Invertible Integers under Multiplication

From ProofWiki
Jump to navigation Jump to search

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 $x y > 0$.

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 Invertible Elements under Natural Number Multiplication 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 Product with Ring Negative, the result follows.

$\blacksquare$


Sources