* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download APPENDIX B EXERCISES In Exercises 1–8, use the
List of important publications in mathematics wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Mathematics and art wikipedia , lookup
Mathematical model wikipedia , lookup
History of mathematics wikipedia , lookup
Philosophy of mathematics wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Large numbers wikipedia , lookup
Infinitesimal wikipedia , lookup
Non-standard calculus wikipedia , lookup
Ethnomathematics wikipedia , lookup
Surreal number wikipedia , lookup
Brouwer–Hilbert controversy wikipedia , lookup
Peano axioms wikipedia , lookup
Foundations of mathematics wikipedia , lookup
Collatz conjecture wikipedia , lookup
Mathematical proof wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Real number wikipedia , lookup
Hyperreal number wikipedia , lookup
Non-standard analysis wikipedia , lookup
APPENDIX B EXERCISES In Exercises 1–8, use the principle of mathematical induction to establish the stated identity involving the integer n. 1. 12 + 2 2 + + n 2 = n(n + 1)(2n + 1) for n ≥ 1 . 6 2. 13 + 2 3 + + n 3 = n 2 (n + 1)2 for n ≥ 1 . 4 3. 2 + 2 2 + + 2 n = 2 n +1 − 2 for n ≥ 1 . 4. If r is a real number and r ≠ 1 , then 1 + r + r 2 + + r n = 5. 1 + 5 + 9 + + (4n + 1) = (n + 1)(2n + 1) for n ≥ 0 . 6. 1⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 + + n(n + 1)(n + 2) = 7. 1⋅1!+ 2 ⋅ 2!+ 3 ⋅ 3!+ + n ⋅ n! = (n + 1)!− 1 for n ≥ 1 . 8. 1 ⎞⎛ 1⎞ ⎛ 1 ⎞ n +1 ⎛ for n ≥ 2 . ⎜⎝ 1 − 2 ⎟⎠ ⎜⎝ 1 − 2 ⎟⎠ ⎜⎝ 1 − 2 ⎟⎠ = 2 3 n 2n r n +1 − 1 for n ≥ 0 . r −1 n(n + 1)(n + 2)(n + 3) for n ≥ 1 . 4 In Exercises 9–12, find a formula for the given expression and then prove that your formula is correct using the principle of mathematical induction. 9. 1 − 2 + 3 − 4 + + (2n − 1) − 2n 10. 1 − 4 + 9 − 16 + + (2n − 1)2 − (2n)2 11. 1 1 1 1 + + ++ 1⋅ 2 2 ⋅ 3 3 ⋅ 4 n(n + 1) 12. 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ ⎜⎝ 1 − ⎟⎠ ⎜⎝ 1 − ⎟⎠ ⎜⎝ 1 − ⎟⎠ 2 3 n In Exercises 13–18, use the principle of mathematical induction to prove the stated assertion involving the integer n. 13. 3n > 2 n +1 for n ≥ 1 . 14. n! > 2 n for n ≥ 4 . 15. If x is a real number, x > −1 and x ≠ 0 , then (1 + x)n > 1 + nx for n ≥ 2 . 16. n 3 − n is a multiple of 3 for n ≥ 1 . 17. 4 n − 1 is a multiple of 3 for n ≥ 1 . 18. 10 2n + 2 ⋅10 2n −1 + 1 is a multiple of 11 for n ≥ 1 . In Exercises 19 and 20, prove the stated property of exponents using the principle of mathematical induction. 19. For all real numbers a and b, (ab)n = a nb n for n ≥ 1 . 20. For all real numbers a and b, (a m )n = a mn for all nonnegative integers m and n. [Hint: Fix m and use induction on n ≥ 1 .] 21. Using the principle of mathematical induction, prove that for any integer n ≥ 1 and any n real numbers x1 , x2 , …, xn , x1 + + xn ≤ x1 + + xn . [Note: The case n = 2 is the Triangle Inequality for real numbers: x1 + x2 ≤ x1 + x2 .] 22. If we glue three unit squares together as shown below, we get an L-tromino. Consider a 2 n × 2 n checkerboard with one corner square removed. Using the principle of mathematical induction, prove that for any integer n ≥ 1 , such a checkerboard can always be completely covered by non-overlapping L-trominoes. (The L-trominoes can be rotated if necessary.) 23. Using the second principle of mathematical induction, prove that every positive integer can be written as a product of an odd integer and a power of 2. 24. Using the second principle of mathematical induction, prove that every positive integer can be written as a sum of distinct powers of 2. [Hint: Every positive integer lies between two consecutive powers of 2 (or is equal to one of these powers).] 