Download The Riemann hypothesis

The real primes and the Riemann hypothesis
Jamel Ghanouchi
Ecole Supérieure de Sciences et Techniques de Tunis
[email protected]
Abstract
In this study, the Riemann problem is presented with highlights on history of
the zeta function.
Thereafter, the real primes, which constitute a novelty, are defined.
It allows to generalize the Riemann hypothesis to the reals. A calculus of
integral solves the problem.
The proof is generalized to the integers by an elementary development.
Keywords
Primes ; Reals ; Riemann hypothesis
Introduction
The Riemann conjecture is a conjecture which has been formulated in 1859 by
Bernard Riemann in the subject of the Riemann function zeta or  . It is called
the zeta Riemann function. This conjecture is one the seven problems of the
millennium.
The first result is the divergence of the harmonic serie. The second consist of
the results of Leonard Euler. Finally, Bernard Riemann has presented the
conjecture.
The problem is reconsidered here as the real primes are defined for the first
time. This definition will allow to generalize the hypothesis to the reals for
which the solution is easier. Thereafter, the conjecture will be solved after an
elementary development.
Materials or methods
1
The main novelty of the study consists on the definition of the real primes. In
fact, it is a generalization of the concept of prime to the reals. The discovery of
the real primes will make easier the resolution of the hypothesis since a
calculus of integral will be sufficient.
The Riemann hypothesis
The zeta function of Riemann is defined as follows
n 
 ( s)   (
n 1
1
1 1
)  1  s  s  ...
s
n
2 3
The first result is the divergence of the harmonic serie
n 
1
n
1
2
1
3
 (1)   ( )  1    ...
n 1
It has been proved in the middle age by Nicole Oresme.
In the XVIII century, Leonard Euler has discovered the main proprieties of the 
function.
In the 1730’s he conjectured after numerical calculus the following equality,
which is often called the Basel problem.
1
1 1
2
)

1



...

n2
22 32
6
n 
 (2)   (
n 1
Euler proved it in 1748 and introduced the  function. He calculated its value
for the positive even numbers.
n 
 (2k )   (
n 1
| B2k | (2 )2 k
1
1
1
)

1



...

n2 k
22 k 32 k
2(2k )!
Where B2k are the Bernoulli numbers.
Thereafter, he proved in 1744 the Euler idendity where prime numbers are
related to the  function.
n 
 ( s)   (
n 1
1
1 1
)  1  s  s  ... 
s
n
2 3
1
 1 p
s
primes
Consequently he deduced the divergence of the serie of the inverse of primes.
2
With Bernard Riemann, s can be complex number. Riemann proved the
following formula
s
 (1 s )
2
s
2
 2 ( ) ( s)  
1 s
(
) (1  s)
2
Where

(s)   t s 1et dt
0
This formula demonstrates that this equation does not change if we replace s
by 1-s. Thus it is symmetric | s 
1
2
Riemann demonstrates that the only zeros in the R( s)  0 are the trivial zeros
negative even numbers and that there is no zero in the R ( s )  1 .
The other zeros are the non trivial zeros. They are in the critical zone 0  R( s)  1
1
2
. Riemann conjectured they are all in the critical line R( s)  .
This conjecture is called the Riemann hypothesis.
They calculated numerically one billion zeros of the Riemann they are all
located in the critical line.
Resolution of the Riemann hypothesis for the reals
Definition
A real number is compound if it can be written as
p
nj
j
where p j are primes
j
and n j are rationals. This decomposition in prime factors is unique. A prime real
number or R-prime can be written only as p=p.1 . Thereafter, there are other
real prime numbers like  , e, ln(2) . Of course, it is a convention, because, if  2
is prime  will be no more prime. It is equivalent in what will follow.
Thus
2i
q
1
q
p  p is compound. Also
q
1
q
p  1  p  1 is prime when p is prime and
p 1  ( p 1)(2 p  1)1 (2 p  1)1...( p 1)1
i
i 1
example.
3
is compound for p prime, for
The approach of the Riemann hypothesis
The Rieman hypothesis states that the non trivial zeros of the Riemann zeta
function

 ( z)  
t 1
1
1
lie on the critical line  iy .
z
2
t

1

z
t 1 t
For t integer, Euler has proved that  ( z )  
the Euler identity.

