A question about limit and sequence

Question

Let $\{p_n\},\{q_n\}$ be sequences of positive integers such that $$ \lim_{n \to \infty}\frac{p_n}{q_n}=\sqrt{2}. $$

Show that $$\lim_{n\to\infty}\dfrac{1}{q_n}=0.$$

I have no clue in proving this. Could somebody give me some hints?

Answer 1

If not, then there is an $\varepsilon>0$, such that $$ \frac{1}{q_n}\ge\varepsilon, $$ for infinitely many $n$. So there are subsequences $p_{n_k},q_{n_k}$, $k\in\mathbb N$, such that $$ \frac{p_{n_k}}{q_{n_k}}\to\sqrt{2}\quad\text{and}\quad q_n\le\frac{1}{\varepsilon}. $$ But $$ q_{n_k}\le\frac{1}{\varepsilon}\quad\Longrightarrow\quad q_{n_k}\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\}, $$ Let $$ d=\min\left\{\left|\sqrt{2}-\frac{p}{q}\right|: p\in\mathbb N\,\,\text{and}\,\, q\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\}\right\}. $$ Clearly $d>0$, as $$ d=\min\left\{\left|\sqrt{2}-\frac{p}{q}\right|: q\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\} \,\,\text{and}\,\, p\in\{1,2,\ldots,2q\} \right\}, $$ since if $p>2q$, then $\big|\sqrt{2}-\frac{p}{q}\big|>\big|\sqrt{2}-\frac{p}{p}\big|$. Thus $$ \left|\frac{p_{n_k}}{q_{n_k}}-\sqrt{2}\right|\ge d. $$ Thus $\frac{p_{n_k}}{q_{n_k}}\not\to\sqrt{2}$.

Answer 2

The idea is to try and approximate $\sqrt{2}$ with a rational number. Because $\sqrt{2}$ is irrational, the integer making up the denominator of the rational approximation must continue to grow endlessly in order for the approximation to improve.