# 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 $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.

Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.

From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$.

Thus we have:

 $\ds x \times y$ $=$ $\ds \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}$ $\ds$ $=$ $\ds \eqclass {\sequence {x_n \times y_n} } {}$ $\ds$ $=$ $\ds \eqclass {\sequence {y_n \times n_n} } {}$ Rational Multiplication is Commutative $\ds$ $=$ $\ds \eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$ $\ds$ $=$ $\ds y \times x$

$\blacksquare$