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Let $a$ and $b$ be positive integers. Show that $\sqrt{2}$ always lies between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$.

Please give the easy solution as possible.

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  • $\begingroup$ @Cure: "Obviously"? If you see a counterexample, it would be helpful to show it. $\endgroup$ – hmakholm left over Monica Sep 18 '16 at 3:18
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Denote $x=\frac{a}{b}$ and $y=\frac{a+2b}{a+b}=\frac{x+2}{x+1}=1+\frac{1}{1+x}$. If $x< \sqrt{2}$ ($x$ cannot be $\sqrt{2}$ because $x$ is a rational), then $1+x< \sqrt{2}+1$, $\frac{1}{1+x}> \frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. Thus $y> \sqrt{2}$.

The other case can be proved similarly.

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    $\begingroup$ Nice. Very clean. $\endgroup$ – S. Y Sep 18 '16 at 3:25

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