Rational Numbers form Subfield of Real Numbers

Theorem
The field $$\left({\Q, +, \times, \le}\right)$$ of rational numbers forms a subfield of the field $$\left({\R, +, \times, \le}\right)$$ of real numbers.

That is, the totally ordered field of real numbers $$\left({\R, +, \times, \le}\right)$$ is an extension of the rational numbers $$\left({\Q, +, \times, \le}\right)$$.

Proof
We already have that the rational numbers form a totally ordered field.

We need to show that $$\Q \subseteq \R$$.

Let $$x \in \Q$$.

Then there exists a Cauchy sequence $$\left[\!\left[{\left \langle{x_n}\right \rangle}\right]\!\right]$$ in $$\Q$$ such that $$x = \left[\!\left[{\left \langle{x_n}\right \rangle}\right]\!\right]$$.

Such a sequence is $$x, x, x \ldots$$ which is trivially Cauchy.

So by the definition of a real number, $$x \in \R$$.