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Old 2011-10-30, 22:20   #1
SarK0Y
 
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Smile 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.
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Old 2011-10-30, 22:34   #2
Mr. P-1
 
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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.
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Old 2011-10-30, 22:45   #3
SarK0Y
 
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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.
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Old 2011-10-31, 02:13   #4
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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
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Old 2011-10-31, 11:00   #5
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Quote:
Originally Posted by SarK0Y View Post
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.
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Old 2011-10-31, 11:02   #6
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Quote:
Originally Posted by Mr. P-1 View Post
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
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Old 2011-10-31, 11:32   #7
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Quote:
Originally Posted by xilman View Post
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)
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Old 2011-10-31, 13:27   #8
R.D. Silverman
 
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Quote:
Originally Posted by SarK0Y View Post
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.
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Old 2011-10-31, 13:30   #9
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Quote:
Originally Posted by xilman View Post
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.
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Old 2011-10-31, 13:32   #10
R.D. Silverman
 
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Quote:
Originally Posted by SarK0Y View Post
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.
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Old 2011-10-31, 14:04   #11
xilman
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Quote:
Originally Posted by R.D. Silverman View Post
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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