To understand bijection between $\mathbb{N^2}$ and
$\mathbb{N}$ I found this pdf on internet. But have couple of confusion.
N.B: Here we take $0 \in \mathbb{N}.$
Confusion:1 Why directly proof of bijection between $f:\mathbb{N^2}\to\mathbb{N}$ is messy ? But we found easily one-to-one correspondence $(0, 0)$ and $0, (0, 1)$ and $1, (0, 2)$ and $2.......$so on. That's may be proofs bijection.
Confusion:2 To prove injective function $f_1:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$, $f_1(n)=(n,0).$ My question is why I define the function in this way,$f_1(n)=(n,0)?$ could I define any function to prove injective like $f_1(n)=(n,1)$ or$f_1(n)=(n,2)$ or $f_1(n)=(1,n).... .....$ and so on
Confusion:2 To prove injective function $f_2:\mathbb{N^2}\to\mathbb{N}$, $f_2(n,m)=(2^n,3^m).$My question is why I define the function in this way $f_2(n,m)=(2^n,3^m).$could I define any function to prove injective like $f_2(n,m)=(1^n,3^m),$ or$f_2(n,m)=(2^n,4^m).............$ and so on.
The map $(n,m) \to (2n+1)2^m - 1$ is a perfectly valid bijection from $\Bbb N^2$ to $\Bbb N$. (The $-1$ at the end is just to get $0$ as the image of $(0,0)$, etc.)
Confusion 1: the function you sketch ($(0,0) \to 0, (0,1) \to 1, (0,2) \to 2$.... ) is not a complete description of a bijection at all, so does not belong in such a proof. It seems more like you're just using the second coordinate of the pair only (a projection in essence) which is not a bijection. Where does $(345, 671)$ map to? You don't specify...
Confusion 2: yes, the $0$ second coordinate is arbitrary , $n \to (n,1)$ or $n \to (n,n)$ also works. The author just wants some injection that is trivial, 1-1 on sight.
Confusion 3: $(n,m)$ maps to a product in $\Bbb N$, not a pair (recall he's defing an injection $\Bbb N^2 \to \Bbb N$ here..) $f(n,m)= 2^n \times 3^m$, which is 1-1 because the number of prime factors $2$ and $3$ that divide a number are uniquely determined (elementary number theory), so we get an injection (not onto as $0$ or $5$ etc are not reached at all as images, just numbers with only factors $2$ or $3$ or both). That's why $1^m \times 2^n$ or $2^m \times 4^n$ do not work as injections. Again, minimal knowledge required to see it's an injection.