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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 \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
 one year ago
 one year 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 \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
 one year ago
 one year ago

This Question is Closed

LolWolfBest ResponseYou've already chosen the best response.0
It 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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Could 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.
 one year ago

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

LolWolfBest ResponseYou've already chosen the best response.0
Yeah, 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:\[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}[i]\;\forall m\in\mathbb{Z}[i], n\equiv m\;(\!\!\!\!\!\!\mod z)\} \]
 one year ago

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

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

LolWolfBest ResponseYou've already chosen the best response.0
Wait, 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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
That's what I thought, but just wanted to verify.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I must go, but I will think about this.
 one year ago

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

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

Herp_DerpBest ResponseYou've already chosen the best response.0
And I'm too anally autistic to not correct it :)
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Haha, I'll retype it later, I guess...
 one year ago

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

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

LolWolfBest 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)
 one year ago
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