rational
  • rational
If \(r\) is a positive rational approximation to \(\sqrt{2}\), then show that \(\frac{r+2}{r+1}\) is always a better rational approximation.
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ikram002p
  • ikram002p
we need to show error in second case is less sqrt2-r/sqrt2 >sqrt2-(that value)/sqrt2 we can reduce that to r>r+2/r+1
ikram002p
  • ikram002p
i ignored absolute value :\
rational
  • rational
Exactly! 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.

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More answers

rational
  • rational
please finish off the proof @ikram002p i think you had the right start
Loser66
  • Loser66
.
ikram002p
  • ikram002p
we need to show that \( \Large |\sqrt2-r|> | \sqrt2 -\frac{r+2}{r+1}| \)
rational
  • rational
Yes, that will do.
rational
  • rational
maybe we can start with right hand side
ikram002p
  • ikram002p
as we are interesting in positive error only \(\Large \sqrt{(\sqrt2-r)^2} > \sqrt {(\sqrt2 -\frac{r+2}{r+1})^2 } \) which reduce it only to show \(\Large r>\frac{r+2}{r+1} \)
rational
  • rational
Ahh I see that looks neat!
ikram002p
  • ikram002p
type it again
rational
  • rational
\[ \begin{align} \left| \sqrt2 -\dfrac{r+2}{r+1}\right|&= \left| \dfrac{\sqrt2(r+1) -r-2}{r+1}\right|\\~\\ &=\left| \dfrac{\sqrt2(r-\sqrt2) -1(r-\sqrt2)}{r+1}\right|\\~\\ &=\left| \dfrac{(\sqrt2-1)(r-\sqrt2) }{r+1}\right|\\~\\ &\le\left| (\sqrt2-1)(r-\sqrt2) \right|\\~\\ &\lt \left|r-\sqrt{2}\right| \end{align}\]
ikram002p
  • ikram002p
:D
Loser66
  • Loser66
Question: why don't we use the condition: r is a positive rational approaximation to \(\sqrt 2\)?
ikram002p
  • ikram002p
how is that would be useful @Loser66 ?
Loser66
  • Loser66
Suppose \(r =\dfrac{a}{b}\) is approaximation of \(\sqrt 2\) that is \(\sqrt 2\approx \dfrac{a}{b}\\b\sqrt 2\approx a\) Now \(\dfrac{r+2}{r-1}=\dfrac{(a/b)+2}{(a/b)-1}=\dfrac{a+2b}{a-b}\)\) Make a comparison between \(\dfrac{r+2}{r-1}=\dfrac{a+2b}{a-b}\) and \(\dfrac{a}{b}\)
Loser66
  • Loser66
By replacing \(a\approx b\sqrt2\) we can get the result, right?
ikram002p
  • ikram002p
interesting in which way this comparison could be ? :D
Loser66
  • Loser66
I don't know :)

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