which number is closer to sqrt(5): (a/b) or (2a+5b)/(a+2b)

**chewitard** · March 28th 2011, 12:25 PM

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

**running-gag** · March 29th 2011, 08:59 AM

Hi

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

Using the same denominator and rearranging a little bit

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

Then you can show that

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

Thread: which number is closer to sqrt(5): (a/b) or (2a+5b)/(a+2b)

LinkBack

Thread Tools

Display

which number is closer to sqrt(5): (a/b) or (2a+5b)/(a+2b)

Similar Math Help Forum Discussions

Using sqrt(d) factor large number

Principal Argument of the Complex Number -1 + i sqrt(3)

Prove the sqrt(8) is not a rational number

sqrt of a complex number

prove sqrt(3) + sqrt (5) is an irrational number

Search Tags