Real Addition is Commutative

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Theorem

The operation of addition on the set of real numbers $\R$ is commutative:

$\forall x, y \in \R: x + y = y + x$


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.


Thus:

\(\displaystyle x + y\) \(=\) \(\displaystyle \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] + \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]\) by definition of real numbers
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {x_n + y_n} \right \rangle}\right]\!\right]\) by definition of real addition
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {y_n + x_n} \right \rangle}\right]\!\right]\) Rational Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] + \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]\) by definition of real addition
\(\displaystyle \) \(=\) \(\displaystyle y + x\)

$\blacksquare$


Sources