I think $\lfloor0.999\dots\rfloor= 1$, as $0.999\dots=1$,but I have doubt, as $\lfloor0.9\rfloor=0$,$\lfloor0.99\rfloor=0$,$\lfloor0.9999999\rfloor=0$, etc.
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4@Ross This is not a duplicate... – user1729 Oct 16 '13 at 13:15
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@user1729, Thanks. – Silent Oct 16 '13 at 13:16
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1(Or rather, the OP is wanting to understand the flaw in their argument rather than just "I want a proof of this fact!".) – user1729 Oct 16 '13 at 13:24
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1This is a good question. Thanks – Rustyn Oct 16 '13 at 13:45
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3Induction allows you to prove a statement about an infinite number of finite cases, it does not (usually) tell you anything about the infinite case itself. – DanielV Oct 16 '13 at 15:14
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Your first assertion is correct. The other observation just says that the function $x\mapsto\lfloor x\rfloor$ is not continuous.
Rasmus
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For one thing, $0.999...$ is exactly equal to $1$.
To prove that $0.999... = 1$, we use sums of infinite geometric sequences. We know that $$\sum_{k=0}^\infty r^k = \frac{1}{1-r} \forall \;\left|r\right|\lt1$$ It is fairly simple to prove this statment, although I won't go into that. For our specific instance, we have $$0.999... = \frac{9}{10}\sum_{k=0}^\infty (\frac{1}{10})^k = \frac{9}{10} \cdot\frac{1}{1-\frac{1}{10}} = \frac{9}{10}\cdot\frac{10}{9}=1$$
Using the substitution property, we can substitute any alternate representation of a number into an equation, and it will yield the same result. Therefore,
$$\lfloor0.999...\rfloor = \lfloor1\rfloor = 1$$
AJMansfield
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