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Please how to prove this:

If a rational sequence $\left(\dfrac{p_n }{q_n }\right)$ (where $p_n\in \mathbb{Z}$ and $q_n\in \mathbb{N}\setminus\{0\}$) converge to an irrational number $l\in \mathbb{R}\setminus \mathbb{Q})$ in $(\mathbb{R},|.|)$ then $q_n\to\infty$ and $p_n\to\pm\infty$

I suppose that $\dfrac{p_n }{q_n }\to l$ that is $$\forall \varepsilon>0, \exists n_0,\forall n\in \mathbb{N}, n\geq n_0\Rightarrow \left|\frac{p_n }{q_n }-l\right|\leq \varepsilon$$ that is $$\forall \varepsilon>0, \exists n_0,\forall n\in \mathbb{N}, n\geq n_0\Rightarrow l-\varepsilon<\frac{p_n }{q_n }\leq l+\varepsilon$$

I don't see how i can find that $p_n\to+\infty, q_n\to\pm\infty$?

Thank you

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Since $\frac{p_n}{q_n}$ converges, it is bounded in a certain interval $[s,t]$. Assume $q_n\not\to\infty$. Then there would be a number $q$ and a subsequence $(n_i)$ such that $q_{n_i}\leq q$ for all $i\in\mathbb{N}$. Then the subsequence $\frac{p_{n_i}}{q_{n_i}}$ also converges to $l$ (every subsequence of a converging sequence converges to the same limit). This subsequence $\frac{p_{n_i}}{q_{n_i}}$ takes values in the set $A:= \{x\in [s,t]\mid \exists a,b\in\mathbb{Z}: \frac ab=x \text{ and } b\leq q\}$ of those values in $[s,t]$ which can be written as $\frac{a}{b}$ with $b\leq q$. But this set $A$ is finite! Since finite subsets of $\mathbb{R}$ are closed, the limit $l$ is also in this finite set $A$. But this is a contradiction since all elements of $A$ are rational by definition.

This proves that $q_n\to\infty$. It follows immediately that $p_n\to\pm\infty$.

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  • $\begingroup$ i don't understand the part where you suppose that A is a set of rational numbers $\endgroup$ – Vrouvrou Jan 13 '18 at 18:56
  • $\begingroup$ $A$ is the set of those rational numbers $\frac {a}{b}\in\mathbb{Q}\cap[s,t]$ where $b\leq q$. I did not fully understand your question, could you reformulate? $\endgroup$ – Tashi Walde Jan 13 '18 at 19:02
  • $\begingroup$ normaly when we say $(p_n/q?n)$ that is $\exists M>0, |p_n/q_n|\leq M,\forall n\in N$, i understand that the subsequence is also bouded , but i don't understand how i introduce the set A $\endgroup$ – Vrouvrou Jan 13 '18 at 19:31
  • $\begingroup$ define A to be the set of those rational numbers x which can be written as a quotient $x=\frac ab$ (of integers) where $b\leq q$ and which lie in the bounding intervalery (as you put it $|x|\leq M$). Now the subsequence $\frac{p_{n_i}}{q_{n_i}}$ I defined lies completely in this set by definition (because every $q_{n_i}\leq q$). $\endgroup$ – Tashi Walde Jan 13 '18 at 19:57
  • $\begingroup$ i have that $|p_k|/q_k< M$ and $q_k<q$ then i can say that $|p_k|< M q$ i don't understand from where you get that $p_k/q_k< a/b$ where $a,b\in Z$ @Tashi $\endgroup$ – Vrouvrou Jan 13 '18 at 21:07
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Ok, let's prove a bit more.

If the sequence $p_n/q_n$ (where $p_n, q_n\in\mathbb {Z}, q_n>0$) tends to a real number $l$ (which may be rational also) and $p_n/q_n\neq l$ for all $n$ greater than some specific positive integer $n_0$ then $q_{n} \to \infty$. If $l$ is non-zero then $p_n \to\pm\infty$ based on sign of $l$. If $l=0$ then the limiting behavior of $p_n$ can not be guaranteed.

We can assume without any loss of generality that $p_n/q_n\neq l$ for all $n$. Let $k=\lfloor l\rfloor$ so that $k\in\mathbb{Z} $ and $k\leq l<k+1$. Let $N$ be any arbitrary positive integer and consider the set $A$ of all rational numbers not equal to $l$ which have denominator less than or equal to $N$ and which lie in interval $[k-1, k+1]$. This set $A$ is finite and thus $$\epsilon=\min\{|l-x|\mid x\in A\} >0 $$ Since $p_n/q_n\to l$ it follows that there is a positive integer $m$ such that $|p_n/q_n-l|<\epsilon$ for $n\geq m$. Then clearly $p_n/q_n\notin A$ and hence $q_n>N$ for all $n\geq m$. Thus $q_n\to\infty$. Further $p_n=(p_n/q_n) q_n$ so $p_n\to\pm\infty$ based on sign of $l$. For $l=0$ consider the sequences $1/n,(-1)^{n}/n,n/(n^2+1),-n/(n^2+1),(-1)^n n/(n^2+1)$ and you can note that the limiting behavior of $p_n$ cannot be guaranteed.

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