Curious about a question on zeta zeros? [closed]

I have Edwards and Titmarsch books on Riemann zeta function with me. I could not find (maybe I did not read through that carefully), but are there results similar to the form like the one given below:

Is there a non-trivial zero $\rho$ starting from which $(\Im(\rho)\log(2)/2\pi )$ runs only through integer values

I would love any elaboration or links on this topic.

Thanks,
Roupam Ghosh

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closed as not a real question by Will Jagy, Yemon Choi, S. Carnahan♦Oct 15 '10 at 13:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I can't imagine that the imaginary parts of the roots are integral multiples of, what, $$\frac{2 \pi}{\log 2}$$ down the road. I think there is no mention because it is not plausible. – Will Jagy Oct 15 '10 at 3:21
Can you make a look at mathoverflow.net/questions/32324 ? As far as I understand the "real" conjecture is that the imaginary parts of the zeros are linearly independent over $\mathbb Q$... – Wadim Zudilin Oct 15 '10 at 3:57
Thanks for that link Zudilin... I'll check it... – Roupam Ghosh Oct 15 '10 at 6:05
Dear Author, I don't wish to make you sceptical about Splotchy's answer below, as it makes use of a plausible conjecture to show that yours is false. Even more, if you assume implicitly in your question that the real parts of zeros are $1/2$, then the answer is unconditional. I would suggest in the future to cite the conjectures more accurate: if I were Edwards or Titmarsch, I would be offended by attributing of such expectations to myself. – Wadim Zudilin Oct 15 '10 at 8:44
Scott, many questions can be resolved in exactly the same way. Your criteria to judge whether a question is trivial or not are completely mysterious. – Wadim Zudilin Oct 15 '10 at 14:00

EDIT: A simpler argument - a more precise asymptotic for the number of zeroes $N(t)$ of imaginary part $\le t$ (counted with multiplicity) is $$N(t) = \frac{t}{ \pi} \log \frac{t}{2 \pi e} + o(\log t),$$ This is enough to show that, for any fixed $\epsilon > 0$, $$N(t + \epsilon) - N(t) \sim \frac{\epsilon}{\pi} \log t,$$ and thus, for sufficiently large $t$, and for any interval of length $\epsilon$, there are zeroes (whose imaginary part lies) in this interval, which also implies the conjecture is false. THIS DOES NOT USE GRH.
PS: Scott Carnahan helpfully remarks that the wikipedia article points out that Littlewood noticed that the difference in the imaginary parts of the zeros tends to zero as $t \rightarrow \infty$ (presumably by exactly using this asymptotic result of von Mangoldt above). Personally I prefer mathematics rather than an appeal to authority, but apparently that is not enough for some.
REMARK: Dear Wadim, please read this again, and realize that it DOESN'T USE GRH. The estimate of zeros (which was basically known by Riemann) is about zeroes in the CRITICAL STRIP (real part in $[0,1]$) not the CRITICAL LINE. Having done this, you can delete all your comments, I'll edit this answer, and we can all pretend it never happened. (In fact, I'll make this community wiki so you can delete this remark yourself.)
Splotchy, you miss the point: the conjecture from OP is about 1/2+in\alpha$for some$n\in\mathbb Z$, not all$n\ge n_0$. – Wadim Zudilin Oct 15 '10 at 5:44 One can obviously choose a set$S\subset\mathbb Z$such that$\{1/2+in\alpha:n\in S\}$satisfies both Riemann's, Montgomery's etc statistics. The asymptotics does not imply arithmetic. – Wadim Zudilin Oct 15 '10 at 6:18 Oh-oh, they are too dense! :-))) Taken. Let me say the truth: I don't believe RH and I am not sure that there are unconditional estimates for number of zeta zeros with real part$=1/2$and imaginary part$<T$(for any$T$). – Wadim Zudilin Oct 15 '10 at 7:05 The original question does not refer to$\Re z$, so you can take for any$n$randomly symmetric$\log n$-set for$\Re\rho$. Of course, this is quite unlikely (so, the original question is irrealistic as well) but your claim "Such a conjecture is false" is true modulo RH. – Wadim Zudilin Oct 15 '10 at 7:31 @Wadim, it's my understanding that it has been known for about 40 years, unconditionally, that at least a third of the zeros up to$T\$ have real part 1/2. – Gerry Myerson Oct 15 '10 at 12:52