Invertible Integers under Multiplication

Theorem
The only invertible elements of $$\Z$$ for multiplication are $$1$$ and $$-1$$.

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.