20111030, 22:20  #1 
Jan 2010
77_{8} Posts 
Just a thought of Quark numbers.
My greetings, Amici.
example of QN's: i^2=1, ij=1, j=1/i, (a1*i+b1*j)*(a2*i+b2*j)= a1*a2+a2*b1+b2*a1b1*b2.  seems quite useful to factorize primes into'em. 
20111030, 22:34  #2 
Jun 2003
7×167 Posts 
i^2 + ij = 1 + 1 = 0
Therefore i(i + j) = 0 Therefore i + j = 0 Therefore j = i Your "quantum numbers" appear to be just the complex numbers with i labeled as j. 
20111030, 22:45  #3 
Jan 2010
3^{2}×7 Posts 
Mr. P1
Thanks for reply. actually, you're right. but perhaps we can design new algorithms of prime factorization with QN's. i cannot confirm it at now, but looks useful. 
20111031, 02:13  #4 
Jan 2010
3F_{16} Posts 
in fact, QN's are ndimensional complex numbers. each prime P= Aqn*Bqn, set of solutions for each prime is infinite, for some solutions, a1, b1, a2, b2 are integers

20111031, 11:00  #5 
Jun 2003
7×167 Posts 
As I said, QNs as you defined them appear to be just plain ordinary complex numbers with i labeled as j. Simply attaching a new label to an element of a structure doesn't make it into a new structure.

20111031, 11:02  #6  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10109_{10} Posts 
Quote:
A fair assumption, given the algebraic manipulations in the post to which you refer, but worth pointing out, IMO. Paul 

20111031, 11:32  #7  
Jun 2003
4706_{10} Posts 
Quote:
Assumes nothing more than associativity of multiplication (not even commutativity is needed) 

20111031, 13:27  #8 
Nov 2003
16100_{8} Posts 
There is no such thing as 'ndimensional complex numbers'. There do exist e.g. higher dimensional division algebras (Quaternions, Octonions) but one loses something. For the Quaternions we lose commutitivity. For the Octonions, we also lose associativity. For (say) Clifford algebras we also lose both. Try as you might you will be unable to construct what you are looking for in a consistent way. 
20111031, 13:30  #9  
Nov 2003
1110001000000_{2} Posts 
Quote:
no zero divisors (i.e. you are working in an Integral domain) For those of you who don't know what a zero divisor is, consider the ring Z/12Z (i.e. the integers mod 12) We have 3*4 = 0 mod 12, but 3 != 0 and 4!=0. Thus, it is possible to have xy = 0 without either x=0 or y=0. 

20111031, 13:32  #10  
Nov 2003
2^{6}·113 Posts 
Quote:
What you are doing is NOT useful. Stop now. Your time would be better spent reading some texts. You need to learn some modern algebra and some algebraic number theory. 

20111031, 14:04  #11  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
11·919 Posts 
Quote:


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