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rational
 one year ago
If \(r\) is a positive rational approximation to \(\sqrt{2}\), then show that \(\frac{r+2}{r+1}\) is always a better rational approximation.
rational
 one year ago
If \(r\) is a positive rational approximation to \(\sqrt{2}\), then show that \(\frac{r+2}{r+1}\) is always a better rational approximation.

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ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1we need to show error in second case is less sqrt2r/sqrt2 >sqrt2(that value)/sqrt2 we can reduce that to r>r+2/r+1

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i ignored absolute value :\

rational
 one year ago
Best ResponseYou've already chosen the best response.1Exactly! for example, if you choose \(r=1\) to approximate \(\sqrt{2}\), then \(\frac{1+2}{1+1} = 3/2\) is more closer to \(\sqrt{2}\) than \(1\) is.

rational
 one year ago
Best ResponseYou've already chosen the best response.1please finish off the proof @ikram002p i think you had the right start

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1we need to show that \( \Large \sqrt2r>  \sqrt2 \frac{r+2}{r+1} \)

rational
 one year ago
Best ResponseYou've already chosen the best response.1maybe we can start with right hand side

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1as we are interesting in positive error only \(\Large \sqrt{(\sqrt2r)^2} > \sqrt {(\sqrt2 \frac{r+2}{r+1})^2 } \) which reduce it only to show \(\Large r>\frac{r+2}{r+1} \)

rational
 one year ago
Best ResponseYou've already chosen the best response.1Ahh I see that looks neat!

rational
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align} \left \sqrt2 \dfrac{r+2}{r+1}\right&= \left \dfrac{\sqrt2(r+1) r2}{r+1}\right\\~\\ &=\left \dfrac{\sqrt2(r\sqrt2) 1(r\sqrt2)}{r+1}\right\\~\\ &=\left \dfrac{(\sqrt21)(r\sqrt2) }{r+1}\right\\~\\ &\le\left (\sqrt21)(r\sqrt2) \right\\~\\ &\lt \leftr\sqrt{2}\right \end{align}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: why don't we use the condition: r is a positive rational approaximation to \(\sqrt 2\)?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1how is that would be useful @Loser66 ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Suppose \(r =\dfrac{a}{b}\) is approaximation of \(\sqrt 2\) that is \(\sqrt 2\approx \dfrac{a}{b}\\b\sqrt 2\approx a\) Now \(\dfrac{r+2}{r1}=\dfrac{(a/b)+2}{(a/b)1}=\dfrac{a+2b}{ab}\)\) Make a comparison between \(\dfrac{r+2}{r1}=\dfrac{a+2b}{ab}\) and \(\dfrac{a}{b}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0By replacing \(a\approx b\sqrt2\) we can get the result, right?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1interesting in which way this comparison could be ? :D
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