Difference between revisions of "Chinese Remainder Theorem"
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The '''Chinese Remainder Theorem''' is a [[number theory | number theoretic]] result.
The '''Chinese Remainder Theorem''' is a [[number theory | number theoretic]] result. is one of the only [[theorem]]s named for an oriental person or place, due to the closed development of mathematics in the western world.
== Theorem ==
== Theorem ==
Let <math>m</math> be [[relatively prime]] to <math>n</math>. Then each [[residue class]] mod <math>mn</math> is equal to the [[intersection]] of a unique residue class mod <math>m</math> and a unique residue class mod <math>n</math>, and the intersection of each residue class mod <math>m</math> with a residue class mod <math>n</math> is a residue class mod <math>mn</math>.
== Proof ==
== Proof ==
Revision as of 21:43, 11 July 2008
|This is an AoPSWiki Word of the Week for July 11-July 16|
Let be relatively prime to . Then each residue class mod is equal to the intersection of a unique residue class mod and a unique residue class mod , and the intersection of each residue class mod with a residue class mod is a residue class mod .
If , then and clearly differ by a multiple of , so and . This is the first part of the theorem. The converse follows because and must differ by a multiple of and , and . This is the second part of the theorem.
Much like the Fundamental Theorem of Arithmetic, many people seem to take this theorem for granted before they consciously turn their attention to it. It ubiquity derives from the fact that many results can be easily proven mod (a power of a prime), and can then be generalized to mod using the Chinese Remainder Theorem. For intance, Fermat's Little Theorem may be generalized to the Fermat-Euler Theorem in this manner. Its application in problem-solving is similar.
General Case: the proof of the general case follows by repeatedly applying to above result (k-1) times.
Extended version of the theorem
Suppose one tried to divide a group of objects into , and parts instead and found , and fish left over, respectively. One can show this to be impossible, because any number giving remainder modulo must be odd and any number giving remainder modulo must be even. Thus, the number of objects must be odd and even simultaneously, which is absurd. Thus, there must be some restrictions on the numbers to ensure that at least one solution exists. It turns out (and this is the extended version of the theorem) that
- Here is an AoPS thread in which the Chinese Remainder Theorem is discussed and implemented.