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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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    $\begingroup$ @Ross This is not a duplicate... $\endgroup$
    – user1729
    Oct 16, 2013 at 13:15
  • $\begingroup$ @user1729, Thanks. $\endgroup$
    – Silent
    Oct 16, 2013 at 13:16
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    $\begingroup$ (Or rather, the OP is wanting to understand the flaw in their argument rather than just "I want a proof of this fact!".) $\endgroup$
    – user1729
    Oct 16, 2013 at 13:24
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    $\begingroup$ This is a good question. Thanks $\endgroup$
    – Rustyn
    Oct 16, 2013 at 13:45
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    $\begingroup$ Induction 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. $\endgroup$
    – DanielV
    Oct 16, 2013 at 15:14

2 Answers 2

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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.

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  • $\begingroup$ Thanks for so quick reply and additional information. $\endgroup$
    – Silent
    Oct 16, 2013 at 13:14
  • $\begingroup$ You're welcome. $\endgroup$
    – Rasmus
    Oct 16, 2013 at 13:15
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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,

$$\lfloor0.999...\rfloor = \lfloor1\rfloor = 1$$

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    $\begingroup$ For more proofs, see here. $\endgroup$
    – user1729
    Oct 17, 2013 at 10:35

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