Question

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

Answer 1

we do not know whether $\frac{a}{b}<\frac{a+2b}{a+b}$ or $\frac{a}{b}>\frac{a+2b}{a+b}$

Therefore, to compare these two numbers, let us compute $\frac{a}{b}-\frac{a+2b}{a+b}$

We have,

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

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

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

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

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

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

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

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

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

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

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

Thus, $\frac{a}{b}>\frac{a+2b}{a+b}$, if $a>\sqrt{2}b$ and $\frac{a}{b}<\frac{a+2b}{a+b}$ if $a<\sqrt{2}b$.

So, we have the following cases:

Case $I$ when $a>\sqrt{2}b$

In this case, we have

$\frac{a}{b}>\frac{a+2b}{a+b}$ i.e., $\frac{a+2b}{a+b}<\frac{a}{b}$

We have to prove that

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

We have,

$a>\sqrt{2b}$

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

$\Rightarrow a^2+a^2>a^2+{2b}^2$ $\left[adding{a}^{2}bothsides\right]$

$\Rightarrow 2a^2+2b^2>(a^2+2b^2)+2b^2$ $\left[adding{2b}^{2}onbothsides\right]$

$\Rightarrow 2(a^2+2ab+b^2)>a^2+4ab+4b^2$ $\left[adding4abbothsides\right]$

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

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

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

Again,

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

From $\left(i\right)$ and $\left(ii\right)$, we get

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

Case $II$ when $a<\sqrt{2}b$

In this case, we have

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

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

We have,

$a<\sqrt{2}b$

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

$\Rightarrow a^2+a^2<a^2+2b^2$ $\left[adding{a}^{2}onbothsides\right]$

$\Rightarrow 2a^2+2b^2<a^2+4b^2$ $\left[adding{2b}^{2}onbothsides\right]$

$\Rightarrow 2a^2+4ab+2b^2<a^2+4ab+4b^2$

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

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

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

From $\left(iii\right)$ and $\left(iv\right)$, we get

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

Hence $\sqrt{2}$ lies between $\frac{a}{b}$ and $\frac{a+2b}{a+b}$