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September 10th, 2017, 06:40 PM  #1 
Newbie Joined: Jul 2014 From: Taiwan Posts: 8 Thanks: 0  Riemann Hypothesis
P: ζ(sᵢ)=0 Q: sᵢ=σ+it R: t≠0 S: σ=1/2 A: P≡[Q≡(R≡S)] It is that ζ(1/2)=0 if the Proposition A doesn't hold, but actually, ζ(1/2)≈1.4603545; therefore RH does hold. 
September 11th, 2017, 06:38 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,402 Thanks: 547 
I am not sure what you are trying to say. However (using your notation) the Riemann hypothesis states that all zeroes for 0<σ<1 are on the line σ=1/2+it, not necessarily for t=0.

October 20th, 2017, 11:44 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,879 Thanks: 766 
I presume you mean "for 0< σ<1".

October 20th, 2017, 02:13 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,402 Thanks: 547  

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hypothesis, riemann 
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