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Lecture

# MAT246 Lecture1.docx

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Department
Mathematics
Course
MAT246H1
Professor
Regina
Semester
Fall

Description
MAT246 Lecture 1 (2012-09-12) Part1. Basic Review The fundamental theorem of Algebra (by Gause): let p(x) be a non-constant polynomial whose coefficients are complex numbers. Then the equation f(x) = 0 has a solution in complex numbers. Direct proof:  Theorem: the sum of two even integers is always even. Proof: Let x and y be even, that is x = 2a, a∈Z; y = 2b, b∈Z then, x+y = 2a+2b = 2(a+b), a+b∈Z Since x+y has a divisor 2, so the theorem is established. Proof by Transposition: Want to prove “if p then q”, where p is the hypothesis, q is the conclusion. This can be achieved by proving an equivalent contra-positive statement “if not q then not p”. Proof by Contradiction: Assume q is not true = show the “hypothesis p” is not true.  Ex: not all prime numbers are odd. Hint: prove 2 is not odd. Proof by Mathematical Induction:  Ex: prove 1 + 2 + 3 + ⋯+ n = n n+1) 2 Non-constructive proof/Proof of Existence: (P1). a+(b+c) = (a+b)+c ----- Associative Law of Addiction (P2). a+0 = 0+a ----- Existence of Additive identity (0) (P3). a+(-a) = (-a)+a = 0 ----- Existence of Additive Inverse (P4). a+b = b+a ----- Commutative Law (P5). a×(b×c) = (a×b)×c ----- Associative for Multiplication (P6). a×1 = 1×a = a ----- Existence of Multiplication Identity (P7). a×a = a ×a = 1, a≠0 ----- Existence of Multiplication Inverse (P8). a×b = b×a ----- Commutative Law for Multiplication (P9). a×(b+c) = a×b+b×a ----- Distributive Law  Theorem: a×0 = 0 Proof: a×0 = a×(0+0) ----- (P2) a×0 = a×0+a×0 ----- (P9) Add –(a×0) to LHS&RHS, then a×0+[-(a×0)] = a×0+a×0+[-(a×0) ----- (P3) 0 = a×0+0 0 = a×0 Thus, the theorem is established. (P10). For every number a, one and only one of the following holds. *denote the collection of positive numbers as p ----- Trichotomy Law i. a = 0 ii. a∈p iii. -a∈p MAT246 Lecture 1 (2012-09-12)  Theorem: for both numbers a and b, we have i. a + b ≤ a + b | | ii. a = a,if a ≥ 0 iii. a = −a,if a ≤ 0 Proof: 2 2 2 2 (|a + b|) = a + b ) = a + 2ab + b = 2 2 2 2 2 |a| + 2ab + b| | ≤ a| | + 2 a b + b | | = (|a + b| |) That is,(|a + b|)2 ≤ (|a + b| |)2 According to theorem: √a = a | | a + b ≤ a + b | | Part2. Prime Numbers (Chapter1.1) Natural numbers/Positive integers/Positive whole numbers: 1,2,3,4 and so on. No largest natural number. Prime numbers: a natural number greater than 1 whose only natural number divisors are 1 and the number itself. Composite numbers: a natural number, other than 1, that is not a prime. Divisibility: the integer m is divisible by the integer n if there is an integer q such that m=nq. Divisors/Factors
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