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Old 2006-10-14, 16:32   #23
S485122
 
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Quote:
Originally Posted by troels munkner View Post
Please, read my original contribution again.
I said that [(6*m)+1)] with the integer m running from - infinity to + infinity
constitute a special group of integers (exactly one third of all integers).
(6*m)+1 gives one sixth of all integers. Unless your definition of multiplication and addition are different from mine.

Quote:
Originally Posted by troels munkner View Post
By the way [(6*(-5)+1] = 29, and [(6*(+5)+1) = 31.
And the line above is evidence for it: 6*(-5) is -30 adding one gives -29 not 29.

Last fiddled with by S485122 on 2006-10-14 at 16:32
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Old 2006-10-15, 05:35   #24
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You are correct, I did mis-understand 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?

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Old 2006-10-15, 09:59   #25
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Quote:
Originally Posted by Jacob Visser View Post
(6*m)+1 gives one sixth of all integers. Unless your definition of multiplication and addition are different from mine.



And the line above is evidence for it: 6*(-5) is -30 adding one gives -29 not 29.


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
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Old 2006-10-16, 16:03   #26
ewmayer
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Quote:
Originally Posted by troels munkner View Post
a lapsus calami.
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.
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Old 2006-10-16, 18:52   #27
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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:
Originally Posted by troels munkner View Post
"Possible primes" were defined as [(6*m)+1], m being an integer from - infinity to + infinity.
Negative possible primes (-5,-11,-17,-23.....) have modules V, II or VIII.
Positive possible primes (1,7,13,19,25,....) have modules I, VII or IV.
So since all primes other than 2 and 3 must be of the form 6*n+-1 (n running over the naturals), basically you catch the ones == +1 (mod 6) with m > 0, and the ones == -1 (mod 6) by way of their arithmetic inverses, using m < 0. This is completely trivial.

Quote:
The integers 2 and 3 cannot be defined as possible primes (6*m +1)
and should not be considered as primes.
Again ignoring your obfuscatory "possible primes" terminology, this is just the statement that "the primes 2 and 3 are the only ones not of the form 6*m + 1." Again, trivially true.

Quote:
The integer 1 is a square (6*0 +1)*(6*0 +1), just as 25 is equal to
[(6*(-1) +1)] * [(6*(-1) +1)] and 49 is equal to [(6*(+1) +1)] * [(6*(+1) +1)].
"The square of a number == 1 (mod 6) is == 1 (mod 6)". Trivial, and in fact a subset of another trivial observation:

Quote:
Products of possible primes remain possible primes 36 * (n*m) + 6* (n+m) +1, n being an integer from - infinity to +infinity.
"The product of two numbers == 1 (mod 6) is == 1 (mod 6)".
Wow - how enlightening.

Quote:
All Mersenne primes are positive possible primes and will be defined
in a later thread.
Actually, that should read "All Mersenne primes with odd exponent." Since you find the actual proof of this to apparently be quite challenging, allow me to help you out here:

M-primes are of the form M(p) := 2p-1, we consider the ones with p an odd prime, the smallest of which is 23-1. now observe that 23 == 2 (mod 6), and thus also that 2*22 == 2 (mod 6). It follows that in fact 2*22*k == 2 (mod 6) for all integter k. Since all odd-prime M(p) must have an exponent of form p = 3+2*k, their power-of-2 component must satisfy 2p = 23+2*k == 2 (mod 6). QED
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Old 2006-10-16, 19:51   #28
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Quote:
Originally Posted by ewmayer View Post
So since all primes other than 2 and 3 must be of the form 6*n+-1 (n running over the naturals), basically you catch the ones == +1 (mod 6) with m > 0, and the ones == -1 (mod 6) by way of their arithmetic inverses, using m < 0. This is completely trivial.
Sorry but Troels Munkner uses the form 6*m+1 not (6*m+1 and 6*m-1) that is why I insist that his third of all integers is really a sixth of all integers. A lot of posters here go to the logical conclusion that to have a third of integers (the integers not divisible by 2 nor by 3) one must use (6*m+1 and 6*m-1) but in this thread Troels Munkner uses only 6*m+1. This means divising Z in 4 classes, m being a member of Z:
- 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:
Originally Posted by troels munkner View Post
C. Odd integers which are not divisible by 3.
Their general form is (6*m +1), m being an integer from - infinity to + infinity.
These integers can be grouped as "possible primes" and comprise real primes and possible prime products.
All possible primes are "located" along a straight line wuith an individual difference of 6, i.e. (6*m +1) --- (-35), (-29), (-23), (-17), (-11), (-5),1,7,13,19,25,31,37 --- (6*m +1)
Possible primes constitute exactly one third of allo integers.
By this definition 5 could be a possible prime and even a real prime since it is not even and not divisble by three, but since 5 is not of the general form 6m+1 it can not be a real prime nor a product or primes (on this last point I am inclined to think that most will agree with him :-) In fact 5 is not a "Troels Munkner integer". Troels Munkner confirms this here:
Quote:
Originally Posted by troels munkner View Post
Let us put it in another way: "possible primes" are located along a straight line with a difference of 6 between the individual integers such as --. (-41), (-35), (-29). (-23), (-17), (-11), (-5), 1,7,13,19,25,31, --.
Note that a THIRD of all primes are separated by a difference of SIX!
A question for Mally: did you READ Troels Munkners's posts ?

Last fiddled with by S485122 on 2006-10-16 at 19:57
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Old 2006-10-16, 20:43   #29
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Quote:
Originally Posted by Jacob Visser View Post
Sorry but Troels Munkner uses the form 6*m+1 not (6*m+1 and 6*m-1) that is why I insist that his third of all integers is really a sixth of all integers.
Yes, but as I pointed out (and this is why is used 'n' in my positive-integer expression 6*n+-1, rather than 'm'), he gets the primes == 5 (mod 6) via a "backdoor" route by allowing m to be negative.

My conclusion that there is nothing new or even remotely interesting here stands.
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Old 2006-10-17, 05:44   #30
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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:
Originally Posted by troels munkner View Post
"Possible primes" were defined as [(6*m)+1], m being an integer from - infinity to + infinity.
Negative possible primes (-5,-11,-17,-23.....) have modules V, II or VIII.
Positive possible primes (1,7,13,19,25,....) have modules I, VII or IV.
If I read well the modules are modulo 9.

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 2006-10-17 at 05:56
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Old 2006-10-17, 19:43   #31
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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
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Old 2006-10-18, 01:51   #32
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I think the best way to find out what mr. Munkner means would perhaps be asking him plain, simple, easy-to-understand 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
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Old 2006-10-18, 12:15   #33
mfgoode
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Quote:
Originally Posted by Jacob Visser View Post

Note that a THIRD of all primes are separated by a difference of SIX!
A question for Mally: did you READ Troels Munkners's posts ?

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
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