If a,b>=0, then a<=b equivalent to sqrt{a}<=sqrt{b}

[princesssarah]

Here is my problem.

Prove that if a,b are real numbers and a > (or equal to) 0 and b> (or equal to) 0 then a < (or equal to) b is equvilant to the sqaure root a < (or equal to) square root b.

I'm stumped! How can i prove this? Can anyone get me started in the rigth direction??
Thank you a ton.

 

[stapel]

Prove that if a and b are real numbers and a > 0 and b > 0 then a < b is equvilant to the sqaure root a < square root b.

I'm stumped! How can i prove this?

If a < b, then a - b < 0, so, applying the "difference of squares" formula in a slightly non-standard way, we get:

. . . . .\(\displaystyle \left(\sqrt{a}\, -\, \sqrt{b}\right)\left(\sqrt{a}\, +\, \sqrt{b}\right)\, \leq\, 0\)

What do you know must be true about the sum in the second factor? (Specifically, think about positivity, negativity, and zero.)

The product of the two factors (one a subtraction and the other a sum) is less than or equal to zero, then either the product is negative (less than zero), or it is not (equal to zero). Consider each case, considering in particular the implications for the difference in the first factor above. :wink:

Eliz.