anonymous
  • anonymous
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 closed-form expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
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 closed-form expression for the number of equivalence classes in \(\mathbb{Z}[i]/z\mathbb{Z}[i]\)
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
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.
KingGeorge
  • KingGeorge
\[\mathbb{Z}[i]=\{a+bi\;|\;a,b\in\mathbb{Z}\} \quad\text{and}\quad i=\sqrt{-1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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:\[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}[i]\;|\forall m\in\mathbb{Z}[i], n\equiv m\;(\!\!\!\!\!\!\mod z)\} \]
KingGeorge
  • KingGeorge
So in the complex numbers, modular arithmetic takes place as follows correct?\[a+bi\pmod{c+di}\equiv\left[ a\mod{c}+bi\mod{d}\right]\]
anonymous
  • anonymous
I don't quite understand your case... But in the Gaussian Integers, the modulus has the same definition as the normal integers: \[ a\equiv b\mod c \iff c\,|\,(b-a)\\ a, b, c \in \mathbb{Z}[i] \]
anonymous
  • anonymous
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.
KingGeorge
  • KingGeorge
That's what I thought, but just wanted to verify.
KingGeorge
  • KingGeorge
I must go, but I will think about this.
anonymous
  • anonymous
So your definition isn't quite correct; it should be:\[\large\mathbb{Z}[i]/z\mathbb{Z}[i]=\{n\in\mathbb{Z}i:\forall m\in\mathbb{Z}[i]~~\text{with}~~|m|<|z|,~n\equiv m\pmod{z}\}\]
anonymous
  • anonymous
Yes, it's not, it'd be easier to show using division algorithm... but, frankly, I'm too lazy to re-type it.
anonymous
  • anonymous
And I'm too anally autistic to not correct it :)
anonymous
  • anonymous
Haha, I'll re-type it later, I guess...
KingGeorge
  • KingGeorge
Just fyi for people reading, how I thought the modular arithmetic worked at first, was not correct.
anonymous
  • anonymous
Here's the better definition: \[ \mathbb{Z}[i]/z\mathbb{Z}[i]=\{r\in\mathbb{Z}[i]\;|\;\forall n\in\mathbb{Z}[i], q\in\mathbb{Z}[i], r=n-qz, \,N(r)<\frac{1}{2}N(z)\} \]
anonymous
  • anonymous
(Quintessentially, you have to remember that these are the sets of a single element (Gaussian Integer), which represents an equivalence class)

Looking for something else?

Not the answer you are looking for? Search for more explanations.