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/07 4:40 PM Page 44 44 Logic A compound sentence may contain both negations and conjunctions at the same time. For example: Let p represent “Ten is divisible by 2.” Let q represent “Ten is divisible by 3.” Then p ∧ q represents “Ten is divisible by 2 and ten is not divisible by 3.” Here p is true, q is false, and q is true. Then for p ∧ q, both parts are true so the conjunction is true. This can be summarized in the following table EXAMPLE 1 Let p represent “Albany is the capital of New York State.” (True) Let q represent “Philadelphia is the capital of Pennsylvania.” (False) For each given sentence: a. Write the sentence in symbolic b. Tell whether the statement is form. true or false. (1) Albany is the capital of New York State and Philadelphia is the capital of Pennsylvania. (2) Albany is the capital of New York State and Philadelphia is not the capital of Pennsylvania. (3) Albany is not the capital of New York State and Philadelphia is the capital of Pennsylvania. (4) Albany is not the capital of New York State and Philadelphia is not the capital of Pennsylvania. (5) It is not true that Albany is the capital of New York State and Philadelphia is the capital of Pennsylvania. Solution (1) The statement is a conjunction. Since p is true and q is false, the statement is false. (2) The statement is a conjunction. Since p is true and q is true, the statement is true Answers: a. p ∧ q b. False Answers: a. p ∧ q b. True 14365C02.pgs 7/9/07 4:40 PM Page 45 Conjunctions 45 (3) The statement is a conjunction. Since p is false and q is false, the statement is false. (4) The statement is a conjunction. Since p is false and q is true, the statement is false Answers: a. p ∧ q b. False Answers: a. p ∧ q b. False (5) The phrase “It is not true that” applies to the entire conjunction. Since p is true and q is false, (p ∧ q) is false and the negation of (p ∧ q) is true. p T q F p ∧ q (p � |
� q) F T Answers: a. (p ∧ q) b. True EXAMPLE 2 Solution Use the domain {1, 2, 3, 4} to find the truth set for the open sentence (x 3) ∧ (x is a prime) Let x 1 Let x 2 Let x 3 Let x 4 (1 3) ∧ (1 is a prime) (2 3) ∧ (2 is a prime) (3 3) ∧ (3 is a prime) (4 3) ∧ (4 is a prime) T ∧ F False T ∧ T True F ∧ T False F ∧ F False A conjunction is true only when both simple sentences are true. This condition is met here when x 2. Thus, the truth set or solution set is {2}. Answer {2} EXAMPLE 3 Three sentences are written below. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. Today is Friday and I have soccer practice. Today is Friday. I have soccer practice. (False) (True) (?) Solution Since the conjunction is false, at least one of the conjuncts must be false. But “Today is Friday” is true. Therefore, “I have soccer practice” must be false. Answer “I have soccer practice” is false. 14365C02.pgs 7/9/07 4:40 PM Page 46 46 Logic EXAMPLE 4 Three sentences are written below. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. Today is Monday and the sun is shining. Today is Monday. The sun is shining. (False) (False) (?) Solution (1) Use symbols to represent the sentences. Indicate their truth values. p: Today is Monday. q: The sun is shining. p ∧ q: Today is Monday and the sun is shining. (False) (?) (False) (2) Construct a truth for the conjunction. Study the truth values of p and p ∧ q. Since p ∧ q is false, the last three rows apply. Since p is false, the choices are narrowed to the last two rows. Therefore q could be either true or false Answer The truth value of “The sun is shining” is uncertain. Exercises |
Writing About Mathematics 1. Is the negation of a conjunction, (p ∧ q), the same as p ∧ q? Justify your answer. 2. What must be the truth values of p, q, and r in order for (p ∧ q) ∧ r to be true? Explain your answer. Developing Skills In 3–12, write each sentence in symbolic form, using the given symbols. Let p represent “It is hot.” Let q represent “It is raining.” Let r represent “The sky is cloudy.” 3. It is hot and it is raining. 4. It is hot and the sky is cloudy. 5. It is not hot. 6. It is not hot and the sky is cloudy. 7. It is raining and the sky is not cloudy. 8. It is not hot and it is not raining. 14365C02.pgs 7/9/07 4:40 PM Page 47 9. The sky is not cloudy and it is not hot. 10. The sky is not cloudy and it is hot. 11. It is not the case that it is hot and it is raining. 12. It is not the case that it is raining and it is not hot. In 13–20, using the truth value for each given statement, tell if the conjunction is true or false. Conjunctions 47 A piano is a percussion instrument. A piano has 88 keys. A flute is a percussion instrument. A trumpet is a brass instrument. (True) (True) (False) (True) 13. A flute is a percussion instrument and a piano is a percussion instrument. 14. A flute is a percussion instrument and a trumpet is a brass instrument. 15. A piano has 88 keys and is a percussion instrument. 16. A piano has 88 keys and a trumpet is a brass instrument. 17. A piano is not a percussion instrument and a piano does not have 88 keys. 18. A trumpet is not a brass instrument and a piano is a percussion instrument. 19. A flute is not a percussion instrument and a trumpet is a brass instrument. 20. It is not true that a piano is a percussion instrument and has 88 keys. In 21–28, complete each sentence with “true” or “false” to make a correct statement. 21. When p is true and q is true, then p ∧ q is _______. 22. When p |
is false, then p ∧ q is _______. 23. If p is true, or q is true, but not both, then p ∧ q is _______. 24. When p ∧ q is true, then p is ______ and q is ______. 25. When p ∧ q is true, then p is ______ and q is ______. 26. When p ∧ q is true, then p is ______ and q is ______. 27. When p is false and q is true, then (p ∧ q) is _______. 28. If both p and q are false, then p ∧ q is _______. Applying Skills In 29–36, three sentences are written. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 29. It is noon and I get lunch. (True) 30. I have the hiccups and I drink some It is noon. (True) I get lunch. (?) water. (False) I have the hiccups. (True) I drink some water. (?) 14365C02.pgs 7/9/07 4:40 PM Page 48 48 Logic 31. I have the hiccups and I drink some 32. I play tennis and Anna plays golf. water. (False) I have the hiccups. (False) I drink some water. (?) (False) I play tennis. (True) Anna plays golf. (?) 33. Pam sees a movie and Pam loves going 34. Jon and Edith like to eat ice cream. to the theater. (True) Pam sees a movie. (True) Pam loves going to the theater. (?) (False) Jon likes to eat ice cream. (True) Edith likes to eat ice cream. (?) 35. Jordan builds model trains and model 36. Bethany likes to chat and surf on the planes. (False) Jordan builds model trains. (False) Jordan builds model planes. (?) internet. (False) Bethany likes to surf on the internet. (True) Bethany likes to chat on the internet. (?) In 37 and 38, a compound sentence is given using a conjunction. Use the truth value of the compound sentence to determine whether each sentence that follows is true or false. 37. In winter I wear a hat and scarf. (True) 38. I do not practice and I know |
that I a. In winter I wear a hat. b. In winter I wear a scarf. c. In winter I do not wear a hat. should. (True) a. I do not practice. b. I know that I should practice. c. I practice. 2-3 DISJUNCTIONS In logic, a disjunction is a compound statement formed by combining two simple statements using the word or. Each of the simple statements is called a disjunct. When p and q represent simple statements, the disjunction p or q is written in symbols as p ∨ q. For example: p: Andy rides his bicycle to school. q: Andy walks to school. p ∨ q: Andy rides his bicycle to school or Andy walks to school. In this example, when is the disjunction p ∨ q true and when is it false? 1. On Monday, Andy rode his bicycle part of the way to school when he met a friend. Then he walked the rest of the way to school with his friend. Here p is true and q is true. The disjunction p ∨ q is true. 2. On Tuesday, Andy rode his bicycle to school and did not walk to school. Here p is true and q is false. The disjunction p ∨ q is true. 3. On Wednesday, Andy did not ride his bicycle to school and walked to school. Here p is false and q is true. The disjunction p ∨ q is true. 14365C02.pgs 7/9/07 4:40 PM Page 49 Disjunctions 49 4. On Thursday, it rained so Andy’s father drove him to school. Andy did not ride his bicycle to school and did not walk to school. Here p is false and q is false. The disjunction p ∨ q is false. The disjunction p or q is true when any part of the compound sentence is true: p is true, q is true, or both p and q are true. The only case in which the disjunction p or qis false is when both p and q are false. The truth values of the disjunction p ∨ q are summarized in the truth table to the right. The possible combinations of the truth values of p and of q, shown in the first two columns, are the same as those used in the truth table of the conjunction. The third column gives the truth values for the |
disjunction, p ∨ q EXAMPLE 1 Use the following statements: Let k represent “Kurt plays baseball.” Let a represent “Alicia plays baseball.” Let n represent “Nathan plays soccer.” Write each given sentence in symbolic form. a. Kurt or Alicia play baseball. b. Kurt plays baseball or Nathan plays soccer. c. Alicia plays baseball or Alicia does not play baseball. d. It is not true that Kurt or Alicia play baseball. e. Either Kurt does not play baseball or Alicia does not play baseball. f. It’s not the case that Alicia or Kurt play baseball. Answers k ∨ a) k ∨ a (a ∨ k) EXAMPLE 2 Symbols are used to represent three statements. For each statement, the truth value is noted. k: “Every line segment has a midpoint.” m: “A line has a midpoint.” q: “A ray has one endpoint.” (True) (False) (True) For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the statement is true or false. 14365C02.pgs 7/9/07 4:40 PM Page 50 50 Logic (1) k ∨ q (2) k ∨ m Answers a. Every line segment has a midpoint or every ray has one endpoint. b. T ∨ T is a true disjunction. a. Every line segment has a midpoint or a line has a midpoint. b. T ∨ F is a true disjunction. (3) m ∨ q a. A line has a midpoint or a ray does not have one endpoint. b. F ∨ T F ∨ F, a false disjunction. (4) (m ∨ q) a. It is not the case that a line has a midpoint or a ray has one endpoint. b. (F ∨ T) T, a false disjunction. EXAMPLE 3 Find the solution set of each of the following if the domain is the set of positive integers less than 8. a. (x 4) ∨ (x 3) b. (x 3) ∨ (x is odd) c. (x 5) ∧ (x 3) Solution The domain is {1, 2, 3 |
, 4, 5, 6, 7}. a. The solution set of x 4 is {1, 2, 3} and the solution set of x 3 is {4, 5, 6, 7}. The solution set of the disjunction (x 4) ∨ (x 3) includes all the numbers that make x 4 true together with all the numbers that make x 3 true. Answer {1, 2, 3, 4, 5, 6, 7} Note: The solution set of the disjunction (x 4) ∨ (x 3) is the union of the solution sets of the individual parts: {1, 2, 3} {4, 5, 6, 7} {1, 2, 3, 4, 5, 6, 7}. b. The solution set of (x 3) is {4, 5, 6, 7} and the solution set of (x is odd) is {1, 3, 5, 7}. The solution set of the disjunction (x 3) ∨ (x is odd) includes all the numbers that make either x 3 true or x is odd true. Answer {1, 3, 4, 5, 6, 7} Note: The solution set of the disjunction (x 3) ∨ (x is odd) is the union of the solution sets: {4, 5, 6, 7} {1, 3, 5, 7} {1, 3, 4, 5, 6, 7}. c. The solution set of (x 5) is {6, 7} and the solution set of (x 3) is {1, 2}. The solution set of (x 5) ∧ (x 3) is {6, 7} {1, 2} or the empty set,. Answer 14365C02.pgs 7/9/07 4:40 PM Page 51 Disjunctions 51 Two Uses of the Word Or When we use the word or to mean that one or both of the simple sentences are true, we call this the inclusive or. The truth table we have just shown uses truth values for the inclusive or. Sometimes, however, the word or is used in a different way, as in “He is in grade 9 or he is in grade 10.” Here it is not possible for both simple sentences to be true at the same time. When we use the word or to mean that one and only one of the simple sentences is true, we call |
this the exclusive or. The truth table for the exclusive or will be different from the table shown for disjunction. In the exclusive or, the disjunction p or qwill be true when p is true, or when q is true, but not both. In everyday conversation, it is often evident from the context which of these uses of or is intended. In legal documents or when ambiguity can cause difficulties, the inclusive or is sometimes written as and/or. We will use only the inclusive or in this book. Whenever we speak of disjunction, p or q will be true when p is true, when q is true, when both p and q are true. Exercises Writing About Mathematics 1. Explain the relationship between the truth set of the negation of a statement and the com- plement of a set. 2. Explain the difference between the inclusive or and the exclusive or. Developing Skills In 3–12, for each given statement: a. Write the statement in symbolic form, using the symbols given below. b. Tell whether the statement is true or false. Let c represent “A gram is 100 centigrams.” Let m represent “A gram is 1,000 milligrams.” Let k represent “A kilogram is 1,000 grams.” Let l represent “A gram is a measure of length.” (True) (True) (True) (False) 3. A gram is 1,000 milligrams or a kilogram is 1,000 grams. 4. A gram is 100 centigrams or a gram is a measure of length. 5. A gram is 100 centigrams or 1,000 milligrams. 6. A kilogram is not 1,000 grams or a gram is not 100 centigrams. 7. A gram is a measure of length or a kilogram is 1,000 grams. 14365C02.pgs 7/9/07 4:40 PM Page 52 52 Logic 8. A gram is a measure of length and a gram is 100 centigrams. 9. It is not the case that a gram is 100 centigrams or 1,000 milligrams. 10. It is false that a kilogram is not 1,000 grams or a gram is a measure of length. 11. A gram is 100 centigrams and a kilogram is 1,000 grams. 12. A gram is not 100 centigrams or is not |
1,000 milligrams, and a gram is a measure of length. In 13–20, symbols are assigned to represent sentences. Let b represent “Breakfast is a meal.” Let s represent “Spring is a season.” Let h represent “Halloween is a season.” For each sentence given in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the sentence is true or false. 13. s ∨ h 17. b ∨ s 14. b ∧ s 18. (s ∧ h) 15. s ∨ h 19. (b ∨ s) 16. b ∧ h 20. b ∧ s In 21–27, complete each sentence with the words “true” or “false” to make a correct statement. 21. When p is true, then p ∨ q is ______. 22. When q is true, then p ∨ q is ______. 23. When p is false and q is false, then p ∨ q is _______. 24. When p ∨ q is false, then p is _______ and q is _______. 25. When p ∨ q is false, then p is _______ and q is _______. 26. When p is false and q is true, then (p ∨ q) is _______. 27. When p is false and q is true, then p ∨ q is ______. Applying Skills In 28–32, three sentences are written. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 28. May is the first month of the year. (False) January is the first month of the year. (True) May is the first month of the year or January is the first month of the year. (?) 29. I will study more or I will fail the course. (True) I will fail the course. (False) I will study more. (?) 30. Jen likes to play baseball and Mason likes to play baseball. (False) Mason likes to play baseball. (True) Jen likes to play baseball. (?) 14365C02.pgs 7/9/07 4:40 PM Page 53 Conditionals 53 31. Nicolette is my friend or Michelle is my friend. (True) Nicolette is my friend. |
