The proof by contradiction of the negation of Riemann Hypothesis
The proof by contradiction of the
negation of Riemann Hypothesis
Riemann Hypothesis s=(+i.t)
All the non trivial zeros of the Zeta function (s) in the critical strip
(0<Real s<1) are located on the critical line (Real s=1/2)
By my effort to solve:
The greatest unsolved problem in Mathematics
I found that:
The Riemann Hypothesis is not valid
Iff there exist p,q,t, (p#q) to match this equation
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
The above equation is correct:
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
http://Mathhighways.blogspot.com/
John Bredakis
Athens Greece 2013
1
A fact for the Zeta function (s) s=(+i.t)
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
Checking the Riemann Hypothesis in the critical strip
s=(p
i.t) or s=(qi.t) 0<p<1 0<q<1 q=1-p
I found a formula for A and B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
And then I made a Working Hypothesis
ie: under what conditions A=B=O
expecting that this zero belongs to (s)
For the proof of the formulas for A and B
see text
2
For the proof of the formulas for A and B
see text
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
-----
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
3
My Working Hypothesis
ie: under what conditions A=B=O
expecting that this zero belongs to (s)
Suppose that:
2 M = Icos
t +p.q 0<p<1
M = = Icos q=1-p
2 2 2 2
t +p.q + (p-q) .t 0<q<1
+oo
p/2 q/2 -1 t (x)-1
Icos = x + x .x .coslog(x)...dx
1 2 2
And suppose also that:
N = Isin
(p-q).t
N = = Isin
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
Isin = x - x .x .sinlog(x)...dx
1 2 2
----
+oo
p/2 q/2 -1 t (x)-1
Icos = x + x .x .coslog(x)...dx
1 2 2
2
(x)-1 +oo -n ..x
= e
2 n=1
+oo
p/2 q/2 -1 t (x)-1
Isin = x - x .x .sinlog(x)...dx
1 2 2
4
By my Working Hypothesis
A=B=0
2
(p-q).t [t + p.q]
Isin = and Icos = 2 2 2 2
=t +p.q +(p-q) .t
.Isin .Icos 0<p<1
ie: = = 1
(p-q).t 2 q=1-p
[t +p.q] 0<q<1
By my Working Hypothesis
The following integral vanishes (becomes zero)
t 2 2 2 2
log(x). = A = t +p.q + (p-q) .t
2
p/2 q/2 p/2 q/2
+oo x -x x +x
-1 (x)-1
..sin(A) - .cos(A).x ..dx
1 (p-q).t 2 2
t +p.q
By the Mean Value Theorem of Integral Calculus
there must be a >1 so that
p/2 q/2 p/2 q/2
- +
t t
.sinlog(). = .coslog().
(p-q).t 2 2 2
t +p.q
r
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
5
So far we have that iff the formulas of A and B are correct
Then the following is also correct
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
The next step is to select a t to satisfy the above
p/2 q/2
2 +
M ± M - 4.(p.q) -1
tanlog(). = .(p-q).M
4 p/2 q/2
-
>1
0<p<1 0<q<1 [2p-1]/2
+ 1
q=1-p -1
Or tan() = .(p-q).M
[2p-1]/2
- 1
Working on the left part of the above equation
>1 0<1
2
log() = 2./+ -4.(p.q) Or log() = 2./ -
-----=-----
2./ + 2./ -
= e r = e
6
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
Working on the right part of the above equation
0<p<1 0<q<1 q=1-p >1
[2p-1]/2 S+1 M.R M.R
= = = S
S-1 (p-q) (2p-1)
S+1 2/[2p-1]
=
S-1
>1 S+1 2/[2p-1]
S+1 2/[2p-1] log
= = e S-1
S-1
>1 0<1
2./ + 2./ -
= e r = e
-----=log()------
±
S+1 2/[2p-1] M.R R=tan(.)
= .log S = 2
2. S-1 (2p-1)
0<1
2
= - 4.(p.q) 0<p<1 0<q<1 q=1-p
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
Those distinct roots are: t1,2 = (M
T)/2
7
Regardless of whether p>q or p<q 0<p<1 0<q<1 q=1-p >1
[p-q]=[2p-1] [p-q]=[2p-1]
[p-q]/2 S+1 M.R+(p-q) S+1 2/[2p-1]
= = =
S-1 M.R-(p-q) S-1
-----------------------------------------------------------
[q-p]/2 S+1 -1 M.R+(p-q) -1 M.R+(q-p)
= = =
S-1 M.R-(p-q) M.R-(q-p)
[q-p]/2 M.R+(q-p) M.R+(q-p) 2/[2q-1]
= =
M.R-(q-p) M.R-(q-p) [2q-1]=[q-p]
Therefore in any case >1 and close (to very close) to 1
As expected
Regardless of whether p>q or p<q
-1<0 for t<0
0<p<1 0<q<1 q=1-p p/2 q/2
+
t t
R = tan(.) = tanlog(). = .(p-q).
