beta + \alpha \gamma$.

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

Suppose that $q' \in \alpha \beta + \alpha \gamma$.

Then there exist $r' \in \alpha \beta$ and $s' \in \alpha \gamma$ such that $q' = r' + s'$.

There exist $u_1 \in \alpha$, $u_1 > 0$, and $v_1 \in \beta$ such that $r' = u_1 v_1$.

There exist $u_2 \in \alpha$, $u_2 > 0$, and $v_2 \in \beta$ such that $s' = u_2 v_2$.

Let $A = \max {\left\{{u_1, u_2}\right\}} \in \alpha$, $A > 0$.

Then:
 * $q' = A \left({\dfrac {u_1} A v_1 + \dfrac {u_2} A v_2}\right)$

If $v_1 \ge 0$, then by the definition of a Dedekind cut, we have that $\dfrac {u_1} A v_1 \in \beta$.

Otherwise, $v_1 < 0$, and so $\dfrac {u_1} A v_1 \in \beta$ because $\beta \ge 0^*$ by hypothesis.

Either way, $\dfrac {u_1} A v_1 \in \beta$; similarly, $\dfrac {u_2} A v_2 \in \gamma$.

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

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

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

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

Thus, because real addition is commutative, we can assume, without loss of generality, that $\beta < 0^*$.

Since $\le$ is compatible with $+$ on $\R$, we have that $-\beta \ge 0^*$, $\gamma \ge 0^*$.

Hence:

Case where $\alpha > 0^*$, $\beta + \gamma < 0^*$
We can use the above result to obtain:
 * $\alpha 0^* = \alpha \left({0^* + 0^*}\right) = \alpha 0^* + \alpha 0^*$

which implies that:
 * $\alpha 0^* = 0^*$

Thus, if either $\beta = 0^*$ or $\gamma = 0^*$, the result follows.

Otherwise, we have $\beta < 0^*$ and $\gamma < 0^*$.

In this case: