The Proof of Hilbert’s Seventh Problem about Transcendence of e+π ()
![](//html.scirp.org/file/69954x4.png)
Subject Areas: Algebra, Algebraic Geometry
![](//html.scirp.org/file/69954x5.png)
1. Introduction
Hilbert’s seventh problem is about transcendental number. The proof of transcendental number is not very easy. We have proved the transcendence of “e” and “π”. However, for over a hundred years, no one can prove the transcendence of “e + π” [1] . The purpose of this article is to solve this problem and prove that e + π is a transcendental number.
2. Proof
1) Assuming
is any one polynomial of degree n.
,
, Let ![](//html.scirp.org/file/69954x9.png)
Now we consider this integral:
. By integrability by parts, we can get the following For- mula (2.1):
(2.1)
2) Assuming
is a algebraic number, so it should satisfy some one algebraic equation with integral coefficients:
,
.
According to Formula (2.1), using
multiplies both sides of Formula (2.1) and let be separately equal to
. We get the following result.
(2.2)
So, all we need to do or the key to solve the problem is to find a suitable
that it doesn’t satisfy the Formula (2.2) above.
3) So we let
[2] ,
,
and b is a prime number Because of
,
, so
can be divisible by
and when
, all of
equal zero.
Furthermore, we consider
whose (p + a)-th derivative (
); when
, the derivative is zero. And when
, the derivative is
. What’s more, the coefficient of
is a multiple of (p + a)!, so it’s alse a multiple of (p − 1)! and p.
By the analysis above, we can know that
are multiples of p.
Now we see
; we know,
![]()
and its the sum of the first p − 1 item is zero (because the degree of each term of
is not lower than
). All from the (p + 1)-th item to the end are multiples of p. But the p-th item
is the (p − 1)-th derivative of
. So,
, and
and
are congruence, written
. Thereby,
, but
,
, and b is a prime number, so
,
(2.3)
4) Next, we need to prove that
when p tends to be sufficiently large.
When x changes from 0 to n, the absolute value of each factor
of
is not more than n, so
,
.
So by integral property: when
,
![]()
Let M equal ![]()
thus, ![]()
When
. So,
(2.4)
Finally, according to (2.3) and (2.4), we know (2.2) is incorrect. So, e + π is a transcendental number.
3. Conjecture
By the proof above, we conclude that e + π is a transcendental number. Besides, I suppose
is also a transcendental number. What’s more, when a and b are two real numbers, and
, I suppose that
is a transcendental number.
Acknowledgements
I am grateful to my friends and my classmates for supporting and encouraging me.