Rational Number Space is not Complete Metric Space

Theorem
Let $\left({\Q, \tau_d}\right)$ be the rational number space under the Euclidean metric $d$.

Then $\left({\Q, \tau_d}\right)$ is not a complete metric space.

Proof
Proof by Counterexample:

First note that Rational Numbers form Metric Space.

It remains to be shown that $\left({\Q, \tau_d}\right)$ is not complete.

Consider the sequence $\left\langle{a_n}\right\rangle$ given by:
 * $\displaystyle a_n := \frac {f_{n+1}} {f_n}$

where $\left\langle{f_n}\right\rangle$ is the sequence of Fibonacci numbers.

By Ratio of Consecutive Fibonacci Numbers:
 * $\displaystyle \lim_{n \to \infty} a_n = \phi := \frac{1 + \sqrt 5} 2$

But Square Root of Prime is Irrational.

That means $\dfrac{1 + \sqrt 5} 2$ is likewise irrational.

Thus $\left\langle{a_n}\right\rangle$ is a Cauchy sequence of rational numbers which converges to a number which is not in $\Q$.

The result follows by definition of complete metric space.

In fact, any sequence of rational numbers that converges to an irrational number (in the metric space $\R$) is a Cauchy sequence that does not converge in $\Q$.