20080611, 07:26  #1 
Oct 2007
22_{16} 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. 
20080611, 12:01  #2 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
I looked for a smilie that expressed my reaction, but the best I can do is

20080611, 12:06  #3 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB_{16} Posts 

20080611, 12:19  #4  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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 socalled argument is nonsense. 

20080611, 14:18  #5  
"William"
May 2003
New Haven
4470_{8} Posts 
Quote:


20080611, 14:55  #6 
Oct 2007
100010_{2} 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. 
20080611, 15:07  #7  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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??? 

20080611, 15:11  #8 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{7}·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.

20080611, 15:35  #9  
"Ben"
Feb 2007
2^{2}·29^{2} Posts 
Quote:


20080611, 15:38  #10 
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. 
20080611, 16:22  #11  
Nov 2003
2^{2}·5·373 Posts 
Quote:
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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