Modulo Addition has Inverses

Theorem

Let $m \in \R$ be a real number.

Then addition modulo $m$ has inverses:

For each element $\left[\!\left[{x}\right]\!\right]_m \in \R_m$, there exists the element $\left[\!\left[{-x}\right]\!\right]_m \in \R_m$ with the property:

$\left[\!\left[{x}\right]\!\right]_m +_m \left[\!\left[{-x}\right]\!\right]_m = \left[\!\left[{0}\right]\!\right]_m = \left[\!\left[{-x}\right]\!\right]_m +_m \left[\!\left[{x}\right]\!\right]_m$

where $\R_m$ is the set of residue classes modulo $m$.

That is:

$\forall a \in \R: a + \left({-a}\right) \equiv 0 \equiv \left({-a}\right) + a \pmod m$

Proof

Follows directly from the definition of addition modulo $m$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_m +_m \left[\!\left[{-x}\right]\!\right]_m\)	\(=\)	\(\displaystyle \left[\!\left[{x + \left({-x}\right)}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{0}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{\left({-x}\right) + x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[\!\left[{-x}\right]\!\right]_m +_m \left[\!\left[{x}\right]\!\right]_m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

As $-x$ is a perfectly good real number, $\left[\!\left[{-x}\right]\!\right]_m \in \R_m$, whatever it may be.

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 19.1$