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We all know that $0.99999...=1$

So does that imply $[0.99999...]=1?$

Or do we consider it as $0?$

My doubt is: any gif of the form $[0.xyz...]=0$. If $[0.99999...]=1$ won't that be contradicting?

There isn't much clear explanation in the previous post.

Here $[.]$ is the greatest integer function. I couldn't find a post containing this query on MSE. Please help :)

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Aditya Kumar
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    Yes, $[0.999\dots]=1$ because $0.999\dots=1$. – Wojowu Jan 16 '16 at 09:00
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    [1] = 1, 0.999… = 1, therefore by the transitivity of equality, [0.999...] = 1. QED. – bjb568 Jan 16 '16 at 09:00
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    Since $0.999...=1$, it’s automatic that $\lfloor0.999...\rfloor=\lfloor 1\rfloor=1$. – Brian M. Scott Jan 16 '16 at 09:00
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    @bjb568 Actually you aren't using transitivity, but rather Leibniz's Law. – Wojowu Jan 16 '16 at 09:02
  • @Wojowu but any gif of the form [0.xyz...] is considered to be equal to 0. So won't this be contradicting? – Aditya Kumar Jan 16 '16 at 09:02
  • Did you link to an answer rather than to the question about $0.99999999999.....=1$ on purpose, or was it a mistake? – Martin Sleziak Jan 16 '16 at 09:03
  • @AdityaKumar The thing is that [0.xyz...]=0 isn't true for all x,y,z,..., and 0.999... is a sole counterexample. – Wojowu Jan 16 '16 at 09:04
  • @AdityaKumar Maybe you could ad you previous comment about [0.xyz.....]=0 to the post. (This could clarify what is the main point that is causing you trouble. And it could also help in deciding whether or not this is actually a duplicate of the previous question.) – Martin Sleziak Jan 16 '16 at 09:05
  • Indeed $[0.99999....]=1$ shows in a certain sense that $[\cdot]$ is not continuous on $\mathbb{N}$ – sinbadh Jan 16 '16 at 09:15
  • @sinbadh: You mean "not continuous on $\mathbb{R}$". And I think this is precisely what the OP's confusion is about; it only appears to be contradition if one believes that all functions must be continuous, so that one can always interchange the order of the operations "applying the function" and "taking the limit". – Hans Lundmark Jan 16 '16 at 10:07

2 Answers2

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Let's recall the definition. $\lfloor x\rfloor$ is the greatest integer $n$ satisfying $n\leq x$. But now we have $1\leq 0.999\dots$, and no integer greater than $1$ doesn't satisfy $n\leq 0.999\dots$, so, from the definition, $\lfloor 0.999\dots\rfloor=1$.

What probably confuses you is the mentioned fact that $\lfloor 0.xyz\dots\rfloor=0$ for any $xyz\dots$. The problem is that this is wrong. It certainly doesn't follow from the definition above, though it might follow from some informal definitions (of the sort "remove everything after decimal point"). Indeed, there is precisely one counterexample to $\lfloor 0.xyz\dots\rfloor=0$, namely $0.xyz\dots=0.999\dots$.

Wojowu
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If $x = y$ then $[x] = [y]$. How could they not?

So as $0.9999.... = 1$ it has to be that $[0.9999...] = [1]$.

Equality means they are the same thing. If they are the same thing whatever you do to them will have the same result.[*]

[*](Provided what you do to them is based on their value. It's possible for things to not be "well-defined" which means they will not have consistent results based on circumstances that aren't about their value. Operations that are not "well-defined" are not considered to be valid. Usually. There are always exceptions.)

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Ah, I didn't see the $[0.xyz...] = 0$ confusion.

Well, that just isn't true (although 0.9999.... is the only exception).

Well.... You have to keep in mind 0.9999.... is an integer even if it doesn't look like one.

Keep in mind math results are based on what thing are; not what they look like. 0.999.... is the only exception to 0.xyz.... < 1. And it's because 0.999.... $\not <$ 1 it doesn't follow that [0.9999... ] < 1.

Believe me. If I could apologize for the confusion, I would.

fleablood
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  • This seems to be convincing. :) – Aditya Kumar Jan 16 '16 at 09:15
  • Please tell me how $1$ and $2$ are the same thing since $1/2=2/4$ and we can take numerators of both sides :) – Wojowu Jan 16 '16 at 09:16
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    @Wojowu I was worried you'd say something like that. I even thought I'd bring it up but thought that just make matters worse. There is a concept of an operation being "well-defined" which means it is defined to give consistent results. "taking the bottom number when you write it" is not well defined. "taking the numerator" isn't either. But "taking the numerator of a rational number express as a ratio m/n where m/n are integers with no common factors and n is positive" is. – fleablood Jan 16 '16 at 09:27
  • "Let x = 49 and y = 16 and z = 37 and so on. Take the numbers 1 to 100 and put them in alphabetical order based on their variables" is utterly NOT well-defined. – fleablood Jan 16 '16 at 09:31
  • .333.... does equal 1/3. But [0.333....] = 0. – fleablood Jan 16 '16 at 09:38