## 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.

1. 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

2. ikram002p

i ignored absolute value :\

3. 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.

4. rational

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

5. Loser66

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6. ikram002p

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

7. rational

Yes, that will do.

8. rational

maybe we can start with right hand side

9. 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 $$\Large r>\frac{r+2}{r+1}$$

10. rational

Ahh I see that looks neat!

11. ikram002p

type it again

12. 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}

13. ikram002p

:D

14. Loser66

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

15. ikram002p

how is that would be useful @Loser66 ?

16. 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 $$\dfrac{a}{b}$$

17. Loser66

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

18. ikram002p

interesting in which way this comparison could be ? :D

19. Loser66

I don't know :)