Real Multiplication is Commutative

Theorem
The operation of multiplication on the set of real numbers $$\R$$ is commutative:
 * $$\forall x, y \in \R: x \times y = y \times x$$

Proof
From the definition, the real numbers are the set of all equivalence classes $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$$ of Cauchy sequences of rational numbers.

Let $$x = \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right], y = \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$$, where $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$$ and $$\left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$$ are such equivalence classes.

From the definition of real multiplication, $$x \times y$$ is defined as $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] = \left[\!\left[{\left \langle {x_n \times y_n} \right \rangle}\right]\!\right]$$.

Thus we have:

$$ $$ $$ $$ $$