Question

If a > 0 and b > 0, and if a ≠ b, prove that sqrt(ab) is strictly between a and b. How would I go about proving this?

Answer 1

can you prove the contrapositive?>

Answer 2

Let a = 1, b = 4 then ab = 4 and sqrt(4) = +2 and -2, so -2 is not between 1 and 4. Are you assuming that only the positive square root of ab is between a and b?

Answer 3

Yes, only positive square root of ab.

Answer 4

You can try proof by contradiction, but you will have 4 cases. let x= sqrt(ab), then if you assume your conclusion is false, then x =a, x=b, x<a, x>a. The first 2 cases are easy. I am not sure about the other 2.

Answer 5

The 4th case was supposed to be x>b, not x>a. I am also assuming b is the larger of a and b.

Answer 6

I think if you assume x< a, then x*x < a*a < a*b which contradicts that x*x =a*b

Answer 7

No, it's much easier to prove directly.

First without loss of generality, let a < b.

Now a < sqrt(ab) < b
if and only if a^2 < ab < b^2 , as a and b are positive numbers.

So let's see if we can prove that.

Well a < b , so multiplying now both sides by a we get
a^2 < ab

Again starting with a < b, multiply both sides by b then
ab < b^2

Hence

a^2 < ab < b^2

As all these numbers are positive, we can take the square root and

a < sqrt(ab) < b.

Answer 8

and if x>b then x*x> b*b >b*a which is a contradiction that x*x = b*a

Answer 9

His way is much more efficient than mine.

Answer 10

Both work nicely. Contradiction might get a bit hairy though. The direct proof is easier in this case. Thanks to both of you.