SOLUTION: Show that: if a > 0, b > 0, and sqrt{a} < sqrt{b}, then a < b. Hint given: b - a = (sqrt{b}

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Question 1208931: Show that: if a > 0, b > 0, and
sqrt{a} < sqrt{b}, then a < b.
Hint given:
b - a = (sqrt{b} - sqrt{a})(sqrt{b} + sqrt{a}

Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

sqrt(a) < sqrt(b)
sqrt(b) > sqrt(a)
sqrt(b)-sqrt(a) > 0
(sqrt(b)-sqrt(a))*(sqrt(b)+sqrt(a)) > 0*(sqrt(b)+sqrt(a)) ....... see note below
b-a > 0
b > a
a < b

Note:
I multiplied both sides by (sqrt(b)+sqrt(a)) to use the hint your teacher gave you.
The output of a square root function is never negative as long as there isn't a negative out front. Eg: sqrt(25) = 5. This means sqrt(a) and sqrt(b) are both positive. Furthermore it means sqrt(b)+sqrt(a) > 0. Multiplying both sides by a positive value will not flip the inequality sign.

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Another approach.

sqrt(a) < sqrt(b) means that sqrt(a) = sqrt(b) - c for some positive real number c.
In other words, sqrt(a) = sqrt(b) - somePositiveNumber

Square both sides of this equation.
sqrt(a) = sqrt(b) - c
( sqrt(a) )^2 = ( sqrt(b) - c)^2
a = ( sqrt(b) )^2 - 2*c*sqrt(b) + c^2
a = b - 2*c*sqrt(b) + c^2
a = b - (2*c*sqrt(b) - c^2)
a = b - c*(2*sqrt(b) - c)
a = b - c*(sqrt(b) + sqrt(b) - c)
a = b - c*(sqrt(b) + sqrt(a))
a = b - positiveNumber*(positiveNumber + positiveNumber)
a = b - positiveNumber*positiveNumber
a = b - positiveNumber
a < b

In the 3rd and 4th steps, notice that the squaring and square roots cancel each other out.
For instance ( sqrt(b) )^2 = b when b > 0.
The terms marked in blue represent a substitution based on the initial equation.

SOLUTION: Show that: if a > 0, b > 0, and sqrt{a} < sqrt{b}, then a < b. Hint given: b - a = (sqrt{b} - sqrt{a})(sqrt{b} + sqrt{a}