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How would I use the Archimedean property to prove that there is no negative real number greater that all negative rational numbers?

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Let $x$ be any negative real number. Then $-x$ is positive, and by the Archimedean property there is a natural number $n$ such that $0<-x<n$.

Then $-\frac 1n$ is a negative rational number, and

$$x<-\frac 1n<0$$

Thus $x$ is not greater than all negative rational numbers. Since $x$ are arbitrary, there is no negative real number greater that all negative rational numbers.

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By archimedian property for all $-x>0\exists n_x\in\mathbb N$ s.t. $0<\frac1{n_x}<-x\implies x<-\frac1{n_x}<0$. I hope you can get the rest of the thing.

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