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???
Best Answer
Your doing things in the wrong order, cf. comments above.
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!), $\frac1{N^3}<x$.