1
$\begingroup$

In this theorem for continued fractions

$\alpha \in \Bbb R /\Bbb Q $ suppose $n >1$ $, 0<q \leq q_n$

$\tfrac{p}{q}\neq\tfrac{p_n}{q_n}$, where $\tfrac{p_n}{q_n}$ is the $n^{th}$ convergent af $\alpha$ then $|\alpha-\tfrac{p_n}{q_n}|<|\alpha-\tfrac{p}{q}|$.

What does $\tfrac{p}{q}$ represent ? Originally I had thought it was $\alpha$ itself but obviously that would make no sense I realised after getting to the inequality at the end.

$\endgroup$
  • 1
    $\begingroup$ Please provide a link, if online, or otherwise a reference, to where the theorem for continued fractions you are asking about comes from. Thanks. $\endgroup$ – John Omielan Jul 20 at 5:17
  • $\begingroup$ @JohnOmielan I can't provide either it's from handwritten lecture notes $\endgroup$ – excalibirr Jul 20 at 5:18
  • 1
    $\begingroup$ Thanks for the very prompt reply. Have you tried asking the lecturer, or perhaps some of your classmates, about this? $\endgroup$ – John Omielan Jul 20 at 5:21
  • $\begingroup$ @JohnOmielan I have a meeting with my lecturer on Wednesday , but I have exams in 2 weeks so I'm trying to get as much covered before I meet her as possible $\endgroup$ – excalibirr Jul 20 at 5:24
  • 1
    $\begingroup$ @excalibirr, no, if $p/q$ was one of the convergents, it would be $p_k/q_k$. It's likely some fraction not belonging to the set of convergents $\endgroup$ – Yuriy S Jul 20 at 6:11
2
$\begingroup$

$\alpha$ is an irrational and so it has a sequence of convergents $\frac{p_n}{q_n}$ (rational numbers) that approximate $\alpha$ closer and closer.

The statement you quote has as its conclusion that $|\alpha - \frac{p_n}{q_n}| < |\alpha-\frac{p}{q}|$, saying that $\frac{p_n}{q_n}$ is a better approximation of $\alpha$ than $\frac{p}{q}$ is (some other rational). But the condition on this other approximating rational $\frac{p}{q}$ makes no sense: what does it mean that "$\frac{p}{q}$ does not divide $\frac{p_n}{q_n}$"? We usually talk about divisibility in proper rings, but in fields (like $\Bbb Q$) every is divisible by any non-zero element). You might want to check what the lecturer means by that statement (or whether you copied it correctly). Check out the "best rational approximations" part of the Wikipedia page as well.

After changing the condition in the question: After adding $\frac{p}{q} \neq \frac{p_n}{q_n}$ and $q \le q_n$ the world is right again and it's the statement in the Wikipedia page I linked to. So it's saying that the $n$-th convergent is the best rational approximation of $\alpha$ among all fractions of smaller or equal denominator.

$\endgroup$
  • $\begingroup$ I wasn't sure if it was an equals or a does not divide sign it was probably an equals, I'll edit it now. But basically the then the $p/q=a_k$ and the $p_n/q_n=n$ for $\alpha=[a_;a_1...a_n….a_k….]$ then right ? $\endgroup$ – excalibirr Jul 20 at 5:54
  • 1
    $\begingroup$ @excalibirr No, that notation makes no sense. $\frac{p_n}{q_n}$ is not $n$, and $\frac{p}{q}$ is probably not one of the convergents $a_k$. $\endgroup$ – Henno Brandsma Jul 20 at 6:03
  • $\begingroup$ sorry that was a typo , I meant to write $p_n/q_n=a_n$, the first line of the proof starts of with let $q=q_n$ by the way that's why I thought maybe $p/q$ was one of the convergents. I'll give everything you said a think anyway and read the Wikipedia article , thanks for your advice :) $\endgroup$ – excalibirr Jul 20 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.