mersenneforum.org Riemann's hypothesis is incorrect a proof
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 2008-06-11, 07:26 #1 Carl Fischbach     Oct 2007 2216 Posts Riemann's hypothesis is incorrect a proof I would be greatful for any feedback on this proof so it is free of mistakes before I take it any further. from Euler's formula e^ipi = -1 let 0 + i0 = 1/1^s +1/2^s + 1/3^s +1/4^s . . . now you choose any term except 1/1^s and move it to the left side of the equation I've choosen 1/2^s -1/2^s = 1/1^s +1/3^s +1/4^s . . . from Euler's formula you now have e^ipi/2^s = 1/1^s +1/3^s +1/4^s. . . substituting you get 0 +i0 = 1/2^s +e^ipi/2^s multiplying both sides by 2^2s you get 0+ i0 = 2^s + 2^s*e^ipi next 0 +i0 = 2^(a +iq)+2^(a+iq)*e^ipi then 0 +i0 = 2^a*2^iq + 2^a*2^iq*e^ipi let 2^q = e^pi therefore 0 +i0 = 2^a*e^ipi + 2^a*e^ipi*e^ipi let -1 = e^ipi 0 + i0 = 2^a*-1 + 2^a*-1*-1 add i0 to both sides 0 +i0 = -2^a + 2^a +i0 therefore 0 +i0 = 0 +i0 Note (a) can be any real value not just 1/2 therefore Riemann's hypothesis is incorrect.
 2008-06-11, 12:01 #2 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22×3×641 Posts I looked for a smilie that expressed my reaction, but the best I can do is
2008-06-11, 12:06   #3
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

10AB16 Posts

Quote:
 Originally Posted by cheesehead I looked for a smilie that expressed my reaction, but the best I can do is
I've got a few:

2008-06-11, 12:19   #4
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Carl Fischbach I would be greatful for any feedback on this proof so it is free of mistakes before I take it any further. from Euler's formula e^ipi = -1 let 0 + i0 = 1/1^s +1/2^s + 1/3^s +1/4^s . . . now you choose any term except 1/1^s and move it to the left side of the equation I've choosen 1/2^s -1/2^s = 1/1^s +1/3^s +1/4^s . . . from Euler's formula you now have e^ipi/2^s = 1/1^s +1/3^s +1/4^s. . . substituting you get 0 +i0 = 1/2^s +e^ipi/2^s multiplying both sides by 2^2s you get 0+ i0 = 2^s + 2^s*e^ipi next 0 +i0 = 2^(a +iq)+2^(a+iq)*e^ipi then 0 +i0 = 2^a*2^iq + 2^a*2^iq*e^ipi let 2^q = e^pi therefore 0 +i0 = 2^a*e^ipi + 2^a*e^ipi*e^ipi let -1 = e^ipi 0 + i0 = 2^a*-1 + 2^a*-1*-1 add i0 to both sides 0 +i0 = -2^a + 2^a +i0 therefore 0 +i0 = 0 +i0 Note (a) can be any real value not just 1/2 therefore Riemann's hypothesis is incorrect.
Your first mistake is that you are unaware that you have no
understanding of mathematics.
Your third mistake is not realizing that the representation that you
are using for the Zeta function ONLY CONVERGES for real(s) > 1.
nonsense.

2008-06-11, 14:18   #5
wblipp

"William"
May 2003
New Haven

44708 Posts

Quote:
 Originally Posted by R.D. Silverman Your first mistake is that you are unaware that you have no understanding of mathematics. Your second mistake is total ignorance about the Zeta function. Your third mistake is not realizing that the representation that you are using for the Zeta function ONLY CONVERGES for real(s) > 1. Your fourth mistake is not realizing that your so-called argument is nonsense.
Who are you and what have you done with the real R.D. Silverman?

 2008-06-11, 14:55 #6 Carl Fischbach     Oct 2007 1000102 Posts an interesting twist I am not really sure of myself with complex math but I thought I would give Riemann's hypothesis a shot, the feedback is useful. But I am sure of this get to the bottom of the year 2012 your really going to need a life boat then.
2008-06-11, 15:07   #7
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Carl Fischbach I am not really sure of myself with complex math but I thought I would give Riemann's hypothesis a shot, the feedback is useful. But I am sure of this get to the bottom of the year 2012 your really going to need a life boat then.
I am curious. You admit that you are unsure of yourself.
Why then would you even try something like this? Don't you
realize it makes you look foolish? I am sure that you are equally
ignorant about (say) neurosurgery. But you wouldn't presume to
suggest a new surgical technique would you?

What is it about mathematics that people who are totally ignorant
(and who know they are ignorant) think they can magically solve
problems that have eluded PhD's for centuries??? We see this phenomenon
constantly on the Internet.

I also suggest that you need to go back to school to learn how
to write cogent English. Not only is your mathematics bad, but so is
your spelling, grammar, punctuation, and sentence structure. Indeed, your
use of the phrase "complex math" doesn't even mean what you
clearly intended. "complex math" is not the same as "mathematics of
complex valued functions". Your public writing makes you appear to be a
total cretin. Do you like to appear foolish???

 2008-06-11, 15:11 #8 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 27·47 Posts We are all fools at some time in our lives. Nothing wrong with that as long as we learn from it and try to improve.
2008-06-11, 15:35   #9
bsquared

"Ben"
Feb 2007

22·292 Posts

Quote:
 Originally Posted by retina We are all fools at some time in our lives. Nothing wrong with that as long as we learn from it and try to improve.

 2008-06-11, 15:38 #10 Carl Fischbach     Oct 2007 2×17 Posts nothing ventured nothing gained I may be an amateur and it may be along shot, but amatuers have cracked problems before and I don't care.
2008-06-11, 16:22   #11
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Carl Fischbach I may be an amateur and it may be along shot, but amatuers have cracked problems before and I don't care.

No, you are not an "amateur". An amateur is someone who knows
at least a little about the subject under discussion. You tried to disprove
R.H. without even knowing what the Zeta function *is*.

You certainly have the right not to care about looking foolish.

Also, your English still hasn't improved. Someone who is deliberately
ignorant and who refuses to learn from clear prior mistakes is a crank.

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