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 2011-10-30, 22:20 #1 SarK0Y     Jan 2010 2×37 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*a1-b1*b2. --------------------- seems quite useful to factorize primes into'em.
 2011-10-30, 22:34 #2 Mr. P-1     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.
 2011-10-30, 22:45 #3 SarK0Y     Jan 2010 4A16 Posts Mr. P-1 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.
 2011-10-31, 02:13 #4 SarK0Y     Jan 2010 2·37 Posts in fact, QN's are n-dimensional complex numbers. each prime P= Aqn*Bqn, set of solutions for each prime is infinite, for some solutions, a1, b1, a2, b2 are integers
2011-10-31, 11:00   #5
Mr. P-1

Jun 2003

7·167 Posts

Quote:
 Originally Posted by SarK0Y in fact, QN's are n-dimensional complex numbers. each prime P= Aqn*Bqn, set of solutions for each prime is infinite, for some solutions, a1, b1, a2, b2 are integers
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.

2011-10-31, 11:02   #6
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

237258 Posts

Quote:
 Originally Posted by Mr. P-1 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.
The second line of your proof assumes that the distributive law holds for the quantities i and j.

A fair assumption, given the algebraic manipulations in the post to which you refer, but worth pointing out, IMO.

Paul

2011-10-31, 11:32   #7
axn

Jun 2003

7·683 Posts

Quote:
 Originally Posted by xilman The second line of your proof assumes that the distributive law holds for the quantities i and j. A fair assumption, given the algebraic manipulations in the post to which you refer, but worth pointing out, IMO. Paul
ij=1 --> i*ij=i*1 --> (i^2)j=i --> -j=i

Assumes nothing more than associativity of multiplication (not even commutativity is needed)

2011-10-31, 13:27   #8
R.D. Silverman

Nov 2003

26×113 Posts

Quote:
 Originally Posted by SarK0Y in fact, QN's are n-dimensional complex numbers.

There is no such thing as 'n-dimensional 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.

2011-10-31, 13:30   #9
R.D. Silverman

Nov 2003

26×113 Posts

Quote:
 Originally Posted by xilman The second line of your proof assumes that the distributive law holds for the quantities i and j. A fair assumption, given the algebraic manipulations in the post to which you refer, but worth pointing out, IMO. Paul
It not only requires the distributive law, it also assumes that there are
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.

2011-10-31, 13:32   #10
R.D. Silverman

Nov 2003

26×113 Posts

Quote:
 Originally Posted by SarK0Y Mr. P-1 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.
Some advice:

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.

2011-10-31, 14:04   #11
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

100111110101012 Posts

Quote:
 Originally Posted by R.D. Silverman It not only requires the distributive law, it also assumes that there are 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.
True. I missed that one.

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