# If a and b are positive integers. Show that root2 always lies between a/b and (a-2b)(a+b).

SHAIK AASIF AHAMED
10 years ago
Hello student,
Iam unable to provide you the exact solution.It seems there is some error in your question.please check whether it is a-2b or a+2b.So please recheck the question and post it again so that i can provide you with a meaningful answer
Ashfaaq Mohammed
35 Points
10 years ago
Sir it is a-2b.

SHAIK AASIF AHAMED
10 years ago
Hello student,
If it is a-2b it is not possible,if it is a+2b you can get
Value of root(2) = 1.414
since a is an integer and b is also an positive integer so ..
a/b is a rational number
If a >b then a/b > 1
also value of a/b can be greater than 1.414 in some cases ...Like (2/1) ( so a>b can not be the option ...a must be less than b)
So condition is if a < b ...then a/b is always less than 1 ...and hence 1.414 will always fall right side of the a/b value...on real axis ...means greater than a/b .
Now (a+2b)/(a+b) can be written as 1 + b/(a+b)
So it is always greater than one ... Critically when a=b ..it will be equal to 1.5 which is greater than 1.414
But once a > b .. the value can fall less than 1.414 ..example 2 and 1 ...value will be 1.333
So Root(2) always lies between a/b and (a+2b)/(a+b) only if a < b ..
And i have shown above that ..how for a<b .. a/b is less than 1.414 and (a+2b)/(a+b) is greater than 1.414