l4 - CS 171 Lecture Outline January 13, 2011 Example. Prove...

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Unformatted text preview: CS 171 Lecture Outline January 13, 2011 Example. Prove that for all positive integers n , n is even 7 n + 4 is even Solution. Let n be a particular but arbitrarily chosen integer. Proof for n is even 7 n + 4 is even. Since n is even, n = 2 k for some integer k . Then, 7 n + 4 = 7(2 k ) + 4 = 2(7 k + 2) Hence, 7 n + 4 is even. Proof for 7 n + 4 is even n is even. Since 7 n + 4 is even, 7 n + 4 = 2 l for some integer l . Then, 7 n = 2 l 4 = 2( l 2) Clearly, 7 n is even. Combining the fact that 7 is odd with the result of the Example 1, we conclude that n is even. We can also prove the latter by proving its contrapositive, i.e., we can prove if n is odd then 7 n + 4 is odd. Since n is odd we have n = 2 k + 1 , for some integer k . Thus we have 7 n + 4 = 7(2 k + 1) + 4 = 14 k + 10 + 1 = 2(7 k + 5) + 1 = 2 k + 1 , where k = 7 k + 5 is an integer. Example. Prove that there are infinitely many prime numbers. Solution. Assume, for the sake of contradiction, that there are only finitely many primes. Let p be the largest prime number. Then all the prime numbers can be listed as 2 , 3 , 5 , 7 , 11 , 13 , ... ,p Consider an integer n that is formed by multiplying all the prime numbers and then adding 1. That is, n = (2 3 5 7 p ) + 1 2 Lecture Outline January 13, 2011 Clearly, n > p . Since p is the largest prime number, n cannot be a prime number. In other words, n is composite. Let q be any prime number. Because of the way n is constructed, when n is divided by q the remainder is 1. That is, n is not a multiple of q . This contradicts the Fundamental Theorem of Arithmetic. Alternate Proof by Filip Saidak. Let n be an arbitrary positive integer greater than 1. Since n and n + 1 are consecutive integers, they must be relatively prime. Hence, the number N 2 = n ( n + 1) must have at least two different prime factors. Similarly, since the integers n ( n + 1) and n ( n + 1) + 1 are consecutive, and therefore relatively prime, the number N 3 = n ( n + 1)[ n ( n + 1) + 1] must have at least three different prime factors. This process can be continued indefinitely, so the number of primes must be infinite. Example. Prove that the sum of the first n positive odd numbers is n 2 . Solution. We want to prove that positive integers n,P ( n ) where P ( n ) is the following property. n 1 summationdisplay i =0 2 i + 1 = n 2 Base Case: We want to show that P (1) is true. This is clearly true as summationdisplay i =0 2 i + 1 = 1 = 1 2 Induction Hypothesis: Assume P ( k ) is true for some k 1. Induction Step: We want to show that P ( k + 1) is true, i.e., we want to show that k summationdisplay i =0 2 i + 1 = ( k + 1) 2 We can do this as follows....
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l4 - CS 171 Lecture Outline January 13, 2011 Example. Prove...

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