🚀 go-pugleaf

In article <491@watshine.waterloo.edu> pfratar@watshine.waterloo.edu (Paul Frattaroli - DCS) writes:
>
>	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.

This works because

 n^2 = 1 + 3 + ... + (2n-1)

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Luiz Henrique de Figueiredo		internet: lhf@aries5.waterloo.edu
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