(True) Michelle is my friend. (?) 32. I practice the cello on Monday or I practice the piano on Monday. (True) I do not practice the piano on Monday. (False) I practice the cello on Monday. (?) 2-4 CONDITIONALS A sentence such as “If I have finished my homework, then I will go to the movies” is frequently used in daily conversation. This statement is made up of two simple statements: p: I have finished my homework. q: I will go to the movies. The remaining words, if... then, are the connectives. In English, this sentence is called a complex sentence. In mathematics, however, all sentences formed using connectives are called compound sentences or compound statements. In logic, a conditional is a compound statement formed by using the words if... then to combine two simple statements. When p and q represent simple statements, the conditional if p then q is written in symbols as p → q. The symbol p → q can also be read as “p implies q” or as “p only if q.” Here is another example: p: It is January. q: It is winter. p → q: If it is January, then it is winter. or It is January implies that it is winter. or It is January only if it is winter. Certainly we would agree that the compound sentence if p then q is true for this example: “If it is January, then it is winter.” However, if we reverse the order of the simple sentences to form the new conditional if q then p, we will get a sentence with a different meaning: q → p: If it is winter, then it is January. When it is winter, it does not necessarily mean that it is January. It may be February, the last days of December, or the first days of March. Changing the order in which we connect two simple statements in conditional does not always give a conditional that has the same truth value as the original. 14365C02.pgs 7/9/07 4:40 PM Page 54 54 Logic Parts of a Conditional The parts of the conditional if p then q can be identified by name: p is the hypothesis, which is sometimes referred to as the premise or the antecedent. It is an assertion or a sentence that begins an argument. The hypothesis usually follows the word if. q is the conclusion, which is |
sometimes referred to as the consequent. It is the part of a sentence that closes an argument. The conclusion usually follows the word then. There are different ways to write the conditional. When the conditional uses the word if, the hypothesis always follows if. When the conditional uses the word implies, the hypothesis always comes before implies. When the conditional uses the words only if, the conclusion follows the words only if. p → q: If it is January, then it is winter. d d hypothesis conclusion p → q: It is January implies that it is winter. d d hypothesis conclusion p → q: It is January only if it is winter. d d hypothesis conclusion All three sentences say the same thing. We are able to draw a conclusion about the season when we know that the month is January. Although the word order of a conditional may vary, the hypothesis is always written first when using symbols. Truth Values for the Conditional p → q In order to determine the truth value of a conditional, we will consider the statement “If you get an A in Geometry, then I will buy you a new graphing calculator.” Let p represent the hypothesis, and let q represent the conclusion. p: You get an A in Geometry. q: I will buy you a new graphing calculator. Determine the truth values of the conditional by considering all possible combinations of the truth values for p and q. CASE 1 You get an A in Geometry. (p is true.) I buy you a new graphing calculator. (q is true.) We both keep our ends of the agreement. The conditional statement is true. p T q T p → q T 14365C02.pgs 7/9/07 4:40 PM Page 55 CASE 2 You get an A in Geometry. (p is true.) I do not buy you a new graphing calculator. (q is false.) I broke the agreement because you got an A in Geometry but I did not buy you a new graphing calculator. The conditional statement is false. Conditionals 55 p T q F p → q F CASE 3 You do not get an A in Geometry. (p is false.) I buy you a new graphing calculator. (q is true.) You did not get an A but I bought you a new graphing calculator anyway. Perhaps I felt that a new calculator would help you to get an A next time. I did not break my promise. My promise only said what I would |
do if you did get an A. The conditional statement is true. p → q T T p q F CASE 4 You do not get an A in Geometry. (p is false.) I do not buy you a new graphing calculator. (q is false.) Since you did not get an A, I do not have to keep our agreement. The conditional statement is true. p F q F p → q T Case 2 tells us that the conditional is false only when the hypothesis is true and the conclusion is false. If a conditional is thought of as an “agreement” or a “promise,” this corresponds to the case when the agreement is broken. Cases 3 and 4 tell us that when the hypothesis is false, the conclusion may or may not be true. In other words, if you do not get an A in Geometry, I may or may not buy you a new graphing calculator. These four cases can be summarized as follows: A conditional is false when a true hypothesis leads to a false conclusion. In all other cases, the conditional is true Hidden Conditionals Often the words “if... then” may not appear in a statement that does suggest a conditional. Instead, the expressions “when” or “in order that” may suggest that the statement is a conditional. For example: 1. “When I finish my homework I will go to the movies.” p → q: If I finish my homework, then I will go to the movies. 14365C02.pgs 7/9/07 4:40 PM Page 56 56 Logic 2. “In order to succeed, you must work hard” becomes p → q: If you want to succeed, then you must work hard. 3. “2x 10; therefore x 5” becomes p → q: If 2x 10, then x 5. EXAMPLE 1 For each given sentence: a. Identify the hypothesis p. b. Identify the conclusion q. (1) If Mrs. Shusda teaches our class, then we will learn. (2) The assignment will be completed if I work at it every day. (3) The task is easy when we all work together and do our best. Solution (1) If Mrs. Shusda teaches our class, then we will learn. i p q e a. p: Mrs. Shusda teaches our class. b. q: We |
will learn. (2) The assignment will be completed if I work at it every day. i f q p a. p: I work at it every day. b. q: The assignment will be completed. (3) Hidden Conditional: If we all work together and do our best, then the task is easy. i p e q a. p: We all work together and we do our best. b. q: the task is easy. Note: In (3), the hypothesis is a conjunction. If we let r represent “We all work together” and s represent “We do our best,” then the conditional “If we all work together and we do our best, then the task is easy” can be symbolized as (r ∧ s) → q. EXAMPLE 2 Identify the truth value to be assigned to each conditional statement. (1) If 4 4 8, then 2(4) 8. (2) If 2 is a prime number, then 2 is odd. 14365C02.pgs 7/9/07 4:40 PM Page 57 Conditionals 57 (3) If 12 is a multiple of 9, then 12 is a multiple of 3. (4) If 2 3 then 2 3 is a positive integer. Solution (1) The hypothesis p is “4 4 8,” which is true. The conclusion q is “2(4) 8,” which is true. The conditional p → q is true. Answer (2) The hypothesis p is “2 is a prime number,” which is true. The conclusion q is “2 is odd,” which is false. The conditional p → q is false. Answer (3) The hypothesis p is “12 is a multiple of 9,” which is false. The conclusion q is “12 is a multiple of 3,” which is true. The conditional p → q is true. Answer (4) The hypothesis p is “2 3,” which is false. The conclusion q is “2 3 is a positive integer,” which is false. The conditional p → q is true. Answer EXAMPLE 3 For each given statement: a. Write the statement in symbolic form using the symbols given below. b. Tell whether the statement is true or false. Let m represent “Monday is the first day of the week.” Let w represent |
“There are 52 weeks in a year.” Let h represent “An hour has 75 minutes.” (True) (True) (False) (1) If Monday is the first day of the week, then there are 52 weeks in a year. (2) If there are 52 weeks in a year, then an hour has 75 minutes. (3) If there are not 52 weeks in a year then Monday is the first day of the week. (4) If Monday is the first day of the week and there are 52 weeks in a year, then an hour has 75 minutes. Answers a. m → w b. T → T is true. a. w → h b. T → F is false. a. ~w → m b. F → T is true. a. (m ∧ w) → h b. (T ∧ T) → F T → F is false. 14365C02.pgs 7/9/07 4:40 PM Page 58 58 Logic Exercises Writing About Mathematics 1. a. Show that the conditional “If x is divisible by 4, then x is divisible by 2” is true in each of the following cases: (1) x 8 (2) x 6 (3) x 7 b. Is it possible to find a value of x for which the hypothesis is true and the conclusion is false? Explain your answer. 2. For what truth values of p and q is the truth value of p → q the same as the truth value of q → p? Developing Skills In 3–10, for each given sentence: a. Identify the hypothesis p. b. Identify the conclusion q. 3. If a polygon is a square, then it has four right angles. 4. If it is noon, then it is time for lunch. 5. When you want help, ask a friend. 6. You will finish more quickly if you are not interrupted. 7. The perimeter of a square is 4s if the length of one side is s. 8. If many people work at a task, it will be completed quickly. 9. 2x 7 11 implies that x 2. 10. If you do not get enough sleep, you will not be alert. In 11–16, write each sentence in symbolic form, using the given symbols. p: The car has a flat tire. q: Danny has a spare tire. r: Danny will change the |
tire. 11. If the car has a flat tire, then Danny will change the tire. 12. If Danny has a spare tire, then Danny will change the tire. 13. If the car does not have a flat tire, then Danny will not change the tire. 14. Danny will not change the tire if Danny doesn’t have a spare tire. 15. The car has a flat tire if Danny has a spare tire. 16. Danny will change the tire if the car has a flat tire. 14365C02.pgs 7/9/07 4:40 PM Page 59 In 17–24, for each given statement: a. Write the statement in symbolic form, using the symbols given below. b. Tell whether the conditional statement is true or false, based upon the truth values given. Conditionals 59 b: The barbell is heavy. t: Kylie trains. l: Kylie lifts the barbell. (True) (False) (True) 17. If Kylie trains, then Kylie will lift the barbell. 18. If Kylie lifts the barbell, then Kylie has trained. 19. If Kylie lifts the barbell, the barbell is heavy. 20. Kylie lifts the barbell if the barbell is not heavy. 21. Kylie will not lift the barbell if Kylie does not train. 22. Kylie trains if the barbell is heavy. 23. If the barbell is not heavy and Kylie trains, then Kylie will lift the barbell. 24. If the barbell is heavy and Kylie does not train, then Kylie will not lift the barbell. In 25–31, find the truth value to be assigned to each conditional statement. 25. If 4 8 12, then 8 4 12. 27. If 1 1 1, then 1 1 1 1. 29. 6 6 66 if 7 7 76. 26. If 9 15, then 19 25. 28. 24 3 8 if 24 8 3. 30. 48 84 if 13 31. 31. If every rhombus is a polygon, then every polygon is a rhombus. In 32–39, symbols are assigned to represent sentences, and truth values are assigned to these sentences. Let j represent “July is a warm month.” Let d represent “I am busy every day.” Let g represent “I work in my garden.” Let f represent “I like |
flowers.” (True) (False) (True) (True) For each compound statement in symbolic form: a. Write a complete sentence in words to show what the symbols represent. b. Tell whether the compound statement is true or false. 32. j → g 36. (j ∧ f ) → d 33. d → g 37. (j ∧ g) → f 34. f → g 38. j → (d ∧ f ) 35. g → j 39. g → (j ∨ d) In 40–45 supply the word, phrase, or symbol that can be placed in the blank to make each resulting sentence true. 40. When p and q represent two simple sentences, the conditional if p then q is written symboli- cally as _______. 41. The conditional if q then p is written symbolically as _______. 14365C02.pgs 7/9/07 4:40 PM Page 60 60 Logic 42. The conditional p → q is false only when p is _______ and q is _______. 43. When the conclusion q is true, then p → q must be _______. 44. When the hypothesis p is false, then p → q must be _______. 45. If the hypothesis p is true and conditional p → q is true, then the conclusion q must be _______. Applying Skills In 46–50, three sentences are written in each case. The truth values are given for the first two sentences. Determine whether the third sentence is true, is false, or has an uncertain truth value. 46. If you read in dim light, then you can strain your eyes. (True) You read in dim light. (True) You can strain your eyes. (?) 47. If the quadrilateral has four right angles, then the quadrilateral must be a square. (False) The quadrilateral has four right angles. (True) The quadrilateral must be a square. (?) 48. If n is an odd number, then 2n is an even number. (True) 2n is an even number. (True) n is an odd number. (?) 49. If the report is late, then you will not get an A. (True) The report is late. (False) You will not get an A. (?) 50. Area 1 2bh if the polygon is a triangle. (True) The polygon is a triangle. |
(True) Area (?) 1 2bh 2-5 INVERSES, CONVERSES, AND CONTRAPOSITIVES The conditional is the most frequently used statement in the construction of an argument or in the study of mathematics. We will use the conditional frequently in our study of geometry. In order to use the conditional statements correctly, we must understand their different forms and how their truth values are related. There are four conditionals that can be formed from two simple statements, p and q, and their negations. The conditional: The converse: p → q q → p The inverse: The contrapositive: p → q q → p 14365C02.pgs 7/9/07 4:40 PM Page 61 Inverses, Converses, and Contrapositives 61 The Inverse The inverse of a conditional statement is formed by negating the hypothesis and the conclusion. For example, the inverse of the statement “If today is Monday, then I have soccer practice” is “If today is not Monday, then I do not have soccer practice.” In symbols, the inverse of (p → q) is (p → q). The following examples compare the truth values of given conditionals and their inverses. 1. A true conditional can have a false inverse. Let p represent “A number is divisible by ten.” Let q represent “The number is divisible by five.” (p → q): (p → q): Conditional If a number is divisible by ten, then it is divisible by five. f i p q Inverse If a number is not divisible by ten, then it is not divisible by five. g i p q We can find the truth value of these two statements when the number is 15. p → q: If 15 is divisible by 10, then 15 is divisible by 5. p: 15 is divisible by 10. q: 15 is divisible by 5. p: 15 is not divisible by 10. q: 15 is not divisible by 5. False True F → T is T True False T → F is F p → q: If 15 is not divisible by 10, then 15 is not divisible by 5. In this case, the conditional and the inverse have opposite truth values. 2. A false conditional can have a true inverse. Let p represent “Two angles are congruent.” Let q represent “Two angles |
