Recent content by Algebracus

We can easily prove it without induction: a_n = 3^{2n+1}+4^{2n+1}\equiv 3^{2n+1} + (-3)^{2n+1} \equiv 3^{2n+1}\cdot(1 + (-1)^{2n+1}) \equiv 0 (mod 7).

It sems like you (Sunshine) are talking about a method for finding rational numbers x_1, x_2, x_3, ... so that the difference between the irrational number x and the rational number x_i is at most half the difference between x and x_{i-1} for every i. Still, there is a difference, and this is...