🚀 go-pugleaf

In article <16075@watdragon.waterloo.edu> afscian@violet.waterloo.edu (Anthony Scian) writes:
>I've tried to find a reference for this technique but
>I can't seem to find any but it is alluded to in Knuth Vol2.
>
>Does anybody remember the algorithm or any references?
>
>Anthony
>//// Anthony Scian afscian@violet.uwaterloo.ca afscian@violet.waterloo.edu ////
>"I can't believe the news today, I can't close my eyes and make it go away" -U2

Hi,
	The method I am familiar with is simply, subtract consecutive odd
numbers starting at 1 ( 1, 3, 5 ...) from the nuber you want to take the 
square root of and count the number of times you subtract.  When you have
a remainder you can use long division to divide the next odd number into
the remainder to get an approximate decimal.  The root is the number of
times you subtract + any decimal from the long division.

Example  square root of 24

sub 1    23
sub 3    20
sub 5	 15
sub 7	 8 
sub 9    nope        8.00000 / 9 = .88888...

square root of 24 is 4.8888888...

Paul F......

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            Paul Frattaroli - Department of Computing Services
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