are both right angles.” Conditional (p → q): If two angles are congruent, g p then the two angles are both right angles. i q Inverse (p → q): If two angles are not congruent, i p then the two angles are not both right angles. i q We can find the truth value of these two statements with two angles A and B when mA 60 and mB 60. 14365C02.pgs 7/9/07 4:40 PM Page 62 62 Logic p: Angle A and B are congruent. q: Angle A and B are both right angles. p → q: If A and B are congruent, then A and B are both right angles. p: Angle A and B are not congruent. q: Angle A and B are not both right angles. p → q: If A and B are not congruent, then A and B are not both right angles. True False T → F is F False True F → T is T Again, the conditional and the inverse have opposite truth values. 3. A conditional and its inverse can have the same truth value. Let r represent “Twice Talia’s age is 10.” Let q represent “Talia is 5 years old.” Conditional If twice Talia’s age is 10, then Talia is 5 years old. f e r s Inverse If twice Talia’s age is not 10, then Talia is not 5 years old. g f r s (r → s): (r → s): When Talia is 5, r is true and s is true. The conditional is true. When Talia is 5, r is false and s is false. The inverse is true. Both the conditional and its inverse have the same truth value. When Talia is 6, r is false and s is false. The conditional is true. When Talia is 6, r is true and s is true. The conditional is true. Again, the conditional and its inverse have the same truth value. These three illustrations allow us to make the following conclusion: A conditional (p → q) and its inverse (p → q) may or may not have the same truth value. This conclusion can be shown in the truth table. Note that the conditional (p → q) and its inverse (p → q) have the same truth value when p and q have the |
same truth value. The conditional (p → q) and its inverse (p → q) have opposite truth values when p and q have opposite truth values. Conditional Inverse 14365C02.pgs 7/9/07 4:40 PM Page 63 Inverses, Converses, and Contrapositives 63 The Converse The converse of a conditional statement is formed by interchanging the hypothesis and conclusion. For example, the converse of the statement “If today is Monday, then I have soccer practice” is “If I have soccer practice, then today is Monday.” In symbols, the converse of (p → q) is (q → p). To compare the truth values of a conditional and its converse, we will con- sider some examples. 1. A true conditional can have a false converse. Let p represent “x is a prime.” Let q represent “x is odd.” Conditional (p → q): If x is a prime, then x is odd. d p q Converse (q → p): If x is odd, then x is a prime. d d d q p When x 9, p is false and q is true. Therefore, for this value of x, the condi- tional (p → q) is true and its converse (q → p) is false. In this example, the conditional is true and the converse is false. The condi- tional and its converse do not have the same truth value. 2. A false conditional can have a true converse. Let p represent “x is divisible by 2.” Let q represent “x is divisible by 6.” Conditional (p → q): If x is divisible by 2, then x is divisible by 6. e p Converse (q → p): If x is divisible by 6, then x is divisible by 2. q e e e q p When x 8, p is true and q is false. Therefore, for this value of x, the condi- tional (p → q) is false and its converse (q → p) is true. In this example, the conditional is false and the converse is true. Again, the conditional and its converse do not have the same truth value. 3. A conditional and its converse can have the same truth value. Let p represent � |
�Today is Friday.” Let q represent “Tomorrow is Saturday.” Conditional (p → q): If today is Friday, then tomorrow is Saturday. e f Converse (q → p): If tomorrow is Saturday, then today is Friday. f e q p p q 14365C02.pgs 7/9/07 4:40 PM Page 64 64 Logic On Friday, p is true and q is true. Therefore, both (p → q) and (q → p) are true. On any other day of the week, p is false and q is false. Therefore, both (p → q) and (q → p) are true. A conditional and its converse may have the same truth value. These three examples allow us to make the following conclusion: A conditional (p → q) and its converse (q → p) may or may not have the same truth value. This conclusion can be shown in the truth table. Note that the conditional (p → q) and its converse (q → p) have the same truth value when p and q have the same truth value. The conditional (p → q) and its converse (q → p) have different truth values when p and q have different truth values. The Contrapositive Conditional p → q Converse We form the inverse of a conditional by negating both the hypothesis and the conclusion. We form the converse of a conditional by interchanging the hypothesis and the conclusion. We form the contrapositive of a conditional by doing both of these operations: we negate and interchange the hypothesis and conclusion. In symbols, the contrapositive of (p → q) is (q → p). 1. A true conditional can have a true contrapositive. Let p represent “Gary arrives late to class.” Let q represent “Gary is marked tardy.” Conditional If Gary arrives late to class, then Gary is marked tardy. f g p q (p → q): (q → p): Contrapositive If Gary is not marked tardy, i q then Gary does not arrive late to class. i p If Gary arrives late to class, p is true and q is true. Therefore, (p → q) is true. Also, if p and q are true, p is false and q is false, so (q → p) is true. If p is false, that is, if Gary does not arrive late to class, q is false |
and (p → q) is true. Similarly, if p and q are false, p is true and q is true, so (q → p) is true. 14365C02.pgs 7/9/07 4:40 PM Page 65 Inverses, Converses, and Contrapositives 65 2. A false conditional can have a false contrapositive. Conditional If x is an odd number, then x is a prime number. f f p q Contrapositive If x is not a prime number, then x is not an odd number. g i (p → q): (q → p): q p → q F → T is true T → F is false T → T is true F → F is true p q → p F → T is true T → F is false F → F is true T → T is true Let x 2 Let x 9 Let x 11 Let x 8 For each value of x, the conditional and its contrapositive have the same truth value. These illustrations allow us to make the following conclusion: A conditional (p → q) and its contrapositive (q → p) always have the same truth value: When a conditional is true, its contrapositive must be true. When a conditional is false, its contrapositive must be false. This conclusion can be shown in the following truth table. Conditional Contrapositive Logical Equivalents A conditional and its contrapositive are logical equivalents because they always have the same truth value. The inverse of (p → q) is (p → q). The contrapositive of (p → q) is formed by negating and interchanging the hypothesis and conclusion of (p → q). The negation of p is p and the negation of q is q. Therefore, the contrapositive of (p → q) is (q → p). For instance: 14365C02.pgs 7/9/07 4:40 PM Page 66 66 Logic Conditional If Jessica likes waffles, then Jessica eats waffles. f f (p → q): p q Inverse If Jessica does not like waffles, i (p → q): p then Jessica does not eat waffles. i q Contrapositive If Jessica eats waffles, then Jessica likes waffles. f f q p of the inverse (q → p): Notice, however, that the contrapositive of the inverse is the same as the converse of the original conditional. Thus, the inverse and the converse of (p → q |
) are contrapositives of each other. Since a conditional and its contrapositive always have the same truth value, the converse and the inverse always have the same truth value. This can be verified by constructing the following truth table EXAMPLE 1 Write the inverse, converse, and contrapositive of the given conditional: If today is Tuesday, then I play basketball. Solution Inverse: If today is not Tuesday, then I do not play basketball. Converse: If I play basketball, then today is Tuesday. Contrapositive: If I do not play basketball, then today is not Tuesday. EXAMPLE 2 Write the inverse, converse, and contrapositive of the given conditional: If a polygon is a square then it has four right angles. Solution Inverse: If a polygon is not a square, then it does not have four right angles. Converse: If a polygon has four right angles, then it is a square. Contrapositive: If a polygon does not have four right angles, then it is not a square. 14365C02.pgs 7/9/07 4:40 PM Page 67 Inverses, Converses, and Contrapositives 67 Write the inverse, converse, and contrapositive of the given conditional: If M is the midpoint of AB, then AM MB. Inverse: If M is not the midpoint of AB, then AM MB. Converse: If AM MB, then M is the midpoint of. AB Contrapositive: If AM MB, then M is not the midpoint of AB. EXAMPLE 3 Solution EXAMPLE 4 Given the true statement, “If the polygon is a rectangle, then it has four sides,” which statement must also be true? (1) If the polygon has four sides, then it is a rectangle. (2) If the polygon is not a rectangle, then it does not have four sides. (3) If the polygon does not have four sides, then it is not a rectangle. (4) If the polygon has four sides, then it is not a rectangle. Solution A conditional and its contrapositive always have the same truth value. The contrapositive of the given statement is “If a polygon does not have four sides then it is not a rectangle.” Answer (3) Exercises Writing About Mathematics 1. Samuel said that if you know that a conditional is true then you know that the converse of |
the conditional is true. Do you agree with Samuel? Explain why or why not. 2. Kate said that if you know the truth value of a conditional and of its converse then you know the truth value of the inverse and the contrapositive. Do you agree with Kate? Explain why or why not. Developing Skills In 3–6, for each statement, write in symbolic form: a. the inverse b. the converse c. the contrapositive. 3. p → q 4. t → w 5. m → p 6. p → q 14365C02.pgs 7/9/07 4:40 PM Page 68 68 Logic In 7–10: a. Write the inverse of each conditional statement in words. b. Give the truth value of the conditional. c. Give the truth value of the inverse. 7. If 6 3, then 6 3. 8. If a polygon is a parallelogram, then the polygon has two pairs of parallel sides. 9. If 3(3) 9, then 3(4) 12. 10. 1f 22 4, then 32 6. In 11–14, write the converse of each statement in words. 11. If you lower your cholesterol, then you eat Quirky oatmeal. 12. If you enter the Grand Prize drawing, then you will get rich. 13. If you use Shiny’s hair cream, then your hair will curl. 14. If you feed your pet Krazy Kibble, he will grow three inches. In 15–18: a. Write the converse of each conditional statement in words. b. Give the truth value of the conditional. c. Give the truth value of the converse. 15. If a number is even, then the number is exactly divisible by 2. 16. If 0.75 is an integer, then it is rational. 17. If 8 1 7, then 82 12 72. 18. If 4(5) 6 20 6, then 4(5) 6 14. In 19–23: a. Write the contrapositive of each statement in words. b. Give the truth value of the conditional. c. Give the truth value of the contrapositive. 19. If Rochester is a city, then Rochester is the capital of New York. 20. If two angles form a linear pair, then they are supplementary. 21. If 3 2 1, then 4 3 2. 22. If all angles of a triangle are equal in |
measure, then the triangle is equiangular. 23. If 0, then 1 2 1 2 is a counting number. In 24–28, write the numeral preceding the expression that best answers the question. 24. When p → q is true, which related conditional must be true? (1) q → p (2) p → q (3) p → q (4) q → p 25. Which is the contrapositive of “If March comes in like a lion, it goes out like a lamb”? (1) If March goes out like a lamb, then it comes in like a lion. (2) If March does not go out like a lamb, then it comes in like a lion. (3) If March does not go out like a lamb, then it does not come in like a lion. (4) March goes out like a lion if it comes in like a lamb. 14365C02.pgs 7/9/07 4:40 PM Page 69 Biconditionals 69 26. Which is the converse of “If a rectangular prism is a cube, then its surface area is 6s2”? (1) If a rectangular prism is not a cube, then its surface area is not 6s2. (2) If the surface area of a rectangular prism is 6s2, then it is a cube. (3) If the surface area of a rectangular prism is not 6s2, then it is not a cube. (4) If the surface area of a cube is 6s2, then it is a rectangular prism. 27. Which is the inverse of “If z 4, then 2z 9”? (1) If 2z 9, then z 4. (2) If 2z 9, then z 4. (3) If z 4, then 2z 9. (4) If z 4, then 2z 9. 28. Which is the contrapositive of “If y is greater than 3, then 2y 10y is not equal to 36”? (1) If 2y 10y is not equal to 36, then y is greater than 3. (2) If 2y 10y equals 36, then y is not greater than 3. (3) If 2y 10y is not equal to 36, then y is not greater than 3. (4) If 2y 10y equals 36, then y is greater than 3. |
Applying Skills In 29–34, assume that each conditional statement is true. Then: a. Write its converse in words and state whether the converse is always true, sometimes true, or never true. b. Write its inverse in words and state whether the inverse is always true, sometimes true, or never true. c. Write its contrapositive in words and state whether the contrapositive is always true, sometimes true, or never true. 29. If Derek lives in Las Vegas, then he lives in Nevada. 30. If a bin contains 3 red marbles and 3 blue marbles, then the probability of picking a red 1 marble from the bin is. 2 31. If a polygon has eight sides, then it is an octagon. 32. If a garden grows carrots, then it grows vegetables. 33. If the dimensions of a rectangle are 8 feet by 6 feet, then the area of the rectangle is 48 square feet. 34. If a number has 7 as a factor, then it is divisible by 7. 2-6 BICONDITIONALS A biconditional is the conjunction of a conditional and its converse. For the conditional (p → q), the converse is (q → p). The biconditional can be written as (p → q) ∧ (q → p) or in the shorter form p ↔ q, which is read p if and only if q. 14365C02.pgs 7/9/07 4:40 PM Page 70 70 Logic Recall that a conjunction is true only when both parts of the compound statement are true. Therefore, (p → q) ∧ (q → p) is true only when (p → q) is true and its converse (q → p) is true. In the last section you learned that a conditional (p → q) and its converse (q → p) are both true when p and q are both true or both false. This is shown in the table belowp → q) ∧ (q → p The biconditional p if and only if q is true when p and q are both true or both false. In other words, p ↔ q is true when p and q have the same truth value. When p and q have different truth values, the biconditional is false. Applications of the Biconditional There are many examples in which the biconditional is always true. Consider the following: 1. |
Every definition is a true biconditional. Every definition can be written in reverse order. Both of the following statements are true: • Congruent segments are segments that have the same measure. • Line segments that have the same measure are congruent. We can restate the definition as two true conditionals: • If two line segments are congruent, then they have the same measure. • If two line segments have the same measure, then they are congruent. Therefore, this definition can be restated as a true biconditional: • Two line segments are congruent if and only if they have the same measure. 2. Biconditionals are used to solve equations. We know that when we add the same number to both sides of an equation or when we multiply both sides of an equation by the same number, the derived equation has the same solution set as the given equation. That is, any number that makes the first equation true will make the derived equation true. 14365C02.pgs 7/9/07 4:40 PM Page 71 Biconditionals 71 For example: p: 3x 7 19 q: 3x 12 p → q: If 3x 7 19, then 3x 12. (7 was added to both sides of q → p: If 3x 12, then 3x 7 19. (7 was added to both sides of the equation.) the equation.) When x 4, both p and q are true and both p → q and q → p are true. When x 1 or when x equals any number other than 3, both p and q are false and both p → q and q → p are true. Therefore, the biconditional “3x 7 19 if and only if 3x 12” is true. The solution of an equation is a series of biconditionals: 3x 7 19 12 3x 4 x 3x 7 19 if and only if 3x 12. 3x 12 if and only if x 4. 3. A biconditional states that two logical forms are equivalent. Two logical forms that always have the same truth values are said to be equivalent. We have seen that a conditional and its contrapositive are logically equivalent and that the converse and inverse of a conditional are logically equivalent. There are many other statements that are logically equivalent. The table below shows that (p ∧ q) and p ∨ q are logically equivalent. We can write |
