 Someone other than RC do this ;) Acolyte Posers and Puzzles 02 Feb '04 11:06
1. 02 Feb '04 11:06
Prove the existence of transcendental numbers (ie, numbers which are not roots of polynomials with integer coefficients).
2. 02 Feb '04 14:59
42
3. 02 Feb '04 19:35
Fine then 😛. I'll try to post an erroneous proof later and make you all find the problem with it...
4. 02 Feb '04 20:581 edit
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 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.
5. 02 Feb '04 21:55
e is also included! WHOOO! No way in hell I am provin' it though! 😀
6. 02 Feb '04 23:33
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.
That works. However, can you prove it without showing or assuming that R is uncountable?
7. 02 Feb '04 23:41
Originally posted by Acolyte
That works. However, can you prove it without showing or assuming that R is uncountable?
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 😉.
8. 03 Feb '04 08:37
Originally posted by royalchicken
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 😉.
You're on
9. 03 Feb '04 09:31
RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:

Signed: Pi-derman
10. 03 Feb '04 23:42
Originally posted by piderman
RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:

Signed: Pi-derman
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-&gt;oo) [h(n) - log n] is transcendental.
11. 04 Feb '04 15:30
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.
Hmm, I've never learned anything about trancendentals, only their definition, but I'll work on it.