Proving TranscendenceDate: 05/15/99 at 20:47:04 From: Tony Johnson Subject: How does one prove transcendence in a number? I've been asked by my mathematics teacher to present a talk about transcendental numbers in class. I've been to nearly every site that is listed by my search engine regarding transcendental numbers. They all seem to be telling me which numbers are transcendental; however, none has shown how to prove transcendence in a number. Can you help me please? Date: 05/17/99 at 13:14:22 From: Doctor Wilkinson Subject: Re: How does one prove transcendence in a number? Hi, Tony! In general it is a very difficult task to prove any particular number is transcendental. In fact, not a single example was known until 1851, when Liouville exhibited a whole class of them, and it was not until 1873, when Hermite proved the transcendence of e, that a number not specifically constructed for the purpose was proved to be transcendental. Although the details are very difficult, all transcendence proofs that I know of are based an a very simple idea. If you want to show that a number is transcendental, you first assume it isn't (indirect proof or proof by contradiction). You then use this assumption to construct a number that (a) is a whole number; and (b) is greater than 0 and less than 1. Since there is no such number, this is a contradiction and the original assumption must have been false. For an example of how such a proof goes, let's prove that e is irrational. We have the formula for e: e = 1 + 1/1! + 1/2! + 1/3! +... Let's suppose e is rational, say e = a/b, where a and b are whole numbers. Multiply the equation for e by b!. This gives b!(a/b) = b!e = b! + b!/1! + b!/2! +...+b!/b! + b!/(b+1)! + b!/(b+2)! +.. Now the left side is an integer, and so are the terms on the right side up to b!/b!. So the difference beween the left side and those terms on the right side is an integer, and it's greater than zero, because it's just the sum of the remaining terms on the right. The sum of those terms, however, is 1/(b+1) + 1/(b+1)(b+2) + 1/(b+1)(b+2)(b+3) +... and this is less than 1/(b+1) + 1/(b+1)^2 + 1/(b+1)^3 +... which is a geometric series with sum 1/b, which is less than 1. So here we have a number that is an integer, and is greater than 0 and less than 1, which is impossible. You might want to fill in all the details of this proof and present it as part of your talk. Transcendence proofs are much more difficult, but they use the same basic idea. Good luck with your talk! - Doctor Wilkinson, The Math Forum http://mathforum.org/dr.math/ |
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