/sci/ - Science & Math

Pic related. Obviously a simple solution, but I can't figure it out. Would the following be sufficient proof?

Conceptually I understand how the cardinality is the same, since for every ordered pair (or 'every element of the set'), there will be a corresponding n in N that it is mapped to. Where I'm lost is actually constructing the bijection, and what that process should look like.

My best attempt so far is below, but it's far from formalized.

Organize the set of all ordered pairs [math] NxN | {[(0,0),(0,1),..(0,a)][(1,0),(1,1),(1,b)]} [\math] such that set (0,0) maps to 1, (0,1) maps to 2, (0,a) maps to i, and, performing the same mapping to the latter half of the sets, (1,0) maps to i+1, (1,1) maps to i+2, (1,b) maps to i+b.