Cauchy Sequences of rationals where none of the rationals are integers.

Question

I am studying from Strichartz's The Way of Analysis, and in section 2.1.3, exercise 2, the following problem is given:

"Show that every real number can be given by a Cauchy sequence of rationals $r_1 , r_2,\ldots$ , where none of the rational numbers $r_1 , r_2,\ldots$ is an integer."

I understand how to do this given any arbitrary Cauchy sequence of rationals, but I'm not sure how the extra condition that none of the rational numbers are integers changes how I approach this problem.

Answer 1

If the real, say $r$, is an integer, take $r_n=r+1/(n+1)$. If it's not, take $r_n$ as $r$ to $n$ decimal places. The latter won't suit the stated need for $r\in\Bbb Z$.

Answer 2

For any Cauchy sequence $\{q_n\}$, define another Cauchy sequence $\{r_n\}$ by:

$$ r_n=\begin{cases} q_n,&q_n \not\in \mathbb{Z}\\ q_n+\frac{1}{n+1},&q_n \in \mathbb{Z}\end{cases} $$

Then $\{r_n\}$ will be Cauchy and equivalent to $\{q_n\}$. So $\{r_n\}$ converges to the same real number as $\{q_n\}$. Additionally, none of the terms $r_n$ are integers, by construction.

Answer 3

If $r$ is not an integer then let $\delta > 0$ be the distance to the nearest integer. You can clearly choose a sequence of rationals within $\delta$ of $r$ than converges to $r$.

If $r$ is an integer then the sequence $r + 1/2, r+ 1/3, r+ 1/4, \ldots$ is a sequence of rationals that converges to $r$.