proving $\big \lfloor xn-\frac 1n \big\rfloor\neq \big\lfloor xn+\frac 1n\big\rfloor$

Question

How can I prove that for any real number $x$ there are infinitely many integers $n$ such:$$\Big \lfloor xn-\frac 1n \Big\rfloor\neq \Big\lfloor xn+\frac 1n\Big\rfloor$$

Answer 1

It is immediately obvious for rational $x$, since there are infinite natural numbers $n$ such that $xn$ is an integer, and therefore $\lfloor xn-\epsilon\rfloor\neq\lfloor xn+\epsilon\rfloor$ for any strictly positive $\epsilon$.

For irrational $x$, we need some extra tools. Specifically, Dirichlet's approximation theorem which says that for any irrational number $x$ there are infinitely many "good" rational approximations $\frac pq$ that satisfy $$\left|x-\frac pq\right|<\frac1{q^2}$$I believe these are precisely the continued fraction estimates for $x$, but this was a long time ago for me.

Anyways, any of those estimates are such that $|xq-\lfloor xq\rfloor|<\frac1q$, so that fits your needs.