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

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.

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 $$r\in\Bbb Z$$.
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.
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 $$r$$.