Author Topic: Imaginary Numbers Question  (Read 9934 times)

0 Members and 1 Guest are viewing this topic.

Offline ruler501

  • Meep
  • LV11 Super Veteran (Next: 3000)
  • ***********
  • Posts: 2475
  • Rating: +66/-9
  • Crazy Programmer
    • View Profile
Re: Imaginary Numbers Question
« Reply #15 on: February 22, 2011, 09:34:26 pm »
I'm happy i understood what you typed and what I am writing about in my book(mostly ;))
Thanks again for the help
I currently don't do much, but I am a developer for a game you should totally try out called AssaultCube Reloaded download here https://assaultcuber.codeplex.com/
-----BEGIN GEEK CODE BLOCK-----
Version: 3.1
GCM/CS/M/S d- s++: a---- C++ UL++ P+ L++ E---- W++ N o? K- w-- o? !M V?
PS+ PE+ Y+ PGP++ t 5? X R tv-- b+++ DI+ D+ G++ e- h! !r y

Offline jnesselr

  • King Graphmastur
  • LV11 Super Veteran (Next: 3000)
  • ***********
  • Posts: 2270
  • Rating: +81/-20
  • TAO == epic
    • View Profile
Re: Imaginary Numbers Question
« Reply #16 on: February 22, 2011, 10:26:20 pm »
I'm happy i understood what you typed and what I am writing about in my book(mostly ;))
Thanks again for the help
Imaginary numbers of the form a+bi where b=0 can be prime. ;-)  Technically, a+bi would have to only have multiples of 1 and itself.  I'm sure you could generalize the multiplication of to complex numbers to get another.  For example with P+Qi and S+Ti:
(P+Qi)(S+Ti)=PS+QSi+PTi+QTi^2=PS+QSi+PTi-QT=PS+PTi-QT+QSi=P(S+Ti)-Q(T+Si)
So, I guess as long as (PS-QT)=a and (PT-QS)=b, and as long as there were no P, Q, T, and S that satisfied that, it would be prime.

This is only in theory.  If the idea of being prime isn't extended to complex numbers, then... yeah.

Offline ruler501

  • Meep
  • LV11 Super Veteran (Next: 3000)
  • ***********
  • Posts: 2475
  • Rating: +66/-9
  • Crazy Programmer
    • View Profile
Re: Imaginary Numbers Question
« Reply #17 on: February 22, 2011, 10:30:33 pm »
Now I'm kondof confused. Must think about this...
Are prime numbers only the natural numbers?
I currently don't do much, but I am a developer for a game you should totally try out called AssaultCube Reloaded download here https://assaultcuber.codeplex.com/
-----BEGIN GEEK CODE BLOCK-----
Version: 3.1
GCM/CS/M/S d- s++: a---- C++ UL++ P+ L++ E---- W++ N o? K- w-- o? !M V?
PS+ PE+ Y+ PGP++ t 5? X R tv-- b+++ DI+ D+ G++ e- h! !r y

Offline jnesselr

  • King Graphmastur
  • LV11 Super Veteran (Next: 3000)
  • ***********
  • Posts: 2270
  • Rating: +81/-20
  • TAO == epic
    • View Profile
Re: Imaginary Numbers Question
« Reply #18 on: February 22, 2011, 10:38:20 pm »
Now I'm kondof confused. Must think about this...
Are prime numbers only the natural numbers?

Well, technically, it is defined as a natural number, so yes.  However, I believe you could extend the pretext of it to other numbers.  For example, above, if you had 0 for Q and T, you would get:
P(S+Ti)-Q(T-Si)=PS=A+Bi=A.  So, the factors of A would be P and S.  So it makes sense.  Take Q=2 and T=5:
P(S+Ti)-Q(T-Si)=P(S+5i)-2(5-Si)=PS+5i-10+2Si=S(P+2i)+5i-10
and P=9, S=3:
3(9+2i)+5i-10=27+6i+5i-10=17+11i
So the factors of 17+11i are (9+2i) and (3+5i) so 17+11i wouldn't be prime by this definition.

But yeah, it's defined as natural (real) numbers only.

Offline phenomist

  • LV4 Regular (Next: 200)
  • ****
  • Posts: 132
  • Rating: +46/-3
    • View Profile
Re: Imaginary Numbers Question
« Reply #19 on: February 23, 2011, 11:32:56 pm »
A complex extension to prime numbers are the Gaussian primes. Here, normal primes like 2 are no longer primes (2 = (1+i)(1-i)), though 3 is still a prime, for instance.

[link: http://en.wikipedia.org/wiki/Gaussian_prime#As_a_unique_factorization_domain]
Level Designer for Graviter

[Disclaimer: I can't program for my life.]