20061014, 16:32  #23  
"Jacob"
Sep 2006
Brussels, Belgium
2^{4}·3·37 Posts 
Quote:
And the line above is evidence for it: 6*(5) is 30 adding one gives 29 not 29. Last fiddled with by S485122 on 20061014 at 16:32 

20061015, 05:35  #24 
Aug 2003
Snicker, AL
7·137 Posts 
You are correct, I did misunderstand part of your proposal. However, my underlying concept looks to be intact. Where you are using 2*3, I suggest using 2*3*5. The manifold is divided into groups of numbers and according to your system, only one out of 6 numbers can possibly be prime. My division suggests that only one out of 30 numbers can be prime. (there are a couple of minor quibbles with this, but it is simplest form) Why is your method valid while mine is not? What about 2*3*5*7? It is conceded that none of your numbers can possibly be divisible by 2 or 3. What about numbers of the form (6*M) + 3?
Fusion 
20061015, 09:59  #25  
May 2006
29 Posts 
Quote:
You are absolutely right. A minus sign was missing in "29", which was a lapsus calami. Later in my text the number was correctly given as (29). Sorry for the inconvenience. Y.s. troels munkner 

20061016, 16:03  #26 
∂^{2}ω=0
Sep 2002
República de California
2^{3}·3·487 Posts 
I tried writing with squid ink several occasions, but the little buggers are indeed exceedingly slippery, as you note. And training them to produce a steady ink flow rather than just "blurting out" whatever is on their little squiddy minds is ... rather challenging.
Oh wait, I'm thinking of a lapsus calamari. Sorry about the confusion, ink on my face, & c. Y'all must've thought I was Kraken up mentally, or something. 
20061016, 18:52  #27  
∂^{2}ω=0
Sep 2002
República de California
2^{3}·3·487 Posts 
OK, so I'm just going to ignore the silly and confusing "possible primes" verbiage invented by Mr. Munkner, and instead use simply "integers of the form 6*m+1" wherever it occurs. So now let's look at the substance of the resulting claims:
Quote:
Quote:
Quote:
Quote:
Wow  how enlightening. Quote:
Mprimes are of the form M(p) := 2^{p}1, we consider the ones with p an odd prime, the smallest of which is 2^{3}1. now observe that 2^{3} == 2 (mod 6), and thus also that 2*2^{2} == 2 (mod 6). It follows that in fact 2*2^{2*k} == 2 (mod 6) for all integter k. Since all oddprime M(p) must have an exponent of form p = 3+2*k, their powerof2 component must satisfy 2^{p} = 2^{3+2*k} == 2 (mod 6). QED 

20061016, 19:51  #28  
"Jacob"
Sep 2006
Brussels, Belgium
2^{4}×3×37 Posts 
Quote:
 the even numbers of the form 6m, 6m+2 and 6m+4;  the odd multiples of three of the form 6m+3;  the "Troels Munkner possible primes" or (real primes and possible prime products" of the form 6m+1 and finally  other numbers of the form 6m+5, the numbers in this last class are not integers by Troels Munkner's definition. Quote:
Quote:
A question for Mally: did you READ Troels Munkners's posts ? Last fiddled with by S485122 on 20061016 at 19:57 

20061016, 20:43  #29  
∂^{2}ω=0
Sep 2002
República de California
2^{3}×3×487 Posts 
Quote:
My conclusion that there is nothing new or even remotely interesting here stands. 

20061017, 05:44  #30  
"Jacob"
Sep 2006
Brussels, Belgium
2^{4}·3·37 Posts 
I agree with you that his theorem is just a set of definitions that do not bring new insights.
But I must be dumb: I did not see 5 in his list only 5. And how could 5 and 7 have a difference of 6? One can see this in his fist port: Quote:
It is true that his arithmetic is a bit shaggy, sometimes one gets the impression that for him [6*(1)+1] might be equal to [1*(6*1+1)]. Last fiddled with by S485122 on 20061017 at 05:56 

20061017, 19:43  #31 
Jan 2006
JHB, South Africa
157 Posts 
I feel that Mr. Munkner is actually not portraying his Mathematical notations correctly and he is actually meaning to imply the abolute values of ((6*m)+1) is one third of all the positive integers, where m can be either a positive or negative integer. That seems the only way I can make sense of how he has written the notations.
Regards Patrick 
20061018, 01:51  #32 
Feb 2006
Brasília, Brazil
3×71 Posts 
I think the best way to find out what mr. Munkner means would perhaps be asking him plain, simple, easytounderstand questions.
So, mr. Munkner, please classify both the statements below as either "true" or "false". Whatever your answers are, we've already understood the logic beneath them, so you don't need to spend time explaining why these statements are true or false. 1) The number 5 (positive five) is a prime number. True or false? 2) The number 5 (negative five) is a prime number. True or false? Please do follow my guidelines strictly, as I personally have a very hard time understanding math which doesn't present itself to me according to them. Thanks a lot, Bruno 
20061018, 12:15  #33  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Quote:
Well jakes, he very generously sent me his book(let) by airmail which he priced at $15 though he has not given me the bill as yet! I must confess it is even more confusing than his posts. and not well connected from section to chapter. Never the less, I personally feel,, that he has a message which he cant explain clearly. Due paucity of time, as Im working on my own theories myself, I have not really studied it. But if you are interested I could Xerox the booklet and send it to you by post, just for the asking. You may PM me at your leisure. Regards, Mally 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Is there any such theorem that states this?  soumya  Miscellaneous Math  17  20130328 10:26 
Fermats theorem and defining a 'full set' for any prime.  David John Hill Jr  Miscellaneous Math  32  20090313 21:45 
New exact theorem  Master Alex  Miscellaneous Math  38  20070305 18:30 
Number Theorem  herege  Math  25  20061118 09:54 
Fermat's Fuzzy Theorem  any good for new prime test?  bearnol  Miscellaneous Math  9  20051116 13:19 