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Prove if that $0<a<b$ where $a,b \in \mathbb{R}$ then there exist some $n \in \mathbb{N}$ such that $\frac{1}{n} < a$ and $b < n $

The question states to use the Archimedean Property; If $a,b \in \mathbb{R}$ where $a < b$, then there exist an $n \in \mathbb{N}$ such that $b <na$.

My guess is to begin with the result of the Archimedean Property ($b<na$) and try to manipulate that and arrive at $\frac1n$ and $b<n$ but I'm having trouble doing so, is this the right idea? Any guidance is appreciated !

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  • $\begingroup$ You could apply the Archimedian Property to $b$ and $1$... $\endgroup$
    – el_tenedor
    Commented Feb 18, 2017 at 19:49
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    $\begingroup$ Show that an $n$ exists for each situation, and then consider the maximum of the two. $\endgroup$
    – The Count
    Commented Feb 18, 2017 at 19:49

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I'll get you started. Using el_tenedor's comments: Consider $b$ and $1$. By the Archimedean Property, there exists $N\in\mathbb{N}$ such that $b<N$. Now consider $a$ and $1$. By the Archimedean property, there exists $M\in\mathbb{N}$ such that $1<aM$, which implies the $\frac{1}{M}<a$. Now, what happens if you take the maximum of $\{M,N\}$?

Here, we've assumed that $a<1$ and $b>1$, by the way. That is also something you'll need to work around.

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  • $\begingroup$ @J.Jones did you find this useful? $\endgroup$
    – The Count
    Commented Feb 20, 2017 at 22:43

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