Why not all Cauchy sequences in $\Bbb Q$ converge to values in $\Bbb Q$?

Question

I recently read about how to construct $\Bbb R$ from $\Bbb Q$ and Cauchy sequences in $\Bbb Q$. My book says that Cauchy sequences built in $\Bbb Q$ do not all have limits in $\Bbb Q$, and therefore by defining an equivalence relation on all sequences in $\Bbb Q$ by Cauchy criterion result in the set of real numbers, $\Bbb R$.

My question is: why do Cauchy sequences in $\Bbb Q$ do not all have a limit in $\Bbb Q$?

Answer 1

We know this because we have a bunch of examples of rational Cauchy sequences which can provably not converge to rational limits.

Example. The proof that there is no rational number $r\in\Bbb Q$ with $r^2=2$ is well know. This is often quoted as "the square root of $2$ is not rational".

However, we can find a Cauchy sequence $(r_n)$ of rational numbers so that $r_n^2$ converges to $2$. What should the limit of this sequence be? It can certainly be no rational number. So you have the following options:

• Not every Cauchy sequence converges.
• It converges, but the limit is not rational.

The first option is disliked because we want to deal with complete spaces.

Answer 2

The rationals are countable amd the reals are not.

Answer 3

Take any real number, written as a decimal. For example, $\sqrt{2}$. The following is a Cauchy sequence: $$1\\1.4\\1.41\\1.414\\1.4142\\1.41421 $$ So this is a Cauchy sequence that approaches $\sqrt{2}$.

Answer 4

Another reason ids that rational numbers have an eventually periodic decimal representation, and this number, for instance: $$x=0.101001000100001000001\dots$$ has a decimal representation which is never periodic – by construction.

Answer 5

Consider the sequence $$\begin{aligned} a_0 &= 1\\ a_{n+1} &= \frac12\left(a_n+\frac{2}{a_n}\right) \end{aligned}$$ This sequence can be shown to be a Cauchy sequence. However its limit, if it exists, must fulfil the condition $$a = \frac12\left(a+\frac{2}{a}\right)$$ which is equivalent to $$a^2=2$$ But there is no rational number whose square is $2$.

Answer 6

Expanding a bit on Winter's argument, we do the following:-

1. For each $n \in \mathbb{N}$, choose (why we can do so?) $m_n \in \mathbb{N}$ such that

$$ \frac{m_n^2}{n^2} < 2 < \frac{(m_n^2 +2m_n + 1)}{n^2} = \frac{(m_n + 1)^2}{n^2} $$

2. Let $r_n := \frac{m_n}{n}$ be a defined sequence in $\mathbb{Q}$.

3. Show that $(r_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{Q}$.

4. Show that $r_n^2$ converges to $2$.

5. Show that the limit of the sequence $(r_n)_{n=1}^{\infty}$ is not a rational number (why?, see Example in Winter's answer).

Conclusion: We obtain a Cauchy sequence in $\mathbb{Q}$ (namely, $(r_n)_{n=1}^{\infty}$) which is not convergent in $\mathbb{Q}$.