the true biconditional (p ∧ q) ↔ (p ∨ q). p ∧ q EXAMPLE 1 Determine the truth value to be assigned to the biconditional. Germany is a country in Europe if and only if Berlin is the capital of Germany. Solution “Germany is a country in Europe” is true. “Berlin is the capital of Germany” is true. Therefore, the biconditional “Germany is a country in Europe if and only if Berlin is the capital of Germany” is also true. Answer 14365C02.pgs 7/9/07 4:40 PM Page 72 72 Logic EXAMPLE 2 The statement “I go to basketball practice on Monday and Thursday” is true. Determine the truth value to be assigned to each statement. a. If today is Monday, then I go to basketball practice. b. If I go to basketball practice, then today is Monday. c. Today is Monday if and only if I go to basketball practice. Solution Let p: “Today is Monday,” and q: “I go to basketball practice.” a. We are asked to find the truth value of the following conditional: p → q: If today is Monday, then I go to basketball practice. On Monday, p is true and q is true. Therefore, p → q is true. On Thursday, p is false and q is true. Therefore, p → q is true. On every other day, p is false and q is false. Therefore, p → q is true. “If today is Monday, then I go to basketball practice” is always true. Answer b. We are asked to find the truth value of the following conditional: q → p: If I go to basketball practice then today is Monday. On Monday, p is true and q is true. Therefore, q → p is true. On Thursday, p is false and q is true. Therefore, q → p is false. On every other day, p is false and q is false. Therefore, q → p is true. “If I go to basketball practice, then Today is Monday” is sometimes true and sometimes false. Answer c. We are asked to find the truth value of the following biconditional: p ↔ q: Today is Monday if and only if I go to basketball practice. The conditionals p → q and q → p |
do not always have the same truth value. Therefore, the biconditional “Today is Monday if and only if I go to basketball practice” is not always true. We usually say that a statement that is not always true is false. Answer EXAMPLE 3 Determine the truth value of the biconditional. 3y 1 28 if and only if y 9. 14365C02.pgs 7/9/07 4:40 PM Page 73 Biconditionals 73 Solution When y = 9, 3y 1 28 is true and y 9 is true. Therefore, 3y 1 28 if and only if y 9 is true. When y 9, 3y 1 28 is false and y 9 is false. Therefore, 3y 1 28 if and only if y 9 is true. 3y 1 28 if and only if y 9 is always true. Answer Exercises Writing About Mathematics 1. Write the definition “A prime number is a whole number greater than 1 and has exactly two factors” as a biconditional. 2. Tiffany said that if the biconditional p ↔ q is false, then either p → q is true or q → p is true but both cannot be true. Do you agree with Tiffany? Explain why or why not. Developing Skills In 3–16, give the truth value of each biconditional. 3. y 7 30 if and only if y 23. 4. B is between A and C if and only if AB AC BC. 5. z 9 13 if and only if z 2 6. 6. A parallelogram is a rhombus if and only if the parallelogram has four sides of equal length. 7. A real number is positive if and only if it is greater than zero. 8. An angle is an acute angle if and only if its degree measure is less than 90. 9. An element belongs to the intersection of sets F and G if and only if it belongs to both F and G. 10. An integer is odd if and only if it is not divisible by 2. 11. Two angles have the same measure if and only if they are right angles. 12. I live in the United States if and only if I live in New York State. 13. A rational number has a multiplicative inverse if and only if it is not zero. 14. An angle is an acute angle if and only if it has a degree measure of 50. 15. |
x 5 if and only if x 3. 16. Today is Friday if and only if tomorrow is Saturday. 14365C02.pgs 7/9/07 4:40 PM Page 74 74 Logic Applying Skills 17. Let p represent “x is divisible by 2.” Let q represent “x is divisible by 3.” Let r represent “x is divisible by 6.” a. Write the biconditional (p ∧ q) ↔ r in words. b. Show that the biconditional is always true for the domain {2, 3, 5, 6, 8, 9, 11, 12}. c. Do you think that the biconditional is true for all counting numbers? Explain your answer. 18. A gasoline station displays a sign that reads “Open 24 hours a day, Monday through Friday.” a. On the basis of the information on the sign, is the conditional “The gasoline station is closed if it is Saturday or Sunday” true? b. On the basis of the information on the sign, is the conditional “If the gasoline station is closed, it is Saturday or Sunday” true? c. On the basis of the information on the sign, is the biconditional “The gasoline station is closed if and only if it is Saturday or Sunday” true? d. Marsha arrives at the gasoline station on Monday and finds the station closed. Does this contradict the information on the sign? e. Marsha arrives at the gasoline station on Saturday and finds the station open. Does this contradict the information on the sign? In 19–22, write a biconditional using the given conditionals and tell whether each biconditional is true or false. 19. If a triangle is isosceles, then it has two congruent sides. If a triangle has two congruent sides then it is isosceles. 20. If two angles are both right angles, then they are congruent. If two angles are congruent, then they are both right angles. 21. If today is Thursday, then tomorrow is not Saturday. If tomorrow is not Saturday, then today is Thursday. 22. If today is not Friday, then tomorrow is not Saturday. If tomorrow is not Saturday, then today is not Friday. 2-7 THE LAWS OF LOGIC We frequently want to combine known facts in order to establish the |
truth of related facts. To do this, we can look for patterns that are frequently used in drawing conclusions. These patterns are called the laws of logic. 14365C02.pgs 7/9/07 4:40 PM Page 75 The Laws of Logic 75 The Law of Detachment A valid argument uses a series of statements called premises that have known truth values to arrive at a conclusion. For example, Cynthia makes the following true statements to her parents: I want to play baseball. If I want to play baseball, then I need a glove. These are the premises of Cynthia’s argument. The conclusion that Cynthia wants her parents to make is that she needs a glove. Is this conclusion valid? Let p represent “I want to play baseball.” Let q represent “I need a glove.” Then p → q represents “If I want to play baseball, then I need a glove.” We know that the premises are true, that is, p is true and p → q is true. The only line of the truth table that satisfies both of these conditions is the first in which q is also true. Therefore, “I need a glove” is a true conclusion The example just given does not depend on the statement represented by p and q. The first line of the truth table tells us that whenever p → q is true and p is true, then q must be a true conclusion. This logical pattern is called the Law of Detachment: If a conditional (p → q) is true and the hypothesis (p) is true, then the con- clusion (q) is true. EXAMPLE 1 If the measure of an angle is greater than 0° and less than 90°, then the angle is an acute angle. Let “mA 40°, which is greater than 0° and less than 90°” be a true statement. Prove that A is an acute angle. Solution Let p represent “mA 40°, which is greater than 0° and less than 90°.” Let q represent “the angle is an acute angle.” Then p → q is true because it is a definition of an acute angle. Also, p is true because it is given. Then by the Law of Detachment, q is true. Answer “A is an acute angle” is true. 14365C02.pgs 7/9/07 4:40 PM Page 76 76 Logic The Law of |
Disjunctive Inference We know that a disjunction is true when one or both statements that make up the disjunction are true. The disjunction is false when both statements that make up the disjunction are false. For example, let p represent “A real number is rational” and q represent “A real number is irrational.” Then p ∨ q represents “A real number is rational or a real number is irrational,” a true statement. When the real number is p, then “A real number is rational” is false. Therefore “A real number is irrational” must be true. When the real number is 7, then “A real number is irrational” is false. Therefore “A real number is rational” must be true. The truth table shows us that when p is false and p ∨ q is true, only the third line of the table is satisfied. This line tells us that q is true. Also, when q is false and p ∨ q is true, only the second line of the table is satisfied. This line tells us that p is true. The example just given illustrates a logical pattern that does not depend on the statements represented by p and q. When a disjunction is true and one of the disjuncts is false, then the other disjunct must be true. This logical pattern is called the Law of Disjunctive Inference If a disjunction (p ∨ q) is true and the disjunct (p) is false, then the other disjunct (q) is true. If a disjunction (p ∨ q) is true and the disjunct (q) is false, then the other disjunct (p) is true. EXAMPLE 2 What conclusion can be drawn when the following statements are true? I will walk to school or I will ride to school with my friend. I do not walk to school. Solution Since “I do not walk to school” is true, “I walk to school” is false. The disjunc- tion “I will walk to school or I will ride to school with my friend” is true and one of the disjuncts, “I walk to school,” is false. By the Law of Disjunctive Inference, the other disjunct, “I ride to school with my friend, |
” must be true. Alternative Solution Let p represent “I walk to school.” Let q represent “I ride to school with my friend.” 14365C02.pgs 7/9/07 4:40 PM Page 77 The Laws of Logic 77 Make a truth table for the disjunction p ∨ q and eliminate the rows that do not apply. We know that p ∨ q is true. We also know that since p is true, p is false. (1) Eliminate the last row of truth values in which the disjunction is false. (2) Eliminate the first two rows of truth values in which p is true. (3) Only one case remains: q is true. Answer I ride to school with my friend Note: In most cases, more than one possible statement can be shown to be true. For example, the following are also true statements when the given statements are true: I do not walk to school and I ride to school with my friend. If I do not walk to school, then I ride to school with my friend. If I do not ride to school with my friend, then I walk to school. EXAMPLE 3 From the following true statements, is it possible to determine the truth value of the statement “I will go to the library”? If I have not finished my essay for English class, then I will go to the library. I have finished my essay for English class. Solution When “I have finished my essay for English class” is true, its negation, “I have not finished my essay for English class” is false. Therefore, the true conditional “If I have not finished my essay for English class, then I will go to the library” has a false hypothesis and the conclusion, “I will go to the library” can be either true or false. Alternative Solution Let p represent “I have not finished my essay for English class.” Let q represent “I will go to the library.” Make a truth table for the disjunction p → q and eliminate the rows that do not apply. We know that p → q is true. We also know that since p is true, p is false. 14365C02.pgs 7/9/07 4:40 PM Page 78 78 Logic (1) Eliminate the second row of truth values in which the conditional is false. (2) |
Eliminate the first row of truth values in which p is true. The second row in which p is true has already been eliminated. (3) Two rows remain, one in which q is true and the other in which q is false Answer “I will go to the library” could be either true or false. EXAMPLE 4 Draw a conclusion or conclusions base on the following true statements. If I am tired, then I will rest. I do not rest. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If I do not rest, then I am not tired” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “I do not rest” is true, “I am not tired” must be true. Alternative Solution Let p represent “I am tired.” Let q represent “I will rest.” Make a truth table for the disjunction p → q and eliminate the rows that do not apply. We know that p → q is true. We also know that since q is true, q is false. (1) Eliminate the second row of truth values in which the conditional is false. (2) Eliminate the first and third rows of truth values in which q is true. (3) Only one case remains: p is false. Since p is false, p is true. “I am not tired” is true Answer I am not tired. 14365C02.pgs 7/9/07 4:40 PM Page 79 The Laws of Logic 79 Exercises Writing About Mathematics 1. Clovis said that when p → q is false and q ∨ r is true, r must be true. Do you agree with Clovis? Explain why or why not. 2. Regina said when p ∨ q is true and q is true, then p ∧ q must be true. Do you agree with Regina? Explain why or why not. Developing Skills In 3–14, assume that the first two sentences in each group are true. Determine whether the third sentence is true, is false, or cannot be found to be true or false. Justify your answer. 3. I save up money or I do not go on the trip. 4. If I speed, then I get a ticket. I go on the trip. I save up money. 5. |
I like swimming or kayaking. I like kayaking. I like swimming. 7. x 18 if x 14. x 14 x 18 I speed. I get a ticket. 6. I like swimming or kayaking. I do not like swimming. I like kayaking. 8. I live in Pennsylvania if I live in Philadelphia. I do not live in Philadelphia. I live in Pennsylvania. 9. If I am late for dinner, then my dinner 10. If I am late for dinner, then my dinner will be cold. I am late for dinner. My dinner is cold. will be cold. I am not late for dinner. My dinner is not cold. 11. I will go to college if and only if I work this summer. I do not work this summer. I will go to college. 13. If I am late for dinner, then my dinner will be cold. My dinner is not cold. I am not late for dinner. 12. The average of two numbers is 20 if the numbers are 17 and 23. The average of two numbers is 20. The two numbers are 17 and 23. 14. If I do not do well in school, then I will not receive a good report card. I do well in school. I receive a good report card. Applying Skills In 15–27, assume that each given sentence is true. Write a conclusion using both premises, if possible. If no conclusion is possible, write “No conclusion.” Justify your answer. 15. If I play the trumpet, I take band. I play the trumpet. 16. 6 " 6 " is rational or irrational. is not rational. 14365C02.pgs 7/9/07 4:40 PM Page 80 80 Logic 17. If 2b 6 14, then 2b 8. If 2b 8, then b 4. 2b 6 14 19. If k is a prime, then k 8. k = 8 18. If it is 8:15 A.M., then it is morning. It is not morning. 20. x is even and a prime if and only if x 2. x 2 21. It is February or March, and it is not 22. If x is divisible by 4, then x is divisible summer. It is not March. by 2. x is divisible by 2. 23. On Saturdays, we go bowling or we 24. I study computer science, and wood shop fly kites. or welding. Last Saturday |
