# How do we prove that $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$?

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 $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ?

• 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. – Najib Idrissi Feb 1 '12 at 7:31
• If $a=a$, then $f(a)=f(a)$. Now let $a=0.999...=1$ and $f(.) = \lfloor .\rfloor$. :-) – Myself Feb 1 '12 at 7:33
• How would you justify the integer part of $0.999\ldots$ not being 1? – Zev Chonoles Feb 1 '12 at 7:35
• 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, $0.9999\dots$). – André Nicolas Feb 1 '12 at 7:35
• @Zev : I can't, I don't know how to justify integer part being 1 without using 1 instead of 0.999.. . – Arjang Feb 1 '12 at 7:39

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.

• +1 Bingo, interchanging limits lurking underneath! Of course that makes sense. – Arjang Feb 1 '12 at 8:03
• 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 $1$. – Srivatsan Feb 1 '12 at 8:20

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 $0.999\cdots = 1$.)

• true, but this also is using the fact that they are equal. – Arjang Feb 1 '12 at 8:01
• 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"? – Jonas Meyer Feb 1 '12 at 19:42
• @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 $1$. – Henry Feb 1 '12 at 21:28