Show set of Gaussian integers is denumberable?

Question

A Gaussian integer is a complex number of the form a+bi where, a,b are integers and i = squareroot -1. Show set of Gaussian integers is denumberable


Any help would be great...thanks

Answer 1

Do you remember how to show it for the rationals? It's exactly the same idea, only simpler.

Line up the integers in some order that has a starting point -- say 0, 1, -1, 2, -2, 3, -3, ...

Label both the rows and columns of an infinite matrix by that list.

Traverse the matrix by zig-zagging back and forth across (finitely long) diagonals.

That's your mapping.

Answer 2

All you have to do is find a one-to-one mapping from the Gaussian integers to the integers.

Lets write out the Gaussian integers in this way:
First list all Gaussian integers where |a| + |b| = 0.
0+0i
Then list all Gaussian integers where |a| + |b| = 1.
1+0i, -1+0i, 0+1i, 0-1i
Then list all Gaussian integers where |a| + |b| = 2.
2+0i, -2+0i, 1+1i, 1 - 1i, -1 +1i, -1-1i, 0+2i, 0-2i
And so on.
This will list all Gaussian integers and above them you can write 1,2,3,4,... and hence you have created a one to one mapping.

Answer 3

Note that there is a bijection between the Gaussian integers and the 2-D lattice points, a+bi<-->(a,b), and since the lattice points are denumerable (countably infinite), so is the set of Gaussian integers. For a proof that the lattice points are countably infinite, consider a path beginning at (0,0) which hits each lattice point exactly once. One example of such a path is (0,0)->(0,1)->(1,1)-> (1,0)->(1,-1)->(0,-1)-> (-1,-1)->(-1,0)->(-1,1)-> (-1,2)->(0,2)->(1,2)-> (2,2)... This path demonstrates that there is a bijection between the lattice points and the positive integers, and the positive integers are by definition denumerable.