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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 $2^{n-1}$?

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  • $\begingroup$ Is this supposed to be in base 2? $\endgroup$ – John Wayland Bales May 17 '16 at 23:11
  • $\begingroup$ @JohnWaylandBales That's what the $(2)$ denotes. $\endgroup$ – user19405892 May 17 '16 at 23:12
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    $\begingroup$ In base ten, when you multiply by a power of ten it moves the decimal place. The same thing happens in base 2. $\endgroup$ – John Wayland Bales May 17 '16 at 23:13
  • $\begingroup$ If $2^{n-1} = 1$ then this equation is true. So this is true if n-1 = 0. so n = 1. That's all. $\endgroup$ – fleablood May 17 '16 at 23:30
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    $\begingroup$ @fleablood this equation is true for all n $\endgroup$ – ASKASK May 18 '16 at 2:32
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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.

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  • $\begingroup$ @user19405892 If my answer is unsatisfactory, please you might want to give me a feedback, so I might improve it. Thanks. $\endgroup$ – Olivier Oloa May 21 '16 at 16:29
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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.

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