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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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    @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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    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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    @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