, we did not go bowling. I do not take woodshop. 25. Five is a prime if and only if five has 26. If x is an integer greater than 2 and x exactly two factors. Five is a prime. is a prime, then x is odd. x is not an odd integer. 27. If a ray bisects an angle, the ray divides the angle into two congruent angles. Ray DF does not divide angle CDE into two congruent angles. 2-8 DRAWING CONCLUSIONS Many important decisions as well as everyday choices are made by applying the principles of logic. Games and riddles also often depend on logic for their solution. EXAMPLE 1 The three statements given below are each true. What conclusion can be found to be true? 1. If Rachel joins the choir then Rachel likes to sing. 2. Rachel will join the choir or Rachel will play basketball. 3. Rachel does not like to sing. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If Rachel does not like to sing, then Rachel will not join the choir” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “Rachel does not like to sing” is true, “Rachel will not join the choir” must be true. Since “Rachel will not join the choir” is true, then its negation, “Rachel will join the choir,” must be false. 14365C02.pgs 7/9/07 4:40 PM Page 81 Drawing Conclusions 81 Since “Rachel will join the choir or Rachel will play basketball” is true and “Rachel will join the choir” is false, then by the Law of Disjunctive Inference, “Rachel will play basketball” must be true. Answer Rachel does not join the choir. Rachel will play basketball. Alternative Solution Let c represent “Rachel joins the choir,” s represent “Rachel likes to sing,” and b represent “Rachel will play basketball.” Write statements 1, 2, and 3 in symbols: 1. c → s 2. c ∨ b 3. s Using Statement 1 c → s is true, so s → c is true. (A conditional and its contrapositive always have the same truth value.) Using Statement 3 s is true and s → c is true, |
so c is true. (Law of Detachment) Also, c is false. Using Statement 2 c ∨ b is true and c is false, so b must be true. (Law of Disjunctive Inference) Answer Rachel does not join the choir. Rachel will play basketball. EXAMPLE 2 If Alice goes through the looking glass, then she will see Tweedledee. If Alice sees Tweedledee, then she will see the Cheshire Cat. Alice does not see the Cheshire Cat. Show that Alice does not go through the looking glass. Solution A conditional and its contrapositive are logically equivalent. Therefore, “If Alice does not see the Cheshire Cat, then Alice does not see Tweedledee” is true. By the Law of Detachment, when the hypothesis of a true conditional is true, the conclusion must be true. Therefore, since “Alice does not see the Cheshire Cat” is true, “Alice does not see Tweedledee” must be true. Again, since a conditional and its contrapositive are logically equivalent, “If Alice does not see Tweedledee, then Alice does not go through the looking 14365C02.pgs 7/9/07 4:40 PM Page 82 82 Logic glass” is also true. Applying the Law of Detachment, since “Alice does not see Tweedledee” is true, “Alice does not go through the looking glass” must be true. Alternative Solution Let a represent “Alice goes through the looking glass.” t represent “Alice sees Tweedledee.” c represent “Alice sees the Cheshire cat.” 1. a → t Then, in symbols, the given statements are: 3. c We would like to conclude that a is true. 2. t → c Using Statement 2 t → c is true, so c → t is true. (A conditional and its contrapositive always have the same truth value.) Using Statement 3 c is true and c → t is true, so t is true. (Law of Detachment) Using Statement 1 a → t is true, so t → a is true. (A conditional and its contrapositive always have the same truth value.) Therefore, by the Law of Detachment, since t is true and t → a is true, a is true. EXAMPLE 3 Three siblings, Ted, Bill, and Mary, each |
take a different course in one of three areas for their senior year: mathematics, art, and thermodynamics. The following statements about the siblings are known to be true. • Ted tutors his sibling taking the mathematics course. • The art student and Ted have an argument over last night’s basketball game. • Mary loves the drawing made by her sibling taking the art course. What course is each sibling taking? Solution Make a table listing each sibling and each subject. Use the statements to fill in the information about each sibling. 14365C02.pgs 7/9/07 4:40 PM Page 83 (1) Since Ted tutors his sibling taking the math course, Ted cannot be taking math. Place an ✗ in the table to show this. (2) Similarly, the second state- Drawing Conclusions 83 Math Art Thermodynamics ✗ Ted Bill Mary Ted ment shows that Ted cannot be taking art. Place an ✗ to indicate this. The only possibility left is for Ted to be taking thermodynamics. Place a to indicate this, and add ✗’s in the thermodynamics column to show that no one else is taking this course. Mary Bill ✗ Math Art Thermodynamics ✗ ✗ ✗ (3) The third statement shows that Mary is not taking art. Therefore, Mary is taking math. Since Ted is taking thermodynamics and Mary is taking math, Bill must be taking art. Math Art Thermodynamics Ted Bill Mary ✗ ✗ ✗ ✗ ✗ ✗ Answer Ted takes thermodynamics, Bill takes art, and Mary takes mathematics. Exercises Writing About Mathematics 1. Can (p ∨ q) ∧ r be true when p is false? If so, what are the truth values of q and of r? Justify your answer. 2. Can (p ∨ q) ∧ r be true when r is false? If so, what are the truth values of p and of q? Justify your answer. Developing Skills 3. When p ∧ q is true and q ∨ r is true, what is the truth value of r? 4. When p → q is false and q ∨ r is true, what is the truth value of r? 5. When p → q is false, what is the truth value of q → r? 6. When p → q and p ∧ q are both true, what are the truth values of p and of q? 14365C |
02.pgs 7/9/07 4:40 PM Page 84 84 Logic 7. When p → q and p ∧ q are both false, what are the truth values of p and of q? 8. When p → q is true and p ∧ q is false, what are the truth values of p and of q? 9. When p → q is true, p ∨ r is true and q is false, what is the truth value of r? Applying Skills 10. Laura, Marta, and Shanti are a lawyer, a doctor, and an investment manager. • The lawyer is Marta’s sister. • Laura is not a doctor. • Either Marta or Shanti is a lawyer. What is the profession of each woman? 11. Alex, Tony, and Kevin each have a different job: a plumber, a bookkeeper, and a teacher. • Alex is a plumber or a bookkeeper. • Tony is a bookkeeper or a teacher. • Kevin is a teacher. What is the profession of each person? 12. Victoria owns stock in three companies: Alpha, Beta, and Gamma. • Yesterday, Victoria sold her shares of Alpha or Gamma. • If she sold Alpha, then she bought more shares of Beta. • Victoria did not buy more shares of Beta. Which stock did Victoria sell yesterday? 13. Ren, Logan, and Kadoogan each had a different lunch. The possible lunches are: a ham sandwich, pizza, and chicken pot pie. • Ren or Logan had chicken pot pie. • Kadoogan did not have pizza. • If Logan did not have pizza, then Kadoogan had pizza. Which lunch did each person have? 14. Zach, Steve, and David each play a different sport: basketball, soccer, or baseball. Zach made each of the following true statements. • I do not play basketball. • If Steve does not play soccer, then David plays baseball. • David does not play baseball. What sport does each person play? 14365C02.pgs 7/9/07 4:40 PM Page 85 Chapter Summary 85 15. Taylor, Melissa, and Lauren each study one language: French, Spanish, and Latin. • If Melissa does not study French, then Lauren studies Latin. • If Lauren studies Latin, then Taylor studies Spanish. • Taylor does not study Spanish. What language does each person study? 16. Three friends, Augustus, Brutus, and Caesar, play a |
game in which each decides to be either a liar or a truthteller. A liar must always lie and a truthteller must always tell the truth. When you met these friends, you asked Augustus which he had chosen to be. You didn’t hear his answer but Brutus volunteered, “Augustus said that he is a liar.” Caesar added, “If one of us is a liar, then we are all liars.” Can you determine, for each person, whether he is a liar or a truthteller? CHAPTER SUMMARY Definitions to Know • Logic is the study of reasoning. • In logic, a mathematical sentence is a sentence that contains a complete thought and can be judged to be true or false. • A phrase is an expression that is only part of a sentence. • An open sentence is any sentence that contains a variable. • The domain or replacement set is the set of numbers that can replace a variable. • The solution set or truth set is the set of all replacements that will change an open sentence to true sentences. • A statement or a closed sentence is a sentence that can be judged to be true or false. • A closed sentence is said to have a truth value, either true (T) or false (F). • The negation of a statement has the opposite truth value of a given statement. • In logic, a compound sentence is a combination of two or more mathematical sentences formed by using the connectives not, and, or, if... then, or if and only if. • A conjunction is a compound statement formed by combining two simple statements, called conjuncts, with the word and. The conjunction p and q is written symbolically as p ∧ q. • A disjunction is a compound statement formed by combining two simple statements, called disjuncts, with or. The disjunction p or q is written symbolically as p ∨ q. • A truth table is a summary of all possible truth values of a logic statement. • A conditional is a compound statement formed by using the words if... then to combine two simple statements. The conditional if p then q is written symbolically as p → q. 14365C02.pgs 7/9/07 4:40 PM Page 86 86 Logic • A hypothesis, also called a premise or antecedent, is an assertion that begins an argument. The hypothesis usually follows the word if. • A conclusion, also |
called a consequent, is an ending or a sentence that closes an argument. The conclusion usually follows the word then. • Beginning with a statement (p → q), the inverse (p → q) is formed by negating the hypothesis and negating the conclusion. • Beginning with a statement (p → q), the converse (q → p) is formed by interchanging the hypothesis and the conclusion. • Beginning with a conditional (p → q), the contrapositive (q → p) is formed by negating both the hypothesis and the conclusion, and then interchanging the resulting negation. • Two statements are logically equivalent––or logical equivalents––if they always have the same truth value. • A biconditional (p ↔ q) is a compound statement formed by the conjunc- tion of the conditional p → q and its converse q → p. • A valid argument uses a series of statements called premises that have known truth values to arrive at a conclusion. Logic Statements Negation: p Conjunction: Disjunction: Conditional: Inverse: Converse: Contrapositive: Biconditional not p p and q p or q if p then q if p then q if q then p if q then p p if and only if q The truth values of the logic connectives can be summarized as follows: n tio tio c n nju o C p ∧ q n tio c n Disju al n ditio e s r e v In sitiv al n ditio n o Bic p ↔ q T F F T 14365C02.pgs 7/9/07 4:40 PM Page 87 Laws of Logic • The Law of Detachment states that when p → q is true and p is true, then q must be true. • The Law of Disjunctive Inference states that when p ∨ q is true and p is false, then q must be true. Review Exercises 87 VOCABULARY 2-1 Logic • Truth value • Mathematical sentence • Phrase • Open sentence • Domain • Replacement Set • Solution set • Truth set • Statement • Closed sentence • Truth value (T and F) • Negation 2-2 Compound sentence • Compound statement • Conjunction • Conjunct • p and q • Tree diagram • Truth table 2-3 Disjunction • Disjunct • p or q • Inclusive or • Exclusive or 2-4 Conditional • If p then |
q • Hypothesis • Premise • Antecedent • Conclusion • Consequent 2-5 Inverse • Converse • Contrapositive • Logical equivalents 2-6 Biconditional 2-7 Laws of logic • Valid argument • Premises • Law of Detachment • Law of Disjunctive Inference REVIEW EXERCISES 1. The statement “If I go to school, then I do not play basketball” is false. Using one or both of the statements “I go to school” and “I do not play basketball” or their negations, write five true statements. 2. The statement “If I go to school, then I do not play basketball” is false. Using one or both of the statements “I go to school” and “I do not play basketball” or their negations, write five false statements. 3. Mia said that the biconditional p ↔ q and the biconditional p ↔ q always have the opposite truth values. Do you agree with Mia? Explain why or why not. In 4 and 5: a. Identify the hypothesis p. b. Identify the conclusion q. 4. If at first you don’t succeed, then you should try again. 5. You will get a detention if you are late one more time. In 6–12, tell whether each given statement is true or false. 6. If July follows June, then August follows July. 7. July follows June and July is a winter month in the northern hemisphere. 14365C02.pgs 7/9/07 4:40 PM Page 88 88 Logic 8. July is a winter month in the northern hemisphere or July follows June. 9. If August follows July, then July does not follow June. 10. July is a winter month if August is a winter month. 11. August does not follow July and July is not a winter month. 12. July follows June if and only if August follows July. 13. Which whole number, when substituted for y, will make the following sen- tence true? (y 5 9) ∧ (y 6) In 14–17, supply the word, phrase, or symbol that can be placed in each blank to make the resulting statement true. 14. (p) has the same truth value as ________. 15. When p is true and q is false, then p ∧ q is _______ |
. 16. When p ∨ q is false, then p is _______ and q is _______. 17. If the conclusion q is true, then p → q must be _______. In 18–22, find the truth value of each sentence when a, b, and c are all true. 18. a 20. b → c 21. a ∨ b 19. b ∧ c 22. a ↔ b In 23–32, let p represent “x 5,” and let q represent “x is prime.” Use the domain {1, 2, 3, 4,..., 10} to find the solution set for each of the following. 23. p 28. p ∧ q 24. p 29. p ∧ q 25. q 30. p → q 26. q 31. p → q 27. p ∨ q 32. p ↔ q 33. For the conditional “If I live in Oregon, then I live in the Northwest,” write: a. the inverse, b. the converse, c. the contrapositive, d. the biconditional. 34. Assume that the given sentences are true. Write a simple sentence that could be a conclusion. • If A is the vertex angle of isosceles ABC, then AB AC. • AB AC 35. Elmer Megabucks does not believe that girls should marry before the age of 21, and he disapproves of smoking. Therefore, he put the following provision in his will: I leave $100,000 to each of my nieces who, at the time of my death, is over 21 or unmarried, and does not smoke. Each of his nieces is described below at the time of Elmer’s death. Which nieces will inherit $100,000? • Judy is 24, married, and smokes. • Diane is 20, married, and does not smoke. 14365C02.pgs 7/9/07 4:40 PM Page 89 Review Exercises 89 • Janice is 26, unmarried, and does not smoke. • Peg is 19, unmarried, and smokes. • Sue is 30, unmarried, and smokes. • Sarah is 18, unmarried, and does not smoke. • Laurie is 28, married, and does not smoke. • Pam is 19, married, and smokes. 36. Some years after Elmer Megabucks prepared his will, he |
