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Title: Irrationality of e Post by william wu on Dec 12th, 2007, 3:28pm Prove that the constant e is irrational. (Apologies if this has been posted already; moderators, feel free to delete this thread if so.) |
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Title: Re: Irrationality of e Post by Aryabhatta on Dec 12th, 2007, 5:16pm I don't remember seeing a thread on this (nor did a search seem to reveal any threads...) Here is a proof which involves the infinite series representation of e [hide] We use the following e = Sum 1/n! Suppose e = p/q where q > 1 (p, q are integers) Then consider q!e = Sum_{n=1 to q} q!/n! + Sum_{n=q+1 to oo} q!/n! The n=1 to q part is an integer. Consider the nth tern of remaining: q!/n! = 1/(q+1)(q+2)...(q+n-q) < 1/qn-q Thus Sum_{n=q+1 to oo} q!/n! < Sum_{n=q+1 to oo} 1/qn-q = 1/(q-1) <= 1 But q!e must be an integer. Hence e is irrational. [/hide] |
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Title: Re: Irrationality of e Post by JiNbOtAk on Dec 12th, 2007, 5:20pm What do you know, Wikipedia (http://en.wikipedia.org/wiki/Proof_that_e_is_irrational) has a topic on this. |
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Title: Re: Irrationality of e Post by cool_joh on Dec 14th, 2007, 2:43am on 12/12/07 at 17:16:34, Aryabhatta wrote:
How can we prove that? |
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Title: Re: Irrationality of e Post by pex on Dec 14th, 2007, 3:13am on 12/14/07 at 02:43:25, cool_joh wrote:
As far as I know, we don't have to; it's the definition of e. |
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Title: Re: Irrationality of e Post by Hippo on Dec 17th, 2007, 12:52pm on 12/14/07 at 02:43:25, cool_joh wrote:
Suppose e is such a constant that (ex)'=ex. Than we can approximate ex in 0 by the "infinite polynomial" such that it agrees in each derivative in 0. The polynomial is \sumk=0\inftyxk/k!. We can show the error of approximation goes to 0 for an arbitrary large x as more and more sum elements are added together. Therefore ex=\sumk=0\inftyxk/k! Special case e=e1=\sumk=0\infty1k/k!. |
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