ppt on conjecture

Conjecture and proof and conjectures and counterexamples worksheets
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Prof.EvanBaros,United Kingdom,Teacher
Published Date:26-07-2017
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Conjecture and proof MATH 1510 Let us start from a simple example. We add more and more Lili Shen of the odd numbers as follows: Mathematical Induction 1 = 1; 1+ 3 = 4; 1+ 3+ 5 = 9; 1+ 3+ 5+ 7 = 16; 1+ 3+ 5+ 7+ 9 = 25: One may observe that the numbers on the right hand sides of these equations are all perfect squares; that is, 2 For 1 n 5, the sum of the first n odd numbers is n . Then it is natural to ask: is the above statement true for every positive integer n?Conjecture and proof MATH 1510 Lili Shen It is not difficult to try more numbers and find that the Mathematical pattern persists for the first 6, 7, 8, 9, and 10 odd numbers. Induction At this point we feel fairly confident that this is always true, so we make a conjecture: 2 The sum of the first n odd numbers is n . Since the nth odd number is 2n 1, we can write this statement more precisely as n X 2 (2i 1) = n . i=1 But this is still a conjecture.Conjecture and proof MATH 1510 Lili Shen Mathematical Induction We cannot conclude by checking a finite number of cases + that a property is true for all n2N since there are infinitely many positive integers. A proof is a clear argument that demonstrates the truth of a statement beyond doubt.Conjecture and proof MATH 1510 Lili Shen Mathematical Induction Conjecture For every positive integers n, n X 2 (2i 1) = n : i=1Conjecture and proof MATH 1510 Lili Shen Mathematical Proof. Induction The odd numbers constitute an arithmetic sequence given by a = 2n 1. Using the formula of partial sums of an n arithmetic sequence, one concludes that 1+(2n 1) 2 S = n = n : n 2 n X 2 Hence (2i 1) = n . i=1Mathematical induction MATH 1510 Now we introduce a special kind of proof called Lili Shen mathematical induction, which is useful for statements Mathematical involving natural numbers n. Induction For example, let P(n) denote the following statement: 2 P(n): The sum of the first n odd numbers is n . Since this statement involves all positive integers, it contains infinitely many statements which will be called P(1), P(2), P(3), ::: : 2 P(1): The sum of the first 1 odd number is 1 . 2 P(2): The sum of the first 2 odd numbers is 2 . 2 P(3): The sum of the first 3 odd numbers is 3 . ::: .Mathematical induction MATH 1510 Lili Shen Mathematical Induction Theorem (Principle of mathematical induction) For each positive integer n, let P(n) denote a statement depending on n. Suppose that the following two conditions are satisfied: 1 P(1) is true; 2 for every positive integer k, if P(k) is true then P(k + 1) is true. Then P(n) is true for all positive integers n.Mathematical induction MATH 1510 Lili Shen Mathematical Induction To apply this principle, there are two steps: Step1. Prove that P(1) is true. Step2. Assume that P(k) is true, and use this assumption to prove that P(k + 1) is true. Notice that in Step 2 we do not prove that P(k) is true. We only show that if P(k) is true, then P(k + 1) is also true. The assumption that P(k) is true is called the induction hypothesis.Proofs by mathematical induction MATH 1510 Lili Shen Mathematical