Is this a proof of the Riemann Hypothesis?

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iCyborg

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Can you provide this chapter 12.3 equivalence verbatim, I don't have that book.

Because it cannot be the way you've stated it: the set of all square-free integers is clearly infinite, and it's quite trivial to see that both those with odd and even number of factors are also infinite. Since all infinite subsets of integers have the same cardinality, you're done. It's hard to believe that something provable by smarter 14-year olds is still unsolved...

I would think that it would be something like for some integer n, the number of odd-squarefree and even-squarefree smaller than n is the same, or something close. In that case, your proof does not work at all, after say the 3rd step where p3=5, your sets would be
Even = {1, 6, 10, 15}
Odd = {2, 3, 5, 30}

Sure they have the same cardinality, but the max values are different, and it does not include everything in range, say 7, 11 aren't in either set. So your construction proves nothing. It's like proving that for every N, the number of integers smaller than N and divisible by 2 is the same as those divisible by 100, by giving a "construction"
Div2 = Div2 U (n*2)
Div100 = Div100 U (n*100)
and claiming that in each step both sets are of the same size. Clearly it has nothing to do with what was supposed to be proven...
 

Rudy Toody

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When applying Mertens from one to X, the ending sum would be within some error amount. That has be shown to be false by finding a counter-example by computer.

Denjoy claimed basically the same thing but used the coin-flip probabilities. That fails because the square-frees are not random. However, my claim is that Denjoy is correct when X==Infinity.

This technique of using infinite sets and describing the step values to get to the infinite value was used by Euler during his Prime Product Function proof.

The equivalence: Scroll down to Denjoy's probabilistic argument

Note: this wiki article states that Riemann predicts randomness to the parities, which is not the case with my theorem. I'm saying that a fixed number of square-free numbers are constructed at each step. We have no way to predict where those numbers land on the number-line. We also can't predict where the even count of factors and the odd count of factors land in relation to each other. All we know is that for all square-free number <= some prime p, there is always an equal number of evens and odds.
 
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iCyborg

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You cannot do that (only) at infinity. At infinity the number of integers is the same as the number of integers divisible by 10^100. But their density is nowhere near, which is what that probabilistic argument needs -> they need to be roughly randomly distributed.

Euler's proof deals with a convergent series, I don't see similarities with this...

All we know is that for all square-free number <= some prime p, there is always an equal number of evens and odds.
I cannot fathom how you see this from your construction. The bolded 'always' would seem to apply this holds for all primes, a simple check of the first couple already fails:
p=5 -> 3 odds (2, 3, 5), 1 even (1)
p=7 -> 4 odds (2, 3, 5, 7) and 2 evens (1, 6)
p=11 -> 5 odds (2, 3, 5, 7, 11) and 3 evens (1, 6, 10)
 

Rudy Toody

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I cannot fathom how you see this from your construction. The bolded 'always' would seem to apply this holds for all primes, a simple check of the first couple already fails:
p=5 -> 3 odds (2, 3, 5), 1 even (1)
p=7 -> 4 odds (2, 3, 5, 7) and 2 evens (1, 6)
p=11 -> 5 odds (2, 3, 5, 7, 11) and 3 evens (1, 6, 10)

I take each prime and multiply it against each set of square-frees---creating even counts from the odds and odd counts from the evens. Then I merge the new odds into the odds and the new evens into the evens. At the end of each step we have two sets of the same sizes.

p=5 -> odds(2,3,5,30), evens(1,6,10,15)
P=7 -> odds{2, 3, 5, 7, 30, 42, 70, 105}, evens{1, 6, 10, 14, 15, 21, 35, 210}
p=11 -> odds{2, 3, 5, 7, 11, 30, 42, 66, 70, 105, 110, 154, 165, 231, 385, 2310}
evens{1, 6, 10, 14, 15, 21, 22, 33, 35, 55, 77, 210, 330, 462, 770, 1155}

The above sets have all the square-frees whose highest factor is p.
 
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iCyborg

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You should be more accurate when you talk about math, saying "for all square-free number <= some prime p" is very different from "all square-free number with highest factor p".

Mertens function looks at the number of odd-sqfree and even-sqfree up to a fixed number x. Look at your sets, they are completely different:
1) the largest number in one of the sets is always 2x than the largest number of the other set
2) Mertens needs all numbers <=x, your sets have a lot of holes in each step (they get filled later)


Let me try to explain what's wrong once more with an example:
k(n) = 1 if n divisible by 2, and k(n)=-1 if n divisible by 101.

You can construct the sets just like you did where in each step you will have the same number of elements for which k(x)=1 as of those with k(x)=-1. And you can do that until infinity.

However Mertens function M for k will be nowhere near that required bound. Obviously you need something more than just constructing sets that have the same size at each step, but you're not giving anything more, you're basing your whole argument on just that.
 

Rudy Toody

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Sep 30, 2006
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You should be more accurate when you talk about math, saying "for all square-free number <= some prime p" is very different from "all square-free number with highest factor p".

Mertens function looks at the number of odd-sqfree and even-sqfree up to a fixed number x. Look at your sets, they are completely different:
1) the largest number in one of the sets is always 2x than the largest number of the other set
2) Mertens needs all numbers <=x, your sets have a lot of holes in each step (they get filled later)


Let me try to explain what's wrong once more with an example:
k(n) = 1 if n divisible by 2, and k(n)=-1 if n divisible by 101.

You can construct the sets just like you did where in each step you will have the same number of elements for which k(x)=1 as of those with k(x)=-1. And you can do that until infinity.

However Mertens function M for k will be nowhere near that required bound. Obviously you need something more than just constructing sets that have the same size at each step, but you're not giving anything more, you're basing your whole argument on just that.

From the paper: At each step we create all the square-free numbers whose factors are <= p_n.
Here is a paper I just found that describes it.
The only thing I stated about Mertens is that it equals zero at infinity.
 
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Rudy Toody

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OK, I'm convinced this is not a proof. I think that what I need to do is show that the distribution of the square-frees is random with regard to the Moebius +1/-1 values.

Thanks for your input, iCyborg.

Note: I will keep the paper at the link in the OP and add a link to a more current paper when I have progress to report.
 
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