# Sequence of Rationals Converging to a Limit

I'm trying to show that for every real number $r$, there exists a sequence of rational numbers $\{q_{n}\}$ such that $q_{n} \rightarrow r$.

Could I get some comments on my proof?

I know that between 2 reals $r, b$ there exists a rational number $m$ such that $r < m < b$. So I can write

$r < q_{1} < b$ ; Now check if $q_{1} = r$ or not. If it does I'm done, and if not, I consider the interval $(a, q_{1})$.

$r < q_{2} < q_{1}$ check if $q_{2} = r$ or not. If it does I'm done, and if not, I consider the interval $(a, q_{3})$

If I continue in this manner, I see that $|r - q_{n+1}| < |r - q_{n}|$. So whether $r$ is rational or irrational, I'm making my the size of the interval $(r, q_{n})$ closer to 0 and as $n \rightarrow \infty$. And so given any $\epsilon > 0$, I know that $|q_{n} - r| < \epsilon$.

Revision

Let $\{q_{n} \}$ be a sequence of rational numbers and $q_{n} \rightarrow r$ where $r \in \mathbb{R}$.

If $r$ is rational, then let every element of $q_{n} = r$.

But if $r$ is irrational, then consider the interval $(r, b)$ where $b \in \mathbb{R}$. Since we can always find a rational number between two reals, consider

$r < q_{1} < b$. Now pick $q_{2}$ such that $q_{2}$ is the midpoint of $r$ and $q_{1}$. So we get that $r < q_{2} < q_{1}$. Then repeat the process so that $r < q_{n} < q_{n-1}$. Note that $|r - q_{n}| = \frac{1}{2}|r - q_{n-1}|$. As we take more values for $q_{n}$, it is clear that $|q_{n} - r| \rightarrow 0$. So given any $\epsilon > 0$, $|q_{n} - r| < \epsilon$.

-
Consider the dumb case where $r=0$ and you pick $q_n = 1+\frac{1}{n}$. The point is that you can have $|r-q_{n+1}|<|r-q_n|$ without eventually having $|q_n-r|<\epsilon$. –  Jason DeVito Mar 6 '12 at 17:10
you's answer is good. You have another option too: Why don't you explicitly choose how fast the intervals shrink? –  Antonio Vargas Mar 6 '12 at 17:12
Idea is reasonable, execution is not so good. What are $a$ and $b$? If $x$ is rational, there is nothing to do. Assume that $x>0$ (changing signs deals with $x<0$). Let $q_1=0$. There is a rational $q_2$ in the interval $(x+q_1)/2, x)$, and so on. Distance from $x$ to $q_2$ is less than $x/2$. Similarly distance from $x$ to $q_3$ is less than $x/4$. Continue. –  André Nicolas Mar 6 '12 at 17:15
The easiest way I know to show that every real is the limit of rationals is by considering the decimal expansion, and the fact that every number with terminating decimal expansion is rational. Although depending on your course/book, rational approximation might be a lemma before decimal expansion –  you Mar 6 '12 at 17:18

You have the right idea but you are handwaving. You are not telling exactly how you will pick the rationals, and how their limit is $r$.

Also $|r - q_{n+1}| \lt |r - q_n|$ does mean $q_n \to r$. For instance, $\left|-1 - \frac{1}{n+1}\right| \lt \left|-1 - \frac{1}{n}\right|$, but $\frac{1}{n} \to 0$, and not $-1$.

Also, you don't have to check if $q_i = r$, as you pick $r \lt q_i$.

(Also, the above writeup contains a possible hint, try to find it :-))

Since you are almost there (based on your Revision) (but note: you still are handwaving and don't have a "proper" proof yet..)

Here is the hint I was referring to:

Since there is at least one rational in $(r, r+\frac{1}{n})$. Pick one and call it $q_n$. What is the limit of $q_n$?

A different proof:

Consider the set $S = \{q: q \ge r, q \in \mathbb{Q}\}$. Show that $\inf S = r$ and show that that implies there is a sequence $q_n \in S$ whose limit is $r$.

-
Using your hint, I can show that $q_{n} \rightarrow r$ by the squeeze theorem right? –  user26139 Mar 8 '12 at 16:12
@user26139: Yes, exactly right! –  Aryabhata Mar 8 '12 at 16:18

Here are some critiques to your proof.

