Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question