It's good that you don't understand that statement, because I was multitasking when I wrote it, and it's wrong. --ms On 2014-02-10 14:18, Eugene Salamin wrote:
Mike, I don't understand this: "If gcd(0,0) = 0, then it is not universally true that gcd(a,b)=gcd(a-b,b)."
-- Gene
________________________________ From: Henry Baker <hbaker1@pipeline.com> To: ms@alum.mit.edu; math-fun <math-fun@mailman.xmission.com> Sent: Monday, February 10, 2014 9:23 AM Subject: Re: [math-fun] More 4th grade math
Ok, what do you propose for gcd(0,0) ?
At 09:20 AM 2/10/2014, Mike Speciner wrote:
And, as I recall, many early computers said n/0 = n.
If gcd(0,0) = 0, then it is not universally true that gcd(a,b)=gcd(a-b,b).
On 2014-02-10 11:06, Henry Baker wrote:
At 07:08 AM 2/10/2014, Mike Speciner wrote:
gcd(0,0) = 0 ???
I always thought a positive integer was prime iff it had exactly two positive integer divisors. Common Lisp & Maxima say gcd(0,0)=0.
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