Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Since the $\lfloor n b \rfloor / \lfloor n a \rfloor$ form a sequence of positive integers converging to $b/a$, it follows that $b = m a$ for some integer $m \ge 2$, and $\lfloor n b \rfloor = m \lfloor n a \rfloor$ for all $n$ large enough. Consequently, if $n$ is large enough, then $\lfloor n m a \rfloor = m \lfloor n a \rfloor$, so $m a < m \lfloor n a \rfloor + 1$; that is, $n a < \lfloor n a \rfloor + 1/m \le \lfloor n a \rfloor + 1/2$. Hence $\{ n a \} = n a - \lfloor n a \rfloor < 1/2$, so the set $\{ \{ n a \} : n \in \mathbb{Z}_+ \}$ is not dense in the closed unit interval $[0, 1]$, and $a$ must be rational, say $a = p/q$, where $p$ and $q$ are coprime positive integers. If $q \ge 2$, choose $n$ large enough such that $n p \equiv -1 \pmod q$, to reach a contradiction: $1/2 > \{ n a \} = \{ n p / q \} = \{ (q-1)/q \} = 1 - 1/q \ge 1/2$. Consequently, $q = 1$ and the conclusion follows.