You're right. As Gene pointed out earlier, and I replied, I was wrong--there's no counterexample. On 2014-02-10 19:07, Tom Karzes wrote:
That's the counterexample for gcd(a,b) = gcd(a-b,b)? It works for a=0, for b=0, for a=b=0, and for a=b:
gcd(0,b) = gcd(-b,b) = |b| gcd(a,0) = gcd(a-0,0) = |a| gcd(0,0) = gcd(0-0,0) = 0 gcd(a,a) = gcd(0,a) = |a|
Seems to work. I don't see this as the same as defining n/0 to be n.
Tom
Mike Speciner writes:
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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