Question

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

Answer 1

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

Answer 2

i ignored absolute value :\

Answer 3

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.

Answer 4

please finish off the proof @ikram002p
i think you had the right start

Answer 5

.

Answer 6

we need to show that
\( \Large |\sqrt2-r|> | \sqrt2 -\frac{r+2}{r+1}| \)

Answer 7

Yes, that will do.

Answer 8

maybe we can start with right hand side

Answer 9

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
\(\Large r>\frac{r+2}{r+1} \)

Answer 10

Ahh I see that looks neat!

Answer 11

type it again

Answer 12

\[ \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}\]

Answer 13

:D

Answer 14

Question: why don't we use the condition: r is a positive rational approaximation to \(\sqrt 2\)?

Answer 15

how is that would be useful @Loser66 ?

Answer 16

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 \(\dfrac{a}{b}\)

Answer 17

By replacing \(a\approx b\sqrt2\) we can get the result, right?

Answer 18

interesting in which way this comparison could be ? :D

Answer 19

I don't know :)