# 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.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983