～～無理數次方～～

Take 2^b =3 .
b is irrational otherwise there exits non-zero m and n such that 2^m = 3^n.
b is irrational and hence b and 1 are linearly independent over rational.
According to the (Theorem 2.4) below, b is not algebraic irrational otherwise the number 3 is transcendental !!!
This means that b is transcendental and hence b is not rational number.

2006-12-17 07:34:06 補充：
SORRY{{ Take 2^b =3 .b is irrational otherwise there exits non-zero m and n such that 2^m = 3^n. }}The above statement is enough. Just forget the nonsense following it and please forgive me about that. I have been confusing it with another question.

2006-12-17 07:36:35 補充：
Please forgive me about not deleting this answer.I just want to keep some record for my work even this is very foolish work.

2006-12-23 20:32:20 補充：
It is very lucky to be selected because I answered the question in a very dreadful way.

有Transcendental Number Theory
的soft copy嗎?

a唔係一就應該沒有了,而且此時的c為超越數,即是不屬於任何一個整係數n次方程的解.

若a=1
b是任何數
c都會是1

2006-12-17 00:52:19 補充：
3^0.630929753571457437....=20.630929753571457437....=log2/log3但log2和log3不是有理數