Let a and b be positive integers show that √2 always lies between a/b and a+2b/ a+b
Please give an easier explanation of this question as it's solution is very complicated in RD Sharma
Let a and b be positive integers show that √2 always lies between a/b and a+2b/ a+b
Please give an easier explanation of this question as it's solution is very complicated in RD Sharma
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 .
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 .
