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September 10th, 2017, 05:40 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, 05:38 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,306 Thanks: 525 
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.


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