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 $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$ 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:

\(\displaystyle x \times y\) \(=\) \(\displaystyle \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {x_n \times y_n} \right \rangle}\right]\!\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {y_n \times x_n} \right \rangle}\right]\!\right]\) by commutativity of $\times$ on $\Q$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]\)
\(\displaystyle \) \(=\) \(\displaystyle y \times x\)

$\blacksquare$


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