Correct $(0.11111111...)_{2}=(10)_{2}$?

Question

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?

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 http://math.stackexchange.com/questions/11/does-99999-1 except with base 2 — lab bhattacharjee, Apr 19 '13 at 17:59
Thank you for the feedback, Josh B., Joe Z. and lab bhattacharjee! — Artem, Apr 20 '13 at 08:06

score 6 · Accepted Answer · edited Apr 19 '13 at 17:58

6

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.

edited Apr 19 '13 at 17:58

Austin Mohr

25,662

answered Apr 19 '13 at 17:52

Joe Z.

6,719

score 4 · Answer 2 · answered Apr 19 '13 at 17:53

4

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

answered Apr 19 '13 at 17:53

Ross Millikan

374,822

Thank you for the feedback, Ross Millikan! – Artem Apr 20 '13 at 08:07

Correct $(0.11111111...)_{2}=(10)_{2}$?

2 Answers2