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 Veränderlichen", Vieweg Studium, 1983