(
R  primes  Q
For
t real, it is
1
 1 p
z
primes
still

1
 ( p
primes
N
z
) , it is
true and it becomes


1
dt
1
t1 z  But But let  1 ( z ) the Riemann function for the
) z 
z
1
p
t
1 z
1
reals and  ( z ) the Riemann function for the integers. We will see below that
 ( z)  0   1 ( z)  0
Thus
 (2k )  0   1 (2k )  lim(
t 
t t
eu ; u  
k  N : t 2 ki
 lim(t
2 ki
t 
t1 2 k  1
)  0  lim(t1 2 k )  1
t 
1  2k
e 2 kiu  (e 2 ki ) u  1u
)  1  lim(t
3 6 ki
t 
)  1  lim(t
1
 ki
2
t 
)e
i 2 m
6
e
i 2 m '
6
lim(t  ki )
t 
If
m  m '  lim(t  ki )  e
i 2( m  m ')
6
t 
; k  N  lim(t  i 2 k )  1  e
i 2( m  m ')
3
t 
 e i 2  e i 2 m ' 
 m  m ' 3  4m '  m '  1; m  4  lim(t ik )  1; k  N
t 
Let k  2k ' 1  1 it is impossible  m  m ' . Another proof :
k  2k '  lim(t ik )  1  e
i( m  m ')
3
t 
e
i 2( m  m ')
3
 e i 2  e
i 6 m '
3
 m  m'
1
2
But lim(t )  1  lim(t )  lim(t )  1  lim(t )  1 and we know there exists
6
t 
5
t 
t 
t 
1
1
 iyI
1
1
yI ;  (  iyI )  0   1 (  iyI )  0  lim(t 2 )  1  lim(t 2 ) it means
t 
t 
2
2
4
1
 iy
lim(t 2
t 
)  1; y  N
Let us suppose there exists m different of 2 verifying
1
1
lim(t m )  lim(t 2 )  1  lim(t
t 
t 
lim(t
m 2
2m
t 
t 
)  1  lim(t
t 
4
 1m
2
4
m2
2m
1
 1  1m
2
4
1
2m
1
1
)  1  lim(t 2 m )  1m  2
t 
1
m2
) 1
1
1
1
 14  lim(t 4 m )  lim(t 8 )
t 
t 
Thus
lim(t
1
2 m ( m  2)
t 
1
) 1
m2  4
 lim(t
1
2 m ( m  2)
t 
)
2
 lim(t
m ( m2  4)
t 
) 1
And
1
m
lim(t )  1  lim(t
t 
1
1
m ( m2  4)
2
t 
)  1  lim(t
1
M1
t 
)
Hence
2M 1  m(m 2  4)  10  22 M 2  2M 1 ( M 12  4)  210
...
2n M n  2M 1 ( M 12  4)...( M n 12  4)  m(m 2  4)( M 12  4)...( M n 12  4)  m.Pn .2n
lim( Pn )  
n 
1
m
And lim(t )  1  ei 2u  lim(t
t 
t 
1
Mn
i 2n1 u
)  1  ei 2v  e ( m
2
 4)...( M n12 4)
, n  N with
2n u  m(m2  4)...( M n 12  4)v  2n Pn v, n  N
It is impossible ! The initial hypothesis is false : m does not exist ! Hence m=2, is the only
value for which lim(t
t 
1
 iy
2
)  1; y  N .
Hypothesis H :
5
n
1
H : n, m | lim(t m )  1  lim(t 2 )
t 
t 
2n  m, m  2,1  n  m
lim(t
2nm
2m
t 
)  1  lim(t
2nm
2m
t 
)  lim(t
1
2 m (2 n  m )
t 
t 
m
 lim(t
n (4 n 2  m 2 )
t 
)  lim(t
1
)  1  lim(t )  lim(t
t 
1
2 m (2 n  m )
4 n2  m2
t 
1
)  14 n
2
 m2
n
m
)  lim(t )  1
t 
n
m
4n 2  m 2  2  lim(t )  1
t 
2n  m, m  2,1  n  m
Let the hypothesis H
n
H : n, m | lim(t m )  a  lim(t  iy )
t 
 lim(t
n
 iy
m
t 
 lim(t
)  1  lim(t
t 
2nm
2m
t 
 lim(t
t 
1
 ik
2
)  lim(t
i ( k  y )
t 
2nm
2n
t 
)  lim(t
t 
)
n
m
)  a  lim(t )
t 
m
1
2n
)  1  lim(t
t 
m
n
m
2n
)
n
m
 lim(t )  1  lim(t )  lim(t )  1  a  a  1 
t 
1
 t2
 iy
t 
t 
2n
1
m
 1; y  R