amended the conditions, by moving a comma, to read: I leave $100,000 to each of my nieces who, at the time of my death, is over 21, or unmarried and does not smoke. Which nieces described in Exercise 35 will now inherit $100,000? 37. At a swim meet, Janice, Kay, and Virginia were the first three finishers of a 200-meter backstroke competition. Virginia did not come in second. Kay did not come in third. Virginia came in ahead of Janice. In what order did they finish the competition? 38. Peter, Carlos, and Ralph play different musical instruments and different sports. The instruments that the boys play are violin, cello, and flute. The sports that the boys play are baseball, tennis, and soccer. From the clues given below, determine what instrument and what sport each boy plays. • The violinist plays tennis. • Peter does not play the cello. • The boy who plays the flute does not play soccer. • Ralph plays baseball. 39. Let p represent “x is divisible by 6.” Let q represent “x is divisible by 2.” a. If possible, find a value of x that will: (2) make p true and q false. (1) make p true and q true. (3) make p false and q true. (4) make p false and q false. b. What conclusion can be drawn about the truth value of p → q? 40. Each of the following statements is true. • Either Peter, Jim, or Tom is Maria’s brother. • If Jim is Maria’s brother, then Peter is Alice’s brother. • Alice has no brothers. • Tom has no sisters. Who is Maria’s brother? 14365C02.pgs 7/9/07 4:40 PM Page 90 90 Logic Exploration 1. A tautology is a statement that is always true. For instance, the disjunction p ∨ p is a tautology because if p is true, the disjunction is true, and if p is false, p is true and the disjunction is true. a. Which of the following statements are tautologies? (1) p → (p ∨ q) (2) Either it will rain or it will snow. (3) (p ∧ q) → p b. Construct |
two tautologies, one using symbols and the other using words. 2. A contradiction is a statement that is always false, that is, it cannot be true under any circumstances. For instance, the conjunction p ∧ p is a contradiction because if p is true, p is false and the conjunction is false, and if p is false, the conjunction is false. a. Which of the following statements are contradictions? (1) (p ∨ q) ∧ p (2) It is March or it is Tuesday, and it is not March and it is not Tuesday. (3) (p ∧ p) ∧ q b. Construct two contradictions, one using symbols and the other using words. 3. One way to construct a tautology is to use a contradiction as the hypothesis of a conditional. For instance, since we know that p ∧ p is a contradiction, the conditional (p ∧ p) → q is a tautology for any conclusion q. a. Construct two conditionals using any two contradictions as the premises. b. Show that the two conditionals from part a are tautologies. c. Explain why the method of using a contradiction as the hypothesis of a conditional always results in a tautology. CUMULATIVE REVIEW CHAPTERS 1–2 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is an undefined term? (3) line (2) angle (1) ray (4) line segment 2. The statement “If x is a prime then x is odd” is false when x equals (1) 1 (2) 2 (3) 3 (4) 4 14365C02.pgs 7/9/07 4:40 PM Page 91 Cumulative Review 91 3. Points J, K, and L lie on a line. The coordinate of J is 17, the coordinate of K is 8, and the coordinate of L is 13. What is the coordinate of M, the midpoint of (1) 8 (3) 2 (2) 4 (4) 2 JL? 4. When “Today is Saturday” is false, which of the following statements could be either true or false? (1) If today is Saturday, then I do not have to go to school. (2) Today is Saturday and I do not have to go to school. (3) Today |
is Saturday or I have to go to school. (4) Today is not Saturday. 5. Which of the following is not a requirement in order for point H to be (3) GH HI GI (4) G, H, and I are distinct points. between points G and I? (1) GH HI (2) G, H, and I are collinear. h UW mTUV? (1) 2.5° (2) 32.5° 6. bisects TUV. If mTUV 34x and mVUW 5x 30, what is (3) 42.5° (4) 85° 7. Which of the following must be true when AB BC AC? AB AC (1) B is the midpoint of (2) AB BC (3) C is a point on (4) B is a point on AC 8. Which of the following equalities is an example of the use of the commu- tative property? (1) 3(2 x) 6 3x (2) 3 (2 x) (3 2) x (3) 3 (0 x) 3 x (4) 3(2 x) 3(x 2) 9. Which of the following must be true when p is true? (1) p ∧ q (2) p → q (3) p ∨ q (4) ~p ∨ q 10. The solution set of x 1 x is (2) {0} (3) {–1} (4) {all real numbers} (1) Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C02.pgs 7/9/07 4:40 PM Page 92 92 Logic 11. The first two sentences below are true. Determine whether the third sentence is true, is false, or cannot be found to be true or false. Justify your answer. I win the ring toss game. If I win the ring toss game, then I get a goldfish. I get a goldfish. 12. On the number line, the coordinate of R is 5 and the coordinate of S is 1. What is the coordinate of T if S is the midpoint of RT? Part III Answer |
all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Let A and B be two points in a plane. Explain the meanings of the sym- bols g AB h, AB h BD 14. The ray perpendicular to AB, and AB., is the bisector of ABC, a straight angle. Explain why. Use definitions to justify your explanation. g ABC h BD is Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The steps used to simplify an algebraic expression are shown below. Name the property that justifies each of the steps. a(2b 1) a(2b) a(1) (a 2) b a(1) (2 a) b a(1) 2ab a 16. The statement “If x is divisible by 12, then x is divisible by 4” is always true. a. Write the converse of the statement. b. Write the inverse of the statement. c. Are the converse and inverse true for all positive integers x? Justify your answer. d. Write another statement using “x is divisible by 12” and “x is divisible by 4” or the negations of these statements that is true for all positive integers x. 14365C03.pgs 7/12/07 2:52 PM Page 93 PROVING STATEMENTS IN GEOMETRY After proposing 23 definitions, Euclid listed five postulates and five “common notions.” These definitions, postulates, and common notions provided the foundation for the propositions or theorems for which Euclid presented proof. Modern mathematicians have recognized the need for additional postulates to establish a more rigorous foundation for these proofs. David Hilbert (1862–1943) believed that mathematics should have a logical foundation based on two principles: 1. All mathematics follows from a correctly chosen finite set of assumptions or axioms. 2. This set of axioms is not contradictory. Although mathematicians later discovered that it is not possible to formalize all |
of mathematics, Hilbert did succeed in putting geometry on a firm logical foundation. In 1899, Hilbert published a text, Foundations of Geometry, in which he presented a set of axioms that avoided the limitations of Euclid. CHAPTER 3 CHAPTER TABLE OF CONTENTS 3-1 Inductive Reasoning 3-2 Definitions as Biconditionals 3-3 Deductive Reasoning 3-4 Direct and Indirect Proofs 3-5 Postulates,Theorems, and Proof 3-6 The Substitution Postulate 3-7 The Addition and Subtraction Postulates 3-8 The Multiplication and Division Postulates Chapter Summary Vocabulary Review Exercises Cumulative Review 93 14365C03.pgs 7/12/07 2:52 PM Page 94 94 Proving Statements in Geometry 3-1 INDUCTIVE REASONING Gina was doing a homework assignment on factoring positive integers. She made a chart showing the number of factors for each of the numbers from 1 to 10. Her chart is shown below. Number Number of factors 10 3 4 She noticed that in her chart, only the perfect square numbers, 1, 4, and 9 had an odd number of factors. The other numbers had an even number of factors. Gina wanted to investigate this observation further so she continued her chart to 25. Number 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Number of factors Again her chart showed that only the perfect square numbers, 16 and 25, had an odd number of factors and the other numbers had an even number of factors. Gina concluded that this is true for all positive integers. Gina went from a few specific cases to a generalization. Was Gina’s conclusion valid? Can you find a counterexample to prove that a perfect square does not always have an odd number of factors? Can you find a non-perfect square that has an odd number of factors? Scientists perform experiments once, twice, or many times in order to come to a conclusion. A scientist who is searching for a vaccine to prevent a disease will test the vaccine repeatedly to see if it is effective. This method of reasoning, in which a series of particular examples leads to a conclusion, is called inductive reasoning. In geometry, we also perform experiments to discover properties of lines, angles, and polygons and to determine geometric relationships. Most of these experiments involve measurements. Because direct measurements depend on the type of instrument used to measure and the care with which the measurement is |
made, results can be only approximate. This is the first weakness in attempting to reach conclusions by inductive reasoning. Use a ruler to draw a pair of parallel line segments by drawing a line segment along opposite edges of the ruler. Turn the ruler and draw another pair of parallel line segments that intersect the first pair. Label the intersections of the line segments A, B, C, and D. The figure, ABCD, is a parallelogram. It appears that the opposite sides of the parallelogram have equal measures. Use the ruler to show that this is true. To convince yourself that this relationship is true in other parallelograms, draw other parallelograms and measure their opposite sides. In each experiment you will find that the opposite sides have the same measure. From these ex- C B D A 14365C03.pgs 7/12/07 2:52 PM Page 95 Inductive Reasoning 95 periments, you can arrive at a general conclusion: The opposite sides of a parallelogram have equal measures. This is an example of inductive reasoning in geometry. Suppose a student, after examining several parallelograms, made the generalization “All parallelograms have two acute angles and two obtuse angles.” Here, a single counterexample, such as a parallelogram which is a rectangle and has right angles, is sufficient to show that the general conclusion that the student reached is false. When we use inductive reasoning, we must use extreme care because we are arriving at a generalization before we have examined every possible example. This is the second weakness in attempting to reach conclusions by inductive reasoning. When we conduct an experiment we do not give explanations for why things are true. This is the third weakness of inductive reasoning. In light of these weaknesses, when a general conclusion is reached by inductive reasoning alone, it can at best be called probably true. Such statements, that are likely to be true but not yet been proven true by a deductive proof, are called conjectures. Then why study inductive reasoning? Simply because it mimics the way we naturally make new discoveries. Most new knowledge first starts with specific cases, then, through careful study, to a generalization. Only afterwards, is a proof or explanation usually sought. Inductive reasoning is therefore a powerful tool in discovering new mathematical facts. SUMMARY 1. Inductive reasoning is a powerful tool in discovering and making conjec- tures. 2. Generalizations arising |
from direct measurements of specific cases are only approximate. 3. Care must be taken when applying inductive reasoning to ensure that all relevant examples are examined (no counterexamples exist). 4. Inductive reasoning does not prove or explain conjectures. Exercises Writing About Mathematics 1. After examining several triangles, Mindy concluded that the angles of all triangles are acute. Is Mindy correct? If not, explain to Mindy why she is incorrect. 2. Use a counterexample to show that the whole numbers are not closed under subtraction. 14365C03.pgs 7/12/07 2:52 PM Page 96 96 Proving Statements in Geometry Developing Skills 3. a. Using geometry software or a pencil, ruler, and protractor, draw three right triangles that have different sizes and shapes. b. In each right triangle, measure the two acute angles and find the sum of their measures. c. Using inductive reasoning based on the experiments just done, make a conjecture about the sum of the measures of the acute angles of a right triangle. 4. a. Using geometry software or a pencil, ruler, and protractor, draw three quadrilaterals that have different sizes and shapes. b. For each quadrilateral, find the midpoint of each side of the quadrilateral, and draw a new quadrilateral using the midpoints as vertices. What appears to be true about the quadrilateral that is formed? c. Using inductive reasoning based on the experiments just done, make a conjecture about a quadrilateral with vertices at the midpoints of a quadrilateral. 5. a. Using geometry software or a pencil, ruler, and protractor, draw three equilateral trian- gles of different sizes. b. For each triangle, find the midpoint of each side of the triangle, and draw line segments joining each midpoint. What appears to be true about the four triangles that are formed? c. Using inductive reasoning based on the experiments just done, make a conjecture about the four triangles formed by joining the midpoints of an equilateral triangle. In 6–11, describe and perform a series of experiments to investigate whether each statement is probably true or false. 6. If two lines intersect, at least one pair of congruent angles is formed. 7. The sum of the degree measures of the angles of a quadrilateral is 360. 8. If a2 b2, then a b. 9. In |
any parallelogram ABCD, AC BD. 10. In any quadrilateral DEFG, DF bisects D and F. 11. The ray that bisects an angle of a triangle intersects a side of the triangle at its midpoint. 12. Adam made the following statement: “For any counting number n, the expression n2 n 41 will always be equal to some prime number.” He reasoned: • When n 1, then n2 n 41 1 1 41 43, a prime number. • When n 2, then n2 n 41 4 2 41 47, a prime number. Use inductive reasoning, by letting n be many different counting numbers, to show that Adam’s generalization is probably true, or find a counterexample to show that Adam’s generalization is false. 14365C03.pgs 7/12/07 2:52 PM Page 97 Definitions as Biconditionals 97 Applying Skills In 13–16, state in each case whether the conclusion drawn was justified. 13. One day, Joe drove home on Route 110 and found traffic very heavy. He decided never again to drive on this highway on his way home. 14. Julia compared the prices of twenty items listed in the advertising flyers of store A and store B. She found that the prices in store B were consistently higher than those of store A. Julia decided that she will save money by shopping in store A. 15. Tim read a book that was recommended by a friend and found it interesting. He decided that he would enjoy any book recommended by that friend. 16. Jill fished in Lake George one day and caught nothing. She decided that there are no fish in Lake George. 17. Nathan filled up his moped’s gas tank after driving 92 miles. He concluded that his moped could go at least 92 miles on one tank of gas. Hands-On Activity STEP 1. Out of a regular sheet of paper, construct ten cards numbered 1 to 10. STEP 2. Place the cards face down and in order. STEP 3. Go through the cards and turn over every card. STEP 4. Go through the cards and turn over every second card starting with the second card. STEP 5. Go through the cards and turn over every third card starting with the third card. STEP 6. Continue this process until you turn over the tenth (last) card. If you played this game correctly, the cards that are face up when you finish should |
