Using Archimedean Property to Prove the following

Question

So I've worked through a few of the properties of Archimedas. That is, I understand that for every real number $x$, there exists a natural number $n$ such that $n>x$

And I've also been able to show that, as a consequence of this fact, that for every positive real number $x$ there exists a natural number $n$ such that $1/n<x$

I'm now trying to (and fairly unsuccessfully) show that, as a consequence of the above that for each positive $x<5$ there exists a natural number $n$ such that $5-1/n>x$

I've started out by considering a positve $x<5$. Then by the results above we can find a natural number $n$ such that $1/n<x$

Do I need to consider cases at this point? That is consider when $0<x<4$ and then when $4<x<5$? I'm a little stuck with how to properly apply Archimedes to reach my proposed result. Help?

From $x<5$, conclude $y:=5-x>0$, then find $n$ with $\frac1n<y$. (We do not need that $x$ is positive). — Hagen von Eitzen, May 14 '13 at 20:23

score 2 · Accepted Answer · answered May 14 '13 at 22:23

Making Hagen's comment above into an answer (until and unless he posts it as an answer and informs me of that), put $y:=5-x$. Note that $y>0$ if (and only if) $x<5$. Using your corollary to the Archimedean property, if $x<5$, we may find a natural number $n$ such that $$\frac1n<y.$$ Recalling the definition of $y$ and adding $x$ to both sides of the above inequality gives us the desired result.

Using Archimedean Property to Prove the following

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