1
1
1 iy 
iy

1
1
1
it means that  (  iy )  
t 2   lim(
(t 2  1))  0
t 
1
1
2
1   iy

1   iy
 2
1
2
Let
1

x
1
1
iy   x


1
1
t 2 1
1
 ( x  iy)  0  
t1 x iy   lim(
(t 2 2  1))  lim(
)0 x
t  1  x  iy
t  1  x  iy
2
1  x  iy
1
Thus the non trivial zeros of the Riemann function for the reals lie in the critical
line ! So the hypothesis is proved for the real numbers. The Riemann
hypothesis is important because it gives information about the zeros of the
6
Riemann function and the distribution of those zeros are related to real primes
!
The generalization to the integers
As




dt
1
1
1
1
1 z 



t






1 t z 1  z  1 t 1 t z Q t ' z Q \ N t ' z

1
  z .B 
t 1 t

(
1
)0
primes  N

1
1
).
B

A

z
z
p
t 1 t
Let now
 ( p
primes
N
( x iy )
If B is finite, the zeta function for the reals will be equal to zero. Else
Let  1 ( z ) the Riemann function for the reals and  ( z ) the Riemann function for
the integers, if B   and if A is finite and
A 1
( A   ( z )) 2  A2  A2   ( z ) 2  2 A ( z )  A2  2 A ( z )  A ( z )  0
( An ( z )  A) 2  A2  A2  ( An ( z )) 2  2 An 1 ( z )  An 1 ( z )  0, n
 | B
A  An  0   1 ( z )  B ( z )  An ( z )  0   1 ( z )  0  1  A
And if
1 A  0
( A1   ( z )) 2  A2  A2   ( z ) 2  2 A1 ( z )  A2  2 A1 ( z )  A1 ( z )  0
( A n ( z )  A1 ) 2  A2  A2  ( A n ( z )) 2  2 A n 1 ( z )  A n 1 ( z )  0, n
 | B
A  A n  0   1 ( z )  B ( z )  A n ( z )  0   1 ( z )  0
Now if
7
0  A  1
( A1   ( z )) 2  A2  A2   ( z ) 2  2 A1 ( z )  A2  2 A1 ( z )  A1 ( z )  0
( A2 n 1 ( z )  A2 ) 2  A4  A4  ( A2 n1 ( z )) 2  2 A2 n3 ( z )  A2 n3 ( z )  0, n
 | B
 A   A2 n 1  0   1 ( z )  B ( z )   A2 n 1 ( z )  0   1 ( z )  0
And if
A  1
( A   ( z )) 2  A2  A2   ( z ) 2  2 A ( z )  A2  2 A ( z )  A ( z )  0
( A2 n 1 ( z )  A2 ) 2  A4  A4  ( A2 n 1 ( z )) 2  2 A2 n 3 ( z )  A2 n 3 ( z )  0, n
 | B
 A   A2 n 1  0   1 ( z )  B ( z )  A2 n 1 ( z )  0   1 ( z )  0  A  1
If now B   and if A is finite and
A 1
( A   ( z )) 2  A2  A2   ( z ) 2  2 A ( z )  A2  2 A ( z )  A ( z )  0
( An ( z )  A) 2  A2  A2  ( An ( z )) 2  2 An 1 ( z )  An 1 ( z )  0, n
 | B
 A   An  0   1 ( z )  B ( z )   An ( z )  0   1 ( z )  0  1  A
And if
1 A  0
( A1   ( z )) 2  A2  A2   ( z ) 2  2 A1 ( z )  A2  2 A1 ( z )  A1 ( z )  0
( A n ( z )  A1 ) 2  A2  A2  ( A n ( z )) 2  2 A n 1 ( z )  A n 1 ( z )  0, n
 | B
 A   A n  0   1 ( z )  B ( z )   A n ( z )  0   1 ( z )  0
And if
0  A  1
( A1   ( z )) 2  A2  A2   ( z ) 2  2 A1 ( z )  A2  2 A1 ( z )  A1 ( z )  0
( A2 n 1 ( z )  A2 ) 2  A4  A4  ( A2 n1 ( z )) 2  2 A2 n3 ( z )  A2 n3 ( z )  0, n
 | B
A  A2 n 1  0   1 ( z )  B ( z )  A2 n 1 ( z )  0   1 ( z )  0
And if
A  1
( A   ( z )) 2  A2  A2   ( z ) 2  2 A ( z )  A2  2 A ( z )  A ( z )  0
( A2 n 1 ( z )  A2 ) 2  A4  A4  ( A2 n 1 ( z )) 2  2 A2 n 3 ( z )  A2 n 3 ( z )  0, n
 | B
A  A2 n 1  0   1 ( z )  B ( z )  A2 n 1 ( z )  0   1 ( z )  0  A  1
8
And if A   : B A 1 hence
A m ( z ) m  1
An  ( z ) n 1  0
Am n 1
B  A 1  A( m  n )  A( m  n )  ( z ) m  n 1  B ( z ) m  n 1  A ( z ) m  n
( 1)( m  n )
A
(m  n)    1
mn2
1

