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Are the floor functions of $0.999\cdots$ and 1 equal?

It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ?

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    If they are the same number, then they obviously have the same integer part. The integer part isn't defined as "what goes before the decimal dot", it's the greatest integer smaller or equal to the number. In this case, $1 \leq 1$ and it's the greatest integer with this property.2012-02-01
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    If $a=a$, then $f(a)=f(a)$. Now let $a=0.999...=1$ and $f(.) = \lfloor .\rfloor$. :-)2012-02-01
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    How would you justify the integer part of $0.999\ldots$ *not* being 1?2012-02-01
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    You gave the full explanation: $0.9999\dots$ **is** $1$. So if I say that the number of bicycles I have is $0.9999\dots$, I am saying, admittedly in a peculiar way, that I have $1$ bicycle. The two expressions $0.9999\dots$ and $1$ denote the same object. So $(0.9999\dots)^2=1$ (or, if you prefer, $0.9999\dots$).2012-02-01
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    @Zev : I can't, I don't know how to justify integer part being 1 without using 1 instead of 0.999.. .2012-02-01
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    Consider [the definition](http://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Definition_and_properties). If you agree that $0.999\ldots=1$, then certainly $1\leq 0.999\ldots$, so it can't be true that $\lfloor 0.999\ldots\rfloor=0$ - this is the point zulon made above.2012-02-01
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    @AndréNicolas : Yes they are the same, but the same thing in the form 0.999.. seems to have no integer part, so how come one representation makes it obvious that 1 is the integer part and the other is not too easy to accept when staring at a 0 before the decimal?2012-02-01
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    Because reals don't have unique decimal expansions (same with other bases).2012-02-01
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    @Arjang: You make a good point, one representation of an object can make a fact much more obvious than another reresentation of the same object. As a simple example, from the decimal representation of an integer $n$, it is trivial to check whether $n$ is divisible by $10$, just look at the last digit. If we write $n$ in base $7$, checking whether it is divisible by (decimal) $10$ is more work. Fairly big chunks of mathematics involve finding representations that make proving a theorem possible.2012-02-01
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    @AndréNicolas : The 0 sitting in the integer part of $0.999\cdot$ was making it annoying to accept that despite the fact of the two being the same thing, it is a 1! , but the anon reminding the fact of non unique representation and that limits can not be interchanged left no unsettling points.2012-02-01
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    @ZevChonoles "How would you justify the integer part of 0.999… not being 1?" : The 0 sitting in the integer position, and that was causing the problem.2012-02-01

2 Answers 2

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The floor function is a discontinuous function. This means that limits cannot be interchanged before and after the action of the function, i.e. $f(\lim\limits_{n\to\infty}x_n)\ne \lim\limits_{n\to\infty} f(x_n)$ generally. In particular,

$$\lim_{n\to\infty}\left\lfloor 0.\underbrace{99\cdots9}_n \right\rfloor=\lim_{n\to\infty}0=0 $$

$\hskip 3.2in$ but

$$\left\lfloor \lim_{n\to\infty} 0.\underbrace{99\cdots9}_n \right\rfloor=\lfloor1\rfloor=1. $$

Now the expression $\lfloor0.999\dots\rfloor$ denotes the latter, which is $1$, but intuitively and at face value one notices that the floor function evaluated at the partial sums always returns $0$, so we might be tempted to accept the first formula above as the real answer, but appearance $\ne$ reality in general.

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    +1 Bingo, interchanging limits lurking underneath! Of course that makes sense.2012-02-01
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    A similar reasoning also explains why $0.\underbrace{9999 \cdots 9}_{n} < 1$ for all $n$, but in the limit $0.999\cdots = 1$. This is because the indicator function $1_{(-\infty, 1)}$ is discontinuous at $1$.2012-02-01
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The floor function $\lfloor x \rfloor$ is càdlàg (continue à droite, limite à gauche, i.e. right continuous with left limits).

Since there is no point between $0.999\ldots$ and $1$, and the real numbers are continuous, this right continuity implies $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$.

(Since the identity function is also càdlàg, you can use a similar argument to show $0.999\cdots = 1$.)

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    true, but this also is using the fact that they are equal.2012-02-01
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    I do not understand this answer. Are you saying that $f(1)=f(1)$ because $f$ is right-continuous and the sequence $(1,1,1,\ldots)$ decreases to $1$? What do you mean by "the real numbers are continuous"?2012-02-01
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    @Jonas: I was trying to play Arjang's game, avoiding saying $0.999\ldots$ and $1$ but instead saying there was nothing between them so a function which is right continuous takes the same value at $0.999\ldots$ as it does at $1$.2012-02-01