The mathematical constant is a transcendental number.
In other words, it is not a root of any polynomial with integer coefficients.
Let us assume that is algebraic, so there exists a polynomial
such that .
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
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.
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.
By part 1, for all we have . Therefore,
By the triangle Inequality we get:
But , hence for sufficiently large we get . A contradiction.