20150915, 18:48  #23  
"William"
May 2003
Near Grandkid
4505_{8} Posts 
Quote:
As pointed out in other posts on this thread, this statement doesn't make sense unless "randomly chosen" is given a definition that is different from the usual "chosen with uniform probability." But this isn't a statement about the probability of choosing something. P is a function from sets of integers to [0,1]. P({m}) is a function from integers to [0,1]. Note that the paragraph before this tells us that this "recourse to the established theory of probability" is NOT a helpful formalization of our intuition. 

20150915, 20:06  #24  
Jul 2014
2·3^{2}·5^{2} Posts 
Quote:
The reason for this is because I've yet to study Measure Theory. Except that apparently in this context, measure and probability mean the same thing. ......nonetheless I don't understand what it means. My confusion is : Since we are talking about probability this sentence is meaningless to me : "the probabilty of one single integer=m" because it seems like if we are talking probabilty, we mean a number on [0 , 1] and the number on [0, 1] of one single integer=m must depend on what the set is it's being chosen from. This was the original motivation for this thread, but since Silverman has turned it offtopic (the ErdosTuran conjecture). /* freudian mistake or something */ I respect Silverman's clarity but I think he makes a far too large generalisation i.e Unclear about maths never before seen taken from postgrad book whilst still undergraduate => deficient on basics of maths => person can't solve ErdosTuran conjecture The only evidence you seem to have that I'm deficient on basic maths is that I don't understand probability. My clear lack of understanding of things concerning the completely seperate issue of the ErdosTuran conjecture (which hardly counts as basic maths), was my actual reason for asking a question about it. I feel I have benefited greatly from your help on that thread but here I feel like I'm just being made fun of. You tell me : that page on the wiki about the ErdosTuran conjecture says that "the case is open for arithmetic progressions of length 3". I claim to have proved that all sets of positive integers with no a.p.'s of length more than 2 when used to form an infinite sum of reciprocals converge i.e the infinite sum is finite. I don't care how many tomatoes you'll throw if I'm mistaken. So, you want to see my proof then? Last fiddled with by wildrabbitt on 20150915 at 20:12 

20150915, 20:19  #25  
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{3}·3·461 Posts 
Quote:
Making mistakes is OK. In fact, it's encouraged. This is how we collectively learn. Please. It's probably wrong, but that's OK. Push the envelope. 

20150915, 21:13  #26 
Jul 2014
2·3^{2}·5^{2} Posts 
Since the last post I made my neighbour asked me to use my internet so he can pay his car insurance. I gave him my wireless key.
Now I'm worried that he's going to do something he shouldn't from his flat. I've since changed my wireless key on the hub but I feel eerie. Regarding my proof. I'm genuinely sorry saying I'd show it and then not showing it but I've been keeping it secret because I don't want someone else to use the method to solve the full conjecture. The fact that I don't know whether or not it's right is telling as I'm sure you'll agree  especially Silverman. It took too much effort for to want to part with it right now. I'm going to continue with studying the problem as Silverman would do. Thanks for the input. 
20150915, 21:23  #27  
"Bob Silverman"
Nov 2003
North of Boston
1110101010100_{2} Posts 
Quote:
It also shows that you do not understand how publishing and exposition work in the public domain. <plonk> 

20150915, 22:25  #28 
If I May
"Chris Halsall"
Sep 2002
Barbados
10101100111000_{2} Posts 

20150916, 04:53  #29 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10273_{10} Posts 
Now, why would you want to solve this conjecture?
Try one for which you can get some money, hehe, there is a million bonus for Beal. I was running around it for half year, without getting nowhere, but at least, this made me read few books and articles about how FLT was proven, so even if I couldn't get anywhere, I am a bit smarter now As said in the parallel thread, no tomato from me. Yet. edit: Well, a little bit of tomato for the "I don't want other guy to use the method". Be sure that if someone does, if the method is new, your merit will be recognized. As said before, this sentence scores hard on the crankometer (believe me, I am kind of a crank myself, hehe). This is what mathematicians do, they provide new methods to be used by other people to provide new methods... And this is how progress comes... Last fiddled with by LaurV on 20150916 at 04:55 
20150916, 07:32  #30 
Jul 2014
2×3^{2}×5^{2} Posts 
are you still game then?
Okay, seeing as you're so sure I'm wrong :
It people want to see it, I'll post the proof bit by bit and at every stage if people let me know whether they want to see more I'll post more. If people tell me there's an oversight or it's wrong somewhere I'll admit defeat and quit. It's not an enourmous proof, but it is quite long and the reason I'm posting it bit by bit is because I have some gaps I left and need to finish if other people can understand it. Now that doesn't mean there's a hole in the proof. Just that I focused on the bits I had to think about, not those I knew were right. So, it'll require a little patience if you want to see the whole thing. Are you game? Be good sports, hear me out. Here is the first small part : the 'proof' I claim to have begins like this http://www.mersenneforum.org/attachm...1&d=1442388405 
20150916, 07:36  #31 
Jul 2014
2×3^{2}×5^{2} Posts 
Back around 10am GMT BTW.

20150916, 08:57  #32 
Jul 2014
2·3^{2}·5^{2} Posts 
Let me know if you want the next part (the proof of that lemma).

20150916, 09:22  #33 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10,273 Posts 
I don't exactly grasp that condition with partial sums (it may be because I am silly, or because I don't exactly know the English terminology). For example, does the sequence of triangular numbers satisfy the condition? 1, 3, 6, 10, 15, 21, .... (I think this has no arithmetic progression in it)
If not, why not? If yes, your lemma is false... Last fiddled with by LaurV on 20150916 at 09:23 
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