August 30th, 2017, 04:38 PM  #31 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  https://en.m.wikipedia.org/wiki/Tabl...factorizations A small sample of the above wiki link follows Note that there are rational primes which are not Gaussian primes. A simple example is the rational prime 5, which is factored as 5=(2+i)(2−i) in the table, and therefore not a Gaussian prime. So there is sense in trying to reason mathematically on the assumption that 5 is not prime. I guess you must be the pope , eh my friend? : 
August 31st, 2017, 12:35 AM  #32  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Quote:
gcd(11, 5) = gcd(5, 1) = gcd(1, 0) = 1. So 11 isn't divisible by x nor by y. I read the of 5 not being prime merely as a change of definition of primes; "A number is prime iff it has exactly 2 divisors and is unequal to 5. " But I imagine there's more than one way to read it. Aww, very kind! A lot of people were leaving, glad to see you back agent! Thought you left too Hm, 6*12 + 5 = 6*13  1 = 6 * (2 * 6 + 2  1)  1 = 2*6*6  1*6 + 2*6  1 = (6*2  1)(6*1 + 1). Right?  
August 31st, 2017, 01:43 AM  #33  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
You could have done gcd(11 , 5) = gcd(6 , 5) = gcd(1 , 5) = gcd(1 , 0) = 1 And other paths are possible to get the right answer , switching order inside parenthesis is ok too and I think you did that in steps 2 and 3 right? Follow up question: Is it possible to write 1 as a linear combination of 11 and 5 using this shortcut or must we go the long way? Quote:
Quote:
$ 6(12) + 5 = 6(11 + 1) + 5 = 6(11)+ 6(1) + 5 = 6(11) + 11 = 7(11) $ Last edited by agentredlum; August 31st, 2017 at 01:52 AM. Reason: boo boo  
August 31st, 2017, 04:36 AM  #34 
Global Moderator Joined: Dec 2006 Posts: 20,926 Thanks: 2205 
6*12 + 5 = 6*12 + 12  6  1 = (6 + 1)(12  1)

August 31st, 2017, 10:32 AM  #35 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
6*12 + 5 = 6*11 + 11 = 7*11 I can skip steps too. 
August 31st, 2017, 01:51 PM  #36 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
I find this thread incompletely specified. If we say that the definition of prime is any positive integer that can be divided evenly by only two distinct integers, provided that 5 is defined as not prime, it has been shown that 11 = 6 * 2  1 is prime. No problem on my part with that though I admit I interpreted the question differently. I interpreted it as saying that if there exist positive integers x and y such that x > 1 and y > 1 and x * y = 5, then 6n  1 is not prime. I have not seen a proof of that conclusion. I have seen a purported proof starting with the assertion that x must equal 2, 3, or 4 and y must equal 2, 3, or 4. That assertion appears to be based on the normal rules of arithmetic, but clearly the normal rules of arithmetic cannot apply because, if so, x * y < 5 or x * y > 5. Therefore, we must select a new set of rules. For example, let's say that 2 * 2 = 5, 2 * 3 = 10, 2 * 4 = 17, and 3 * 3 = 13. Lo and behold, 11 must be prime. This goes back to mashcke's comment that we can prove anything from a false premise. So we can prove that 11 is prime or 11 is not prime if we start from 5 is not prime understood in its normal sense. Unless we are told the rules of an alternative arithmetic in which 5 is not prime, there can be neither proof nor disproof on whether 11, 17, 23, 29, 41 etc are primes. I suspect, that there are sets of arithmetic rules where 5 is composite and 11 is prime and different sets of arithmetic rules where both 5 and 11 are composite. But the rules used in the purported proof are clearly inconsistent. 
August 31st, 2017, 11:31 PM  #37  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
1) in mod 11 , 4*4 = 5 so we can say 5 is composite. Note that 4 < 5 and this ties in nicely with the 'purported' 2 , 3 , 4 'proof' which you found objectionable. To show 11 is composite you need to find integers $ 2 \ \le \ a \ , \ b \ \le \ 10 \ \ $ such that $ (a \times b) \equiv 0 \ \ (\text{mod} \ 11)$ Let me know if you find any 2) In the Gaussian Integers 5 is composite and 11 is prime $5 \ \ $ is factorable as $ \ \ (2 + i)(2i) $ There are slightly more complicated reasons why $ \ \ 11 \ \ $ is prime (not factorable) in the Gaussian Integers and it would be best for you to look it up and post questions for discussion if you have any difficulties understanding.  
September 1st, 2017, 04:11 AM  #38  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry  Quote:
By the way, I'd recommend looking up the definitions of "prime" and "irreducible" in integral domains. The definition of prime you're used to actually corresponds to "irreducible"; "prime" is a stronger condition in general (but in the ring of integers and all other UFDs, prime $\iff$ irreducible). Last edited by cjem; September 1st, 2017 at 04:15 AM.  
September 1st, 2017, 06:14 AM  #39  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
cjem ... look closely at my post. I excluded $ \ 0 \ , \ 1 \ , \ 11 $ from the inequality. It seems like a convention to say there are no primes in modular arithmetic. Modular arithmetic has a great deal in common with regular arithmetic , stripped to it's bare bones mechanics it's just regular arithmetic on a clock! For example , in mod 8 a,b can be found such that $(a \times b) \equiv 0$ $(2 \times 4) \equiv 0$ note $ 2 \ \le \ 2 \ , \ 4 \ \le 7 $ But I do believe from what I have read that a,b cannot be found if the modulus is prime. For me , nothing is etched in stone , I question everything.  
September 1st, 2017, 06:50 AM  #40  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry  Quote:
Quote:
Correct. If the modulus $p$ is prime, $p$ does not divide $ab$ for any integers $1 < a,b < p$, so $ab \not \cong 0 \, (mod \, p)$. In other words, there are no zero divisors in the ring of integers mod $p$, so it is an integral domain. However, when the modulus $n$ is composite, you can always find zero divisors (so we never get an integral domain). Indeed, there exist integers $1 < a,b < n$ such that $n = ab$, so $ab = n \cong 0 \, (mod \, n)$. Last edited by cjem; September 1st, 2017 at 07:13 AM.  

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