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My question is if it is possible to claim that $(0.11111111...)_{2}=(10)_{2}$

Here is my approach. I started out by trying to convert this recurring number to base 10. This can be expressed as a sum, i.e., $$\sum{\frac{1}{2^n} } = 1 + \frac{1}{2} +\frac{1}{4}+\frac{1}{8}+...\frac{1}{2^n}$$ This turns out to be a convergent sum, hence, $$\sum{\frac{1}{2^n} } = \frac{1}{1-0.5}=2$$ This means that the recurring number in base 2 is equal to 2 in base 10. However, 2 in base 10 can be expressed as 10 in base 2, or with other words, $$(0.11111111...)_{2}=(2)_{10}$$ $$(2)_{10}=(10)_2$$ $$(0.11111111...)_{2}=(10)_2$$ $$QED$$ Do you agree with my reasoning?

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In your first sum, the 1 should not be there; it should start with $\frac{1}{2}$. –  Josh B. Apr 19 '13 at 17:52
    
This is like math.stackexchange.com/questions/11/does-99999-1 except with base 2 –  lab bhattacharjee Apr 19 '13 at 17:59
    
Not quite. It basically asks if 1 = 2. –  Joe Z. Apr 19 '13 at 19:08
    
Thank you for the feedback, Josh B., Joe Z. and lab bhattacharjee! –  Artem Apr 20 '13 at 8:06

2 Answers 2

up vote 6 down vote accepted

You have an off-by-one error. Your $\displaystyle \sum \frac{1}{2^n}$ is counting from $n = 0$ when it should be counting from $n = 1$.

So in reality, the convergent sum is $\displaystyle \sum_{n = 1}^\infty \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1$, as expected.

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It would be correct except that the sum starts with $1$, so your claim should be $(1.111\ldots)_2=(10)_2$

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Thank you for the feedback, Ross Millikan! –  Artem Apr 20 '13 at 8:07

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