What Else Can I Do With Modulus?
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From: William Elliot on 1 Jul 2010 02:35 On Thu, 1 Jul 2010, Virgil wrote: > William Elliot <marsh(a)rdrop.remove.com> wrote: >> On Thu, 1 Jul 2010, Gerry Myerson wrote: >>> Yimin Rong <yiminrong(a)yahoo.ca> wrote: >>> >>>> In the integer domain, given c = pq, where p and q are prime and > 0, >>>> and given integer modulus m > 1, the following is true: >>>> >>>> c mod m = (pq) mod m = (p mod m)(q mod m) >> >> False. Here's a counter example: >> 1 = 21 mod 5 = 3*7 mod 5 = (3 mod 5)(7 mod 5) = 3 * 2 = 6. >> >>> What do primes have to do with it? If a, b, c, and d are any integers, >>> and a = b mod m, and c = d mod m, then a c = b d mod m. >> >> Are you two on the same wave length? >> Are you thinking the equivalence relation >> a = b (mod m) when m | a - b, >> >> while he's nurdling with that binary operator thing >> a mod m = min{ r in N \/ {0} | some q in Z with a = qm + r } ? >> >> Notice the lack of symmetry >> 5 mod 3 = 2; 2 mod 3 /= 5. > > Using "==" for the standard three line congruence symbol, > 5 == 2 mod 3 and 2 == 5 mod 3, where "mod" is NOT an operator. > Using the old fashion notation of a == b (mod m) instead of the quick type stuff a == b mod m distinguishs the binary operator a = b mod m from the equivalence relation a = b (mod m). which is never written b (mod m) = a while b mod m = a is indication that mod is that nurd think thing.
From: Gerry Myerson on 1 Jul 2010 03:11 In article <20100630221808.V91726(a)agora.rdrop.com>, William Elliot <marsh(a)rdrop.remove.com> wrote: > On Thu, 1 Jul 2010, Gerry Myerson wrote: > > Yimin Rong <yiminrong(a)yahoo.ca> wrote: > > > >> In the integer domain, given c = pq, where p and q are prime and > 0, > >> and given integer modulus m > 1, the following is true: > >> > >> c mod m = (pq) mod m = (p mod m)(q mod m) > > False. Here's a counter example: > > 1 = 21 mod 5 = 3*7 mod 5 = (3 mod 5)(7 mod 5) = 3 * 2 = 6. > > > What do primes have to do with it? If a, b, c, and d are any integers, > > and a = b mod m, and c = d mod m, then a c = b d mod m. > > Are you two on the same wave length? > Are you thinking the equivalence relation > a = b (mod m) when m | a - b, > > while he's nurdling with that binary operator thing > a mod m = min{ r in N \/ {0} | some q in Z with a = qm + r } ? Yes, I am thinking of the equivalence relation. I assumed OP was, too, since, if OP was thinking of the binary operator, it would be strange of OP to write that something manifestly false (as per your example) is true. So at least one of us has, presumably, written something that should help OP. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Bill Dubuque on 1 Jul 2010 10:58 William Elliot <marsh(a)rdrop.remove.com> wrote: >On Thu, 1 Jul 2010, Gerry Myerson wrote: >> Yimin Rong <yiminrong(a)yahoo.ca> wrote: >> >>> given primes p,q, and integer m>1, the following is true: >>> >>> (pq) mod m = (p mod m)(q mod m) > > False. Here's a counter example: > > 1 = 21 mod 5 = 3*7 mod 5 = (3 mod 5)(7 mod 5) = 3 * 2 = 6. Indeed, it should be ab mod m = (a mod m)(b mod m) mod m or (ab, m) = ( (a, m) (b, m), m) as ideals which is frequently handy when computing gcd's, e.g. the special case (b,m) = 1 yields Euclid's Lemma. --Bill Dubuque
From: William Elliot on 2 Jul 2010 06:15 On Thu, 1 Jul 2010, Bill Dubuque wrote: > William Elliot <marsh(a)rdrop.remove.com> wrote: >> On Thu, 1 Jul 2010, Gerry Myerson wrote: >>> Yimin Rong <yiminrong(a)yahoo.ca> wrote: >>> >>>> given primes p,q, and integer m>1, the following is true: >>>> (pq) mod m = (p mod m)(q mod m) >> >> False. Here's a counter example: >> 1 = 21 mod 5 = 3*7 mod 5 = (3 mod 5)(7 mod 5) = 3 * 2 = 6. > > Indeed, it should be > ab mod m = (a mod m)(b mod m) mod m > Yuck, what an ugly algebraic system (Z,+,*,mod) makes. Compare for example, (Z,+,*,min,max). > or (ab, m) = ( (a, m) (b, m), m) as ideals > > which is frequently handy when computing gcd's, e.g. > the special case (b,m) = 1 yields Euclid's Lemma. >
From: Bill Dubuque on 2 Jul 2010 11:57
William Elliot <marsh(a)rdrop.remove.com> wrote: > On Thu, 1 Jul 2010, Bill Dubuque wrote: >> William Elliot <marsh(a)rdrop.remove.com> wrote: >>> On Thu, 1 Jul 2010, Gerry Myerson wrote: >>>> Yimin Rong <yiminrong(a)yahoo.ca> wrote: >>>> >>>>> given primes p,q, and integer m>1, the following is true: >>>>> (pq) mod m = (p mod m)(q mod m) >>> >>> False. Here's a counter example: >>> 1 = 21 mod 5 = 3*7 mod 5 = (3 mod 5)(7 mod 5) = 3 * 2 = 6. >> >> Indeed, it should be >> >> ab mod m = (a mod m)(b mod m) mod m >> >> or (ab, m) = ( (a, m) (b, m), m) as ideals >> >> which is frequently handy when computing gcd's, e.g. >> the special case (b,m) = 1 yields Euclid's Lemma. >> > Yuck, what an ugly algebraic system (Z,+,*,mod) makes. > Compare for example, (Z,+,*,min,max). You shouldn't be surprised. That's typically the case when one imposes unneeded computational structure in a purely mathematical context. It was Dedekind who championed the methodology of eliminating obfuscational calculations in favor of conceptual proofs. In fact there's an interesting discussion about such currently flowing on MathOverflow, cf. http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313 --Bill Dubuque |