The Irrationality of eDate: 09/04/98 at 00:17:10 From: Shawn Brady Subject: How is it possible to prove a number is transcendental? Hi, I was wondering if you could help me. I'm reading through a book, and it's talking about transcendental numbers. I know that a transcendental number is a number such that it can't be put in normal algeabraic forms, such as pi. However, I was wondering how this is proven. Thanks in advance for any response. - Shawn Brady Date: 09/04/98 at 11:57:49 From: Doctor Wilkinson Subject: Re: How is it possible to prove a number is transcendental? In spite of the fact that most numbers are transcendental (a fact proved by Cantor), proving that any particular number is transcendental is usually quite difficult. There is a large class of such numbers known as Liouville numbers, after the French mathematician Liouville who defined them and showed they were transcendental. A typical Liouville number is: 0.10100100000010000000000000000000000001... Here the successive groups of zeros are of length 1!, 2!, 3!, 4!, and so on. As you mentioned, we know that pi is transcendental, but this was proved only in 1882 after being studied for more than 2000 years. The number e is also transcendental, and so is a^b, if a and b are algebraic and b is irrational (like 2 to the square root of 2 power, for example). But all these theorems are very hard. The basic idea, however, in all these proofs is the same: you assume that the number in question is algebraic, and you arrive at a contradiction. The contradiction always takes the form of showing that the assumption implies the existence of an integer between 0 and 1. I do not know any of these proofs, although I have studied the proof for e, but I can give you some of the idea by showing that e is irrational. e is defined as 1 + 1/1! + 1/2! + 1/3! + 1/4! + .... Suppose e is rational, say e = a/b, where a and b are integers. Then b! e is certainly an integer. So multiply the equation for e by b!. This gives us: b! e = b! + b!/1! +... + b!/b! + b!/(b+1)! + b!/(b+2)! + ... Now all the terms up to b!/b! on the right are integers, and the left side is an integer, so the sum of the rest of the terms on the right must be an integer. This is: 1/(b+1) + 1/(b+1)(b+2) + 1/(b+1)(b+2)(b+3) + ... Now if we replace b+2, b+3, ... by b+1 we make the terms on the right bigger, so the right side is less than: 1/(b+1) + 1/(b+1)^2 + 1/(b+1)^3 + ... which is a geomtric series. Its sum is: 1/(b+1) (1 - 1/(b+1)) = 1/b. Note that b is certainly bigger than 1, since e is not an integer, and we have arrived at the promised contradiction: an integer between 0 and 1. - Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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