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This question already has an answer here:

For a ceiling and floor function, the number is taken to 0 decimal places. Does this process mean that 0.9 recurring inside a floor function would go to 0? Or would the mathematician take 0.9 recurring to be equal to 1, thus making the answer 1?

And if 0.9 recurring does equal 1, does that mean (by definition) that ⌊1⌋ = 0?

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marked as duplicate by user26486, Hans Engler, Nate Eldredge, N. F. Taussig, Gabriel Romon Mar 16 '15 at 19:19

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    $\begingroup$ $0.\overline 9$ is equal to one, so the floor of it is also equal to one. $\endgroup$ – Gregory Grant Mar 16 '15 at 11:58
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    $\begingroup$ The definition of the floor function is not "truncate all digits after the decimal point". Indeed, as you've just shown, such a function is not even well-defined - it depends on the particular decimal representation you choose. $\endgroup$ – Jack M Mar 16 '15 at 14:46
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$$\lim_{n\to\infty}\left\lfloor\sum_{k=1}^n\frac9{10^k}\right\rfloor=0\\ \left\lfloor\lim_{n\to\infty}\sum_{k=1}^n\frac9{10^k}\right\rfloor=1\\$$

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  • $\begingroup$ Thank you!, It makes much more sense now you put it like that. $\endgroup$ – user3344560 Mar 16 '15 at 12:02
  • $\begingroup$ @user3344560: I'm glad to be of help. $\endgroup$ – Regret Mar 16 '15 at 12:03
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    $\begingroup$ In other words, the floor function is (left-)discontinuous at 1, which hopefully surprises nobody :-) $\endgroup$ – Steve Jessop Mar 16 '15 at 13:46
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You have to look at the $.9$ recurring as a sum... then you'll know the answer.

$$\bar{.9} = \sum_{i=1}^{\infty} \frac{9}{10^{i}}$$

So, $$\lfloor \bar{.9}\rfloor = \left\lfloor \sum_{i=1}^{\infty}\frac{9}{10^{i}}\right\rfloor=\lfloor 1 \rfloor = 1.$$ You cannot split up the floor function over a sum, i.e. $\lfloor a+b\rfloor \neq \lfloor a\rfloor + \lfloor b\rfloor$.

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  • $\begingroup$ Travis, I am not sure you have the exponent correct.$$0.999...=0.9+0.09+0.009+...\ne 0.9+0.81+0.729+...$$ $\endgroup$ – Regret Mar 16 '15 at 12:09
  • $\begingroup$ @Regret, thank you. You are absolutely correct. I attempted to make an edit to your answer (removing the first limit, and re-parenthesizing... that was a mistake on my part) $\endgroup$ – TravisJ Mar 16 '15 at 12:11
  • $\begingroup$ @Regret, is there anyway to "un-edit" an answer (I'm new to the site). $\endgroup$ – TravisJ Mar 16 '15 at 12:12
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    $\begingroup$ I have rejected the edit you proposed to my answer, if that is the edit you are talking about. $\endgroup$ – Regret Mar 16 '15 at 12:13
  • $\begingroup$ @Regret, yes. Thanks. I misunderstood the limit. I thought you intended to be a finite sum (top) and the infinite sum on bottom, but you really pointed out the same phenomena (the limit cannot be brought in and out of a floor). $\endgroup$ – TravisJ Mar 16 '15 at 12:15
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As Gregory Grant says, $\lfloor0.9\overline{9}\rfloor = 1$; your phenomenon illustrates the jump discontinuity in the floor function at each integer.

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$\lfloor x\rfloor$ is defined as the unique integer $n$ such that $n\leq x<n+1$. Because $0.999...=1$ we have $1\leq 0.999...$ and obviously $0.999...<2$, so $\lfloor 0.999...\rfloor=1$.

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The question embeds an alternate question of what is the division rule that can produce $0.\bar{9} = 0.999999$ (recurring)?

A normal (long) division rule produces an integer (complete division) and a remainder (aka "floor"). The number $0.\bar{9} = 0.999999$ (recurring) does not exist as the result of a classic division, hence the confusion.

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Because $0.\bar{9} = 0.999999$ (recurring) is equal to $1$, the floor function of it is equal to the floor of $1$, and that is $1$.

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