Invertible Integers under Multiplication

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Theorem

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


Corollary 1

Let $a, b \in \Z$ such that $a b = 1$.

Then $a = b = \pm 1$.


Corollary 2

The integers $\struct {\Z, +, \times}$ do not form a field.


Proof

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

Aiming for a contradiction, suppose first that $y \le 0$.

Then from Multiplicative Ordering on Integers and Ring Product with Zero:

$x y \le x \, 0 = 0$

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


Let $x > 0$ and $x y = 1$.

Then:

$y > 0$

and by Natural Numbers are Non-Negative Integers:

$y \in \N$

Hence by Invertible Elements under Natural Number Multiplication:

$x = 1$

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

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

$\blacksquare$


Sources

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