Question

let a and b be positive integers . show that √2 always lie between a/b and a+2b/a+b.

Answer 1

Answer:

Step-by-step explanation:

let a and b be positive integers . show that √2 always lie between a/b and a+2b/a+b.

There can be two cases

a/b > √2    &  (a+2b)/(a+b) < √2

or

a/b < √2    &  (a+2b)/(a+b) > √2

Case 1

(a+2b)/(a+b) < √2

=> (a + 2b) < √2 a + √2 b

=> a - √2 a + 2b - √2 b < 0

=> a(1 - √2) - √2b(1 - √2) < 0

=> (a - √2b)(1 - √2) < 0

1 - √2 is -ve

=> a - √2b > 0

=> a > √2b

=> a/b > √2

=> √2 lies between (a+2b)/(a+b) & ab

Case 2

(a+2b)/(a+b) > √2

=> (a + 2b) > √2 a + √2 b

=> a - √2 a + 2b - √2 b > 0

=> a(1 - √2) - √2b(1 - √2) > 0

=> (a - √2b)(1 - √2) > 0

1 - √2 is -ve

=> a - √2b < 0

=> a < √2b

=> a/b < √2

=> √2 lies between  ab & (a+2b)/(a+b)