Invertible Integers under Multiplication

Theorem
The only invertible elements of $\Z$ for multiplication (that is, units of $\Z$) are $1$ and $-1$.

Corollary
$\left({\Z, +, \times}\right)$ is not a field.

Proof
Let $x > 0$ and $x y > 0$.

Suppose $y \le 0$.

Then from Multiplicative Ordering on Integers and Ring Product with Zero:
 * $x y \le x \, 0 = 0$

From this contradiction we deduce that $y > 0$.

Now, if we have $x > 0$ and $x y = 1$, then $y > 0$ and hence $y \in \N$ by Natural Numbers are Non-Negative Integers.

Hence by Ordering on Naturally Ordered Semigroup Product it follows that $x = 1$.

Thus $1$ is the only element of $\N$ that is invertible for multiplication.

Therefore by Natural Numbers are Non-Negative Integers and Negative Product, the result follows.

Proof of Corollary
For $\left({\Z, +, \times}\right)$ to be a field, it would require that all elements of $\Z$ have an inverse.

Take $2$, for example.

In the Field of Rational Numbers, we have that $2 \times \dfrac 1 2 = 1$ and so $2$ has an inverse in $\Q$.

But that inverse is not in $\Z$.