Integers are Countable

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Theorem

Define the inclusion mapping $i: \N \to \Z$.

Thus there exists an injection from $\N$ to $\Z$.

Hence $\Z$ is infinite.

Next, let us arrange $\Z$ in the following order:

$\Z = \left\{{0, 1, -1, 2, -2, 3, -3, \ldots}\right\}$

Then we can directly see that we can define a mapping $\phi: \Z \to \N$ as follows:

$\forall x \in \Z: \phi \left({x}\right) = \begin{cases} 2x - 1 & : x > 0 \\ - 2x & : x \le 0 \end{cases}$

It is straightforward to show that this is an injection:

Let $\phi \left({x}\right) = \phi \left({y}\right)$. Then one of the following applies:

$(1): \quad -2x = -2y$ in which case $x = y$

$(2): \quad 2x - 1 = 2y - 1$ in which case $2x = 2y$ and so $x = y$

$(3): \quad 2x - 1 = -2y$ in which case $y = -x + \frac 1 2$ and therefore $y \notin \Z$

$(4): \quad 2y - 1 = -2x$ in which case $x = -y + \frac 1 2$ and therefore $x \notin \Z$.

So $2x - 1 = -2y$ and $2y - 1 = -2x$ can't happen and so $x = y$.

Thus $\phi$ is injective.

$\blacksquare$