Math135Lecture14StudentNotes

Math135Lecture14StudentNotes - Math 135: Lecture 14:...

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Unformatted text preview: Math 135: Lecture 14: Chinese Remainder Theorem Example 14.1. Solve n 2 (mod 5) n 9 (mod 11) The first congruence is equivalent to . Substituting this into the second congruence we get Now is equiv. to . Substi- tuting Equation (2) into Equation (1) gives the solution which is equivalent to Check We can check by substitution that the solution is Theorem 14.2 (Chinese Remainder Theorem ( CRT )) . If gcd( m 1 ,m 2 ) = 1, then for any choice of integers a 1 and a 2 , there exists a solution to the simultaneous congruences n a 1 (mod m 1 ) n a 2 (mod m 2 ) Moreover, if n = n is one integer solution, then the complete solution is n n (mod m 1 m 2 . ) Analysis: There are two things to prove. First, that a solution exists and second, what a complete solution looks like. For the first part: Hypothesis: Conclusion: 1 Math 135: Lecture 14: Chinese Remainder Theorem Constructing a Solution A1 : The integer n satisfies n a 1 (mod m 1 ) if and only if Substitute this expression into the second congruence.Substitute this expression into the second congruence....
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Math135Lecture14StudentNotes - Math 135: Lecture 14:...

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