To be honest, my algorithm didn't account for numbers less than one either. I had to add a couple of lines after I saw your post. But in the original problem I was working on, this wasn't a consideration.

>Sorry. This only works with numbers >= 1
>
>Here is some documentation:
>pOrig = number for which you want the square root
>pLower = the best lower bound to the square root
>pUpper = the best upper bound to the square root
>
>In the example below I assure that for any number >= 1 that the square root is between 1 and the number. Pretty crappy guess but the only interval that you can be sure of.
>
>>If I didn't have access to my numerical analysis books and no connection to the internet, and a no pressure to optimize the function, it would look something like this.
>>
>>

>>nAnswer = sr(100,1,100)
>>? "THE ANSWER:", nAnswer
>>
>>FUNCTION sr(pOrig,pLower,pUpper)
>>* attempt to determine the square root by dividing the interval
>>? pOrig, pLower, pUpper
>>WAIT
>>nMiddle = ((pUpper - pLower) / 2) + pLower
>>
>>DO CASE
>>	CASE pOrig = nMiddle*nMiddle
>>		RETURN nMiddle
>>	CASE pOrig > nMiddle*nMiddle
>>		RETURN sr(pOrig,nMiddle,pUpper)
>>	CASE pOrig < nMiddle*nMiddle
>>		RETURN sr(pOrig,pLower,nMiddle)
>>	OTHERWISE
>>		? "ERROR: an impossible result!"
>>		RETURN 0
>>ENDCASE
>>RETURN
>>
>>

>>
>>>Write a function to calculate the square root of a number to a given number of decimal places. (Yes. I know, VFP has the SQRT function, but what would you do without it?)