 A1( m  n ) 
A
B
1
B 2 ( z ) m  n 1 A1  1
1
1
1 
B 2 Am  n
BA
2  m  n 1 m  n




A

B
A

B
A
B mn
B
1
 ( A1 B) 
1
 A1  A2  B  A 1  1  
It is impossible !
 A  , A  
Thus A=0 And

t
1
dt
x  iy


1
t1 x iy  
1
1  x  iy
 ( p
primes
N
1
( x  iy )
).B 
 ( p
primes
N
1
( x  iy )
) A0 x 
1
2
Thus the non trivial zeros of the Riemann funtion zeta lie in the critical line like for the
reals ! It is the proof of the Riemann hypothesis !
Results
The concept of prime has been generalized to the reals. It allowed to
reconsider the Riemann hypothesis under a new point of view.
Then, it has been proved for the reals after a calculus of integral and it has
been generalized to the integers after an elementary calculs. Finally, the
conjecture has been proved.
Discussion
As it has been said, the Riemann hypothesis is one of the most important
problems of number theory. It is one of the seven problems of the millennium.
Until now, the approaches never reconsidered the concept of prime as it has
been done in this study.
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In fact, this new approach will generalize the concept of prime to the reals. It
constitutes a novelty. Therefore, the resolution will depend on a simple integral
and an elementary calculus.
Conclusion
The generalization of the concept of prime is the main novelty of this study. It
allows to generalize the hypothesis to the reals and to solve the conjecture.
The Bibliography
[1]R. J. Backlund, « Sur les zéros de la fonction ζ(s) de Riemann », CRAS, vol. 158,
1914, p. 1979–1981.
[2]X. Gourdon, « The 1013 first zeros of the Riemann zeta function, and zeros
computation at very large height »
[3]J.P.Gram, « Note sur les zéros de la fonction ζ(s) de Riemann », Acta Mathematica,
vol. 27, 1903, p. 289–304.
[4]J.I.hutchinson « On the Roots of the Riemann Zeta-Function », Trans. AMS, vol.
27, no 1, 1925, p. 49–60.
[5]A. M. Odlyzko, The 1020-th zero of the Riemann zeta function and 175 million of
its neighbors, 1992.
[6]J.Barkley Rosser, J. M. Yohe et Lowell Schoenfeld, « Rigorous computation and
the zeros of the Riemann zeta-function.», Information Processing 68 (Proc. IFIP
Congress, Edinburgh, 1968), Vol. 1: Mathematics, Software, Amsterdam, NorthHolland, 1969, p. 70–76.
[7]http://fr.wikipedia.org/wiki/Edward_Charles_TitchmarshE.C.Titchmarsh, « The Zeros
of the Riemann Zeta-Function », Proceedings of the Royal Society, Series A,
Mathematical and Physical Sciences, vol. 151, no 873, 1935, p. 234–255.
[8]E. C. Titchmarsh, « The Zeros of the Riemann Zeta-Function », Proceedings of the
Royal Society, Series A, Mathematical and Physical Sciences, The Royal Society, vol.
157, no 891, 1936, p. 261–263.
[9]A.M.Turing, « Some calculations of the Riemann zeta-function », Proceedings of
the LMS, Third Series, vol. 3, 1953, p. 99–117.
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[10]J. van de Lune, H.te Riele et D. T. Winter, « On the zeros of the Riemann zeta
function in the critical strip. IV », Mathematics of Computation, vol. 46, no 174, 1986,
p. 667–681.
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