Let `a and b` be positive integers. Show that `sqrt2` always lies between and `a/b and (a+2b)/(a +b)`.

Explanation

`frac{a+2 b}{a+b}>sqrt{2} `


`=> a+2 b>sqrt{2}(a+b), `


` => sqrt{2} b(sqrt{2}-1)>a(sqrt{2}-1), `


` => frac{a}{2} b>a `


` => sqrt{2} .`



So, `frac{a}{b} < sqrt{2} < frac{a+2 b}{a+b}`.


Similarly by assuming `frac{a+2 b}{a+b } < sqrt{2}`, we will get `frac{a+2 b}{a+b} < sqrt{2} < frac{a}{b}`.


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