A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
Here's a little brain teaser, that I feel some would enjoy:
Given some Gaussian Integer \(z\in \mathbb{Z}[i]\) find and prove a closedform expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
 2 years ago
Here's a little brain teaser, that I feel some would enjoy: Given some Gaussian Integer \(z\in \mathbb{Z}[i]\) find and prove a closedform expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)

This Question is Closed

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0It might seem easy to show, but it's not as easy as you might think to prove. I will say, though, there *is* a pretty neat proof.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0Could you give me a brief explanation of what \(\mathbb{Z}[i]/z\mathbb{Z}[i]\) means? Specifically what it means to do the mod operation.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0\[\mathbb{Z}[i]=\{a+bi\;\;a,b\in\mathbb{Z}\} \quad\text{and}\quad i=\sqrt{1}\]

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, sorry, I was out. But, here: So, we have, like @KingGeorge said, \[ \mathbb{Z}[i]=\{a+bi\;\;a,b\in \mathbb{Z}\}\text{ where }i^2=1 \]While we have:\[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}[i]\;\forall m\in\mathbb{Z}[i], n\equiv m\;(\!\!\!\!\!\!\mod z)\} \]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0So in the complex numbers, modular arithmetic takes place as follows correct?\[a+bi\pmod{c+di}\equiv\left[ a\mod{c}+bi\mod{d}\right]\]

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0I don't quite understand your case... But in the Gaussian Integers, the modulus has the same definition as the normal integers: \[ a\equiv b\mod c \iff c\,\,(ba)\\ a, b, c \in \mathbb{Z}[i] \]

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Wait, why'd my previous reply not make it? I forgot to add to the previous other post that the norm of the integer must be less than that of the modulus.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0That's what I thought, but just wanted to verify.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0I must go, but I will think about this.

Herp_Derp
 2 years ago
Best ResponseYou've already chosen the best response.0So your definition isn't quite correct; it should be:\[\large\mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}i:\forall m\in\mathbb{Z}[i]~~\text{with}~~m<z,~n\equiv m\pmod{z}\}\]

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, it's not, it'd be easier to show using division algorithm... but, frankly, I'm too lazy to retype it.

Herp_Derp
 2 years ago
Best ResponseYou've already chosen the best response.0And I'm too anally autistic to not correct it :)

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Haha, I'll retype it later, I guess...

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0Just fyi for people reading, how I thought the modular arithmetic worked at first, was not correct.

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Here's the better definition: \[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{r\in\mathbb{Z}[i]\;\;\forall n\in\mathbb{Z}[i], q\in\mathbb{Z}[i], r=nqz, \,N(r)<\frac{1}{2}N(z)\} \]

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0(Quintessentially, you have to remember that these are the sets of a single element (Gaussian Integer), which represents an equivalence class)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.