Proving the archimidean property and a variant of it

Question

We want to prove that the set of natural numbers $\{1,2,3,... \}$ in $\mathbb{R}$. Use this to prove also that for any $x,y \in \mathbb{R}^+$($x,y$ both positive) there exist $n \in \mathbb{N}$ such that $nx \geq y$

Attempt: If we consider $A = \{ n \in \mathbb{N} : n \leq x \}$ for fixed $x \in \mathbb{R}$. Then , we see that $A$ is bounded above by $x$ so $\sup A = \alpha $ exists. And $\alpha - 1$ is not upper bound so can find some $m \in A$ such that $m > \alpha - 1$ so that $m+1 > \alpha$. so $m+1$ is not in $A$. but this is a contradiction since we assumed that $A = \mathbb{N}$.

The only part I still not sure how to prove is that $A \neq \varnothing$. I argued that because $[x-1] \leq x $ then $A \neq \varnothing$ but it was marked incorrect. How to show $A$ is nonempty?

As for second part, I did the following: By previous exercise, for any $x \in \mathbb{R}$ we can always find $n > x$ so if we make $x = a/b$ then we have $bn > a$. How do we get $\geq $?

Answer 1

You can make 2 cases here

1) $A=\phi$ then you have all of $\mathbb N $ outside $A$

2)$A\not = \phi$ after which your argument follows.