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I saw a post on r/learnmath about this, but I wanted to hear what you guys think.

If I think in terms of sequences, I believe that it would be equal to zero, since 0.9999...9 (with n '9's) would always be less than 1 and greater than 0 for all natural n and hence, the floor would be zero (making the limit as n goes to infinity 0). However, if we first evaluate what's inside, then we'd get the floor of 1, which is just 1.

I personally think the first 'method' yielding 0 is more consistent with the idea of sequences.

Aqeel
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    You’ve just proven that the floor function is discontinuous at $x=1$ – Yalikesifulei May 18 '23 at 11:02
  • So, you think that $\left\lfloor\lim_{n\to\infty}1-\frac1n\right\rfloor=0$, right?! – Another User May 18 '23 at 11:02
  • @AnotherUser yes, that was what I was thinking of, but Lee Mosher's answer has corrected me by clarifying that 0.999.... is not a sequence and hence, the floor is 1 – Aqeel May 18 '23 at 11:08
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    @MartinR that was not my question, I already knew that it equals 1 because if we define it as the limit of a sequence to infinity, we simply obtain 1 because it's an infinite geometric series with starting term 0.9 and common ratio 0.1.

    My question has already been answered, but thanks for taking the time to refer me to the link. I'll check it out

     – Aqeel May 18 '23 at 11:11
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    I notice that you have a lot of questions but none of them have an accepted answer. Please keep in mind: When people answer your question, you can choose to accept one of the answers by clicking on the check mark. – Lee Mosher May 18 '23 at 11:12
  • @LeeMosher thanks, I'll do that from now on. – Aqeel May 18 '23 at 11:14
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    You ask for the floor of a single number (not a sequence). So , the first approach does not apply. And $0.\bar 9$ is a number , I disagree with the claim it is not. – Peter May 18 '23 at 11:34
  • I don't know what is 0.999999.... There is several notations for writing periodical decimal fractions, if you mean this. – Ivan Kaznacheyeu May 18 '23 at 14:06

1 Answers1

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Although (like every number) the number $0.9999\ldots = \lim_{n \to \infty} \left( 1 - \frac{1}{10^n}\right)$ is the limit of a sequence, the number $0.9999\ldots$ is not itself a sequence. It is a number. And it is equal to $1$. So its floor is $1$.

Lee Mosher
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  • Thanks, I get what you mean now. – Aqeel May 18 '23 at 11:07
  • Nitpicking, a "decimal number" isn't a number at all, but rather a representation of a number. Now for the sake of uniqueness of that representation it must not finally end in a sequence of infinite-many nines. Agree in this, and the (mathematical) world would be a (slightly) better one ... – Michael Hoppe May 18 '23 at 11:14
  • Uniqueness is not all its cracked up to be. – Lee Mosher May 18 '23 at 11:17