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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: 6,821 Thanks: 723 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, 10:44 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I presume you mean "for 0< |σ|<1".
October 20th, 2017, 01:13 PM   #4
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Quote:
 Originally Posted by Country Boy I presume you mean "for 0< |σ|<1".
No, sorry. I meant all σ = x + iy where 0 < x < 1.

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