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Lecture

MATH135 lecture 2.docx

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Department
Mathematics
Course Code
MATH 135
Professor
Janelle Resch

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MATH135 week 2 Sets (ctd), Quantifiers, Nested Quantifiers, Induction Sept. 16-20, 2013 Quantifiers In math, we have a couple of key words/phrases used in small statements. Depending on the quantifier used certain proof techniques can be used to prove the statement. Existential quantifiers: There are, there is, there exists These are existential quantifiers and we denote them wWe used ∃ when we are looking for math objects. Basic Structure: There is some ‘object’ in the ‘set where the object was taken’ with a ‘certain property’ such that ‘something happens’. For ‘every object’ with a ‘certain property’, something happens. Universal Quantifiers: For all, for each, for any These are universal quantifiers and we denoted the. When ‘ ∀ ’ is used we are looking for a set of objects that all share the same property. Ex: a) There exists an x in the set S such that P(x) is true ∃ xES : ∃ P(x) is true b) ∃ x ∀ y, (x≥ y), x, y EZ There is an integer. That integer is greater than or equal to all integers. There exists an x that is greater than or equal to all integers y. QUESTION: How do we communicate statements with quantifiers? Negation A ~A, ¬A T F F T So the negation of is ∀ , the negation of is ∃ Therefore ~∀ is ∃ Ex: An integer n divides an integer m (n|m), if m=kn, kEZ ∃ If n|m, then kEZ : m=kn By definition n|m means m=kn, for some kEZ MATH135 week 2 Sets (ctd), Quantifiers, Nested Quantifiers, Induction Sept. 16-20, 2013 Before doing another example it is important to know the following types of compound statements Implication: A  B OR (~A)V(B) Converse of Implication: B  A Contrapositive: (~B)(~A) logically equivalent to implication Negation of Implication: ~(AB) OR (A)Λ(~B) Ex: Recall A: ∃ xES : P(x) is true ~A: ∀ xES : P(x) is false B: ∃ x, ∀ y (x≥ y) ~B: ∀ x, ∃ y (x or ≤ instead of < An example of a statement with ‘nested quantifiers’ is… DEFINITION: The LIMIT of f(x) as x -> a equals L means that ε>0, ∃ δ>0 NOTE: ε is epsilon δ is delta 0 < |x-a|ax)=L) |x-a|ax+b)=ma+b Let ε>0 be a real number We want to show ∀ ε>0 ∃ δ >0 -> 0a Define: the limit of f(x) as x->a=x->ax)=L) means that ∀ ε0 ∃ δ>0 MATH135 week 2 Sets (ctd), Quantifiers, Nested Quantifiers, Induction Sept. 16-20, 2013 0amx+b)=ma+b Let ε>0 be a real number We want to show ∀ε >0 ∃ δ>0 : 0im f(x)=0 Let ε>0 be a real number We want to show that ∀ ε>0 ∃ δ>0 : 0
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