What Else Can I Do With Modulus?
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From: Yimin Rong on 30 Jun 2010 14:36 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) Are there any other non-equivalent equalities involving p, q, c and m? Any references, on-line or otherwise, are welcome. Thanks for reading. Regards, Yimin
From: Gerry Myerson on 30 Jun 2010 19:12 In article <152d93b9-44f3-4af3-b0e5-68f588c01ac6(a)j8g2000yqd.googlegroups.com>, 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) 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 there any other non-equivalent equalities involving p, q, c and m? It works for addition, too: under the previous hypotheses, a + c = b + d mod m. And the distributive law holds: u (v + w) = u v + u w (mod m). What's happening is that the integers form a ring, Z, the multiples of m form an ideal, mZ, in this ring, and when you work mod m, you're working in the set of cosets of mZ in Z, which set itself has a ring structure as the quotient ring, Z / mZ. > Any references, on-line or otherwise, are welcome. Any intro Number Theory text will give you the basics on the arithmetic of congruences. Some of them will also give you the ring-theoretic setting, as will most intro abstract algebra texts. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: cwldoc on 30 Jun 2010 15:09 > 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) > > Are there any other non-equivalent equalities > involving p, q, c and m? > > Any references, on-line or otherwise, are welcome. > > Thanks for reading. > > Regards, > > Yimin > Well, there's always the theorem that if p is prime and k is an integer that does not have p as a factor, then k^(p - 1) = 1 (mod p)
From: William Elliot on 1 Jul 2010 02:03 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. >> Are there any other non-equivalent equalities involving p, q, c and m? > > It works for addition, too: under the previous hypotheses, > a + c = b + d mod m. > > And the distributive law holds: > > u (v + w) = u v + u w (mod m). > > What's happening is that the integers form a ring, Z, the multiples > of m form an ideal, mZ, in this ring, and when you work mod m, > you're working in the set of cosets of mZ in Z, which set itself has > a ring structure as the quotient ring, Z / mZ. > >> Any references, on-line or otherwise, are welcome. > > Any intro Number Theory text will give you the basics on > the arithmetic of congruences. Some of them will also give you > the ring-theoretic setting, as will most intro abstract algebra texts. His notation indicates he's not interested in number theory but in the binary computer operator thing mod.
From: Virgil on 1 Jul 2010 02:18
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 } ? > > 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. > > >> Are there any other non-equivalent equalities involving p, q, c and m? > > > > It works for addition, too: under the previous hypotheses, > > a + c = b + d mod m. > > > > And the distributive law holds: > > > > u (v + w) = u v + u w (mod m). > > > > What's happening is that the integers form a ring, Z, the multiples > > of m form an ideal, mZ, in this ring, and when you work mod m, > > you're working in the set of cosets of mZ in Z, which set itself has > > a ring structure as the quotient ring, Z / mZ. > > > >> Any references, on-line or otherwise, are welcome. > > > > Any intro Number Theory text will give you the basics on > > the arithmetic of congruences. Some of them will also give you > > the ring-theoretic setting, as will most intro abstract algebra texts. > > His notation indicates he's not interested in number > theory but in the binary computer operator thing mod. |