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37+PS4+Solutions

# 37+PS4+Solutions - Handout#37 May 2 2011 CS103 Robert...

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Handout #37 CS103 May 2, 2011 Robert Plummer Problem Set #4 Solutions 1. Prove by induction that for any natural number n, 3 ) 2 )( 1 ( ) ) 1 ( ( 1 n n n i i n i Proof: Let P n be the proposition that the equality above holds. We will prove the general case by induction on n. BASE CASE: P 1 states that 1(1+1) = 1(1+1)(1+2)/2. Since LHS = RHS = 2, the base case is true. INDUCTIVE STEP: Assume that for some k 1, P k is true; i.e., that I.H. We will show that P k+1 must be true; i.e., that Adding (k + 1)(k + 1 + 1) to both sides of the I.H. and doing simple algebra gives Applying the definition of on the left and factoring on the right gives as desired. Thus P k P k+1 , and by the Principle of Mathematical Induction, P n is true for all n 0. 3 ) 2 )( 1 ( ) ) 1 ( ( 1 k k k i i k i 3 ) 3 )( 2 )( 1 ( 3 ) 2 1 )( 1 1 )( 1 ( ) ) 1 ( ( 1 1 k k k k k k i i k i 3 ) 2 )( 1 ( 3 ) 2 )( 1 ( ) 1 1 )( 1 ( )) ) 1 ( ( ( 1 1 k k k k k k k i i k i 3 ) 3 )( 2 )( 1 ( ) ) 1 ( ( 1 1 k k k i i k i

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2 2. Let x be an integer. Show that for every integer n 2, x n is even if and only if x is even. Proving the biconditional will naturally involve two subproofs. You are required to use induction in at least one of them, but otherwise, do not use induction where a very simple non-inductive proof will suffice. This is a biconditional proof. ( direction) Show the following is true for all n 2: P n : If x n is even, then x is even. BASE CASE: P 2 states that x 2 even x is even. The contrapositive of that is: x odd x 2 is odd. If x is odd, x = 2k + 1 for some integer k, so x 2 = (2k + 1)(2k + 1) = 4k 2 + 4k + 1, which is odd. So P 2 is true. INDUCTIVE STEP: Assume P k : x k even x is even for some k 2. Show P k+1: x k+1 even x is even. Proof of the inductive step: Suppose x k+1 is even. Thus x k x is even. This means that either (case 1) x k is even or (case 2) x is even, since the product of two odd numbers is odd. But x k even x is even by the IH, so in both cases x is even, and P k+1 is true. By the Principle of Mathematical Induction, P n is true for n 2. ( direction) Show that for n 2, x even x n even. Direct proof: If x is even, x = 2m for some integer m. So x n = (2m) n = 2 n m n , which is even since it has 2 as a factor. 3. Prove by induction that for every integer n 2, the complement of the union of any n sets equals the intersection of the complements of these sets. Using set notation, we could state this as: If A 1 , A 2 , ..., A n are sets and n 2, then (A 1 A 2 ... A n )' = A 1 ' A 2 ' ... A n ' . Proof by induction. P n : (A 1 A 2 ... A n )' = A 1 ' A 2 ' ... A n ' BASE CASE: P 2 states that (A 1 A 2 )' = A 1 ' A 2 '. This is true by De Morgans's Law. INDUCTIVE STEP: Assume P k : (A 1 A 2 ... A k )' = A 1 ' A 2 ' ... A k ' for some k 2.
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