Question

let a and b be positive integers show that root 2 always lies between a/b and a+2b/a+b

please answer this question fastly

I'll mark it as the BRAINLIEST ONE

Answer 1

1.let a and b be any two positive integers.show that n√2always lies between 1⁄band a+2b/a+b.please give me the answer in a simple way.2.let a ,b,c,d be positive rationals such that a+√b=c+√d,then either a=c and b=c or b and d are squares of the rationals.

Answer 2

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

Step-by-step explanation:

We are given that a and b be are two positive integers and we have to show that $\sqrt{2}$ always lies between $\frac{a}{b}$ and $\frac{a+2b}{a+b}$ .

Firstly, assume that $\frac{a+2b}{a+b}$ is greater than $\sqrt{2}$ ;

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

Cross multiplying we get;

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

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

$\sqrt{2}b(\sqrt{2} -1)>a(\sqrt{2}-1)$

Now, canceling $\sqrt{2}-1$ from both sides, we get;

$\sqrt{2}b>a$

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

This means that $\sqrt{2}$ is greater than $\frac{a}{b}$ and less than $\frac{a+2b}{a+b}$, that is;

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

Similarly, is we assume that  $\frac{a+2b}{a+b}$ is less than $\sqrt{2}$ ;

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

Cross multiplying we get;

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

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

$\sqrt{2}b(\sqrt{2} -1)<a(\sqrt{2}-1)$

Now, canceling $\sqrt{2}-1$ from both sides, we get;

$\sqrt{2}b<a$

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

This means that $\sqrt{2}$ is less than $\frac{a}{b}$ and greater than $\frac{a+2b}{a+b}$, that is;

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

This shows that $\sqrt{2}$ always lies between $\frac{a}{b}$ and $\frac{a+2b}{a+b}$ .