Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think $\lfloor0.999\dots\rfloor= 1$, as $0.999\dots=1$,but I have doubt, as $\lfloor0.9\rfloor=0$,$\lfloor0.99\rfloor=0$,$\lfloor0.9999999\rfloor=0$, etc.

share|cite|improve this question
@Ross This is not a duplicate... – user1729 Oct 16 '13 at 13:15
@user1729, Thanks. – Silent Oct 16 '13 at 13:16
(Or rather, the OP is wanting to understand the flaw in their argument rather than just "I want a proof of this fact!".) – user1729 Oct 16 '13 at 13:24
This is a good question. Thanks – Rustyn Oct 16 '13 at 13:45
Induction allows you to prove a statement about an infinite number of finite cases, it does not (usually) tell you anything about the infinite case itself. – DanielV Oct 16 '13 at 15:14
up vote 47 down vote accepted

Your first assertion is correct. The other observation just says that the function $x\mapsto\lfloor x\rfloor$ is not continuous.

share|cite|improve this answer
Thanks for so quick reply and additional information. – Silent Oct 16 '13 at 13:14
You're welcome. – Rasmus Oct 16 '13 at 13:15

For one thing, $0.999...$ is exactly equal to $1$.

To prove that $0.999... = 1$, we use sums of infinite geometric sequences. We know that $$\sum_{k=0}^\infty r^k = \frac{1}{1-r} \forall \;\left|r\right|\lt1$$ It is fairly simple to prove this statment, although I won't go into that. For our specific instance, we have $$0.999... = \frac{9}{10}\sum_{k=0}^\infty (\frac{1}{10})^k = \frac{9}{10} \cdot\frac{1}{1-\frac{1}{10}} = \frac{9}{10}\cdot\frac{10}{9}=1$$

Using the substitution property, we can substitute any alternate representation of a number into an equation, and it will yield the same result. Therefore,

$$\lfloor0.999...\rfloor = \lfloor1\rfloor = 1$$

share|cite|improve this answer
For more proofs, see here. – user1729 Oct 17 '13 at 10:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.