# Let $a$ and $b$ be positive integers. Show that $\sqrt{2}$ is always between $\frac{2}{b}$ and $\frac{a+2 b}{a+b}$.

Given: $a$ and $b$ be positive integers.

To do: To show that $\sqrt{2}$ is always between $\frac{a}{b}$ and $\frac{a+2 b}{a+b}$.

Solution:

To compare each number, let us find $\frac{a}{b}-\frac{a+2b}{a+b}$ first.

$\frac{a}{b}-\frac{a+2b}{a+b}$

$=\frac{a( a+b)-b( a+2b)}{b( a+b)}$

$=\frac{a^2+ab-ab-2b^2}{b( a+b)}$

$=\frac{a^2-2b^2}{b( a+b)}$

$\therefore,\ \frac{a}{b}-\frac{a+2b}{a+b}>0$

$\Rightarrow \frac{a^2-2b^2}{b( a+b)}>0$

$\Rightarrow a^2-2b^2>0$

$\Rightarrow a^2>2b^2$

$\Rightarrow a>\sqrt{2}b$

If $\frac{a}{b}-\frac{a+2b}{a+b}<0$

$\Rightarrow \frac{a^2-2b^2}{b( a+b)}<0$

$\Rightarrow a^2-2b^2<0$

$\Rightarrow a^2<2b^2$

$\Rightarrow a<\sqrt{2}b$

Therefore, If $a>\sqrt{2}b$, then $\frac{a}{b}>\frac{a+2b}{a+b}$

If $a<\sqrt{2}b$, then $\frac{a}{b}<\frac{a+2b}{a+b}$

Now , we have two case:

Case I :

$a>\sqrt{2}b$

$\Rightarrow \frac{a}{b}>\frac{a+2b}{a+b}$

$\Rightarrow \frac{a+2b}{a+b}<\frac{a}{b}$

And we have to prove that

$\frac{a+2 b}{a+b}<\sqrt{2}<\frac{a}{b}$

We know, $a>\sqrt{2}b$

$\Rightarrow a^2>2b^2$      [On squaring both sides]

$\Rightarrow a^2+a^2>a^2+2b^2$     [On adding $a^2$ both sides]

$\Rightarrow 2a^2+2b^2>a^2+2b^2+2b^2$   [On adding $2b^2$ both sides]

$\Rightarrow 2( a^2+2ab+b^2)>a^2+4ab+4b^2$   [On adding $4ab$ both sides]

$\Rightarrow 2( a+b)^2>( a+2b)^2$

$\Rightarrow \sqrt{2}( a+b)>( a+2b)$

$\Rightarrow \sqrt{2}>\frac{a+2b}{a+b}$

Again, $a>\sqrt{2}b$

$\Rightarrow \frac{a}{b}>\sqrt{2}$

Therefore, $\frac{a+2 b}{a+b}<\sqrt{2}<\frac{a}{b}$

Case II

$a<\sqrt{2}b$

We have to prove that $\frac{a}{b}<\sqrt{2}<\frac{a+2b}{a+b}$

As $a<\sqrt{2}b$

$\Rightarrow a^2<2b^2$  [On squaring both sides]

$\Rightarrow a^2+a^2$\Rightarrow 2a^2+2b^2

$\Rightarrow 2( a^2+2ab+b^2)<(a^2+4ab+4b^2$    [On adding $4ab$ both sides]

$\Rightarrow 2( a+b)^2<( a+2b)^2$

$\Rightarrow \sqrt{2}( a+b)<( a+2b)$

$\Rightarrow \sqrt{2}<\frac{a+2b}{a+b}$

Again, $a<\sqrt{2}b$

$\Rightarrow \frac{a}{b}<\sqrt{2}$

Therefore,  $\frac{a}{b}<\sqrt{2}<\frac{a+2b}{a+b}$

Thus, in each case $\sqrt{2}$ lies between  $\frac{a}{b}$ and $\frac{a+2 b}{a+b}$.

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Updated on: 10-Oct-2022

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