Prove that given a real number $x$, there exists a rational sequence $r_n$ such that $r_n \to x$ as $n$ grows.
Proof: Suppose $x$ is a real number. Then we know by definition, there exists a rational number such that $x < q < x + \frac1n$. Using the same argument. Then, $x < r_n < x + \frac1n$.
Can I say $x\to x$, and $x + \frac1n \to x$ as $n$ grows. Thus by the sandwich theorem, $r_n \to x$?
Or should I start with, let $\varepsilon>0$. Then we need to show $|r_n – x | < \varepsilon$?
Please any feedback/hint or anything to make it better would be really appreciated.
Thank you.
Best Answer
Well, you can simply show an example of such sequence for any $r\in\mathbb{R}$:
$r_n=\dfrac{\lfloor{r\cdot10^n}\rfloor}{10^n}$, i.e., the rational number that is given by the first $n$ digits of $r$ on base $10$.