MAT 300 RECITATIONS WEEK 7 SOLUTIONS. Exercise #1. Use induction to prove that for every natural number n 4, n! > 2 n. 4! = 24 > 16 = 2 4 = 2 n

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1 MAT 300 RECITATIONS WEEK 7 SOLUTIONS LEADING TA: HAO LIU Exercise #1. Use induction to prove that for every natural number n 4, n! > 2 n. Proof. For any n N with n 4, let P (n) be the statement n! > 2 n. (i) For n = 4,. 4! = 24 > 16 = 2 4 = 2 n (ii) Let n N be such that n 4, and suppose n! > 2 n. Note that Then So P (n + 1) is true. n 4 n n + 1 > 2. (n + 1)! = (n + 1)(n)! > (n + 1) 2 n > 2 2 n = 2 n+1. Therefore, by mathematical induction, for every natural number n 4, n! > 2 n. Date: October 7, Key words and phrases. Induction, Critique of Proof. 1

2 2 LEADING TA: HAO LIU Exercise #2. Let r be a real number with r 1. r be any number in the set R \ {1}.Prove that for every positive integer n, r 0 + r 1 + r n = n+1. Proof. For all n N, let P (n) be the statement r 0 + r r n = n+1, for any real number r. (i) For n = 1, r 0 + r 1 = 1 + r = So P (1) is true. ()(1 + r) = r. (ii) Let n N. Assume P (n) is true. Then, for any real number r, r 0 + r r n = n+1. We must show P (n + 1) is true. We can compute r 0 + r r n + r n+1 = n+1 So P (n + 1) is true. + r n+1 = n+1 + r n+1 () = n+1 + r n+ r n+1 = rn+1 = n+2. By mathematical induction, we have shown that for every r R with r 1 and for all positive integers n, r 0 + r 1 + r n = n+1..

3 MAT 300 RECITATIONS WEEK 7 SOLUTIONS 3 Exercise #3. Prove that for all n N, k 3 = 1 4 n2 (n + 1) 2. Proof. (i) For n = 1, k 3 = 1 k 3 = 1 3 = = (ii) Let n N. Suppose k 3 = 1 4 n2 (n + 1) 2. Then, n+1 k 3 = k 3 + (n + 1) 3 = 1 4 n2 (n + 1) 2 + (n + 1) 3 = 1 4 (n4 + 2n 3 + n 2 ) + (n 3 + 3n 2 + 3n + 1) = 1 4 (n4 + 2n 3 + n 2 ) (4n3 + 12n n + 4) = 1 4 (n4 + 6n n n + 4) = 1 4 (n + 1)(n3 + 5n 2 + 8n + 4) = 1 4 (n + 1)2 (n 2 + 4n + 4) = 1 4 (n + 1)2 (n + 2) 2 Therefore, by mathematical induction, for any n N, k 3 = 1 4 n2 (n + 1) 2.

4 4 LEADING TA: HAO LIU Exercise #4. Determine if the proof below is a legitimate proof. If not, identify what is wrong with it. If the original claim is true, prove it correctly. Recall: An integer a is even if there exists b Z such that a = 2 b. Theorem: For each n N, n 2 + 7n + 3 is an even integer. Proof. Let P (n) be the statement n 2 + 7n + 3 is an even integer. Suppose that P (n) is true for some n N. Then there exists m Z such that n 2 + 7n + 3 = 2m. Let k = n + 4 Then k Z since n N hence n Z and since Z is closed under addition. Then (n + 1) 2 + 7(n + 1) + 3 = (n 2 + 2n + 1) + 7n = (n 2 + 7n + 3) + 2(n + 4) = 2m + 2k = 2(m + k). m, k Z, m + k Z. Thus P (n + 1) is true. Therefore, P (n) is true for all n N. Solution: This is not a legitimate proof. In order to use mathematical induction, it is necessary to consider the initial step. This proof demonstrates how important that initial step is. Since it was not included here, the student was able to conclude that for each n N, n 2 + 7n + 3 is an even integer. However, this is a false statement. Indeed, if we consider n = 1, we see that n 2 + 7n + 3 = = 11, which is an odd number. This provides a counterexample to the claim and disproves the theorem. Another mistake was not stating what n is when writing Let P (n) be the statement n 2 +7n+3 is an even integer. This should say For any n N, let P (n) be the statement n 2 + 7n + 3 is an even integer. Exercise #5. Determine if the proof below is a legitimate proof. If not, identify what is wrong with it. If the original claim is true, prove it correctly. Theorem: For any n, m N {0} with m 0, m + n 0. Proof. We argue by mathematical induction on n. Let m N {0} be such that m 0. For any n N {0} let P (n) be the statement

5 MAT 300 RECITATIONS WEEK 7 SOLUTIONS 5 m + n 0. (i) Let n = 0. Since we defined m to be non-zero, And so P (0) is true. m + n = m + 0 = m 0. (ii) Let n N {0} be arbitrary. Suppose P (n) is true and P (n+1) is not true. Because n was chosen arbitrarily to be such that P (n) is true, and P (0) is true by the initial step, we can let n = 0. Then, hence m + 1 = 0. Since m N, 0 = m + (n + 1) = m + (n + 1) = (m + n) + 1 = (m + 0) + 1 = m + 1, m 1 m m Hence m and m + 1 = 0. This is a contradiction. Thus P (n + 1) is true. Therefore, by the mathematical induction, for any n, m N {0} with m 0, m + n 0. Solution: This is not a legitimate proof. In the second step of the proof, he states Let n N {0} be arbitrary. Suppose P (n) is true and P (n + 1) is not true. However, he then proceeds to choose n = 0. Since n was arbitrary, it is not possible to then choose it. The theorem is true, so we prove it below: Proof. We argue by mathematical induction on n. Let m N {0} be such that m 0. For any n N {0} let P (n) be the statement m + n 0. (i) Let n = 0. Since we defined m to be non-zero, And so P (0) is true. m + n = m + 0 = m 0.

6 6 LEADING TA: HAO LIU (ii) Let n N {0} be arbitrary. Suppose P (n) is true and that P (n + 1) is not true. Then 0 = m + (n + 1) = (m + n) + 1. Hence, m+n = 1, and so m+n < 0. However, m, n N {0}, therefore, m + n 0. Then, m + n < 0 and m + n 0. This is a contradiction. Thus, P (n + 1) is true. Therefore, by mathematical induction, for any n, m N {0} with m 0, m + n 0. Exercise #6. Determine if the proof below is a legitimate proof. If not, identify what is wrong with it. If the original claim is true, prove it correctly. Theorem: All horses are the same color. Proof. For all n N, let P (n) be the statement Any collection of n horses are the same color. P (1) is true since any collection of 1 horse is the same color. Suppose P (n) is true. Let S be a collection of n + 1 horses. Call one of the horses Charlie. Consider the collection Let S = S \ {Charlie} Then S is a set of n horses. Since P (n) is true by the assumption, they are all the same color, say color C. Now, add Charlie back into the set and remove any other horse. Now this new set also has n horses so all of them must be the same color. Since Charlie is now included in the set, Charlie must be the same color as the rest. Thus, Charlie must also be colored C. Thus, every horse in the collection S is the same color, so P(n+1) is true. Therefore, by the mathematical induction all horses are the same color. Solution: This is not a legitimate proof. The flaw in the logic occurs when we have only 2 horses. At this point, removing Charlie leaves only 1 other horse. When we add Charlie back into the set and remove the other horse, we state that Charlie must be the same color as the rest. However, there is nothing else in this new set.