- Sep 30, 2006
- 4,267
- 421
- 126
I have created a brief proof of the Riemann Hypothesis. I would like some input on its completeness.
http://math.rudytoody.us/RiemannProof.pdf
http://math.rudytoody.us/RiemannProof.pdf
Last edited:
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: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)
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.