Prove that $$\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$$ where $\sqrt 3 = 1.b_1b_2b_3 \dots _{(2)}$. (Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)
I am not sure how to prove this result because of the fractional part. Should I expand $\sqrt{3}$ and then multiply by $2^{n-1}$?
On
This is a rather pointlessly complicated expression of something basically trivial.
Let $w = A.b_1b_2b3....._d = A + \sum_{i=1}^{\infty} b_i*d^{-i}$ be any number in any base $d$.
Then $d^m*w = Ab_1b_2..b_m.b_{m+1}b_{m+2}.... = A*d^m + \sum_{i=1}^{\infty}b_i*d^{m-i}$.
So $\{w*d^m\} = 0.b_{m+1}b_{m+2}.....$ for all $w$ (including $\sqrt 3$) for all $d$ (including 2) and for all $m$ (including $n -1$).
This says absolutely nothing more than "if you multiply by Base to the $m$th power, you shift the decimals over by $m$ places" which is something so utterly basic I'd expect every junior high school student to know it.
However the way this question was written was a masterwork in intimidating obfuscation. If our goal is to scare people away from mathematics so that we can enjoy a private elitist club, this was a very good question for that goal.
From $$ \sqrt 3 = 1.b_1b_2b_3 \cdots _{\,(2)} \tag1 $$ one gets, using $b_n \in \{0,1\}$, $$ \begin{align} \left\{2^{n-1}\sqrt 3\right\} &=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}+\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\} \\\\&=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}\right)+2^{n-1}\left(\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\} \\\\&=\left\{\underbrace{2^{n-1}+2^{n-2}b_1+2^{n-3}b_2+\cdots+b_{n-1}}_{\text{integer part}} +\overbrace{\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots}^{\text{fractional part}}\right\} \\\\&=\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots \\\\&=0.b_nb_{n+1}\cdots_{\,(2)} \end{align} $$ as announced.