# A follow-up problem once the Riemann hypothesis has considered proven to be truth? [closed]

What will happen if a mathmatician have prove that 99.999....% of the solution stay on the critical line and receive the prize but after that another mathmatician find finite numbers of interesting solution out of the line?

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## closed as not a real question by BenjaLim, Old John, Asaf Karagila, William, ArgonAug 30 '12 at 1:55

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I would imagine the mathematician who found the counter examples would demand that the mathematician who received the prize retract his paper and hand over the money. :) –  Bill Cook Aug 29 '12 at 22:53
99.999...%=100% (if ... means repeating forever) –  Graphth Aug 29 '12 at 23:07
@dotdot - 4 over infinity equal to 0... –  Victor Aug 29 '12 at 23:15
A proof that "99.999%" of the zeros are on the critical line, or any similar statement about "most" or "almost all" of the zeros, would not, strictly speaking, constitute a proof of the Riemann hypothesis, and I presume it would not be eligible for the Clay prize. –  Nate Eldredge Aug 31 '12 at 0:57

In theory, one could prove that 100% of the zeros on the line, that is a set of density $1$, which does not neccesarily imply the Riemann Hypothesis. There still could exist a very sparse set of zeros off the line, a set of density $0$. To be precise, let $$N(T)=\left|\left\{s=\sigma+i\gamma:\ \zeta(s)=0,\ 0\leq \sigma\leq 1\ \text{and} |\gamma|\leq T \right\}\right|$$ be the number of zeros up to height $T$, and let $$N_0(T)=\left|\left\{s=\frac{1}{2}+i\gamma:\ \zeta(s)=0,\ \text{and} |\gamma|\leq T \right\}\right|$$ be the number of zeros on the critical line up to height $T$. Then, proving that $$\lim_{T\rightarrow \infty}\frac{N_0(T)}{N(T)}=1$$ means that almost all of the zeros lie on the critical line, i.e. that the density is $1$, but it does not imply the Riemann hypothesis.