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$

  • $\begingroup$ Looks correct to me. $\endgroup$ – Ryan Oct 11 '13 at 15:41
  • $\begingroup$ 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 :) $\endgroup$ – Arek Krawczyk Oct 11 '13 at 15:43

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 $\lfloor an \rfloor - \lceil an \rceil = -1$.

  • $\begingroup$ Ah yes, thank you :) $\endgroup$ – Arek Krawczyk Oct 11 '13 at 15:49
  • $\begingroup$ @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. $\endgroup$ – Stefan Smith Oct 11 '13 at 23:48
  • $\begingroup$ @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 $n > 0$). $\endgroup$ – Vedran Šego Oct 11 '13 at 23:59
  • $\begingroup$ @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. $\endgroup$ – Stefan Smith Oct 12 '13 at 1:18
  • $\begingroup$ @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. $\endgroup$ – Stefan Smith Oct 12 '13 at 1:28

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