2 2 p/2 q/2 2
- t + p.q
0<1 >1
0<1
2 . log()
For t>0 and M.t = t +p.q M = + (p.q).
log() .
2
Those distinct t's (Roots of t -M.t+p.q=0) exist , but cannot
be defined , since the exact value of (>1) is not known
Those distinct roots are: t1,2 = (M
T)/2
8
The proof of the formulas for A and B
Checking the Riemann Hypothesis in the critical strip
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
A.
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2
q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
---------------------------------------------------------------
2
-t +p.q-i.(p-q).t
-1 -1
= =
[p+i.t].[q-i.t] 2 2 2 2 2
t +p.q-i.(p-q).t t +p.q +(p-q) .t
=
+oo
(p+i.t)/2 (q-i.t)/2 -1 (x)-1
+ x + x .x ..dx
1 2
9
Checking the Riemann Hypothesis in the critical strip
-s/2 -(1-s)/2
.(s/2).(s) = .[(1-s)/2].(1-s)
+oo
-1 s/2 (1-s)/2 -1 (x)-1
= + x + x .x ..dx
s.(1-s) 1 2
2 2
n=+ -n ..x (x)-1 +oo -n ..x (x) 1
(x) = e = e =
n=- 2 n=1 (1/x)
x
B.
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2
q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
---------------------------------------------------------------
2
-t +p.q+i.(p-q).t
-1 -1
= =
[p-i.t].[q+i.t] 2 2 2 2 2
t +p.q+i.(p-q).t t +p.q +(p-q) .t
=
+oo
(p-i.t)/2 (q+i.t)/2 -1 (x)-1
+ x + x .x ..dx
1 2
10
Checking the Riemann Hypothesis in the critical strip
A+B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
A+B -2.t +p.q +oo
(p+i.t)/2 (p-i.t)/2 -1 (x)-1
= + x +x .x ..dx
2 2 2 2 1 2
t +p.q+(p-q) .t
+oo
(q+i.t)/2 (q-i.t)/2 -1 (x)-1
+ x +x .x ..dx
1 2
2
A+B -2.t +p.q +oo
p/2 q/2 -1 t (x)-1
= + x +x .x .coslog(x)...dx
2 2 2 2 1 2 1
t +p.q+(p-q) .t
11
Checking the Riemann Hypothesis in the critical strip
A-B
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
A-B +oo
-2.i.(p-q).t (p+i.t)/2 (p-i.t)/2 -1 (x)-1
= + x -x .x ..dx
2 2 2 2 1 2
t +p.q+(p-q) .t
+oo
(q+i.t)/2 (q-i.t)/2 -1 (x)-1
- x -x .x ..dx
1 2
A-B +oo
-2.i.(p-q).t p/2 q/2 -1 t (x)-1
= + i. x -x .x .sinlog(x)...dx
2 2 2 2 1 2 1
t +p.q+(p-q) .t
12
Checking the Riemann Hypothesis in the critical strip
By Combination
A=
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
13
Checking the Riemann Hypothesis in the critical strip
By Combination
B=
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
14
Checking the Riemann Hypothesis in the critical strip
By Combination Summary
p + i.t
- p + i.t 0<p<1
= 2 . .(p + i.t) = A
2 q=1-p
q - i.t
- q - i.t 0<q<1
= 2 . .(q - i.t) = A
2
----------------------------------------------------------------
p - i.t
- p - i.t 0<p<1
= 2 . .(p - i.t) = B
2 q=1-p
q + i.t
- q + i.t 0<q<1
= 2 . .(q + i.t) = B
2
2
-t +p.q - i.(p-q).t
A =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
+ i. x - x .x .sinlog(x)...dx
1 2 2
2
-t +p.q + i.(p-q).t
B =
2 2 2 2
t +p.q + (p-q) .t
+oo
p/2 q/2 -1 t (x)-1
+ x + x .x .coslog(x)...dx
1 2 2
+oo
p/2 q/2 -1 t (x)-1
- i. x - x .x .sinlog(x)...dx
1 2 2
15
Some general remarks on the Zeta function (s)
And a reference to my pdf
Understanding the Zeta function and the Riemann Hypothesis
16
For further details see my pdf:
Understanding the Zeta function
and the Riemann Hypothesis
http://Mathhighways.blogspot.com/
John Bredakis
Athens Greece 2013
17
References
And a lot of personal work
------
I like to express my special thanks to Professor of Mathematics
Elias Kastanas
for his advise and check up
18
Additional
References From The Internet
-----
And my pdf:
Big Bang in Math,John Bredakis Method & the Gamma function
19
20
The reason to deal with the Zeta function
(s),
despite my shallow knowledge of complex analysis
-------
The common things between those pdf of mine
Big Bang in Math , John Bredakis method & the Gamma function
and
The proof by contradiction of the negation of Riemann Hypothesis
The Gamma function and the tranfer of i
from denominator to numerator
Trying to understand the difficult to comprehend topic of complex
analysis I ended up to the calculus of residues.
