Attempt to prove that every real number is a limit of a sequence of rational numbers

Question

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.

I think either of those approaches is reasonable, although you've not written up complete details. "Can I say...?": Can you give full justification for those things you want to say? If so, yes. Otherwise, the second approach of using the definition directly might be a better idea. — Jonas Meyer, Sep 11 '14 at 05:44
Your answer will depend on your definition of the real numbers. What is your definition of $\mathbb{R}$? — D Wiggles, Oct 02 '16 at 02:34

barak manos · Answer 1 · 2014-09-11T06:52:40.787

2

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 $10$.

edited Sep 11 '14 at 06:52

answered Sep 11 '14 at 05:45

barak manos

43,109

@Jonas Meyer: Can you please elaborate on the down-vote? – barak manos Sep 11 '14 at 05:52

Attempt to prove that every real number is a limit of a sequence of rational numbers

1 Answers1