be 1, 4, and 9. a. Play this same game with cards numbered 1 to 20. What cards are face up when you finish? What property do the numbers on the cards all have in common? b. Play this same game with cards numbered 1 to 30. What cards are face up when you finish? What property do the numbers on the cards all have in common? c. Make a conjecture regarding the numbers on the cards that remain facing up if you play this game with any number of cards. 3-2 DEFINITIONS AS BICONDITIONALS In mathematics, we often use inductive reasoning to make a conjecture, a statement that appears to be true. Then we use deductive reasoning to prove the conjecture. Deductive reasoning uses the laws of logic to combine definitions and general statements that we know to be true to reach a valid conclusion. 14365C03.pgs 7/12/07 2:52 PM Page 98 98 Proving Statements in Geometry Before we discuss this type of reasoning, it will be helpful to review the list of definitions in Chapter 1 given on page 29 that are used in the study of Euclidean geometry. Definitions and Logic Each of the definitions given in Chapter 1 can be written in the form of a conditional. For example: A scalene triangle is a triangle that has no congruent sides. Using the definition of a scalene triangle, we know that: 1. The definition contains a hidden conditional statement and can be rewrit- ten using the words If... then as follows: t: A triangle is scalene. p: A triangle has no congruent sides. t → p: If a triangle is scalene, then the triangle has no congruent sides. 2. In general, the converse of a true statement is not necessarily true. However, the converse of the conditional form of a definition is always true. For example, the following converse is a true statement: p → t: If a triangle has no congruent sides, then the triangle is scalene. 3. When a conditional and its converse are both true, the conjunction of these statements can be written as a true biconditional statement. Thus, (t → p) ∧ (p → t) is equivalent to the biconditional (t ↔ p). Therefore, since both the conditional statement and its converse are true, we can rewrite the above definition as a biconditional statement, using the |
words if and only if, as follows: A triangle is scalene if and only if the triangle has no congruent sides. Every good definition can be written as a true biconditional. Definitions will often be used to prove statements in geometry. EXAMPLE 1 A collinear set of points is a set of points all of which lie on the same straight line. a. Write the definition in conditional form. b. Write the converse of the statement given in part a. c. Write the biconditional form of the definition. 14365C03.pgs 7/12/07 2:52 PM Page 99 Definitions as Biconditionals 99 Solution a. Conditional: If a set of points is collinear, then all the points lie on the same straight line. b. Converse: If a set of points all lie on the same straight line, then the set of points is collinear. c. Biconditional: A set of points is collinear if and only if all the points lie on the same straight line. Exercises Writing About Mathematics 1. Doug said that “A container is a lunchbox if and only if it can be used to carry food” is not a definition because one part of the biconditional is false. Is Doug correct? If so, give a counterexample to show that Doug is correct. 2. a. Give a counterexample to show that the statement “If true. a b, 1, then a b” is not always b. For what set of numbers is the statement “If a b, 1, then a b” always true? Developing Skills In 3–8: a. Write each definition in conditional form. b. Write the converse of the conditional given in part a. c. Write the biconditional form of the definition. 3. An equiangular triangle is a triangle that has three congruent angles. 4. The bisector of a line segment is any line, or subset of a line that intersects the segment at its midpoint. 5. An acute angle is an angle whose degree measure is greater than 0 and less than 90. 6. An obtuse triangle is a triangle that has one obtuse angle. 7. A noncollinear set of points is a set of three or more points that do not all lie on the same straight line. 8. A |
ray is a part of a line that consists of a point on the line, called an endpoint, and all the points on one side of the endpoint. In 9–14, write the biconditional form of each given definition. 9. A point B is between A and C if A, B, and C are distinct collinear points and AB BC AC. 10. Congruent segments are segments that have the same length. 11. The midpoint of a line segment is the point of that line segment that divides the segment into two congruent segments. 14365C03.pgs 7/12/07 2:52 PM Page 100 100 Proving Statements in Geometry 12. A right triangle is a triangle that has a right angle. 13. A straight angle is an angle that is the union of opposite rays and whose degree measure is 180. 14. Opposite rays are two rays of the same line with a common endpoint and no other point in common. Applying Skills In 15–17, write the biconditional form of the definition of each given term. 15. equilateral triangle 16. congruent angles 17. perpendicular lines 3-3 DEDUCTIVE REASONING A proof in geometry is a valid argument that establishes the truth of a statement. Most proofs in geometry rely on logic. That is, they are based on a series of statements that are assumed to be true. Deductive reasoning uses the laws of logic to link together true statements to arrive at a true conclusion. Since definitions are true statements, they are used in a geometric proof. In the examples that follow, notice how the laws of logic are used in the proofs of geometric statements. Using Logic to Form a Geometry Proof Let ABC be a triangle in which angle. In this proof, use the following definition: AB'BC. We can prove that ABC is a right • Perpendicular lines are two lines that intersect to form right angles. This definition contains a hidden conditional and can be rewritten as follows: • If two lines are perpendicular, then they intersect to form right angles. Recall that this definition is true for perpendicular line segments as well as for perpendicular lines. Using the specific information cited above, let p repre,” and let r represent “ABC is a right angle.” sent “ AB'BC The proof is shown by the reasoning that follows: p: ⊥ BC AB p is true because it is given. p → r: If AB � |
� BC, then ABC is a right angle. p → r is true because it is the definition of perpendicular lines. r: ABC is a right angle. r is true by the Law of Detachment. 14365C03.pgs 7/12/07 2:52 PM Page 101 A B A B Deductive Reasoning 101 In the logic-based proof above, notice that the Law of Detachment is cited as a reason for reaching our conclusion. In a typical geometry proof, however, the laws of logic are used to deduce the conclusion but the laws are not listed among the reasons. Let us restate this proof in the format used often in Euclidean geometry. We write the information known to be true as the “given” statements and the conclusion to be proved as the “prove”. Then we construct a two-column proof. In the left column, we write statements that we know to be true, and in the right column, we write the reasons why each statement is true. Given: In ABC,. Prove: ABC is a right angle. AB'BC C Proof: Statements Reasons AB'BC 1. 2. ABC is a right angle. 1. Given. 2. If two lines are perpendicular, then they intersect to form right angles. Notice how the Law of Detachment was used in this geometry proof. By combining statement 1 with reason 2, the conclusion is written as statement 2, just as p and p → r led us to the conclusion r using logic. In reason 2, we used the conditional form of the definition of perpendicular lines. Often in a proof, we find one conclusion in order to use that statement to find another conclusion. For instance, in the following proof, we will begin by proving that ABC is a right angle and then use that fact with the definition of a right triangle to prove that ABC is a right triangle. Given: In ABC, Prove: ABC is a right triangle. AB'BC. C Proof: Statements Reasons AB'BC 1. 2. ABC is a right angle. 1. Given. 2. If two lines are perpendicular, then they intersect to form right angles. 3. ABC is a right triangle. 3. If a triangle has a right angle, then it is a right triangle. Note that in this proof, we used the conditional form of the definition of perpendicular lines in reason 2 and the converse form of the definition of a right triangle in reason 3. The proof can also be |
written in paragraph form, also called a paragraph proof. Each statement must be justified by stating a definition or another 14365C03.pgs 7/12/07 2:52 PM Page 102 102 Proving Statements in Geometry statement that has been accepted or proved to be true. The proof given on page 101 can be written as follows: Proof: We are given that AB'BC. If two lines are perpendicular, then they intersect to form right angles. Therefore, ABC is a right angle. A right triangle is a triangle that has a right angle. Since ABC is an angle of ABC, ABC is a right triangle. EXAMPLE 1 Given: DEF with DE DF Prove: DEF is an isosceles triangle. Proof We will need the following definitions: • An isosceles triangle is a triangle that has two congruent sides. D E F • Congruent segments are segments that have the same measure. We can use these two definitions to first prove that the two segments with equal measure are congruent and then to prove that since the two segments, the sides, are congruent, the triangle is isosceles. Statements Reasons 1. DE = DF 2. DE > DF 3. DEF is isosceles. 1. Given. 2. Congruent segments are segments that have the same measure. 3. An isosceles triangle is a triangle that has two congruent sides. The converse of the definition of congruent segments states that if two segments have the same measure, then they are congruent. The given statement, DE DF, means that have the same measure. Therefore, these two sides of DEF are congruent. The converse of the definition of an isosceles triangle states that if a triangle has two congruent sides, then it is isosceles. Therefore, since are congruent sides of DEF, DEF is isosceles. and and DE DF DE DF Alternative Proof EXAMPLE 2 h BD is the bisector of ABC. Given: Prove: mABD mDBC Proof We will need the following definitions: A D C B 14365C03.pgs 7/12/07 2:52 PM Page 103 Deductive Reasoning 103 • The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. • Congruent angles are angles that have the same measure. |
First use the definition of an angle bisector to prove that the angles are congruent. Then use the definition of congruent angles to prove that the angles have equal measures. Statements is the bisector of ABC. h BD 1. 2. ABD DBC 3. mABD mDBC Exercises Writing About Mathematics Reasons 1. Given. 2. The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. 3. Congruent angles are angles that have the same measure. 1. Is an equilateral triangle an isosceles triangle? Justify your answer. 2. Is it possible that the points A, B, and C are collinear but AB BC AC? Justify your answer. Developing Skills In 3–6, in each case: a. Draw a diagram to illustrate the given statement. b. Write a definition or definitions from geometry, in conditional form, that can be used with the given statement to justify the conclusion. 3. Given: h SP bisects RST. Conclusion: RSP PST 5. Given: g BCD g'ACE Conclusion: mACD 90 4. Given: ABC is a scalene triangle. Conclusion: AB BC 6. Given: AB BC AC with ABC Conclusion: B is between A and C. In 7–12, in each case: a. Draw a diagram to illustrate the given statement. b. Write a definition from geometry, in conditional form, that can be used with the given statement to make a conclusion. c. From the given statement and the definition that you chose, draw a conclusion. 7. Given: LMN with 9. Given: PQ QR PR with LM'MN PQR 11. Given: M is the midpoint of LN. 8. Given: g AB h 10. Given: ST 12. Given: 0 mA 90 bisects h SR and DE at F. are opposite rays. 14365C03.pgs 7/12/07 2:52 PM Page 104 104 Proving Statements in Geometry Applying Skills In 13–15: a. Give the reason for each statement of the proof. b. Write the proof in paragraph form. 13. Given: M is the midpoint of. AMB 14. Given: RST with RS ST. Prove: AM MB A M B Prove: RST is an isosceles triangle. |
S Statements Reasons Statements Reasons R T 1. M is the midpoint of. AMB AM 2. MB 3. AM MB 1. 2. 3. 1. RS ST RS > ST 2. 3. RST is isosceles. 1. 2. 3. 15. Given: In ABC, h CE bisects ACB. Prove: mACE mBCE Statements Reasons B E h CE bisects ACB. 1. 2. ACE ECB 3. mACE mECB 1. 2. 3. C A 16. Complete the following proof by writing the statement for each step. Given: DEF with DE EF. Prove: E is the midpoint of DEF. D E F Statements Reasons 1. 2. 3. 1. Given. 2. Congruent segments are segments that have the same measure. 3. The midpoint of a line segment is the point of that line segment that divides the segment into congruent segments. 14365C03.pgs 7/12/07 2:52 PM Page 105 17. In ABC, mA 90 and mB 90. If ABC is an obtuse triangle, why is mC 90? Direct and Indirect Proofs 105 Justify your answer with a definition. 18. Explain why the following proof is incorrect. Given: Isosceles ABC with A as the vertex angle. Prove: BC AC Statements 1. ABC is isosceles. BC > AC 2. 3. BC AC Reasons 1. Given. 2. An isosceles triangle has two congruent sides. 3. Congruent segments are segments that have the same measure. 3-4 DIRECT AND INDIRECT PROOFS A proof that starts with the given statements and uses the laws of logic to arrive at the statement to be proved is called a direct proof. A proof that starts with the negation of the statement to be proved and uses the laws of logic to show that it is false is called an indirect proof or a proof by contradiction. An indirect proof works because when the negation of a statement is false, the statement must be true. Therefore, if we can show that the negation of the statement to be proved is false, then we can conclude that the statement is true. Direct Proof All of the proofs in Section 3-3 are direct proofs. In most direct proofs we use definitions together with the Law of Detachment to arrive at the desired conclusion. Example 1 uses direct |
proof. Given: ABC is an acute triangle. Prove: mA 90 In this proof, we will use the following definitions: • An acute triangle is a triangle that has three acute angles. • An acute angle is an angle whose degree measure is greater than 0 and less than 90. C A B In the proof, we will use the conditional form of these definitions. EXAMPLE 1 14365C03.pgs 7/12/07 2:52 PM Page 106 106 Proving Statements in Geometry Proof C Statements 1. ABC is an acute triangle. 2. A, B, and C are all acute. A 3. mA 90 B Reasons 1. Given. 2. If a triangle is acute, then the triangle has three acute angles. 3. If an angle is acute, then its degree measure is greater than 0 and less than 90. In this proof, the “and” statement is important for the conclusion. In Note: statement 2, the conjunction can be rewritten as “A is acute, and B is acute, and C is acute.” We know from logic that when a conjunction is true, each conjunct is true. Also, in Reason 3, the conclusion of the conditional is a conjunction: “The degree measure is greater than 0 and the degree measure is less than 90.” Again, since this conjunction is true, each conjunct is true. Indirect Proof In an indirect proof, let p be the given and q be the conclusion. Take the following steps to show that the conclusion is true: 1. Assume that the negation of the conclusion is true. 2. Use this assumption to arrive at a statement that contradicts the given statement or a true statement derived from the given statement. 3. Since the assumption leads to a contradiction, it must be false. The nega- tion of the assumption, the desired conclusion, must be true. Let us use an indirect proof to prove the following statement: If the mea- sures of two segments are unequal, then the segments are not congruent. EXAMPLE 2 Given: AB and CD such that AB CD. Prove: AB and CD are not congruent segments A C B D Proof Start with an assumption that is the negation of the conclusion. 14365C03.pgs 7/12/07 2:52 PM Page 107 Direct and Indirect Proofs 107 Statements Reasons CD 1. and AB segments. 2. AB CD are congru |
