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}
Algebra.Com
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.
--------------------------------------------------------------------------------------
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.
RELATED QUESTIONS
simplify
a-b/sqrt a-sqrt... (answered by edjones)
{{{ (a-b)/(sqrt ( a )+sqrt ( b ))... (answered by josgarithmetic)
find the value of A and B if
sqrt(A) x sqrt(B) = 4
sqrt(A) - sqrt(B) =... (answered by stanbon)
Rationalize the denominator..... (answered by Fombitz)
if a, b belong to R+ such that
{{{system(a*sqrt(a) + b*sqrt(b) = 183,
a*sqrt(b) +... (answered by Edwin McCravy)
Given that {{{sqrt((a/b)sqrt((b/a)sqrt(b/a))) = (a/b)^x}}}, where a,b are integers > 0,... (answered by MathLover1,math_tutor2020,MathTherapy)
Simplify: sqrt (21) (sqrt(7) + sqrt(3))
a) 7 sqrt (3) + 3 sqrt(7)
b) 7 sqrt(3) + sqrt (answered by Fombitz)
simplify: sqrt()27 - sqrt()8 + 2sqrt()2
a)3 sqrt() 3
b) 0
c)3 sqrt () 3 - 4 sqrt()2
(answered by Fombitz)
If 0 < a < b, show that a < sqrt{a•b} < b. The number sqrt{a•b} is called the... (answered by math_tutor2020)