Skip to main content
MathOverflow

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

8
  • 13
    $\begingroup$ It's extremely unsurprising, because you put more than 70 digits of information into that long number. $\endgroup$
    – Reid Barton
    Commented Nov 9, 2009 at 16:57
  • 5
    $\begingroup$ Agreed. That number is about 90 digits long (I might have miscounted a few.) The spacing between 5th roots of 90 digit numbers is about 10^{-72}. So it's hardly a surprise that you can find some number in that range whose 5th root agrees with a given real number to 70 decimal places. $\endgroup$
    – David E Speyer
    Commented Nov 9, 2009 at 17:05
  • 8
    $\begingroup$ Somewhat surprisingly, exp(Pisqrt(163))^5 is within 10^(-5) of an integer. And this integer is not N^5 where N is the integer closest to exp(Pisqrt(163)). $\endgroup$
    – Michael Lugo
    Commented Nov 9, 2009 at 17:08
  • 1
    $\begingroup$ Either prod Anton to re-open, or post another question. I can prove that c^2 is close to an integer, where c is Ramanujan's constant, and this isn't a formal consequence of c being close to an integer (because it's not close enough, as it were). But my proof doesn't stretch to c^5. $\endgroup$
    – Kevin Buzzard
    Commented Nov 9, 2009 at 18:39
  • 2
    $\begingroup$ Seconding that Michael should ask this as a separate question, unless he wants to work it out himself. $\endgroup$
    – JSE
    Commented Nov 9, 2009 at 19:58