Originally posted by TheMaster37That works. However, can you prove it without showing or assuming that R is uncountable?
Pi is transcendental, but the proof for it is hard
A non-transcedental number, eg algebraic number, is the root of a polynomial of finite order with integer (positive and negative whole numbers) coefficients, right?
If that is so then;
given a certain order O, there is a countable number of polynomials of that order with integer co ...[text shortened]... The reals form an over-countable set of numbers, and therefore some reals must be transcedental.
Originally posted by AcolyteWhy don't you issue a formal challenge, from the Cult of Maths to those mathematical usurpers, the RUNning Pawn clan, asking them to show e is transcendental? We could taunt them mercilessly .
That works. However, can you prove it without showing or assuming that R is uncountable?
Originally posted by pidermanHah! The ''e'' bit of that proof is almost the same as mine.
RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:
Originally posted by royalchickenHmm, I've never learned anything about trancendentals, only their definition, but I'll work on it.
Hah! The ''e'' bit of that proof is almost the same as mine.
However, since the RUNning Pawn clan is very skilled, I challenge them to solve the following:
Let h(n) be defined as 1+ 1/2 + 1/3 + 1/4 +...+1/n
Now prove that lim(n->oo) [h(n) - log n] is transcendental.