1. When you write "So I can write $r<q_1<b$ ", what is $b$?

2. If $q_1 = r$, how are you done? (I agree that you are, but what would $q_2, q_3,...$ be in this case?)

3. "So whether $r$ is rational or irrational, I'm making my the (sic) size of the interval $(r,q_n)$ closer to $0$ and as $n\rightarrow\infty$." As $n\rightarrow \infty$, what happens?

4. As noted in the comments, the condition $|r-q_{n+1}|<|r-q_n|$ does not guarantee that $|q_n-r|\rightarrow 0$.

Your idea for the proof is great, you just need a slight adjustment to force $|q_n-r|\rightarrow 0$. You could, for example, pick $q_n$ to be between $r$ and the midpoint of $(r,q_{n+1})$. This would force $|q_n-r|$ to be less than half of $|q_{n-1}-r|$, which will force $|q_n-r|\rightarrow 0$. Fix this, and fix up the general presentation a bit and you'll have a great proof.

-
Thanks for the helpful comments. Could you take a look at my revision? –  user26139 Mar 6 '12 at 18:08
For your revision: if $r$ is irrational you can't take $q_{n+1}=\frac{r+q_n}{2}$, since this will not be rational if $q_n$ is rational. Instead, choose $q_{n+1}$ to be any rational number in the interval $(r,\frac{r+q_n}{2})$ –  you Mar 6 '12 at 18:30
@user26139: As "you" said, your choice of $q_{n+1}$ need not be rational, so you must pick it in the interval $(r, \frac{r+q_n}{2})$. Also, I think you should delete the first line of your revision since you're not letting $q_n$ be a sequence, and you don't know if converges to $r$ (yet). Finally, while it is true that the choice of $q_{n+1}$ I gave above will force $q_n\rightarrow r$, it may not be that easy to show. You should also consider Aryabhatta's first hint since, depending on what you've covered so far, it may be easier to show $1/n \rightarrow 0$ than $1/2^n\rightarrow 0$. –  Jason DeVito Mar 6 '12 at 19:15

Your question set me thinking, and I came up with an argument inspired by it, but a bit different from your solution attempt. I do not imagine that the idea is new, but I do not recall seeing it done in this way (this may only mean that my education is incomplete- no pun intended). The idea is to take an irrational real number $r,$ and to recursively construct two sequences $(x_n)$ and $(y_n)$ of rational numbers which approach $r$ respectively from below and above (and are respectively non-decreasing and non-increasing). We start with a rational number $x_1 < r$ and a rational number $y_1 >r.$ Having found rational $x_n <r$ and rational $y_n >r,$ we set $z_n = \frac{x_n +y_n}{2}.$ Then $x_n < z_n < y_n.$ If $z_n <r,$ we set $x_{n+1} = z_n$ and $y_{n+1} = y_n,$ while if $z_n > r,$ we set $x_{n+1} = x_n$ and $y_{n+1} = z_n.$ (Note that $z_n \neq r$ as $z_n$ is rational, but $r$ is irrational). In the first case, we have $y_{n+1}-x_{n+1} = \frac{y_n - x_n}{2}$ and in the second case, we also have $y_{n+1} -x_{n+1} = \frac{y_n - x_n}{2}.$ Hence for each $n > 1,$ we have $y_n -x_n = \frac{y_1 - x_1}{2^{n-1}}.$ For large enough $n,$ we can make this less than any chosen real positive quantity $\varepsilon$. Then for large enough $n,$ we have both $0 < y_n -r < \varepsilon$ and $0 < r - x_n < \varepsilon.$ Hence the sequences $(x_n)$ and $(y_n)$ both have limit $r.$

-
To make your proof a little more formal, it is obvious that $a_{n},b_{n}$ converge to $a,b \in \mathbb{R}$ by Monotone Convergence Theorem. Next show by induction that $\forall n\in \mathbb{N}\ \left| {{a}_{n}}-{{b}_{n}} \right|\le \frac{\left| {{a}_{1}}+{{b}_{1}} \right|}{{{2}^{n}}}$ and using the squezze theorem that $a=b=r$ –  nick Mar 27 '12 at 17:57