- 02 Feb '04 20:58 / 1 editPi is transcendental, but the proof for it is hard

EDIT:

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 coefficients (integers are countable, and for a polynomial of order O you require O+1 integers).

The order is an element of the natural numbers (positive integers) so there is only a countable set of polynomials of finite order with integer coefficients.

This means there is only a countable set of algebraic numbers.

The reals form an over-countable set of numbers, and therefore some reals must be transcedental. - 02 Feb '04 23:33

That works. However, can you prove it without showing or assuming that R is uncountable?*Originally posted by TheMaster37***Pi is transcendental, but the proof for it is hard**

EDIT:

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. - 02 Feb '04 23:41

Why 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 .*Originally posted by Acolyte***That works. However, can you prove it without showing or assuming that R is uncountable?** - 03 Feb '04 23:42

Hah! The ''e'' bit of that proof is almost the same as mine.*Originally posted by piderman***RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:**

http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes6.pdf

Signed: Pi-derman

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. - 04 Feb '04 15:30

Hmm, I've never learned anything about trancendentals, only their definition, but I'll work on it.*Originally posted by royalchicken***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.

Have you allready thought about Q(sqrt(6))?