# I Rational sequence converging to irrational

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1. Dec 10, 2017

### Delta²

In the textbook I have (its a textbook for calculus from my undergrad studies, written by Greek authors) some times it uses the lemma that

"for any irrational number there exists a sequence of rational numbers that converges to it",

and it doesn't have a proof for it, just saying that it is a consequence of the fact that $\mathbb{Q}$ is dense in $\mathbb R$.

Any ideas how to proceed for a rigorous proof?

My idea is that if $x=x_0.x_1x_2...x_n,...$ is the representation of the irrational x in the decimal system with $x_i \in {0...9}$ then the sequence

$\sigma_n=\sum\limits_{k=0}^{n}\frac{x_k}{10^k}$ is rational and converges to the number but something tells me this is not a rigorous proof.

Last edited: Dec 11, 2017
2. Dec 11, 2017

### Staff: Mentor

What does it mean, that $\mathbb{Q}$ is dense in $\mathbb{R}$?

3. Dec 11, 2017

### Delta²

That for any $a,b$ reals there exists $c\in \mathbb Q$ such that $a<c<b$.

4. Dec 11, 2017

### Staff: Mentor

Can you use this to find a sequence of rational numbers that converge to $a$, if you keep $a$ fixed here?

Side remark: you should require a<b in the definition, otherwise a=2, b=1 is a counterexample.

5. Dec 11, 2017

### Staff: Mentor

It might be easier to use the topological approach: given $a \in \mathbb{R}$ then every open neighborhood of $a$ contains a number $c \in \mathbb{Q}$. Now all is left, is to define the sequence $(c_n)_{n \in \mathbb{N}}$.

6. Dec 11, 2017

### Delta²

That's where all my problem is to define the sequence, you got to give me as hint something more than the definition of dense. It seems there are infinitely many such sequences, got to find a way to choose one...

7. Dec 11, 2017

### Staff: Mentor

You certainly have a lot of choice. Just pick an element, then think about how to find another element that is closer.

8. Dec 11, 2017

### Staff: Mentor

Take open balls as neighborhoods and narrow down the diameter. The difficulty here is, that a hint and a solution is the same thing.

9. Dec 11, 2017

### Delta²

sorry cant come up with any ideas (regarding the topological approach i was really bad at Topology) other than the one based in the decimal(or any other base) representation .

$x_0=[x]$

$x_n=\frac{[10^nx]}{10^n}$ where [x] denotes the integer part of x.

10. Dec 11, 2017

### Delta²

I think we can prove kind of easily that $\lim x_n=x$ (because $[10^nx]=10^nx-\epsilon_n(x)$ with $0<\epsilon_n(x)<1$ so that $\frac{\epsilon_n(x)}{10^n}$ converges to zero easily...

BUT the million dollar question,

"where do I need that $\mathbb Q$ is dense in $\mathbb R$""???

(ok probably we need it for the rigorous definition of function [x]=the unique integer z such that z<x<z+1)

Last edited: Dec 11, 2017
11. Dec 11, 2017

### Staff: Mentor

O.k. let's take your definition. We have two real numbers $a$ and $b=a+\dfrac{1}{n}$ for $n \in \mathbb{N}$.
Now you said, there is a rational number $c_n$ with $a < c_n < a+\dfrac{1}{n}$.

So does the sequence $(c_n)_{n \in \mathbb{N}}$ converge, and if, what is the limit?

12. Dec 11, 2017

### Delta²

I wonder why I didn't think of that, I was trying to uniquely determine $c_n$ maybe that's why.

13. Dec 11, 2017

### Math_QED

If you want to learn more about it, you can read about the cauchy construction of $\mathbb{R}$. Let me give you a brief sketch. Let's assume $\mathbb{Q}$ is given. I.e., we know such an ordered field exists.

Define the set $F := \{(u_n)_n \in \mathbb{Q}^{\mathbb{N}} \mid \forall d \in \mathbb{Q^{+}: \exists n_0: \forall p,q > n_0: |u_p - u_q| < d}\}$

and the set $N := \{(u_n)_n \in F \mid \forall d \in \mathbb{Q}^{+}: \exists n_0: \forall n > n_0: |a_n| < d\}$

While this may seem abstract, you can think about $F$ as the cauchy sequences and $N$ as the sequences converging to $0$.

Now, you can define the equivalence relation $\approx$ on $F$

such that $(u_n) \approx (v_n) : \iff (u_n-v_n)_n \in N$

We then define $\mathbb{R} := F/\approx$.

This means, we define the real numbers as the equivalence classes of sequences whose difference 'converges' to 0.

So in this context, real numbers are sets! One can prove (using denseness), that if we take any element out of this set (and such a set is never empty because equivalence classes are never empty, by reflexivity), that this sequence of rationals will converge to the real number it represents, as desired.

14. Dec 11, 2017

### Staff: Mentor

I would rather call it using $\mathbb{R}$ is the completion of $\mathbb{Q}$ by definition, and thus the result follows immediately from the convergence of Cauchy sequences. That was my first thought, too, but density is an unnecessary step here.