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# lecnotes4.1 - Lecture Notes Concepts of Mathematics(21-127...

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Unformatted text preview: Lecture Notes, Concepts of Mathematics (21-127) Lecture 1, Recitation A–D, Spring 2008 4 Induction and the Binomial Theorem 4.1 Induction We begin the study of induction by describing some properties of the natural numbers N : 1. Successor properties: 1 is not the successor of any x ∈ N , and any x ∈ N that is not 1 has a unique successor x + 1. 2. Closure properties: for all x, y ∈ N , x + y ∈ N and xy ∈ N . 3. Associativity properties: For all x, y, z ∈ N , x +( y + z ) = ( x + y )+ z and x ( yz ) = ( xy ) z . 4. Commutativity properties: For all x, y ∈ N , x + y = y + x and xy = yx . 5. Distributivity properties: For all x, y, z ∈ N , x ( y + z ) = xy + xz and ( y + z ) x = yx + zx . 6. Cancellation properties: For all x, y, z ∈ N , if x + z = y + z , then x = y . 7. Order properties: For all x, y, z ∈ N , • x < y iff there is a w ∈ N such that x + w = y . • x ≤ y iff x < y or x = y . • x < y and y < z implies x < z . • x ≤ y and y ≤ x implies x = y . • if x < y , then x + z < y + z and xz < yz . These properties will be assumed as we focus on another characteristic property of the nat- ural numbers: Inductive property of the positive integers: Let S be a subset of N . If (i) 1 ∈ S , and (ii) k ∈ S ⇒ k + 1 ∈ S , then S = N . 1 Definition 4.1 An inductive definition is a means to define a set of objects that can be indexed by the natural numbers. Inductive definitions follow from the inductive property above: there is a first object, and the ( n + 1) st object is defined in terms of the n th object. Example 4.1 The factorial of a natural number n , given by n ! = n · ( n- 1) ··· 2 · 1 , can be defined inductively as: (i) Define 1! = 1 (ii) For n ∈ N , define ( n + 1)! = ( n + 1) n ! . Example 4.2 Suppose we let O be the set of all odd natural numbers. Then we can define O by: (i) 1 ∈ O (ii) If n ∈ O , then n + 2 ∈ O . Then O = { n | n = 2 k- 1 for some k ∈ N } . Theorem 4.1 (Principle of Mathematical Induction) Let the universe be N and let P ( n ) be an open proposition. If (i) P (1) is true, and (ii) If P ( n ) is true, then P ( n + 1) is true, then P ( n ) is true for all n ∈ N . Proof: Assume P (1) is true and assume that if P ( n ) is true, then P ( n + 1) is true. Let S be the set of all natural numbers n such that P ( n ) is true, i.e., S = { n ∈ N | P ( n ) is true } . We must show that S = N . By definition, S ⊆ N . The assumptions (i) and (ii) here imply that S satisfies the two conditions of the inductive property of the positive integers. Thus S = N , or P ( n ) is true for all n ∈ N ....
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lecnotes4.1 - Lecture Notes Concepts of Mathematics(21-127...

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