Showing the set of 2$\times$2 matrices with integer entries are countable.

Question

Just had an analysis test, wasn't too sure if $\mathbb{Z} \times \mathbb{Z} \times\mathbb{Z}\times \mathbb{Z} \text{ was countable (it is), so I defined an injection onto }\mathbb{N}, \text{a countable set.}$

Given $M=\{ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} : a,b,c,d \in \mathbb{Z} \}$

My injection was given by: $f: M \to \mathbb{N} \text{ by}$

$\begin{equation} f(a,b,c,d) = \begin{cases} 2^{a}3^{b}5^{c}7^{d} & \text{if } a,b,c,d \geq 0 \\ 11^{|a|}13^{|b|}17^{|c|}19^{|d|} & \text{if } a<0 \text{ or } b<0 \text{ or } c<0\text{ or } d<0. \end{cases} \end{equation}$

Thus, as prime factorisations are unique, f is an injection from M onto a countable set, and thus M is countable.

Answer 1

Not a bad start, but as Bram28 has pointed out, not quite correct. To save this I'd start by defining an injection from the integers to the nonnegative integers. For example define $g: \mathbb{Z} \to \mathbb{N}$ by $g(n) = 2n$ if $n \ge 0$ and $2(-n) - 1$ if $n < 0$. That is:

• $g(0) = 0$
• $g(1) = 2, g(2) = 4, \ldots$
• $g(-1) = 1, g(-2) = 3, \ldots$

Then let $f(a, b, c, d) = 2^{g(a)} 3^{g(b)} 5^{g(c)} 7^{g(d)}$ which is an injection from the 2 by 2 matrices onto $\mathbb{N}$.

Answer 2

Good try, but it doesn't work: $(-1,1,1,1)$ gets mapped to the same number as $(1,-1,1,1)$