Gaussian Integers Proof?

Question

The Gaussian Integers Zi = {a + bi : a,b are elements of Z} are a subset of the complex numbers, where i is the imaginary constant. Show that Zi is countable. (A full proof. No idea where to start!)

(Hint: the Gaussian integers use pairs of integers. So does Q, one in the numerator, and one in the denominator, and it's countable. Try the ideas in those proofs for this one, making a systematic 'list' for counting.)

Answer 1

The simplest idea is to enumerate the points on a Gaussian grid by traveling (outward from the origin) on a "square spiral" like the one pictured at the address given below.



At that webpage, the square spiral is used to "count" rational numbers by discarding the grid points whose coordinate ratios have already been encountered. Here you are "counting" the grid points themselves, so the square spiral gives you directly a one-to-one correspondence between the natural integers and the Gaussian integers.

Answer 2

For a gaussian integer z = a + bi (the place a and b are the two integers and that i is the sq. root of -a million), the norm is defined as: norm(z) = a^2 + b^2 it will be straight forward to instruct why that won't be able to probably be 3.

Answer 3

F(a + bi) = (a + b - 2)(a + b -1) + b is a one-to-one function from the "positive" Gaussian integers onto the positive integers. You know how to extend the argument to all Gaussian integers and all integers.

Answer 4

" No idea where to start!)"



you can start with the definition of countable set?

what means countable?