
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:48:39)
Re: Riemann hypothesis
The Beginning Of All Things To End. The End Of All Things To Come.
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:49:26)
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:52:30)
 Ricky
 Moderator
Re: Riemann hypothesis
 means divides. a  b means that a divides b.
 is one of the first things anyone learns in an introduction to proofs class. I have a hard time believing you have any idea what you are talking about.
So I ask the question again, first, write:
If a  b and b  c, then a  c.
In terms of AND, NOT, OR, etc. Then prove the statement the traditional way (is it litterally a 1 line proof).
Edit: You can ignore the first part about writing it in terms of AND, NOT, OR, etc, but I bet you have no idea how you would prove it the traditional way.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Riemann hypothesis
"EDIT: Btw, I'm almost certain that even = can be written in terms of the logical operations"
lol, please explain what you are even on about here! because that doesnt really make any sense at all to say that, in terms of computing, all = does is copy the data from one memory address to another memory address for primary data like characters, numbers, boolean data and pointers

so a  b means that b/a is an integer?
well in that case, logically, if a divides into b whatever times it might be (lets say n), and b divides c, then (an) divides c, so a divides c
would that be valid as a proof ricky?
The Beginning Of All Things To End. The End Of All Things To Come.
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:50:09)
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:50:53)
 Ricky
 Moderator
Re: Riemann hypothesis
a  b means that a divides b, or rather as you put it, a goes into b a number of times. Not the other way around.
A formal definition of such is that:
b = ka, where k is some integer.
So b = ka for for some integer k and c = lb for some integer l, so c = l(ka) = lka = (lk)a. Since the integers are closed under multiplication, lk is an integers and so it must be the case that a  c.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Riemann hypothesis
Heh, I could have proved that.
Really, I know what I'm talking about, but I can see why people would doubt me. Anyway I didn't come here to try and make people believe me, I just wanted information about the Riemann hypothesis.
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:51:50)
 Ricky
 Moderator
Re: Riemann hypothesis
se7en, what your posting here is just boolean algebra. These have been known for 100+ years and are really nothing new. Furthermore, you are defining "=" by using "="? Surely you can see why this is a problem.
That, along with the fact you have never had any sort of introduction to proofs, makes me seriously doubt that you could ever prove something such as the Riemann hypothesis. And the fact that you refuse to show these proofs, for whatever excuse, elevates that doubt exponentially.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Riemann hypothesis
yeh, plus what he posted, just made no sense in any way, apart from one bit, which did make sense, but had nothing to do with checking equality between numbers, and was overly complicated anyways.
The Beginning Of All Things To End. The End Of All Things To Come.
 numen
 Full Member
Re: Riemann hypothesis
If I had such a revolutionary algoritm in my hands, and only in my hands, I sure wouldn't keep it for myself, it doesn't matter if I'd give up the million dollars. Go talk to a mathematical doctor about it, maybe you're up to something, but most likely you're not.
If you had been reading up on university math, as you claim, the proof Ricky asked you about would've been easy the traditional way. It was one of the first proofs they introduced when I started my first year at university. Actually, a whole lot has been about proving this and that, you'd surely know that math at university level is a whole different game (it's more fun too! )
Bang postponed. Not big enough. Reboot.
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:48:07)
 Ricky
 Moderator
Re: Riemann hypothesis
"Furthermore, you are defining "=" by using "="? Surely you can see why this is a problem."  Ricky
I don't know what you're talking about, I did nothing of the sort.
What do you mean? Take a look at your post:
a = b is equivalent to...
(IF a = 1 THEN b = 0) XOR (IF a = 0 THEN b = 1)
You are defining "equals" with equals signs.
Anyway, I've probably constructed just as many proofs on my own as any university graduate (it's a long story, but I basically rediscovered most high school mathematics on my own), so although you have tremendous doubt in me coming from your standpoint, knowing what I know about my studies thus far, I have very little doubt in myself.
Any math undergraduate knows what  means, let alone any graduate. Please do share, what kind of proofs have you done? Abstract Algebra? Number Theory? Real Analysis? Vector Calculus? Complex Analysis?
I know it can't be the first two because divisibility is key to the vast majority of their central tenants.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Riemann hypothesis
you want to know how equality between data is found via boolean operators? its simple, each bit of data is checked against the other data's corresponding bit using AND, and as soon as it results in low, the loop is cut > not equal, if the loop ends, it must be equal, because every bit of data is the same, but ofcourse this is for data stored in binary on a computer... hardly what you are saying
but to say that equality between numbers can be found with (IF a = 1 THEN b = 0) XOR (IF a = 0 THEN b = 1) is just retarded
why dont you explain to me exactly what you mean by IF a = 1 THEN b = 0, because to me, that is saying in terms of programming, which is the only time ive EVER seen this kind of structure, is that you are checking equality between a and 1 (or rather, you would be if it were ==, and not = ) and then if a == 1, you are assigning the value of '0' to b
and apart from that, since you are only defining equality between boolean values, theres no need for all that cr.ap, all that is required is a single AND operation a == b if a&b == 1, but how would this apply to any type of mathematical structure, or even just real numbers, even integers!
Last edited by lucadeltodesco (20060731 06:04:38)
The Beginning Of All Things To End. The End Of All Things To Come.
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:47:06)
 Ricky
 Moderator
Re: Riemann hypothesis
Mostly I've proved stuff at high school level, such as that the product of the slopes of two perpendicular lines equals 1. (My proof was much longer and more complex than a proof I saw much later on in a book I was reading, but for a long time it was the proof I was most proud of.) That includes algebra and Euclidian geometry.
And so I would take it that you wouldn't really understand that a 2d line is just a mapping of the the real line to the reals. So you wouldn't even begin to understand what a homomorphism is, and so you wouldn't even have a clue as to what the Poincare conjecture is (I barely do). Yet, somehow, you were able to say, "A demonstration of this for the Poincare conjecture would be insanely hard to do, but I suspect (and am almost certain really) that it could be done..."
It's really starting to seem like you are just making stuff up as you go along.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Riemann hypothesis
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Last edited by se7en (20060801 03:46:08)
 Ricky
 Moderator
Re: Riemann hypothesis
So I take it you admit to saying that you are almost certain you can solve it, with out having any idea whatsoever as to what it is?
Sorry, but your credibility has gone from extremely low to rockbottom.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
 Zhylliolom
 Real Member
Re: Riemann hypothesis
Wow, this thread grew substantially overnight.
Ok se7en, this here isn't a question as to whether or not your thesis works, so I hope you may choose to reply to it: when do you think your algorithm will be sufficiently used by you so you'd feel comfortable publishing some proofs and thus releasing it to the mathematical community? Are you going to wait until you're done with Riemann, or do a few more problems, or wait until you actually present your master's thesis?
I found it interesting that Ricky brought up the Poincaré conjecture... I came here just to ask if you could do that, but I guess it's already been talked about.
Last edited by Zhylliolom (20060731 07:19:51)
 Patrick
 Real Member

Re: Riemann hypothesis
It all comes down to this I guees  se7en still needs to show some of the proofs he's made. If your skills in maths are so great as you say, you should have some work somewhere that will convince us. I find it hard to believe you know how to prove a thing you don't know what is.
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 George,Y
 Super Member
Re: Riemann hypothesis
hehe, Ricky's got a conservative mind.
I suggest you to Let it be let it be Let it be let it be Let it be~
Last edited by George,Y (20060731 11:25:38)
X'(yXβ)=0
 mikau
 Super Member
Re: Riemann hypothesis
what exactly is the riemann hypothesis?
A logarithm is just a misspelled algorithm.
