For any positive real number $x>0$, there exist a positive integer N such that $x>1/N>0$.

Question

I am trying to prove the following: For any positive real number $x>0$, there exist a positive integer N such that $x>1/N>0$.

What I know is: since x is a real number, it can be expressed as the formal limit of a Cauchy sequence $$x:=LIM (b_n)_{n=1}^\inf$$ since $x>0$, the sequence $(b_n)$ is bounded away from zero, i.e. $|b_n|>c$, $c>0$ Also, since $(b_n)$ is Cauchy then its bounded by M ($|b_n|\leq M$)

However, I dont know how to further proceed. Would appreciate your help.

Answer 1

Hint: $$x>0$$ There will be infinitely many positive integers $A$ such that $$A>\frac{1}{x}.$$

Answer 2

Let x and $\epsilon$ be any positive real numbers, there exist M such that $M\epsilon>x$. choose $\epsilon=1$, then $$\frac{1}{M}<x$$ $$x>\frac{1}{M}>0$$