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An exercise from my book is as follows:

Assume that $x > 0$ for $x$ in $\mathbb{R}$ (Real numbers) then there is an $y$ in $\mathbb{N}$ (Natural Numbers) such that $1/y^3 < x$.

By the archimedean property, there exists an y in N such that $1/y < x$. How exactly would I continue on from here?

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    $y^3>y$. no!... – 2017-02-02
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    Have you proven that for any positive real x and any natural n that there exists a real y so that y^n =x? If not, prove it with s= sup {y|y^n < x}. You can prove that s^n=y. It's a bit of a pain but can be done. – 2017-02-02
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    Oh, never mind my comment. I misread. – 2017-02-02
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    0 <1/y < 1. So 1/y.1/y . 1/y < 1 . 1/y.1/y <1.1.1/y. – 2017-02-02

2 Answers 2

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By Archimedean property there exists a natural number $y$ such that $xy >1$. Then $xy^3 >xy >1$. So $1/(y^3)

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The indirect way is to remember for $a>0$ and $s

So if $y > 1$ then $ \frac 1 y < 1$ and $1/y^3 = 1/y*1/y^2=1/y*1/y*1/y$. And $1/y > 0$.

So if you know $1/y

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    In your first line, if $a>0$ and $s – 2017-02-02
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    Yep. That's exactly what it should be. – 2017-02-02