Inverse for Rational Addition

Theorem
Each element $x$ of the set of rational numbers $\Q$ has an inverse element $-x$ under the operation of rational number addition:
 * $\forall x \in \Q: \exists -x \in \Q: x + \left({-x}\right) = 0 = \left({-x}\right) + x$

Proof
Let $x = \dfrac a b$ where $b \ne 0$.

We take the definition of rational numbers as the quotient field of the integral domain $\left({\Z, +, \times}\right)$ of integers.

From Existence of Quotient Field, we have that the inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:

From Negative of Divided By, we have that:
 * $\displaystyle -\frac a b = \frac {-a} b = \frac a {-b}$

So $\dfrac a b$ has a unique and unambiguous inverse for $+$.