The proper definition of prime, gcd, and lcm is in terms of ideals. It is only because Z is a PID that an ideal (n), the set of multiples of n, gets confused with the element n. But of course, we can't teach ring theory to 4th graders, at least not in the US public schools. An ideal A divides an ideal B if A contains B. In Z, if a|b then every multiple of b is a multiple of a. "To contain is to divide", quote from Harvey Cohn, Advanced Number Theory. The gcd of ideals is the smallest ideal containing their union. In Z, gcd((m),(n)) = (m,n). In general, gcd((a,b,...), (c,d,...)) = (a,b,c,d,...), which to me justifies the (m,n) notation. The lcm of ideals is their intersection, which is also an ideal. A product AB of ideals A and B is the smallest ideal containing all xy with x in A and y in B. The ideal (0) consists of the single element 0. In taking the gcd of a list of ideals, (0) can be omitted from the list. In Z, gcd((n),(0)) = (n), or gcd(n,0) = n. The gcd of the empty list is (0), since this is the smallest ideal containing the empty set. Thus gcd(0,0) = gcd() = 0. The lcm of any list of ideals containing (0) is (0). In Z, lcm(n,0) = lcm(0,0) = 0. The ideal (1) is the entire integral domain. The gcd of any list of ideals containing (1) is (1). In Z, gcd(n,1) = 1. Similarly, (1) can be omitted from any list in taking lcm, and lcm() = (1). A prime P is a maximal ideal. That is, P is a proper ideal of the entire integral domain R, but there is no ideal lying strictly between P and R. It is the "maximal" condition that denies primality to 1. 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.