# Show that either there exists an $N$ so that $r_n=r$ for $n \ge N,$ or the set of numbers $\{q_n\}$ is unbounded.

Let $(r_n)$ be a sequence of positive rational numbers that converges to a rational number $r$. Suppose that $r_n = \frac{p_n}{q_n}$, is in lowest terms.

Show that either there exists an $N$ such that $r_n=r$ for $n \ge N,$ or the set of numbers $\{q_n\}$ is unbounded.

If we suppose that $\{q_n\}$ is bounded, then $\{q_n\}$ is a finite set of integers.

I think we want to show that if $\{q_n\}$ is bounded and $r_n\to r$, then $r_n=r$ after $n \ge N$.

But, I can't see how this alone along with $r_n \to r$ would tell me $r_n =r$ for $n \ge N$.

• Well, if there is a finite number of possible denominators, how close could you come to $r$ without hitting it exactly? – spaceisdarkgreen Dec 17 '17 at 22:53
• @spaceisdarkgreen I don't know how to answer that rigorously. – Al Jebr Dec 17 '17 at 22:59
• For any given $q,$ there will be a $p$ that gets you closest. So running through the finite number of q's, for each one, you can find a corresponding $p$ that either hits $r$ dead on or is the closest possible. The closest $p/q$ you find here is the closest you will ever get. Either it's dead on, or you do not come arbitrarily close. – spaceisdarkgreen Dec 17 '17 at 23:12
• @spaceisdarkgreen Ahh, I see now. Got it – Al Jebr Dec 18 '17 at 0:25