2

$$ \sqrt{5} > \frac {13 + 4\pi}{24 - 4\pi} $$

Thomas Andrews
  • 177,126
GohP.iHan
  • 1,376
  • @Simon S: I've tried The one by (published by Nilakantha in the 15th century). Other than calculators, all techniques are permitted. I'm looking for the best / shortest solution. – GohP.iHan Nov 28 '14 at 14:40
  • Nice almost integer! From the observation by @JackDAurizio, $$\frac{\left(133-37\sqrt{5}\right)}{\pi}\approx 16.0000007560086$$

    Maybe this interesting question could be reopened by adding as a context two other approximations to $\pi$ that use $\sqrt{5}$, such as http://math.stackexchange.com/questions/1092215/how-was-this-approximation-of-pi-involving-sqrt5-arrived-at and http://math.stackexchange.com/questions/146831/approximation-for-pi

     – Jaume Oliver Lafont Apr 15 '16 at 07:05
  • The rational multiple of $\frac{1}{\pi}$ can be related to two convergents to $\pi$: $$\frac{133}{\pi}=\frac{111}{\pi}+\frac{22}{\pi}\approx \frac{106}{3}+7=\frac{127}{3}=\frac{2^7-1}{3}$$ – Jaume Oliver Lafont Apr 15 '16 at 07:12

1 Answers1

6

This is the same as proving that $$\frac{1}{16}(133-37\sqrt{5})>\pi$$ and it follows from the fact that the continued fraction of the LHS is: $$ [3;7,15,1,660,\ldots] $$ while the continued fraction of $\pi$ is: $$ [3;7,15,1,292,\ldots].$$

Jack D'Aurizio
  • 353,855