Induction Proposition For all positive integers n, n X 2 (2i 1) = n : i=1Proofs by mathematical induction MATH 1510 Proof. Lili Shen n X 2 Mathematical Let P(n) denote the statement (2i 1) = n . Induction i=1 2 Step1. P(1) is true since 1 = 1 . k X 2 Step2. Suppose P(k) is true, i.e., (2i 1) = k . We i=1 k+1 X 2 show that P(k + 1) is true, i.e., (2i 1) = (k + 1) . i=1 Indeed, k+1 k h i X X 2 2 (2i1) = (2i1) +(2k+1) = k +2k+1 = (k+1) : i=1 i=1 Thus P(k + 1) follows from P(k), completing the proof.Sum of powers MATH 1510 Lili Shen Proposition Mathematical Induction n X (1) 1 = n. i=1 n X n(n + 1) (2) i = . 2 i=1 n X n(n + 1)(2n + 1) 2 (3) i = . 6 i=1 n 2 2 X n (n + 1) 3 (4) i = . 4 i=1Sum of powers MATH 1510 Proof. Lili Shen n X Mathematical (1) Let P(n) denote the statement 1 = n. Induction i=1 Step1. P(1) is true since 1 = 1. k X Step2. Suppose P(k) is true, i.e., 1 = k. We show i=1 k+1 X that P(k + 1) is true, i.e., 1 = k + 1. Indeed, i=1 k+1 k h i X X 1 = 1 + 1 = k + 1: i=1 i=1 Thus P(k + 1) follows from P(k), completing the proof.Sum of powers MATH 1510 Lili Shen n X n(n + 1) (2) Let P(n) denote the statement i = . Mathematical Induction 2 i=1 1(1+ 1) Step1. P(1) is true since 1 = . 2 Step2. Suppose P(k) is true, i.e., k X k(k + 1) i = : 2 i=1 We show that P(k + 1) is true, i.e., k+1 X (k + 1)(k + 2) i = : 2 i=1Sum of powers MATH 1510 Lili Shen Indeed, Mathematical Induction k+1 k h i X X i = i + k + 1 i=1 i=1 k(k + 1) = + k + 1 2 k(k + 1)+ 2(k + 1) = 2 2 k + 3k + 2 = 2 (k + 1)(k + 2) = : 2 Thus P(k + 1) follows from P(k), completing the proof.Sum of powers MATH 1510 Lili Shen (3) Let P(n) denote the statement n X n(n + 1)(2n + 1) Mathematical 2 i = . Induction 6 i=1 1(1+ 1)(2 1+ 1) 2 Step1. P(1) is true since 1 = . 6 Step2. Suppose P(k) is true, i.e., k X k(k + 1)(2k + 1) 2 i = : 6 i=1 We show that P(k + 1) is true, i.e., k+1 X (k + 1)(k + 2)(2k + 3) 2 i = : 6 i=1Sum of powers MATH 1510 Lili Shen Indeed, Mathematical k+1 k Induction h i X X 2 2 2 i = i +(k + 1) i=1 i=1 k(k + 1)(2k + 1) 2 = +(k + 1) 6 2 k(k + 1)(2k + 1)+ 6(k + 1) = 6 2 (k + 1)(2k + 7k + 6) = 6 (k + 1)(k + 2)(2k + 3) = : 6 Thus P(k + 1) follows from P(k), completing the proof.Sum of powers MATH 1510 Lili Shen n 2 2 X n (n + 1) 3 (4) Let P(n) denote the statement i = . Mathematical 4 Induction i=1 2 2 1 (1+ 1) 3 Step1. P(1) is true since 1 = . 4 Step2. Suppose P(k) is true, i.e., k 2 2 X k (k + 1) 3 i = : 4 i=1 We show that P(k + 1) is true, i.e., k+1 2 2 X (k + 1) (k + 2) 3 i = : 4 i=1Sum of powers MATH 1510 Lili Shen Indeed, Mathematical k+1 k Induction h i X X 3 3 3 i = i +(k + 1) i=1 i=1 2 2 k (k + 1) 3 = +(k + 1) 4 2 2 3 k (k + 1) + 4(k + 1) = 4 2 2 (k + 1) (k + 4k + 4) = 4 2 2 (k + 1) (k + 2) = : 4 Thus P(k + 1) follows from P(k), completing the proof.Proofs by mathematical induction MATH 1510 Lili Shen Mathematical Induction Example n Prove that 4n 2 for all n 5.Proofs by mathematical induction MATH 1510 Lili Shen Mathematical Induction Proof. n Let P(n) denote the statement 4n 2 . 5 Step1. P(5) is true since 4 5 = 20 32 = 2 . k Step2. Suppose P(k) is true, i.e., 4k 2 , where k 5. k+1 We show that P(k + 1) is true, i.e., 4(k + 1) 2 . Indeed, k k k+1 4(k + 1) = 4k + 4 4k + 4k 2 + 2 = 2 : Thus P(k + 1) follows from P(k), completing the proof.

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