# 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 $\epsilon > 0$, $|q_{n} – r| < \epsilon$.

#### Solutions Collecting From Web of "Sequence of Rationals Converging to a Limit"

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 $r$.

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.

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 $r.$