Proof
(related to Proposition: Uniqueness of Negative Numbers)
 For a given real number \(x\in\mathbb R\), let \(y\) be any real number with
\begin{align}
x+y=0. \label{eq:60}\
\end{align}
 Using the existence of negative numbers, we can add to both sides of the equation the number \((x)\) from the left side, resulting in $$(x)+(x+y)=(x) + 0.$$
 From the existence of zero and associative law of addition, it follows that $$((x)+x)+y=(x)+0.$$
 This leads to the equation $$0 + y=(x).$$
 Using the existence of zero once again, we find that $y=(x).$
 By applying the commutative law of addition, we get the same result, if we add \((x)\) to both sides of the equation \eqref{eq:60} from the right side.
 Thus, the there is only one such real number $y$ for which $x + y=0$, namely $y=x$.
 Therefore, negative real numbers are unique.
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983