# $\dfrac{-1}{n} < x$ for all $n\in \mathbb{N}$ $\Rightarrow x\geq 0$

I want to show

$\dfrac{-1}{n} < x$ for all $n\in \mathbb{N}$ $\Rightarrow x\geq 0$

And I'am not allowed to use limits. Any ideas? I have tried to use contraposition, but with no luck.

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Could you explain what you are allowed to use? If yo have a textbook, for example, it would be helpful to let us know what it is. – Carl Mummert Oct 2 '11 at 12:47

The idea is clear: $x$ cannot be a negative real number since you can find a sufficiently large $n$ to squeeze $-\frac1n$ between it and zero. So it has to be non-negative. I guess the point of such an exercise is to write this in formal symbols.

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This seems like circular reasoning. I thought the point was essentially to prove that for every $x>0$, there is some $n\in\mathbb{N}$ such that $1/n<x$, or equivalently, for every $y>0$, there is some $n\in\mathbb{N}$ such that $n>y$. This merely asserts it. – Michael Hardy Oct 2 '11 at 13:22
@MichaelHardy: At first I was inclined to agree with you, but, looking at the algebra-precalculus tag, I suppose it may well be that the Archimedean property can be assumed. – Niels Diepeveen Oct 2 '11 at 18:22

This is equivalent to saying that if $x<0$ then for some $n\in\mathbb{N}$, $-1/n\ge x$. That is in turn equivalent to saying that for every $x>0$, for some $n\in\mathbb{N}$, $1/n\le x$. So let $c=\inf\{1/n : n\in\mathbb{N}\}$. Since $1/n\ge 0$ for all $n\in\mathbb{N}$, it must be that $c\ge 0$. If we can prove that $c=0$, we're done, since that means every $x>0$ fails to be a lower bound.

So suppose $c>0$. Then $0<c\le 1/n$ for every $n\in \mathbb{N}$. Since $c$ is the greatest lower bound, $2c$ is not a lower bound. Therefore, for some $n\in\mathbb{N}$, $1/n < 2c$. But then $1/(2n) < c$, and since $2n\in\mathbb{N}$, this contradicts the fact that $c$ is a lower bound.

Later edit: I find it being asserted in other answers that it's enough to observe that for every real $y>0$, there is some $n\in\mathbb{N}$ such that $n>y$. But I took that to be the thing to be proved. I take $\mathbb{N}$ to be a ring and assume the reals have the least-upper-bound property. The problem is to prove that $\mathbb{N}$ has no upper bound in $\mathbb{R}$. If it does, then it has a least upper bound $a$. The smaller number $a/2$ is then not an upper bound. Therefore for some $n\in\mathbb{N}$, $n>a/2$. Since $2n\in\mathbb{N}$, and $2n>a$, the number $a$ is not an upper bound of $\mathbb{N}$ and we have a contradiction.

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I am going to write a little more improvised proof which is by @Micheal Hardy

The statement $\frac{\displaystyle -1}{\displaystyle n}< x$ for all $n \in \mathbb{N}$ $\implies x \geq 0$ is equivalent to for every $x\geq0$ $\exists\ n \in \mathbb{N}$ such that $x \geq \frac{\displaystyle 1}{\displaystyle n}$ i.e $n\geq\frac{\displaystyle 1}{\displaystyle x}$ . Let us assume that $n<\frac{\displaystyle 1}{\displaystyle x}$ $\implies$ $\mathbb{N}$ has an upper bound which is absurd. So the given statement is proved.

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