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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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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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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$\lfloor x\rfloor$ 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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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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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.
$$\lim_{n\to\infty}\left\lfloor\sum_{k=1}^n\frac9{10^k}\right\rfloor=0\\ \left\lfloor\lim_{n\to\infty}\sum_{k=1}^n\frac9{10^k}\right\rfloor=1\\$$