Chapter 1 Tools of Geometry Sec 1 1

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Chapter 1 Tools of Geometry Sec 1 – 1 Patterns and Inductive Reasoning Objectives:

Chapter 1 Tools of Geometry Sec 1 – 1 Patterns and Inductive Reasoning Objectives: 1) To use inductive reasonings to make conjectures.

Inductive Reasoning – Is reasoning that is based on patterns you observe. – If

Inductive Reasoning – Is reasoning that is based on patterns you observe. – If you observe a pattern in a sequence you can use inductive reasoning to find the next term. Ex. 1: Find the next term in the sequence: 48 96 Rule: x 2 A) 3, 6, 12, 24, ___ Rule: +1, +2, +3, +4, … B) 1, 2, 4, 7, 11, 16, 22, ___ 29 37 C)

Inductive Reasoning assumes that an observed pattern will continue. This may or may not

Inductive Reasoning assumes that an observed pattern will continue. This may or may not be true. – Ex: x = x • x – This is true only for x = 0 and x = 1 Conjecture – A conclusion you reach using inductive reasoning.

Ex. 2: Make a conjecture about the sum of the first 30 odd numbers.

Ex. 2: Make a conjecture about the sum of the first 30 odd numbers. 1 1 + 3 + 5 + 7 + 9 1 + 3 + 5 +. . . + 30 = 12 = 4 = 22 = 9 = 32 = 16 = 42 = 25 = 52 = 900 = 302

To show that a conjecture is always true, you must prove it. To show

To show that a conjecture is always true, you must prove it. To show that a conjecture is false, you have to find only one example in which the conjecture is not true. This case is called a counterexample. A counterexample can be a drawing, a statement, or a number.

Counter Example – To a conjecture is an example for which the conjecture is

Counter Example – To a conjecture is an example for which the conjecture is incorrect. Ex. 1 -3: The first 3 odd prime numbers are 3, 5, 7. Make a conjecture about the 4 th. 11 – 3, 5, 7, ___ – One would think that the rule is add 2, but that gives us 9 for the fourth prime number. Is that true? – What is the next odd prime number?

GEOMETRY LESSON 1 -1 The price of overnight shipping was $8. 00 in 2000,

GEOMETRY LESSON 1 -1 The price of overnight shipping was $8. 00 in 2000, $9. 50 in 2001, and $11. 00 in 2002. Make a conjecture about the price in 2003. Write the data in a table. Find a pattern. 2000 2001 2002 $8. 00 $9. 50 $11. 00 Each year the price increased by $1. 50. A possible conjecture is that the price in 2003 will increase by $1. 50. If so, the price in 2003 would be $11. 00 + $1. 50 = $12. 50. 1 -1

Inductive Reasoning 1. Look for a pattern. 2. Make a conjecture. 3. Prove the

Inductive Reasoning 1. Look for a pattern. 2. Make a conjecture. 3. Prove the conjecture or find a counterexample.

Homework Page 6 1 -53 odd and 2, 26, 40, 44, 48

Homework Page 6 1 -53 odd and 2, 26, 40, 44, 48

GEOMETRY LESSON 1 -1 Pages 6– 9 Exercises 1. 80, 160 1 1 5

GEOMETRY LESSON 1 -1 Pages 6– 9 Exercises 1. 80, 160 1 1 5 6 12. , 2. 33, 333; 333, 333 13. James, John 3. – 3, 4 14. Elizabeth, Louisa 4. 15. Andrew, Ulysses 1 1 16 32 , 5. 3, 0 16. Gemini, Cancer 6. 17. 1 1, 3 7. N, T 8. J, J 19. The sum of the first 6 pos. even numbers is 6 • 7, or 42. 20. The sum of the first 30 pos even numbers is 30 • 31, or 930. 21. The sum of the first 100 pos. even numbers is 100 • 101, or 10, 100. 18. 9. 720, 5040 10. 64, 128 1 1 36 49 11. , 1 -

GEOMETRY LESSON 1 -1 22. The sum of the first 100 odd numbers is

GEOMETRY LESSON 1 -1 22. The sum of the first 100 odd numbers is 1002, or 10, 000. 28. ÷ = and is 1 2 1 3 3 2 improper. 29. 75°F 23. 555, 555 24. 123, 454, 321 25– 28. Answers may vary. Samples are given. 25. 8 + (– 5 = 3) and 3 > 8 / 26. 1 1 1 • > and • > / 3 2 27. – 6 – (– 4) < – 6 and – 6 – (– 4) < – 4 31. 31, 43 32. 10, 13 33. 0. 0001, 0. 00001 30. 40 push-ups; answers may vary. Sample: Not very confident, Dino may reach a limit to the number of push-ups he can do in his allotted time for exercises. 1 -1 34. 201, 202 35. 63, 127 31 63 32 64 36. , 37. J, S 38. CA, CO 39. B, C

GEOMETRY LESSON 1 -1 40. Answers may vary. Sample: In Exercise 31, each number

GEOMETRY LESSON 1 -1 40. Answers may vary. Sample: In Exercise 31, each number increases by increasing multiples of 2. In Exercise 33, to get the next term, divide by 10. 42. 43. 44. 41. You would get a third line between and parallel to the first two lines. 45. 46. 102 cm 1 -1 47. Answers may vary. Samples are given. a. Women may soon outrun men in running competitions. b. The conclusion was based on continuing the trend shown in past records. c. The conclusions are based on fairly recent records for women, and those rates of improvement may not continue. The conclusion about the marathon is most suspect because records date only from 1955.

GEOMETRY LESSON 1 -1 48. a. b. about 12, 000 radio stations in 2010

GEOMETRY LESSON 1 -1 48. a. b. about 12, 000 radio stations in 2010 c. Answers may vary. Sample: Confident; the pattern has held for several decades. 49. Answers may vary. Sample: 1, 3, 9, 27, 81, . . . 1, 3, 5, 7, 9, . . . 50. His conjecture is 52. 21, 34, 55 probably false because most 53. people’s growth slows by 18 until they stop growing somewhere between 18 and 22 years. 51. a. b. H and I c. a circle a. Leap years are years that are divisible by 4. b. 2020, 2100, and 2400 c. Leap years are years divisible by 4, except the final year of a century which must be divisible by 400. So, 2100 will not be a leap year, but 2400 will be.

GEOMETRY LESSON 1 -1 54. Answers may vary. Sample: 55. (continued) d. 100 +

GEOMETRY LESSON 1 -1 54. Answers may vary. Sample: 55. (continued) d. 100 + 99 + 98 + … + 3 + 2 + 1 1 + 2 + 3 + … + 98 + 99 + 100 101 + … + 101 56. B The sum of the first 100 numbers is 57. I 100 • 101 , or 5050. 2 The sum of the first n numbers is n(n+1). 2 55. a. 1, 3, 6, 10, 15, 21 b. They are the same. c. The diagram shows the product of n and n + 1 divided by 2 when n = 3. The result is 6. 1 -1 58. [2] a. 25, 36, 49 b. n 2 [1] one part correct

GEOMETRY LESSON 1 -1 59. (continued) 59. [4] a. The product of 11 60

GEOMETRY LESSON 1 -1 59. (continued) 59. [4] a. The product of 11 60 -67. and a three-digit [3] minor error in number that begins explanation and ends in 1 is a four-digit number [2] incorrect description that begins and ends in part (a) in 1 and has middle digits that are each [1] correct products for one greater than the (151)(11), (161)(11), middle digit of the and (181)(11) three-digit number. (151)(11) = 1661 (161)(11) = 1771 68. B b. 1991 69. N c. No; (191)(11) = 2101 70. G