# Proving the archimidean property and a variant of it

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 $$\geq$$?

• May be you need to edit your question a bit in the first sentence . May 10, 2020 at 7:12

1) $$A=\phi$$ then you have all of $$\mathbb N$$ outside $$A$$
2)$$A\not = \phi$$ after which your argument follows.
• Thanks for replying. So, if $A = \empty$, then for every $x$ we have an $n > x$ so that theorem is true right? May 10, 2020 at 7:22
• If $A=\emptyset$ that means every natural number is greater than x and you are done. May 10, 2020 at 7:29
• For every $n$ we have $n>x$. Yes the theorem is true. May 10, 2020 at 7:30
• how can we make $bn \geq a$ instead of $>$? May 10, 2020 at 7:35
• $>\implies \geq$ but not the other way round hence you are ok here May 10, 2020 at 8:25