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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 $?

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

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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.

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

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