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

Given and irrational $a$ and a natural number $n$ prove that $\lfloor an \rfloor +\lfloor (1-a)n \rfloor = n-1 $.

Is this solution correct?

$\lfloor an \rfloor +\lfloor (1-a)n \rfloor = \lfloor an \rfloor +\lfloor n-na \rfloor =$ (we take out $ n $ because it's an integer) $ \lfloor an \rfloor +n - \lfloor - an \rfloor =$ (because floor of a negative number is a negative of the ceiling of it's positive equivalent) $ \lfloor an \rfloor +n - \lceil an \rceil = n-1$

share|cite|improve this question
Looks correct to me. – Ryan Oct 11 '13 at 15:41
Thanks for the verification, I doubted it's correctness because I thought that nowhere did I use the fact that one of the numbers is irrational, but after posting I realised I did :) – Arek Krawczyk Oct 11 '13 at 15:43
up vote 1 down vote accepted

This is wrong, but probably just a typo: $\lfloor an \rfloor +n \color{red}{\bf -} \lfloor - an \rfloor$.

I suggest adding a minor comment that $an$ is not an integer (because $a$ is irrational). Without that, you cannot conclude that $\lfloor an \rfloor - \lceil an \rceil = -1$.

share|cite|improve this answer
Ah yes, thank you :) – Arek Krawczyk Oct 11 '13 at 15:49
@Vedran : I don't understand your answer. Are you saying that you have to add the assumption that $an$ is not an integer? Note that the given equation is false if you take $n=0$ (which most people consider a natural number). Of course, if $n=0$, then $an$ is an integer. – Stefan Smith Oct 11 '13 at 23:48
@StefanSmith No, many people (in areas of mathematics I work with, that would be all people) consider natural numbers to be $\mathbb{N} := \{1,2,\dots\}$. If you want zero, then it is $\mathbb{N}_0 := \{0\} \cup \mathbb{N}$. For example, sequences (which are $\mathbb{N} \to S$ functions) are indexed from $1$, not from $0$. Here, it is obvious that zero has to be excluded. Also, I didn't say that one has to make an assumption that $an$ is not an integer, but that one should emphasize that it is not (the reason is that $a$ is irrational; I am keeping the assumption that $n > 0$). – Vedran Šego Oct 11 '13 at 23:59
@VedranSego: Thank you. I also always begin sequences with $n=1$. It is unfortunate that there seems not be universal agreement on what a "natural number" is. Given that there is not universal agreement on what a natural number is, the problem should have specified that they start with $1$ here. It is not obvious that $0$ should be excluded: it is obvious that either $0$ should be excluded or there is a bug in the problem. I see bugs in problems here all the time. – Stefan Smith Oct 12 '13 at 1:18
@VedranSego : I admit I can't prove that "most" mathematicians consider zero a natural number (I haven't taken a poll of all of them), that's just the impression I've picked up over the years. Sorry for nitpicking, I just wanted to emphasize that the problem could be stated better. Nice solution. By the way, I use $\mathbb{N}^+$ for the positive integers. – Stefan Smith Oct 12 '13 at 1:28

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.