Re: [math-fun] Fwd: [CASE:423927] GCD suggestions
27 Nov
2013
27 Nov
'13
6:37 p.m.
So gcd(sqrt(2),1)=gcd(%pi,%e)=(+)0, no? At 03:33 PM 11/27/2013, Bill Gosper wrote:
New wording: GCD(a,b):= the limit of the Euclidean process of iteratively subtracting the smaller from the larger.
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Henry Baker