Proof that e is transcendental

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The first proof that the base of the natural logarithms, e, is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following:

Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients c_{0},c_{1},\ldots,c_{n}, satisfying the equation:


and such that c_0 and c_n are both non-zero.

Depending on the value of n, we specify a sufficiently large positive integer k (to meet our needs later), and multiply both sides of the above equation by \int^{\infty}_{0}, where the notation \int^{b}_{a} will be used in this proof as shorthand for the integral:


We have arrived at the equation:

c_{0}\int^{\infty}_{0}+c_{1}e\int^{\infty}_{0}+\cdots+c_{n}e^{n}\int^{\infty}_{0} = 0

which can now be written in the form




The plan of attack now is to show that for k sufficiently large, the above relations are impossible to satisfy because

\frac{P_{1}}{k!} is a non-zero integer and \frac{P_{2}}{k!} is not.

The fact that \frac{P_{1}}{k!} is a nonzero integer results from the relation


which is valid for any positive integer j and can be proved using integration by parts and mathematical induction.

To show that

\left|\frac{P_{2}}{k!}\right|<1 for sufficiently large k

we first note that x^{k}[(x-1)(x-2)\cdots(x-n)]^{k+1}e^{-x} is the product of the functions [x(x-1)(x-2)\cdots(x-n)]^{k} and (x-1)(x-2)\cdots(x-n)e^{-x}. Using upper bounds for |x(x-1)(x-2)\cdots(x-n)| and |(x-1)(x-2)\cdots(x-n)e^{-x}| on the interval [0,n] and employing the fact

\lim_{k\to\infty}\frac{G^k}{k!}=0 for every real number G

is then sufficient to finish the proof.

A similar strategy, different from Lindemann's original approach, can be used to show that the number π is transcendental. Besides the gamma-function and some estimates as in the proof for e, facts about symmetric polynomials play a vital role in the proof.

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