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I should suppose that it has a small element and find a contradiction, but I did not know how to construct a number that smaller than $ r $ and its square is greater than $ \frac{1}{2} $.

As I am stuck also in the second of determining the greatest lower bound if it exists.

Thank you for your help.

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    From what set are you allowed to choose greatest lower bounds? All real numbers, or only rational? – 2017-01-23
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    Only rational numbers. – 2017-01-23

1 Answers 1

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Suppose $r^2>\frac12$, $r=\frac ab$, $a,b\in\Bbb N$. As suitable number just slightly smaller than $r$, we can try $s=\frac{na-1}{nb}$ with $n$ large enough. But will $s^2$ be $>\frac12$? $$ s^2=\frac{n^2a^2-2na+1}{n^2b^2}=r^2-\frac{2a}{nb^2}+\frac1{nb^2}>r^2-\frac{2a}{nb^2},$$ hence it suffices to pick $n$ so large that $\frac{2a}{nb^2}\frac{2a}{b^2(r^2-\frac12)}$$

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    What about the greatest lower bound? thank you. – 2017-01-23
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    How can I show that the greatest lower bound does not exist? Thanks. – 2017-01-24