SOLUTION: Here is another one, 5/SQRT(8)

  5
————— 
 Ö8

The idea is to rationalize the denominator; that is get the
denominator to be free of roots (radicals).  First we break 
8 into primes like this 2·2·2

   5
————————
 Ö2·2·2
  
Two of those 2's will pair up as 22 which will come out of the 
square root radical since it is a perfect square.

   5
————————
 Ö22·2
  
This will leave one 2 under the radical.  So we need another 2
under that radical so it will pair with the other 2, so we will be able
to take it out too.  To accomplish this we multiply numerator and 
denominator by Ö2
      _
   5·Ö2
——————————
 Ö22·2·Ö2
  
Now we multiply under the radical on the bottom
      _
   5·Ö2
—————————
 Ö22·2·2
 
or 

     _
  5·Ö2
————————
 Ö22·22

  
Now we can take individual square roots on the bottom
     _
  5·Ö2
—————————
 Ö22·Ö22

and get 2·2
      _
   5·Ö2
—————————
   2·2
  
or

   _
 5Ö2
—————
  4

Edwin
AnlytcPhil@aol.com