limit of rational sequence Let $a$ and $b$ be two real number. Assume that there exits two real sequences $a_n, b_n$ such that $$ \lim a_n=1, \lim b_n=1$$ and all  $$\frac{a- a_n}{b-b_n} $$are rational numbers. 
Is it true that $\displaystyle \frac{a-1}{b-1}$  is a rational number?
 A: This is false. Try with the following:
$a=\pi+1$, $b=2$, $b_n=1\ \forall n\in\mathbb{N}$, and
$$
a_n = 1 + \left(\pi-\frac{\lfloor 10^n\pi\rfloor}{10^n}\right)
$$
A: No it is not true 
Suppose $c_n$ is a rational sequence with $\lim c_n=\sqrt{2}$,  {for example take the sequence $a_n$ where $a_n$ is defined inductively by $a_1=2$ and $a_{n+1}=\frac{1}{2}(a_n+\frac{1}{a_n})$}
, and $d_n=2c_n$. Now, set $a_n=\frac{c_n}{\sqrt{2}}$ and $b_n=\frac{d_n}{2\sqrt{2}}$, then $\lim a_n=\lim b_n=1$.
Also let  $a=\sqrt{2}, b=2\sqrt{2}$
$\frac{\sqrt{2}-a_n}{2\sqrt{2}-b_n}=\frac{2(2-c_n)}{8-d_n}$ is alway rational (I mean for any $n$) but $\frac{a-1}{b-1}$ is not.
A: Since you didn't require $b\not=1,$ we can just take $a-1=b=1, a_n=b_n=1+1/n,$ as $\infty\not\in\mathbb Q.$ :P
If it is required that $b\not=1,$ then take $a_n=b_n=(\sum_{k=0}^n\frac{1}{k!}/e), a=\frac{5}{e}, b=\frac{4}{e}.$ Thus $$\lim a_n=\lim b_n=1,$$
$$\frac{a-a_n}{b-b_n}=\frac{5-\sum_{k=0}^n\frac{1}{k!}}{4-\sum_{k=0}^n\frac{1}{k!}}\in\mathbb Q,$$
while $$\frac{a-1}{b-1}=\frac{5-e}{4-e}=1+\frac{1}{4-e}\not\in\mathbb Q.$$
This avails only of the fact that $e=\sum_{k=0}^n\frac{1}{k!}.$
Hope this helps. :)
