$\displaystyle a, b$ are positive integers and $\displaystyle \sqrt5$ lies inbetween $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{2a+5b}{a+2b}$. How would I go about showing which number is closer to $\displaystyle \sqrt5$?

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- Mar 28th 2011, 12:25 PMchewitardwhich number is closer to sqrt(5): (a/b) or (2a+5b)/(a+2b)
$\displaystyle a, b$ are positive integers and $\displaystyle \sqrt5$ lies inbetween $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{2a+5b}{a+2b}$. How would I go about showing which number is closer to $\displaystyle \sqrt5$?

- Mar 29th 2011, 08:59 AMrunning-gag
Hi

You must compare $\displaystyle \frac{2a+5b}{a+2b}-\sqrt{5}$ and $\displaystyle \sqrt{5}-\frac{a}{b}$

Using the same denominator and rearranging a little bit

$\displaystyle \frac{2a+5b}{a+2b}-\sqrt{5} = \frac{b}{a+2b}(\sqrt{5}-2) \left(\sqrt{5}-\frac{a}{b}\right)$

Then you can show that

$\displaystyle \frac{2a+5b}{a+2b}-\sqrt{5} \leq \frac{1}{8} \left(\sqrt{5}-\frac{a}{b}\right)$