Better approximation than convergent of continued fraction
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Let $alpha = [a_0;a_1,a_2,ldots] in mathbb{R}$ and $(p_n/q_n)$ be the convergents to the continued fraction of $alpha$. Prove that, if $q_n leq q < q_{n+1}$, gdc($p$,$q$)$=1$ and $p/q not = p_n/q_n$ then $|alpha - p/q| leq |alpha - p_n/q_n|$ only if $frac{p}{q} = frac{p_{n+1}-rp_n}{q_{n+1}-rq_n}$, where $r in mathbb{N}$.
This is a problem I can't solve. I was only able to show that the continiued fraction of $p/q$ begins with $a_0, a_1, ldots, a_n$. My guess is that the next step is showing that it actually ends at the $(n+1)$th term, and that, I think, would solve it.
number-theory continued-fractions
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
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add a comment |
$begingroup$
Let $alpha = [a_0;a_1,a_2,ldots] in mathbb{R}$ and $(p_n/q_n)$ be the convergents to the continued fraction of $alpha$. Prove that, if $q_n leq q < q_{n+1}$, gdc($p$,$q$)$=1$ and $p/q not = p_n/q_n$ then $|alpha - p/q| leq |alpha - p_n/q_n|$ only if $frac{p}{q} = frac{p_{n+1}-rp_n}{q_{n+1}-rq_n}$, where $r in mathbb{N}$.
This is a problem I can't solve. I was only able to show that the continiued fraction of $p/q$ begins with $a_0, a_1, ldots, a_n$. My guess is that the next step is showing that it actually ends at the $(n+1)$th term, and that, I think, would solve it.
number-theory continued-fractions
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
$endgroup$
$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56
add a comment |
$begingroup$
Let $alpha = [a_0;a_1,a_2,ldots] in mathbb{R}$ and $(p_n/q_n)$ be the convergents to the continued fraction of $alpha$. Prove that, if $q_n leq q < q_{n+1}$, gdc($p$,$q$)$=1$ and $p/q not = p_n/q_n$ then $|alpha - p/q| leq |alpha - p_n/q_n|$ only if $frac{p}{q} = frac{p_{n+1}-rp_n}{q_{n+1}-rq_n}$, where $r in mathbb{N}$.
This is a problem I can't solve. I was only able to show that the continiued fraction of $p/q$ begins with $a_0, a_1, ldots, a_n$. My guess is that the next step is showing that it actually ends at the $(n+1)$th term, and that, I think, would solve it.
number-theory continued-fractions
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
$endgroup$
Let $alpha = [a_0;a_1,a_2,ldots] in mathbb{R}$ and $(p_n/q_n)$ be the convergents to the continued fraction of $alpha$. Prove that, if $q_n leq q < q_{n+1}$, gdc($p$,$q$)$=1$ and $p/q not = p_n/q_n$ then $|alpha - p/q| leq |alpha - p_n/q_n|$ only if $frac{p}{q} = frac{p_{n+1}-rp_n}{q_{n+1}-rq_n}$, where $r in mathbb{N}$.
This is a problem I can't solve. I was only able to show that the continiued fraction of $p/q$ begins with $a_0, a_1, ldots, a_n$. My guess is that the next step is showing that it actually ends at the $(n+1)$th term, and that, I think, would solve it.
number-theory continued-fractions
number-theory continued-fractions
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
asked Nov 26 '18 at 17:50
Raul Philip SchwarzRaul Philip Schwarz
1
1
$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56
add a comment |
$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56
$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56
$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56
add a comment |
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$begingroup$
As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides.
$endgroup$
– Richard Martin
Nov 26 '18 at 17:56