What is $\left\lfloor0.\overline{9}\right\rfloor$?

Question

We know that $0.\overline{9} = 1$ but then what is $\left\lfloor0.\overline{9}\right\rfloor$?

My thought process went: $0.\overline{9} = 1$ so therefore $\left\lfloor0.\overline{9}\right\rfloor = \left\lfloor1\right\rfloor$

But also $\left\lfloor0.9\right\rfloor = 0$ and it shouldn't change no matter how many numbers you add on to the back of it.

So what is the right answer, if there is one, and why?

Answer 1

As you say, $0.\overline9=1$. Thus, $0.\overline9$, like $2-1$, $\frac33$, etc., is just a different notation for the number $1$, and consequently $\left\lfloor0.\overline9\right\rfloor=\lfloor1\rfloor=1$.

Added: The fact that $\lfloor 0.9\ldots9\rfloor=0$ for any finite string of $9$s is irrelevant: the floor function is not continuous from the left at integers. In effect you’re saying that $\lfloor 1\rfloor$ ought to be $0$ simply because $\left\lfloor 1-10^{-n}\right\rfloor=0$ for each $n\in\Bbb Z^+$.

Answer 2

You said:

But also $\left\lfloor0.9\right\rfloor = 0$ and it shouldn't change no matter how many numbers you add on to the back of it.

That is a common mistake when dealing with repeating decimals, one tends to think, that you just add more and more $9$'s and get closer and closer to $1$. In my oppinion, this is because we normally think only in some "finite sense". To a certain degree, everything in our everyday world is finite, so it might be difficult, to think of $0.\overline{9}$ as really infinite many $9$'s. The representation $0.999\dots$ does not help either, as there are only a finite number of $9$'s displayed.

So: for $\lfloor 0.9\rfloor=0$ it does not matter, if you add "some $9$'s" on to the back of it, but once you get to $0.\overline{9}$, it does matter.

Answer 3

To me the intuitive explanation is that if you put any number of 9s after the 0, $\lfloor 0.999...99 \rfloor =0$. But $0.\overline{9}$ has infinite 9s after the 0, and infinity isn't a number. So the pattern doesn't hold.

Cameron Williams pointed out the rigorous answer, which is that you're implicitly assuming that because a sequence of numbers (0.9,0.99,0.999...) gets closer to 1, then if you apply a function to them (in this case, $\lfloor x \rfloor$), the the resulting sequence ($\lfloor 0.9\rfloor$,$\lfloor 0.99\rfloor$,$\lfloor 0.999\rfloor$...) should also get closer to $\lfloor 1\rfloor$. But functions don't always do this. If a certain function does have this property, we say that the function is continuous at that point, but $\lfloor x\rfloor$ is not continuous at 1. In fact, you've more or less given a proof that $\lfloor x\rfloor$ is discontinuous at 1.