I am trying to solve the following problem:
For what real numbers x is: $\lfloor{2x}\rfloor = 3\lfloor{x}\rfloor$?
I'm not sure how to deal with the floor functions, so I have no idea where to start. If someone could walk me through the process that would great!
Best Answer
HINT: Let $n=\lfloor x\rfloor$, so that $n\le x<n+1$. Let $\alpha=x-n$, the fractional part of $x$, so that $x=n+\alpha$. You’re looking for those $x$ such that $\lfloor 2x\rfloor=3\lfloor x\rfloor$, i.e., such that $\lfloor 2(n+\alpha)\rfloor=3n$.
Clearly $\lfloor 2(n+\alpha)\rfloor=\lfloor 2n+2\alpha\rfloor$, and because $2n$ is an integer, $\lfloor 2n+2\alpha\rfloor=2n+\lfloor 2\alpha\rfloor$. Can you finish it from here?