20140616, 18:10  #34  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
what's being pointed out is that without knowing if the PRP is prime or not, we can not know with 100% certainty that it is not completely factored Gordon made that claim. As did you by saying
Quote:
edit: I just realized you could use modular arithmetic to limit what it is and if the factorization fits the required number of each remainder. maybe that'll lead me to your logic. Last fiddled with by science_man_88 on 20140616 at 18:21 

20140616, 20:48  #35  
"William"
May 2003
New Haven
2·3^{2}·131 Posts 
Quote:
I thought a number was prime or not prime independent of our ability to prove that. The number has been represented as a product. It's possible that it is a product of primes, so it's possible the number has been completely factored. 

20140616, 22:06  #36 
Nov 2003
2^{2}·5·373 Posts 

20140616, 22:14  #37 
If I May
"Chris Halsall"
Sep 2002
Barbados
3×7×443 Posts 

20140616, 22:16  #38 
"Kieren"
Jul 2011
In My Own Galaxy!
9991_{10} Posts 

20140616, 22:32  #39 
If I May
"Chris Halsall"
Sep 2002
Barbados
3·7·443 Posts 

20140617, 00:29  #40 
P90 years forever!
Aug 2002
Yeehaw, FL
1BED_{16} Posts 

20140617, 00:56  #41 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×7×409 Posts 
Probably not.
Last fiddled with by retina on 20140617 at 00:57 Reason: Who is "we" anyway? 
20140617, 03:07  #42  
"William"
May 2003
New Haven
100100110110_{2} Posts 
Allopurinol. Baby aspirin. I'm supposed to be taking Simvastatin, but my gout won't tolerate it. I don't think any of these affect my mathematical reasoning.
Let's see if we can figure out what you are trying to say, and where our disagreement. We all agree that: Quote:
Quote:
Quote:
Quote:
But it is also true the nobody here can PROVE the given factorization is incomplete  yet you have asserted that this true. If you disagree, please try to explain why. 

20140617, 03:50  #43 
"William"
May 2003
New Haven
100100110110_{2} Posts 
Perhaps Bob MEANT to give this definition:
A number is completely factored when it is proven to be represented as the product of primes. Everything he said would have made sense if "is proven" had been included in his definition of "fully factored." 
20140617, 10:51  #44  
Nov 2003
2^{2}·5·373 Posts 
Quote:
that it is complete must do so. You are really having trouble with this. The SET of primes is a strict SUBSET of the set of probable primes. A prime is a probable prime. The converse is not true in general. You can not assert that a number is the product of primes when all we know is that it is the product of some primes and an integer that may or may not be prime. Asserting that the number is a product of primes is a CONJECTURE. We have good evidence that the conjecture is correct, but we do not know it to be true. 

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