Show set of Gaussian integers is denumberable?

Question

Show set of Gaussian integers is denumberable?

Answer 1

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 2

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.

Answer 3

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.