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I would like a proof that $e$ isn't a fraction $\frac{a}{b}$, for $a,b \in Z$ and $mdc(a,b)=1$.

Just a observation =) I'd like a proof with fractions, not about $e$ irrationality or if $e$ is transcendental. In other words, prove $e$ is not a fraction prove it is irrational, but, I need a prove to this irrationality using fractions, is there any?

Thx.

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9  
This is a proof for the irrationality that you are seeking. –  Phira Oct 19 '11 at 11:09
    
Yes but^^, If someone proof using other attempts, withou fractions, didn't work to me =//, unfortunally. –  GarouDan Oct 19 '11 at 11:11
    
Please see the edit. –  GarouDan Oct 19 '11 at 11:13
2  
The usual, standard proof is to assume $e=a/b$ and derive a contradiction. I don't see how you're asking for anything "alternative" here. –  anon Oct 19 '11 at 11:16

3 Answers 3

up vote 12 down vote accepted

The Wikipedia proof of the irrationality of $e$ is a little more complicated than necessary, because one has to do a bit of work to estimate the "tail." (We assume familiarity with the Wikipedia proof, which is the usual one.) We prove the same result, using the same basic idea, but with a little twist.

From the series for $e^x$, evaluated at $x=-1$, we can see that $$e^{-1}=\sum_2^\infty (-1)^n\frac{1}{n!}=\frac{1}{2!}-\frac{1}{3!} +\frac{1}{4!}-\frac{1}{5!}+\cdots.$$

Suppose that $e$ is rational. Then $e^{-1}$ is also rational, say $e^{-1}=a/b$, and therefore $b!\, e^{-1}$ is an integer $N$. But $N=b!\,e^{-1}=H+T$, where $$H=b!\:\sum_{2}^b (-1)^n \frac{1}{n!}$$ and $$T=b!\:\sum_{b+1}^\infty (-1)^n\frac{1}{n!}.$$ Note that $H$ is an integer. If we can show that the "tail" $T$ is not an integer, we will have a contradiction.

By a standard fact about alternating series, the error when we truncate an alternating series has absolute value less than the absolute value of the first neglected term, and the same sign as the first neglected term.

If we truncate the series for $e^{-1}$ at the term $(-1)^b \frac{1}{b!}$, the error is therefore non-zero and has absolute value less than $\frac{1}{(b+1)!}$. It follows that $$0<|T| < \frac{b!}{(b+1)!}<1,$$ and therefore $T$ cannot be an integer.

Comment: In the proof on Wikipedia, we bound the tail by comparing with a geometric series. This is certainly not difficult! But in the setting of a standard calculus course, the basic facts about alternating series are always discussed, and one might as well exploit that.

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Thx. It's a good proof. I will change my answer to this post becuase it's a local proof to math.SE, so, if other people search about that they will find this. And it's a bit simple too. Asked too in Meta.SE how to proceed. Thx. –  GarouDan Oct 19 '11 at 17:42

What is wrong with the wikipedia proof?

http://en.wikipedia.org/wiki/Proof_that_e_is_irrational

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Lol, I didn't see that, I'll read and return. Thx. –  GarouDan Oct 19 '11 at 11:18

http://en.wikipedia.org/wiki/Continued_fraction will help you. There is obvious theorem: $x$ is rational if and only if it has finite continued fraction. Then try to find continued fraction to $e$, you'll get infinite fraction. It means $e$ is irrational. IMHO very simple. The main problem is to find this fraction. $[2,1,1,2,1,1,4,1,1,6,1,1...]$ and so on corresponds to $e$

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Thx...I think I have a strange process to find if a number is transcendental, using this fractions...if it works i will post later. Thx. –  GarouDan Oct 19 '11 at 11:34

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