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

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

Grade:10

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find my response to your question below
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
7 years ago
Sir it is a-2b.
 
SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your question below
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

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