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I was trying to prove that e is irrational without using the typical series expansion, so starting off
$e = a/b $
Take the natural log so
$1 = \ln(a/b)$
Then
$1 = \ln(a)-\ln(b)$
So unless I did something horribly wrong showing the irrationality of $e$ is the same as showing that the equation $c = \ln(a)-\ln(b)$ or $1 = \ln(a) - \ln(b)$ (whichever one is easiest) has no solutions amongst the natural numbers. I feel like this would probably be easiest with infinite descent, but I'm in high school so my understanding of infinite descent is pretty hazy. If any of you can provide a proof of that, that would be awesome.

EDIT: What I mean by "typical series expansion" is Fourier's proof http://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Proof

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Cool question! +1. –  ncmathsadist Jan 6 '13 at 22:31
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A Diophantine equation is a polynomial equation, not just any equation looking for integer solutions. –  Thomas Andrews Jan 6 '13 at 22:32
    
So now which properties of $\ln$ do you want to exploit? –  Hagen von Eitzen Jan 6 '13 at 22:43
    
@ThomasAndrews I didn't know that Diophantine equations were limited to polynomials, I had assumed they were just ones that looked for integer solutions but thank you for clarifying. –  Simon Means Jan 6 '13 at 22:50
    
@HagenvonEitzen I have thought of 1 or 2 methods of doing it but they all implicitly assumed e was irrational and circular reasoning is generally to be avoided. Would a taylor series be of any use here? I'm genuinely stumped. –  Simon Means Jan 6 '13 at 22:51

4 Answers 4

Disclaimer: Below I present many reasons why the approach you offer is not likely to work. I do not intend this to be discouraging but rather to indicate why I think it is not a good attempt, trying to convey my intuition.

Usually solving Diophantine equations, or more generally solving equations where the solutions are only allowed to be integers is a very difficult task. If you were to randomly choose an equation and try to find its solutions in the integers it is highly likely that it will be an intractable problem. Only very few families of Diophantine equations have general methods of solution. Thus, it is generally not a very promising idea to restate a given problem as an arbitrary Diophantine equation. If the restatement yields a familiar type of Diophantine (or generally, integer solution problem) then that's good. But if the equation you get doesn't look like anything familiar from number theory (and this is the case for your attempt at an alternative proof above) then you most likely should not pursue that path.

In particular, you should always be suspicious of superficial restatements of a problem that transforms a problem stated in analysis, by applying $\ln$ (an analytical function), into a problem stated in a mixed form of number theory (Diophantine, or more generalized, equations) and analytical terms (e.g., $\ln$). If the transformation would have resulted in a clean number theoretic equation this would have been more interesting and promising.

Since the problem, either in its original statement (show that $e$ is irrational) as well as the transformed statement involve the same analytical concept (since $\exp$ and $\ln$ are each other's inverse) any solution, in each formulation, will have to use the definition of $\exp$ or $\ln$. These definitions will either directly involve a series expansion or some other analytical gadget for which a series expansion will be very useful (just because series expansion are extremely useful).

So, while your curiosity is trying to take you to interesting places, I hope that you understand from this answer the reasons why this kind of curiosity might kill the cat. Solving problems in mathematics by translating them between realms is very powerful (and cool) but it works well when the restatement is complete and natural (for instance, in the proof that $\sqrt 2$ is irrational a Diophantine-like proof does work because the square root quickly disappears and what one is left with are divisibility arguments) and when the restatement takes you from a place where you have little tools to tackle the problem to a place where more tools are available.

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Yep! I voiced a similar opinion in my comment above. :) –  Haskell Curry Jan 6 '13 at 23:32
    
After laboring over this for a while, you're definitely right. I had been hoping there was a simple proof that no solution existed that I was missing but that seems very unlikely. Thank you both for your help. :) –  Simon Means Jan 7 '13 at 3:16

One situation in which the existence of a solution to a Diophantine equation implies an irrationality result is this:

If, for a positive integer $n$, there are positive integers $x$ and $y$ satisfying $x^2 - n y^2 = 1$, then $\sqrt n$ is irrational.

I find this amusing, since this proof is more complicated than any of the standard proofs that $\sqrt n$ is irrational and also, since it does not assume the existence of solutions to $x^2 - n y^2 = 1$, requires the "user" to supply a solution. Solutions are readily supplied for 2 and 5, but are harder to find for 61.

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Pell's equation, there are ways of finding a solution involving continued fractions. –  vonbrand Jan 23 '13 at 2:21

There is a fairly simple setup that shows that $e$ is irrational, which is that the simple continued fraction is infinite, see FRACTION. See also OEIS. The simple continued fraction is $$ [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,\ldots ] $$

A short and accessible proof is available at COHN

Give me a few minutes to work out the first few convergents. The infinite but never repeating s.c.f. says that $e$ is irrational and is not of the form $(a + \sqrt b)/c$ for integer $a$ and positive integers $b,c.$

Meanwhile, simple continued fractions should be a fine object of study for a high school student. Long ago, they were standard curriculum at a certain level (I'm not sure at what stage) but have fallen through the cracks. Anyway, clear relations to Pell's equation and variants, therefore indefinite binary quadratic forms, real quadratic fields.

Convergents: $$ \frac{2}{1}, \frac{3}{1}, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \frac{106}{39}, \frac{193}{71}, \frac{1264}{465}, \frac{1457}{536}, \frac{2721}{1001}, \frac{23225}{8544}, \frac{25946}{9545}, \frac{49171}{18089}, $$ $$ \frac{517656}{190435}, \frac{566827}{208524}, \frac{1084483}{398959}, \frac{13580623}{4996032}, \frac{14665106}{5394991}, \frac{28245729}{10391023}, \frac{410105312}{150869313}, \ldots $$

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Please point to a proof of this fraction. –  marty cohen Jan 9 '13 at 4:13

There is the classic proof by Fourier: For the sake of contradiction, suppose $e = \frac{a}{b}$ for $a, b \in \mathbb{N}$ in lowest terms. As $e$ isn't an integer, $b > 1$. Now: $$ e = \sum_{n \ge 0} \frac{1}{n!} \\ b! e = \sum_{n \ge 0} \frac{b!}{n!} = \sum_{0 \le n \le b} \frac{b!}{n!} + \sum_{n \ge b} \frac{b!}{n!} $$ The first sum is an integer, we'll show that the second sum, call it $S$, satisfies $0 < S < 1$, and so can't be an integer. Firstly, $S$ is a sum of positive numbers, so $0 < S$. Now: $$ S = \sum_{n > b} \frac{1}{(b + 1) (b + 2) \ldots n} = \sum_{k \ge 1} \frac{1}{(b + 1)(b + 2) \ldots (b + k)} < \sum_{k \ge 1} b^{-k} $$ The inequality follows because each term in the last sum is larger than the corresponding term of the second one; the second one is just restating the first. But: $$ \sum_{k \ge 1} b^{-k} = b^{-1} \sum_{k \ge 0} b^{-k} = b^{-1} \cdot \frac{1}{1 - b^{-1}} = \frac{1}{b - 1} \le 1 $$ This is just an infinite geometric series, the last inequality follows from $b > 1$. So $0 < S < 1$, as promised; and $e$ isn't rational.

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That proof is already in the OP - "I was trying to prove that $e$ is irrational without using the typical series expansion" –  L. F. Jan 23 '13 at 2:42

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