lecnotes4.1 - Lecture Notes, Concepts of Mathematics...

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Unformatted text preview: Lecture Notes, Concepts of Mathematics (21-127) Lecture 1, Recitation AD, 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...

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