Since we have already shown f(m,n) is equal to the product of the prime numbers, f(m,n)= (not sure how to finish it but I think this setup should go somewhere).....
Since we have already shown f(m,n) is equal to the product of the prime numbers, f(m,n)= (not sure how to finish it but I think this setup should go somewhere).....
You've just proved that $\displaystyle f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}m,n)\mapsto 2^{m-1}(2n-1)$ is a bijection. So, by definition $\displaystyle \text{card }\mathbb{N}\times\mathbb{N}=\text{card }\mathbb{N}=\aleph_0$
You've just proved that $\displaystyle f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}m,n)\mapsto 2^{m-1}(2n-1)$ is a bijection. So, by definition $\displaystyle \text{card }\mathbb{N}\times\mathbb{N}=\text{card }\mathbb{N}=\aleph_0$
Oh, boy, I had a lot fun watching Drexel Math Show.