# Prove that $x$ is any positive real number greater than $0$, $x>0$, then exists $N$ in the natural numbers such that $\frac{1}{N^3}<x$

Prove that $x$ is any positive real number greater than $0$, $x>0$, then exists $N$ in the natural numbers such that $\frac{1}{N^3}<x$

My steps:

Well I begin with $N\in\mathbb{N}$ and $x\in\mathbb{R}$ such that $N$ could be any integer number begining in 1; and $x$ could be any real number.

Then if $\frac{1}{N^3}<x$ and $N$ always be and integer number... and $x$ could be real, I will analyze a few number:

With $N=1$ and $x=1$, then $\frac{1}{1^3}<1$ its totally false. Well if $N=2$ and $x=2$, then $\frac{1}{2^3}<2$ it's true. Then I will check that $\frac{1}{N^3}<x$ is true if and only if $N>0$, and $x>\frac{1}{N^3}$

I am in the correct way???

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We're saying that if someone gives you any $x>0$ you will be able to find $N$ such that $N^{-3}<x$. The point is you pick $N$ after $x$ has been given. Think about Archimedeanity with $y=x^{1/3}$. – Pedro Tamaroff Oct 9 '13 at 16:26
I'm not entirely sure I understand, this statement is false for any $x\ge1$ to start with. Secondly, I don't get what you want to prove in your last sentence, isn't it the same since you already know that $N\in\mathbb N$ ie $N>0$? – user88595 Oct 9 '13 at 16:32

Since $x\ne0$ we can consider $\frac1x$ and since $x>0$ we also have that $\frac1x>0$. By the Archimedean property of $\mathbb R$, there exists some $N\in\mathbb N$ with $\frac1x<N$. Since $N\ge 1$, we have $N^3\ge N>\frac1x$ and by taking reciprocals (all numbers involved are positive!), $\frac1{N^3}<x$.
By taking reciprocals then $\frac{1}{N^3}\leq\frac{1}{N}<x$ and finally prove that $\frac{1}{N^3}<x$ right?? Thank you!!! – Salvattore Oct 9 '13 at 16:53