Question

let a and b be positive integers,show that ✓2 always lies between (a/b) and (a+2b) (a+b)​

Answer 1

Appropriate Question :-

Let a and b be positive integers,show that ✓2 always lies between (a/b) and (a+2b)/ (a+b).

$\large\underline{\sf{Solution-}}$

Since, we have to show that,

✓2 always lies between (a/b) and (a+2b)/(a+b)

So, two cases arises.

Case :- 1

$\red{\rm :\longmapsto\:When \: \dfrac{a}{b} > \sqrt{2}}$

Now,

$\rm :\longmapsto\:a > \sqrt{2}b$

On squaring both sides, we get

$\rm :\longmapsto\: {a}^{2} > {2b}^{2}$

can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {a}^{2} > {2b}^{2} + {a}^{2}$

$\rm :\longmapsto\: 2{a}^{2} > {2b}^{2} + {a}^{2}$

can be rewritten as

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} > {2b}^{2} + {a}^{2} + {2b}^{2}$

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} > {4b}^{2} + {a}^{2}$

Now, Adding 4ab on both sides, we get

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} + 4ab > {4b}^{2} + {a}^{2} + 4ab$

$\rm :\longmapsto\: 2({a}^{2} + {b}^{2} + 2ab) > {(2b)}^{2} + {a}^{2} + 2a(2b)$

$\rm :\longmapsto\:2 {(a + b)}^{2} > {(a + 2b)}^{2}$

$\rm :\longmapsto\: \sqrt{2} {(a + b)}> {(a + 2b)}$

$\bf\implies \: \sqrt{2} > \dfrac{a + 2b}{a + b}$

6

Hence,

$\bf\implies \: \dfrac{a + 2b}{a + b} < \sqrt{2} < \dfrac{a}{b}$

Case :- 2

$\red{\rm :\longmapsto\:When \: \dfrac{a}{b} < \sqrt{2}}$

Now,

$\rm :\longmapsto\:a < \sqrt{2}b$

On squaring both sides, we get

$\rm :\longmapsto\: {a}^{2} < {2b}^{2}$

can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {a}^{2} < {2b}^{2} + {a}^{2}$

$\rm :\longmapsto\: 2{a}^{2} < {2b}^{2} + {a}^{2}$

can be rewritten as

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} < {2b}^{2} + {a}^{2} + {2b}^{2}$

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} < {4b}^{2} + {a}^{2}$

On adding 4ab on both sides, we get

$\rm :\longmapsto\: 2{a}^{2} + {2b}^{2} + 4ab < {4b}^{2} + {a}^{2} + 4ab$

$\rm :\longmapsto\: 2({a}^{2} + {b}^{2} + 2ab )< {(2b)}^{2} + {a}^{2} + 2a(2b)$

$\rm :\longmapsto\:2 {(a + b)}^{2} < {(a + 2b)}^{2}$

$\rm :\longmapsto\: \sqrt{2} {(a + b)} < {(a + 2b)}$

$\bf\implies \: \sqrt{2} < \dfrac{a + 2b}{a + b}$

Hence,

$\bf\implies \:\dfrac{a}{b} < \sqrt{2} < \dfrac{a + 2b}{a + b}$

Hence, from both the two cases, we concluded that

$\bf\implies \: \sqrt{2} \: always \: lies \: between \: \dfrac{a}{b} \: \: and \: \: \dfrac{a + 2b}{a + b}$