Real Multiplication is Commutative

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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
\(\ds \) \(=\) \(\ds \eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}\)
\(\ds \) \(=\) \(\ds y \times x\)

$\blacksquare$


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