Define f: Z+ X Z+ -> Z+ by
f(a,b) = 2^(a-1)(2b-1) for all a,b in Z+
where Z+ is the set of all positive integers,
and X is the Cartesian product
The Attempt at a Solution
If we assume (a,b) as ordered pairs and write them as follows:
(1,1) (1,2) (1,3) (1,4)...(a,b+n)
(2,1) (2,2) (2,3) (2,4)...(a+1,b+n)
Using diagonal processing, we deduce that f is one-one.
Now can we also assume it is onto because f(a,b) maps to exactly one ordered pair?