Proof:
For ever number k there exists -k which has the same parity as k (that is, if k is even, -k is even as well)
For every k there is (k + 1) with opposite parity (same thing with -k and -(k + 1)
Now, if we count the numbers, from 1 to infinite, the number of even and odd numbers is the same
from 1 to -infinite the number of even and odd numbers is the same as well
And you get an extra zero, an even number
Curious what is possible in Integer numbers
> That's what happens when you try to reason with infinity ;)
Ah yes! Fascinating subject
The "infinity" referenced here, aleph-null, is provably both the cardinality of the set of even integers, and the set of odd integers. (And also the set of integers, and any other countably infinite set.)
So, unintuitive as it might be when thinking about finite subsets of the integers, there are not only as many positive integers as negative integers, but as many of either as there are integers.
(There are, however, bigger infinite sets, there are, for instance, more reals than integers.)
There are as many prime numbers as there are integers.
There are as many fractions between 10 and 11 as there are integers.
There are more real numbers between 41.00001 and 41.0002 than there are integers from -infinity to infinity. Even though both are "infinite."