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

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 $\Bbb Q$?

• That pesky square root of two! – Lord Shark the Unknown Dec 9 '17 at 13:55
• Can you explain more precisely ? – toto Dec 9 '17 at 13:57
• Find a limit of rationals that converge to an irrational. – user223391 Dec 9 '17 at 13:59
• It will be an infinite sequence of rational which converge to this irrational number ? – toto Dec 9 '17 at 14:02

We know this because of 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.

• Ok, i see the idea, pretty abstract – toto Dec 9 '17 at 14:14

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 $\sqrt{2}$.

The rationals are countable amd the reals are not.

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.

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 $2$.