3
$\begingroup$

I came across the following question across a math contest and was wondering how to solve it.

Let a be a positive real number that is not an integer and let

$$ n= \left\lfloor \frac {1}{ a- \lfloor a \rfloor } \right\rfloor $$

Prove that $\lfloor (n+1)a \rfloor -1 $ is divisible by $n+1$.

So I played around some values and got that that the quotient would be $\lfloor a \rfloor$. Would it be rigorous enough to prove that $\lfloor a \rfloor (n+1) = \lfloor (n+1)a \rfloor -1 $ if we have the above definition of $n$. Or would you recommend another approach?

Thanks.

|cite|improve this question
$\endgroup$
  • $\begingroup$ A good title for this might be, "Prove that $\lfloor(n+1)a\rfloor-1$ is divisible by $n+1$ if $n= \left\lfloor\frac{1}{a-\lfloor a\rfloor}\right\rfloor$." $\endgroup$ – David K Sep 13 '16 at 13:57
  • $\begingroup$ Thanks for the suggestion, I made the fix, however, I don't know how to format the title as nice as how it is displayed in the question. $\endgroup$ – Jade Sep 14 '16 at 3:34
  • $\begingroup$ It's the same markup, just copied from one box to the other while editing. I went ahead and did that step, hoping it would be OK with you. $\endgroup$ – David K Sep 14 '16 at 6:07
0
$\begingroup$

The doubt seems to be due to the fact that the quotient $\lfloor a \rfloor $ comes from trying a few examples with literal numbers.

This procedure for coming up with $\lfloor a \rfloor $ is of course not a rigorous way of finding the quotient of the division by $n+1$, rather just a way to conjecture (really just to guess) what the quotient should be. But once you have such a guess, no matter how you guessed it, if you can then prove rigorously that $\lfloor a \rfloor (n+1) = \lfloor (n+1)a \rfloor -1 $ in all cases, not just in some cases, that is a perfectly rigorous proof that $\lfloor (n+1)a \rfloor - 1$ is divisible by $n + 1$.

Since your conjectured fact is true (I checked), I hope you will not have too much difficulty proving it.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ So the process is right, i'ts just a matter of actually doing it now. So I rearranged to get $1 = \lfloor (n+1)a \rfloor− ⌊a⌋(n+1)$. At this point I'm hoping to factor n on the right side and divide by the other factor to get my definition of $n$. However, how do I do that? I can't just pull terms out of the floor function without changing it's value. $\endgroup$ – Jade Sep 14 '16 at 3:44
  • $\begingroup$ I've already misplaced my notes, but it's often useful to assign names to the whole and fractional parts of some things. If you choose names wisely you can convert your equation to one without floor functions, which I think is easier to work with. $\endgroup$ – David K Sep 14 '16 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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