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#1 |
"Bob Silverman"
Nov 2003
North of Boston
22×1,877 Posts |
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In another thread, it was witten:
"also,one of the definitions of a mersenne number is that p is prime.another is that p is a positive integer.i could easily accuse you of not knowing your definitions. " It is FALSE that another definition allows p to be any positive integer. I have heard this false assertion many times. Repeating it does not make it true. If you will not accept me as an authority, perhaps you will accept: P. Ribenboim: "The Book of Prime Number Records" 1st edition. p.75 "The numbers M_q = 2^q-1 (with q prime) are called Mersenne numbers Or maybe the discussion in Hardy & Wright, sect. 2.4 will convince you. Mersenne himself only concerned himself with 2^q-1 for q prime. Or maybe definition 5 in D. Shanks' book will convince you? etc. etc. If anyone writes 2^n-1 for general n as a Mersenne number, they are being sloppy and ignoring mathematical history. Go to any number theory conference. Ask any number theorist there. Mersenne was concerned with the question of when 2^q-1 is prime. And this ONLY happens when q is prime. |
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#2 |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts |
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The following claim n must be an integer, or a positive integer: (and don't require n to be prime)
http://mathworld.wolfram.com/MersenneNumber.html http://mersennewiki.org/index.php/Mersenne_number The following claim n must be prime: (they're all from Silverman's post because I can't find any web sites that directly claim n must be prime) Silverman P. Ribenboim: "The Book of Prime Number Records" 1st edition. p.75 definition 5 in D. Shanks' book the discussion in Hardy & Wright, sect. 2.4 "Mersenne was concerned with the question of when 2^q-1 is prime." http://en.wikipedia.org/wiki/Mersenne_prime says that definitions vary http://primes.utm.edu/mersenne/ only discusses Mersenne primes without particularly defining Mersenne numbers and whether n may be composite or not. http://www.mersenne.org/various/math.php is unclear as it too is focused on Mersenne primes, but in my opinion it implies that n need not be prime. "Mersenne numbers are of the simple form 2P-1, where P is the exponent to be tested. It is easy to prove that if 2P-1 is prime, then P must be a prime. Thus, the first step in the search is to create a list of prime exponents to test." If Mersenne numbers required a prime n, why would it even bother to suggest that the first step is to find Mersenne numbers with a prime n? Isn't this all really a bit pointless, though? It is purely a question of definition. It simply does not matter whether n must be prime or not to be a "Mersenne number". It's almost always quite clear whether the author is speaking only of ns that are prime or of all ns. And we all know that in order to be a Mersenne prime, it must first be a Mersenne number. Besides, it seems to me it should be understood by the fact that you're using p, since p and q are usually be used for primes, while n and k are usually used for any integer. (at least from my observations, this seems to be true) Last fiddled with by TimSorbet on 2009-09-18 at 15:22 |
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#3 | |
"Bob Silverman"
Nov 2003
North of Boston
165248 Posts |
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On-line sources are unreliable, notoriously error-prone, and should not be trusted. The sources that I quoted are all from highly respected NUMBER THEORISTS. Trying to quote Wikipaedia as an authority on anything is a joke. Although mathworld is somewhat more reliable, I doubt whether their post on the subject was written by a number theorist. And if you still don't believe me, then I suggest you ask Hugh Williams. He is a world recognized authority on the history of computational (and other) number theory. Check to see what Ore's History of Theory of Numbers has to say... This is mathematics. Having a precise definition DOES matter. |
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#4 | |
Cranksta Rap Ayatollah
Jul 2003
641 Posts |
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You're arguing that having a consistent convention matters in mathematics, which is patently false. If someone were to present a brilliant proof with regard to Mersenne primes and referred to Mersenne numbers as all numbers of the form 2^n - 1 with n a positive integer, nobody would care. If in 50 years the convention shifts to where leading number theorists use Mersenne number in the broader sense, the mathematics will not change. |
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#5 |
"Brian"
Jul 2007
The Netherlands
2·11·149 Posts |
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Are you sure? If mathematicians took this attitude to extremes and used their own pet conventions for all sorts of ideas, their contributions would be hard to read and understand no matter how carefully they defined everything first. It seems to me that we ought to respect established conventions. These may change over time but there is no need to hasten the process.
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#6 |
Jun 2003
2·2,719 Posts |
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I guess, many times you need to refer to numbers of the form 2^n-1 (n primes/composites). Instead of inventing a new term, people just conveniently re-use Mersenne numbers as a shortcut for it, the context making it clear whether it is all n's or prime n's.
Last fiddled with by axn on 2009-09-18 at 16:55 |
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#7 | |
Cranksta Rap Ayatollah
Jul 2003
64110 Posts |
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Pretty good advice for most things in life too. |
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#8 | |
Cranksta Rap Ayatollah
Jul 2003
641 Posts |
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I think "Mersenne Numbers" and "Mersenne Primes" together make for an aesthetically satisfying definition of numbers of the form 2^n - 1 and the subset of them that are prime. |
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#9 | |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts |
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![]() Um...not that I know of. They'd probably just be referred to as [Blank] Primes, even if there are no [Blank] Numbers that aren't [Blank] Primes. Last fiddled with by TimSorbet on 2009-09-18 at 17:50 |
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#10 |
Jul 2006
Calgary
52·17 Posts |
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It seems one common convention is to use the name of the variable to indicate its type. 2^n-1 implies n is any (positive) integer. 2^p-1 implies p is a prime.
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#11 |
"Lucan"
Dec 2006
England
2·3·13·83 Posts |
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Bob deserves all the stick he is getting after
circumnavigating the temporary closure of the thread where he was taking the piss out of someone (possibly stupider than he is). Lord Lucan |
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