Point 9 recurring: This equals 1, right?

Yes. There are many, many proofs of this, but one I prefer is to think of it in terms of a series.

\(\displaystyle 0.\overline{99} = 0.9 + 0.09 + 0.009 + \cdots = \sum\limits_{k = 1}^{\infty} \frac{9}{10^k} = 9 \sum\limits_{k = 1}^{\infty} \frac{1}{10^k} \)

This, then, is a geometric series with parameter \(r = \frac{1}{10}\). The infinite sum of any geometric series is:

\(\displaystyle \sum\limits_{k = 1}^{\infty} r^k = -\frac{r}{r-1}\)

So the infinite sum of our series must be:

\(\displaystyle 9 \sum\limits_{k = 1}^{\infty} \frac{1}{10^k} = -9 \cdot \frac{\frac{1}{10}}{-\frac{9}{10}} = -9 \cdot -\frac{1}{9} = 1\)
 
Cat, I know you knew this fact. So what exactly are you really saying or trying to do? Is it just to annoy Denise? I thought that was my job.
 
Jomo said:
Cat, I know you knew this fact. So what exactly are you really saying or trying to do? Is it just to annoy Denise? I thought that was my job.
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I think Denis was trying to disallow \(\displaystyle .\dot{9} = 1 \) in his little number game on another post. (You know the one!)
 
Seriously I do not know the one. I thought that it was a recent one. Please give me the link
 
OK, I found it. Not sure why Denis' little number game did not ring a bell.
 
ksdhart2 said:
Yes. There are many, many proofs of this, but one I prefer is to think of it in terms of a series.

\(\displaystyle 0.\overline{99} = 0.9 + 0.09 + 0.009 + \cdots = . . . \)
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\(\displaystyle 0.\overline{9} \ \ \) is sufficient for the number of nines.

Actually, what you wrote should translate to "0.99 + 0.0099 + 0.000099 + ..."

And that is equivalent to 1 as well.
 
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