# Prove that $\lfloor(n+1)a\rfloor-1$ is divisible by $(n+1)$ if $n= \left\lfloor \frac {1}{ a- \lfloor a \rfloor } \right\rfloor$

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.

• A good title for this might be, "Prove that $\lfloor(n+1)a\rfloor-1$ is divisible by $n+1$ if $n= \left\lfloor\frac{1}{a-\lfloor a\rfloor}\right\rfloor$." – David K Sep 13 '16 at 13:57
• 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. – Jade Sep 14 '16 at 3:34
• 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. – David K Sep 14 '16 at 6:07

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 $n + 1$.
• 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. – Jade Sep 14 '16 at 3:44