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a symbolic phrasing of the Riemann Hypothesis

Name: furfag 2005-01-17 4:36

Not to solve the million dollar problem, but just to look at the strongest way of phrasing it.

The Riemann Hypothesis says that the non-trivial (I'll define that in a bit) zeroes of the zeta function all lie on the same line, that line in the complex plane having real part 1/2.  the riemann zeta function is analytically extended in such a way (which I don't even understand yet) that the zeta of any even negative integer is zero.  zeta(-2)=0, zeta(-4)=0, zeta(-6)=0, and so forth.  To date, then, all points s of the complex plane for which zeta(s)=0  have either been the aforementioned trivial zeroes, or been complex numbers of the form a + bi, with a=1/2. they have 'real part' 1/2.  Let C denote the complex numbers, and Z denote the integers (never mind double struck lettering!)  Let C-Z be their set difference, and let є mean 'is an element of'.  So, a naive first attempt at phrasing the hypothesis itself might be:

1)   s  є  C - Z  =>  ( zeta(s) = 0  =>  Re(s) = 1/2)

Now to my question.  Could this symbolic phrasing of the question be strengthened, by changing conditionals to biconditionals?  For example:

2)   s  є  C - Z  <=>  ( zeta(s) = 0  =>  Re(s) = 1/2)     ,

3)   s  є  C - Z  =>  ( zeta(s) = 0  <=>  Re(s) = 1/2)     ,  or

4)   s  є  C - Z  <=>  ( zeta(s) = 0  <=>  Re(s) = 1/2)    .

Which of 1)-4) is the strongest, correct logical phrasing?  give counterexamples for the others.

Name: Anonymous 2005-01-19 14:34

2.

<=> in the parentheses in 3 and 4 is wrong since Re(s) = 1/2 doesn't imply that zeta(s) = 0; It would be sufficient to show one number c for which zeta(1/2 + ci) != 0 as a quick counterexample.

Name: Anonymous 2008-06-29 4:00

bump

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