Re: closest integer value

Here is my solution: If there are mistakes could you please let me know? Thank you so much (Clapping)

If we look at $\displaystyle x=0.b_{1} b_{2} \cdots b_{m}$ then

$\displaystyle 2^{m} \cdot x = 2^{m} \sum_{i=1}^{m} b_{i} 2^{-i} = \sum_{i=1}^{m} b_{i} 2^{m-i} = b_{1} 2^{m-1} + b_{2} 2^{m-2} \cdots + b_{m} 2^{m-m}$

will be an integer.

When we look at $\displaystyle x=\overline{0.b_{1} b_{2} \cdots b_{m}}$ we get

$\displaystyle 2^{m} \cdot \overline{0.b_{1} b_{2} \cdots b_{m}} = 2^{m} \cdot \left(\sum_{i=1}^{m} b_{i} 2^{-i} + \underbrace{0.0 \cdots 0}_{m \ zeros} \overline{b_{1} \cdots b_{m}}\right) = \\ 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + 2^{m} \cdot \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i-m} = 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i}$

Thus

$\displaystyle \{ 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \} = \{\sum_{i=1}^{m} \overline{b_{i}} \} $

We can repeat such a procedure k times and that is why we get the result.

I want to ask: What do you think of that solution? Thanks! :)

Re: closest integer value

Hello guys!

I donīt want to bother but I really wonder what you think of my solution? I just want to make sure I donīt have any big mistakes in there :)

Thank you!