Just a thought of Quark numbers.
 

[B]My greetings, Amici. [/B]

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.

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.

[B]Mr. P-1[/B]
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. :smile:

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 :smile:

[QUOTE=SarK0Y;276397]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 :smile:[/QUOTE]

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.

[QUOTE=Mr. P-1;276379]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.[/QUOTE]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

[QUOTE=xilman;276437]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[/QUOTE]

ij=1 --> i*ij=i*1 --> (i^2)j=i --> -j=i

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

[QUOTE=SarK0Y;276397]in fact, QN's are n-dimensional complex numbers.

[/QUOTE]


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.

[QUOTE=xilman;276437]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[/QUOTE]

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.

[QUOTE=SarK0Y;276383][B]Mr. P-1[/B]
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. :smile:[/QUOTE]

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.

[QUOTE=R.D. Silverman;276457]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.[/QUOTE]True. I missed that one. :doh!: