2 (a)
let f(n) = 2n where n is any integer
Assuming that f(a) = f(b)
Then, f(a) = 2a = f(b) = 2b
2a = 2b
a = b by cancellation
Also, f(n) always equals to 2i for some int i
Therefore, the set of even numbers is countable.
(Edited: 2020-03-09)
2 (a)
let f(n) = 2n where n is any integer
Assuming that f(a) = f(b)
Then, f(a) = 2a = f(b) = 2b
2a = 2b
a = b by cancellation
Also, f(n) always equals to 2i for some int i
Therefore, the set of even numbers is countable.