Timeline for Can $N^2$ have only digits 0 and 1, other than $N=10^k$?
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2009 at 20:39 | comment | added | Ilya Nikokoshev | No, sometimes the probability arguments provide the solid intuition and may lead to a formal proof. | |
| Oct 23, 2009 at 20:19 | comment | added | Kevin P. Costello | The key isn't so much that the probability are tending to 0 as that that it does so very quickly. The general rule is this: We know (from the Borel-Cantelli lemmas) that if we consider a set of independent events with probabilities p_1, p_2, ... and the sum of the p_i's converge, then almost surely only finitely many events hold. Conversely, if the sum diverges, almost surely infinitely many. In a crude sense, the probability type argument would say that P(N is a square)=1/Sqrt(N) and P(N is prime)=1/Log N correspond to divergent sequences, while the all (0,1) probability (4/10)^n doesn't | |
| Oct 23, 2009 at 19:30 | history | answered | Per Alexandersson | CC BY-SA 2.5 |