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Lecture

# MAT246 Lecture7.pdf

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School
University of Toronto St. George
Department
Mathematics
Course
MAT246H1
Professor
Regina
Semester
Fall

Description
MAT246 Lecture 7 (2012-10-31) Chinese remainder theorem/The mathematical classic of Sunzi: If , then , where and are relatively primes. Case of two equations: , are relatively primes is multiplicative inverse of , , that is Bezout’s identity: let a and b be integers, not both zero, and . Then there exist integers x and y such that: Since, , then => Since then Since , then  Ex1: solve Known that => Thus, , then Thus, the general solution is 19+56k  Ex2: solve Known that Thus, , then Since Since since That is, Thus, the general solution is 233+105k If k=-2, the minimum solution is 233-2 105=23 Theorem: two simultaneous congruences are only solvable when . The solution is unique module if and are relatively primes Proof: by the generalization of Euclidean algorithm. Since => That is, Suppose there exist integers s and t satisfying , then => =>  Ex3. solve Since , 8-2=6 is divisible by 3,is solvable. Known that MAT246 Lecture 7 (2012-10-31) Step1: Suppose there exist integers s and t satisfying , then , thus That is, Step2: , then Thus, Step3: That is, the general solution is -16+36k=20+26k Theorem: are solvable only when . The solution is unique modulo  Ex4. Solve From Ex3, the general solution for is n=20+36k. Then the problem is solving Since , 20-12=8 is divisible by 4, is solvable Known that Step1. Suppose there exist integers s and t satisfying , then Thus, Step2. : , then Thus, Step3. That is, the general solution is 236+252k  Ex5. Solve Known that Thus, the general solution to the first five equations is . Then, solve Since , 1-0=1 is divisible by 1,is solvable Known that Step1. Suppose there exist integers s and t satisfying , then Thus, Step2. Thus, Step3. That is, the general solution is -119+420k MAT246 Lecture 7 (2012-10-31)  Ex6. Solve Known that Since , 3-7=-4 is divisible by 1, is solvable Step1: Suppose there exist integers s and t satisfying , then Thus, Step2. , then Thus, Step3. That is, the general solution is  Ex7. Solve From Ex6, the general solution for is Then the problem is solving Known that Since , 4-40=-36 is divisible by 1, is solvable Step1: Suppose there exist integers s and t satisfying , then Thus, Step2. , then Thus, Step3. That is, the general solution is  Ex8. Solve Known that Since , 7-1=6 is divisible by 3, is solvable Step1. Suppose there exist integers s and t satisfying , then Step2. , then Thus, Step3. That is, the general solution is MAT246 Lecture 7 (2012-10-31) The complex numbers Definition 9.1.1: a polynomial is an expression of the form , where n is a natural number and the are numbers with . The are called the coefficients of the polynomial. The natural number n, the highest power to which x occurs in the polynomial, is called the degree of the polynomial. If , the polynomial is simply , which is called a constant polynomial. Definition 9.
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