Z+ times Z+ Solution First, note that you can define an injective map f: Z+ x Z+ --> Z+ by taking f(m, n) = (2^m)(3^n) (or choose any other two primes for bases). To show this is injective, suppose f(m, n) = f(i, j). Then (2^m)(3^n) = (2^i)(3^j), whence 2^(m - i)3^(n - j) = 1. Thus m - i = 0, so m = i, and n - j = 0, from which n = j. Naturally, you can define an injective map g: Z+ --> Z+ x Z+ by g(n) = (n, n). This is clearly an injection: g(m) = g(n) --> (m, m) = (n, n) --> m = n. Now, you can invoke the Schroeder-Bernstein theorem [1] to conclude that there exists a bijection between the two sets, and since Z+ is countable (in fact, it is the canonical countable set), so too is Z+ x Z+..