Lecture_11 - MA1100 Lecture 11 Mathematical Induction Axiom...

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1 MA1100 Lecture 11 Mathematical Induction Axiom of Induction Principle of Mathematical Induction Base Case Inductive Step
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MA1100 Lecture 11 2 Example 1 Cardinality of power set Let A be a finite set with n element. How many elements does P (A) have? |A| 1 2 3 | P (A)| A = {a} P (A) = { « , {a} } P (A) = { « , {a}, {b}, {a, b} } A = {a, b, c} P (A) = { « , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } A = {a, b}
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MA1100 Lecture 11 3 Example 2 Regions in a circle Place n points on a circle and connect each pair of points with a line. Count the number of distinct regions partitioned by the lines. Number of points 2 3 4 Number of regions
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MA1100 Lecture 11 4 Example 2 Number of points 2 3 4 5 6 Number of regions Regions in a circle
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MA1100 Lecture 11 5 Axiom of Induction If T is a subset of Z + such that: 1. 1 œ T 2. for every k œ Z + , if k œ T, then k+1 œ T then T = Z + This is called the Axiom of Induction T 1 k œ T Ø k+1 œ T Z + = +
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MA1100 Lecture 11 6 Axiom of Induction If T is a subset of Z such that: 1. 1 œ T 2. for every k œ Z , if k œ T, then k+1 œ T then T = Z Z does not have a smallest number does not hold Counter-example: T = {-1, 0, 1, 2, 3, …} Axiom of Induction is a characteristic of Z + It does not hold for Z.
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MA1100 Lecture 11 7 Principle of Mathematical Induction Let P(n) be a predicate such that: 1.P(1) is true 2.For all k œ Z + , if P(k) is true, then P(k+1) is true Then P(n) is true for all n œ Z + Theorem (Principle of Mathematical Induction) Proof Given P(1) is true Given: If P(k) is true, then P(k+1) is true Let T be the set: T = {n œ Z + | P(n) is true} P(n) is true for all n œ Z +
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MA1100 Lecture 11 8 Principle of Mathematical Induction PMI can be used to prove ( " n œ Z + ) P(n) is true Base Case: Prove P(1) is true Inductive Step: Prove ( " k œ Z + ) P(k) Ø P(k+1) is true inductive hypothesis We can then conclude that P(n) is true for all n œ Z +
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