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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 $\frac{a+2 b}{a+b}$. 31 Views
