Constructing Series : Alternating between Rational Number and Irrational Number In the range of $[0,1] \in \Bbb R$, I'd like to construct a series which is strictly increasing and rational number and irrational numbers are alternatively appearing, like $0< \sqrt{1\over 3} < 2/3 < \sqrt{1 \over 2} < ... $
I want to make the range being partitioned in this way, with "infinitely many terms".
Any recommendation?
 A: You may use numbers of the form $\frac{n}{n+1}$ such as $\frac{2}{3}$ for rational numbers and their geometric mean $\sqrt{\frac{n}{n+1}\frac{n+1}{n+2}}=\sqrt{\frac{n}{n+2}}$ in between. 
This way, your sequence would start
$$\frac{1}{2}<\sqrt{\frac{1}{3}}<\frac{2}{3}<\sqrt{\frac{1}{2}}<\frac{3}{4}<\sqrt{\frac{3}{5}}<\frac{4}{5}<\sqrt{\frac{2}{3}}<...$$
Starting at $n=0$, with rational $a(2n)$ and irrational $a(2n+1)$, the closed form is
$$a(n)=\frac{1+(-1)^n}{2}\frac{n+2}{n+4}+\frac{1-(-1)^n}{2}\sqrt{\frac{n+1}{n+5}}$$
Equivalently,
$$\begin{align}
a(n)&=\sqrt{\frac{(2n+3+(-1)^n)(2n+5-(-1)^n)}{(2n+7+(-1)^n)(2n+9-(-1)^n)}}\\
&=\sqrt{\frac{2n^2+8n+7+(-1)^n}{2n^2+16n+31+(-1)^n}}\\
\end{align}$$
A: More abstractly let  $(A_n)_{n\in \mathbb N}$ and $(B_n)_{n\in \mathbb N}$ be strictly increasing sequences in  $[0,1]$ with $\lim_{n\to \infty}A_n=\lim_{n\to \infty}B_n,\;$ with $A_n$ rational and $B_n$ irrational. For example $A_n=1-2^{-n}$ and $B_n=1-2^{-n}/\sqrt 2\;. $ 
Construct the desired sequence  $(C_n)_{n\in \mathbb N}$ recursively:
Let $C_1=A_1.$ 
Let $f(2n)=\min \{m:B_m>C_{2n-1}\}.$ Let $C_{2n}=B_{f(2n)}.$
Let $f(2n+1)=\min \{m:A_m>C_{2n}\}.$ Let $C_{2n+1}=A_{f(2n+1)}.$
