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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