Rationals form a Subfield of Reals

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, $x \in \R$.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 4.16$: Example $21$