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Famous Theorems of Mathematics/e is transcendental

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The mathematical constant is a transcendental number.

In other words, it is not a root of any polynomial with integer coefficients.

Proof

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Let us assume that is algebraic, so there exists a polynomial

such that .

Part 1

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Let be a polynomial of degree . Let us define . Taking its derivative yields:

Let us define . Taking its derivative yields:

Since is differentiable, we shall apply the mean Value Theorem on the interval for . So there exists a such that

Now let:

Summing all the terms yields

Part 2

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Let be a polynomial with a root of multiplicity . We will show that for all we get .

Let us write , such that is a polynomial with .

with all polynomials.

Part 3

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Let us now define a polynomial

for prime such that and . We get:

hence for all , the function is a polynomial with integer coefficients all divisible by .


By part 2, for all we get:

Therefore is also an integer divisible by .

On the other hand, for we get:

but , and are not divisible by . Therefore is not divisible by .

In other words, the sum is an integer not divisible by , and particularly non-zero.

Part 4

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By part 1, for all we have . Therefore,

By the triangle Inequality we get:

But , hence for sufficiently large we get . A contradiction.