25. A chemist has an unlimited supply of 3 gram and 5 gram weights that she uses on a 2-pan balance scale. Using the second principle of mathematical induction, prove that she can accurately weigh any amount greater than or equal to 8 grams. [Hint: k + 1 = (k − 2) + 3 ] 26. A unit fraction is the reciprocal 1 n of a positive integer n. An Egyptian fraction is any rational number that can be written as a sum of distinct unit fractions. For example, 2 1 1 = + 3 2 6 and 4 1 1 1 = + + 5 2 4 20 are Egyptian fractions. Prove that every rational number between 0 and 1 is an Egyptian fraction. [Hint: If 0 < n m < 1 , let 1 d be the largest unit fraction such that 1 d ≤ n m . Use the second principle of mathematical induction on the numerator and consider n m − 1 d for the induction step.] 27. A sequence x1 , x2 , …, xn of real numbers satisfies x1 = 3 , x2 = 5 , and xn +1 = 3xn − 2xn −1 for n ≥ 2 . Using the second principle of mathematical induction, prove that xn = 2 n + 1 for all integers n ≥ 1 . 28. A sequence x1 , x2 , …, xn of real numbers satisfies x1 = 2 , x2 = 3 , and xn +1 = xn xn −1 for n ≥ 2 . Using the second principle of mathematical induction, prove that xn + 6 = xn for all integers n ≥ 1 . In Exercises 29–32, find the flaw in the induction “proof”. 29. Let’s prove that n + 1 < n , where n is a positive integer. For n = k , we assume that k + 1 < k . Then for n = k + 1 we have (k + 1) + 1 < k + 1 . So the result follows by the principle of mathematical induction. 30. All real numbers are equal. More specifically, in any set of n real numbers, all numbers in the set are equal. Clearly if we have only one real number, it is equal to itself. Now assume that in any set of k real numbers, all numbers in the set are equal. Consider a set {x1 , x2 , …, xk , xk +1 } of k + 1 real numbers. By the induction hypothesis, x1 = x2 = = xk and also x2 = = xk = xk +1 . Hence x1 = x2 = = xk = xk +1 and so all of the numbers are equal. The result follows by the principle of mathematical induction. 31. For all positive integers n, if the maximum of the positive integers a and b is n, then a = b . Let max(a, b) denote the maximum of a and b . If max(a, b) = 1 , then a = b = 1 . Assume that if max(a, b) = k , then a = b . Now let max(a, b) = k + 1 . Then max(a − 1, b − 1) = k and so a − 1 = b − 1 by the induction hypothesis. Therefore, a = b . This completes the induction step and so the result is true for all integers n ≥ 1 . 32. If a is a nonzero real number, then a n = 1 for all n ≥ 0 . Certainly a 0 = 1 . Now assume that a k = 1 for all k ≤ n . Then we may write n + 1 = p + q for p, q ≤ n . Hence, a n +1 = a p + q = a p a q = 1⋅1 = 1 and so the result follows by the second principle of mathematical induction. 33. Professor Snarf’s linear algebra class meets five days a week, Monday through Friday. One Friday, he said to the class, “There will be a surprise quiz next week. You won’t know which day it will occur until the actual day of the quiz.” Annie Zog, one of the students in the class, whispered to Bert, the student sitting beside her, “Ha! That means we can’t have a quiz at all next week!” “What are you talking about?” replied Bert, perplexed. “Induction!” Annie announced. “I can prove by induction that Snarfie can’t give us a quiz next week.” “You’re kidding,” said Bert. “Convince me!” Annie continued, “Number the days next week like this: Monday=1, Tuesday=2, …, Friday=5. Let S(n) be the statement ‘There can’t be a quiz on day 5 − n .’ Now S(0) says ‘There can’t be a quiz on day 5.’ That’s Friday and if we haven’t had the quiz by Thursday, then we’ll know it’s on Friday and it won’t be a surprise. So S(0) is true. Next, assume that S(k) is true for 0 ≤ k ≤ 3 . That is, there can’t be a surprise quiz on days 5 − k through 5. But then day 5 − k − 1 is the latest possible day for the quiz and, if we haven’t already had the quiz by then, we’ll know it’s coming and it won’t be a surprise. So there can’t be a quiz on day 5 − k − 1 = 5 − (k + 1) either. That’s the induction step and so we can’t have a quiz any day next week!” “Beautiful!” exclaimed Bert. “You’ve caught Snarf at his own game! Now we don’t have to study for a quiz.” It certainly came as a surprise to Annie and Bert when the quiz happened on Wednesday! Can you find a flaw in Annie’s reasoning? Is there a flaw?