And then I realized that the complex analysis can be bypassed to
a large part by the classical advanced analysis , hence my pdf
An attempt to bypass the calculus of residues
Searching very carefully in the Internet , I found the one and only pdf
by Theodore Yoder (Introduction to Riemann Hypothesis)
unique in the sense of handling the
(s),with minimal complex analysis
and epecially with the introduction of the idea of analytic continuation
Thereafter I needed only few lines from the pdf by Lorentzo Menici
to comprehend the Riemann-Siegel formula , hence my pdf:
Understanding the Zeta function and the Riemann Hypothesis
This is the whole story
Handling Mathematics for many years
I must admitt that my mentor in Mathematics was my beloved
Uncle Fotis , a medical doctor , brilliant mathematical brain
and great teacher of Mathematics
Sincerelly:
John Bredakis MD
http://Mathhighways.blogspot.com/
Athens Greece 2013
21
Frequently Asked Questions About "The proof by contradiction of the negation of Riemann Hypothesis"
What is the main claim of this document?
The document claims to present a proof by contradiction that the Riemann Hypothesis is not valid.
What equation does the author present as key to the proof?
The author states that the Riemann Hypothesis is not valid if there exist p, q, and t (p ≠ q) to match a specific equation (provided in the document). The correctness of this equation is stated.
What is the author's Working Hypothesis?
The author's Working Hypothesis involves finding conditions under which A=B=0, expecting that these zeros belong to the Zeta function.
What formulas are used for A and B in the Working Hypothesis?
The document provides formulas for calculating A and B based on p, q, and t, where s=(p + i.t) or s=(q + i.t) and 0 < p < 1, 0 < q < 1, q = 1-p. These formulas involve the Zeta function and the Gamma function.
What integrals are discussed?
The document discusses integrals related to cosine and sine functions, with respect to a variable x, over the interval from 1 to positive infinity. These integrals are related to the formulas for A and B and the Working Hypothesis.
What does the author say about selecting 't' values?
The author mentions the next step is to select a 't' value to satisfy an equation. These ‘t’ values, roots of the polynomial, exist but cannot be defined because the exact value of a variable (>1) is not known.
What are the distinct roots, t1,2, found?
The author states that those roots exist and are equal to (MT)/2, however the T cannot be calculated. Also related expressions for these roots are shown.
What is the Mean Value Theorem of Integral Calculus used for?
The Mean Value Theorem is applied to find a value greater than 1 so that expressions involving sine and cosine of log() are considered, where the values are part of calculations related to proving the hypothesis.
What is the relationship to the Gamma function?
The formulas for A and B involve the Gamma function, denoted by (s/2) and [(1-s)/2].
Where can I find more information about the author's work?
The author references a blog: http://Mathhighways.blogspot.com/, and indicates further details are in a PDF titled "Understanding the Zeta function and the Riemann Hypothesis".
What resources did the author use to prove the Riemann Hypothesis?
The document mentions the book "Big Bang in Math , John Bredakis method & the Gamma function, a pdf by Theodore Yoder (Introduction to Riemann Hypothesis) , Lorentzo Menici.
Who is the author and when was it written?
The document is by John Bredakis, written in Athens, Greece in 2013.
Who is Elias Kastanas?
Elias Kastanas is a Professor of Mathematics who provided advice and checked the work.
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- Prof. Dr. med. John Bredakis (Author), 2013, The proof by contradiction of the negation of Riemann Hypothesis, Munich, GRIN Verlag, https://www.grin.com/document/213699
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