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### CMSC 56 | Lecture 11: Mathematical Induction

• 1. Mathematical Induction Lecture 11, CMSC 56 Allyn Joy D. Calcaben
• 2. Find the sum and the product of each pairs of numbers. Express your answers as a binary expansion. 1. (100 0111)2 , (111 0111)2 2. (1110 1111)2 , (1011 1101)2 3. (10 1010 1010)2 , (1 1111 0000)2 Assignment (1 whole Sheet Paper)
• 3. Suppose that we have an infinite ladder, as shown in the figure, and we want to know whether we can reach every step on this ladder. We know two things: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung.
• 4. Can we conclude that we are able to reach every rung of this infinite ladder?
• 5. The answer is yes, something we can verify using an important proof technique called mathematical induction.
• 6. To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps: Principle of Mathematical Induction
• 7. To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps: Basis Step: We verify that P(1) is true. Inductive Step: We show that the conditional statement P(k) → P(k + 1) is true for all positive integers k. Principle of Mathematical Induction
• 8. The assumption that P(k) is true is called the inductive hypothesis. When we use mathematical induction to prove a theorem, we first show that P(1) is true. Then we know that P(2) is true, because P(1) → P(2). Further, we know that P(3) is true, P(2) → P(3). Continuing along these lines, we see that P(n) is true for every positive integer n. Principle of Mathematical Induction
• 9. 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we reach the next rung. Infinite Ladder
• 10. 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we reach the next rung. Basis Step: Verify that P(1) is true. Inductive Step: P(k) → P(k + 1) is true for all positive integers k Therefore, P(n) is true for all positive integers n, where P(n) is the statement that we can reach the nth rung of the ladder. Consequently, we can invoke mathematical induction to conclude that we can reach every rung. Infinite Ladder
• 12. “Domino Effect” Step 1. The first domino falls Step 2. When any domino falls, the next domino falls
• 13. “Domino Effect” Step 1. The first domino falls Step 2. When any domino falls, the next domino falls ….. all dominos will fall! This is how Mathematical Induction works.
• 14. Example There are infinitely many stations on a train route. Suppose that the train stops at the first station and suppose that if the train stops at a station, then is stops at the next station. Show that the train stops at all stations.
• 15. Solution Let P(n) be the statement that the train stops at station n. Basis Step: We are told that P(1) is true. Inductive Step: We are told that P(k) implies P(k+1) for each n ≥ 1. Therefore by the principle of the mathematical induction, P(n) is true for all positive integers n.
• 16. Proving Summation Formulae Mathematical Induction is particularly well suited for proving that such formulae are valid. However, summation formulae can be proven in other ways. major disadvantage of using mathematical induction: 1. you cannot use it to derive this formula.
• 17. Example Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2
• 18. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Let P (n) be the proposition that the sum of the ﬁrst n positive integers, 1 + 2 + ··· n = 𝑛 (𝑛+1) 2 , is 𝑛 (𝑛+1) 2 . We must do two things to prove that P (n) is true for n = 1, 2, 3,.... Namely, we must show that P(1) is true and that the conditional statement P(k)implies P(k+1) is true for k = 1, 2, 3,....
• 19. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2
• 20. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: We show that the conditional statement P(k) → P(k + 1) is true for all positive integers k.
• 21. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: First, assume P(k) is true for all positive integers k.
• 22. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: First, assume P(k) is true for all positive integers k. (Inductive Hypothesis)
• 23. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Inductive Hypothesis: 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
• 24. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true.
• 25. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true. 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘+1 [(𝑘+1)+1] 2 = 𝑘+1 (𝑘+2) 2
• 26. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
• 27. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
• 28. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 (𝑘+1) 2
• 29. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Right Side: P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 (𝑘+1) 2 + (𝑘 + 1)
• 30. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 𝑘+1 + 2 𝑘+1 2
• 31. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘+1 𝑘+2 2
• 32. Example Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛.
• 33. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Let P(𝑛) be the proposition that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for the integer 𝑛.
• 34. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1
• 35. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k.
• 36. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k. (Inductive Hypothesis)
• 37. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Inductive Hypothesis: 1 + 2 + 22 + . . . + 2 𝑘 = 2 (𝑘 + 1) − 1
• 38. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true.
• 39. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true. 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) + 1 −1 = 2 (𝑘 + 2) −1
• 40. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 = 2 (𝑘 + 1) − 1
• 41. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) − 1
• 42. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) Right Side: P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) − 1 + 2 (𝑘 + 1)
• 43. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 ⋅ 2 (𝑘 + 1) − 1
• 44. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 2) − 1
• 45. 1. Prove that 12 + 32 + 52 + . . . + (2𝑛 + 1) 2 = ( 𝑛 + 1) (2 𝑛 + 1)(2 𝑛 + 3) 3 whenever 𝑛 is a nonnegative integer. 2. Prove that 3 + 3 ⋅ 5 + 3 ⋅ 52 + . . . + 3 ⋅ 5n = 3(5 𝑛 +1 − 1) 4 whenever 𝑛 is a nonnegative integer. Challenge (1/2 Sheet Paper)

### Editor's Notes

1. That is, we can show that P (n) is true for every positive integer n, where P (n) is the statement that we can reach the nth rung of the ladder.
2. We assume that P(k)is true for an arbitrary positive integer k and show that under this assumption, P(k+1) must also be true.
3. (The left-hand side of this equation is 1 because 1 is the sum of the ﬁrst positive integer. The right-hand side is found by substituting 1 for n in n(n+1)/2.)
4. This last equation shows that P(k + 1) is true under the assumption that P(k) is true. This completes the inductive step. We have completed the basis step and the inductive step, so by mathematical induction we know that P (n) is true for all positive integers n. That is, we have proven that 1 + 2 +···+n = n(n+1)/2 for all positive integers n.
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