ent 1. Assumption. 2. Congruent segments are segments that have the same measure. 3. AB CD 3. Given. 4. CD and AB segments. are not congruent 4. Contradiction in 2 and 3. Therefore, the assumption is false and its negation is true. In this proof, and most indirect proofs, our reasoning reflects the contra- positive of a definition. For example: Definition: Congruent segments are segments that have the same measure. Conditional: If segments are congruent, then they have the same measure. Contrapositive: If segments do not have the same measure, then they are not congruent. Note: To learn how the different methods of proof work, you will be asked to prove some simple statements both directly and indirectly in this section. Thereafter, you should use the method that seems more efficient, which is usually a direct proof. However, in some cases, only an indirect proof is possible. EXAMPLE 3 Write a direct and an indirect proof of the following: C Given: mCDE 90 Prove: CD is not perpendicular to DE. In this proof, we will use two definitions: D E • If an angle is a right angle, then its degree measure is 90. • If two intersecting lines are perpendicular, then they form right angles. Direct Proof We will use the contrapositive form of the two definitions: • If the degree measure of an angle is not 90, then it is not a right angle. • If two intersecting lines do not form right angles, then they are not per- pendicular. 14365C03.pgs 7/12/07 2:52 PM Page 108 108 Proving Statements in Geometry C Statements Reasons 1. mCDE 90 2. CDE is not a right angle. D E 1. Given. 2. If the degree measure of an angle is not 90, then it is not a right angle. 3. CD is not perpendicular to. DE 3. If two intersecting lines do not form right angles, then they are not perpendicular. Indirect Proof We will use the negation of the statement that is to be proved as an assumption, and then arrive at a contradiction using the conditional form of the two definitions. Statements Reasons CD 1. is perpendicular to 2. CDE is a right angle.. DE 3. mCDE 90 1. Assumption. 2. If two intersecting lines are perpendicular, then they form right angles. 3 |
. If an angle is a right angle, then its degree measure is 90. 4. mCDE 90. 4. Given. 5. CD is not perpendicular to. DE 5. Contradiction in 3 and 4. Therefore, the assumption is false and its negation is true. Exercises Writing About Mathematics 1. If we are given ABC, is it true that intersects g BC? Explain why or why not. g AB g ABC 2. Glen said that if we are given line, then we know that A, B, and C are collinear and B is between A and C. Do you agree with Glen? Justify your answer. Developing Skills In 3–8: a. Draw a figure that represents the statement to be proved. b. Write a direct proof in twocolumn form. c. Write an indirect proof in two-column form. 14365C03.pgs 7/12/07 2:52 PM Page 109 3. Given: LM MN Prove: LM > MN 5. Given: PQR is a straight angle. Prove: h QP and h QR are opposite rays. 7. Given: PQR is a straight angle. Postulates,Theorems, and Proof 109 4. Given: PQR is a straight angle. Prove: mPQR 180 6. Given: h QP and h QR are opposite rays. Prove: P, Q, and R are on the same line. bisects DEF. h EG 8. Given: Prove: P, Q, and R are on the same line. Prove: mDEG mGEF 9. Compare the direct proofs to the indirect proofs in problems 3–8. In these examples, which proofs were longer? Why do you think this is the case? 10. Draw a figure that represents the statement to be proved and write an indirect proof: h EG does not bisect DEF. Given: Prove: DEG is not congruent to GEF. Applying Skills 11. In order to prove a conditional statement, we let the hypothesis be the given and the con- clusion be the prove. h BD If is perpendicular to g ABC, then h BD is the bisector of ABC. a. Write the hypothesis of the conditional as the given. b. Write the conclusion of the conditional as the prove. c. Write a direct proof for the conditional. 12. If mEFG 180, then h |
FE and h FG are not opposite rays. a. Write the hypothesis of the conditional as the given. b. Write the conclusion of the conditional as the prove. c. Write an indirect proof for the conditional. 3-5 POSTULATES, THEOREMS, AND PROOF A valid argument that leads to a true conclusion begins with true statements. In Chapter 1, we listed undefined terms and definitions that we accept as being true. We have used the undefined terms and definitions to draw conclusions. At times, statements are made in geometry that are neither undefined terms nor definitions, and yet we know these are true statements. Some of these statements seem so “obvious” that we accept them without proof. Such a statement is called a postulate or an axiom. 14365C03.pgs 7/12/07 2:52 PM Page 110 110 Proving Statements in Geometry Some mathematicians use the term “axiom” for a general statement whose truth is assumed without proof, and the term “postulate” for a geometric statement whose truth is assumed without proof. We will use the term “postulate” for both types of assumptions. DEFINITION A postulate is a statement whose truth is accepted without proof. When we apply the laws of logic to definitions and postulates, we are able to prove other statements. These statements are called theorems. DEFINITION A theorem is a statement that is proved by deductive reasoning. The entire body of knowledge that we know as geometry consists of undefined terms, defined terms, postulates, and theorems which we use to prove other theorems and to justify applications of these theorems. The First Postulates Used in Proving Statements Geometry is often concerned with measurement. In Chapter 1 we listed the properties of the number system. These properties include closure, the commutative, associative, inverse, identity, and distributive properties of addition and multiplication and the multiplication property of zero. We will use these properties as postulates in geometric proof. Other properties that we will use as postulates are the properties of equality. When we state the relation “x is equal to y,” symbolized as “x y,” we mean that x and y are two different names for the same element of a set, usually a number. For example: 1. When we write AB CD, we mean that length of AB and the length of CD are the same number. 2 |
. When we write mP mN, we mean that P and N contain the same number of degrees. Many of our definitions, for example, congruence, midpoint, and bisector, state that two measures are equal. There are three basic properties of equality. The Reflexive Property of Equality: a a The reflexive property of equality is stated in words as follows: Postulate 3.1 A quantity is equal to itself. 14365C03.pgs 7/12/07 2:52 PM Page 111 Postulates,Theorems, and Proof 111 For example, in CDE, observe that: C 1. The length of a segment is equal to itself: CD CD DE DE EC EC 2. The measure of an angle is equal to itself: D E mC mC mD mD mE mE The Symmetric Property of Equality: If a b, then b a. The symmetric property of equality is stated in words as follows: Postulate 3.2 An equality may be expressed in either order. For example: 1. If LM NP, then NP LM. 2. If mA mB, then mB mA. L A M N P B The Transitive Property of Equality: If a b and b c, then a c. This property states that, if a and b have the same value, and b and c have the same value, it follows that a and c have the same value. The transitive property of equality is stated in words as follows: Postulate 3.3 Quantities equal to the same quantity are equal to each other. The lengths or measures of segments and angles are numbers. In the set of real numbers, the relation “is equal to” is said to be reflexive, symmetric, and transitive. A relation for which these postulates are true is said to be an equivalence relation. Congruent segments are segments with equal measures and congruent angles are angles with equal measures. This suggests that “is congruent to” is also an equivalence relation for the set of line segments. For example: 1. AB > AB A line segment is congruent to itself. 2. If AB > CD, then CD > AB. Congruence can be stated in either order. 3. If AB > CD and CD > EF, then AB > EF. Segments congruent to the same segment are congruent to each other. 14 |
365C03.pgs 7/12/07 2:52 PM Page 112 112 Proving Statements in Geometry Therefore, we can say that “is congruent to” is an equivalence relation on the set of line segments. We will use these postulates of equality in deductive reasoning. In con- structing a valid proof, we follow these steps: 1. A diagram is used to visualize what is known and how it relates to what is to be proved. 2. State the hypothesis or premise as the given, in terms of the points and lines in the diagram. The premises are the given facts. 3. The conclusion contains what is to be proved. State the conclusion as the prove, in terms of the points and lines in the diagram. 4. We present the proof, the deductive reasoning, as a series of statements. Each statement in the proof should be justified by the given, a definition, a postulate, or a previously proven theorem. EXAMPLE 1 If AB > BC and BC > CD, then AB = CD. Given: AB > BC and BC > CD Prove: AB = CD A B C D Proof Statements Reasons 1. AB > BC 2. AB = BC 3. BC > CD 4. BC = CD 5. AB = CD 1. Given. 2. Congruent segments are segments that have the same measure. 3. Given. 4. Congruent segments are segments that have the same measure. 5. Transitive property of equality (steps 2 and 4). Alternative Proof Statements Reasons 1. AB > BC 2. BC > CD 3. AB > CD 4. AB = CD 1. Given. 2. Given. 3. Transitive property of congruence. 4. Congruent segments are segments that have the same measure. 14365C03.pgs 7/12/07 2:52 PM Page 113 EXAMPLE 2 AB'BC If mABC mLMN. and LM'MN, then Given: AB'BC and LM'MN Prove: mABC = mLMN Proof Statements AB'BC 1. 2. ABC is a right angle. 3. mABC 90 LM'MN 4. 5. LMN is a right angle. 6. mLMN 90 7. 90 mLMN 8. mABC mLMN Postulates,Theorems, and Proof 113 A B M L C N Reasons 1. Given. 2. Perpendicular lines are two lines that |
intersect to form right angles. 3. A right angle is an angle whose degree measure is 90. 4. Given. 5. Perpendicular lines are two lines that intersect to form right angles. 6. A right angle is an angle whose degree measure is 90. 7. Symmetric property of equality. 8. Transitive property of equality (steps 3 and 7). Exercises Writing About Mathematics 1. Is “is congruent to” an equivalence relation for the set of angles? Justify your answer. 2. Is “is perpendicular to” an equivalence relation for the set of lines? Justify your answer. Developing Skills In 3–6, in each case: state the postulate that can be used to show that each conclusion is valid. 3. CD CD 4. 2 3 5 and 5 1 4. Therefore, 2 3 4 1. 5. 10 a 7. Therefore a 7 10. 6. mA 30 and mB 30. Therefore, mA mB. 14365C03.pgs 7/12/07 2:52 PM Page 114 114 Proving Statements in Geometry Applying Skills In 7–10, write the reason of each step of the proof. 7. Given: y x 4 and y 7 8. Given: AB BC AC and AB BC 12 Reasons Prove: x 4 7 Statements 1. y x 4 2. x 4 y 3. y 7 4. x 4 7 1. 2. 3. 4. Reasons Prove: AC 12 Statements 1. AB BC AC 2. AC AB BC 3. AB BC 12 4. AC 12 1. 2. 3. 4. 9. Given: M is the midpoint of LN and N is the midpoint of MP. Prove: LM NP Statements Reasons 1. M is the midpoint of LN. LM > MN 2. 3. LM MN 4. N is the midpoint of MP. MN > NP 5. 6. MN NP 7. LM NP 1. 2. 3. 4. 5. 6. 7. 10. Given: mFGH mJGK and mHGJ mJGK is the bisector of FGJ. Prove: h GH Reasons G Statements 1. mFGH mJGK 2. mHGJ mJGK 3. mFGH mHGJ 4. FGH HGJ h GH 5. is the bisector of FGJ. |
1. 2. 3. 4. 5. F H J K 14365C03.pgs 7/12/07 2:52 PM Page 115 11. Explain why the following proof is incorrect. B Given: ABC with D a point on Prove: ADB ADC. BC The Substitution Postulate 115 A D C Statements 1. ADB and ADC are right angles. 2. mADB 90 and mADC 90. 3. mADB mADC 4. ADB ADC Reasons 1. Given (from the diagram). 2. A right angle is an angle whose degree measure is 90. 3. Transitive property of equality. 4. If two angles have the same measure, then they are congruent. Hands-On Activity Working with a partner: a. Determine the definitions and postulates that can be used with the given statement to write a proof. b. Write a proof in two-column form. Given: ABC and BCD are equilateral. C Prove: AB 5 CD A D B 3-6 THE SUBSTITUTION POSTULATE The substitution postulate allows us to replace one quantity, number, or measure with its equal. The substitution postulate is stated in words as follows: Postulate 3.4 A quantity may be substituted for its equal in any statement of equality. From x y and y 8, we can conclude that x 8. This is an example of the transitive property of equality, but we can also say that we have used the substitution property by substituting 8 for y in the equation x y. 14365C03.pgs 7/12/07 2:52 PM Page 116 116 Proving Statements in Geometry From y x 7 and x 3, we can conclude that y 3 7. This is not an example of the transitive property of equality. We have used the substitution property to replace x with its equal, 3, in the equation y x 7. We use this postulate frequently in algebra. For example, if we find the solution of an equation, we can substitute that solution to show that we have a true statement. 4x 1 3x 7 x 8 Check 4x 1 3x 7 4(8) 1 3(8) 7 5? 31 31 ✔ In a system of two equations in two variables, x and y, we can solve one equation for y and substitute in the other equation. For example, we are using the substitution postulate in the third line of this solution: |
3x 2y 13 y x 1 3x 2(x 1) 13 3x 2x 2 13 5x 15 x 3 y 3 1 2 EXAMPLE 1 Given: CE 2CD and CD DE Prove: CE 2DE Proof Statements 1. CE 2CD 2. CD DE 3. CE 2DE Reasons 1. Given. 2. Given. 3. Substitution postulate. (Or: A quantity may be substituted for its equal in any expression of equality.) 14365C03.pgs 7/12/07 2:52 PM Page 117 The Substitution Postulate 117 EXAMPLE 2 Given: mABD mDBC 90 and mABD mCBE Prove: mCBE mDBC 90 Proof Statements 1. mABD mDBC 90 2. mABD mCBE 3. mCBE mDBC 90 Reasons 1. Given. 2. Given. 3. Substitution postulate. A B D C E Exercises Writing About Mathematics 1. If we know that PQ > RS and that RS > ST, can we conclude that PQ > ST? Justify your answer. 2. If we know that clude that PQ'ST PQ > RS, and that? Justify your answer. RS'ST, can we use the substitution postulate to con- Developing Skills In 3–10, in each case write a proof giving the reason for each statement in your proof. and RM MT Prove: RM 3. Given: MT 1 2RT 1 2RT 5. Given: ma mb 180 and ma mc Prove: mc mb 180 7. Given: 12 x y and x 8 Prove: 12 8 y 9. Given: AB CD, 1 2GH EF, and " CD 5 EF " Prove: AB 1 2GH 4. Given: AD DE AE and AD EB Prove: EB DE AE 6. Given: y x 5 and y 7 2x Prove: x 5 7 2x 8. Given: BC 2 AB2 AC 2 and AB DE Prove: BC 2 DE 2 AC 2 10. Given: mQ mR mS 75, mQ mS mT, and mR mT mU Prove: mU 75 14365C03.pgs 7/12/07 2:52 PM Page 118 118 Proving Statements in Geometry 3-7 THE ADDITION AND SUBTRACTION POSTULATES The Partition Postulate |
When three points, A, B, and C, lie on the same line, the symbol of indicating the following equivalent facts about these points: g ABC is a way • B is on the line segment. AC • B is between A and C. • AB 1 BC 5 AC Since A, B, and C lie on the same line, we can also conclude that into two segments whose sum is. This fact is expressed in the following postulate called the partition postulate. AB BC AC. In other words, B separates AC AC Postulate 3.5 A whole is equal to the sum of all its parts. This postulate applies to any number of segments or to their lengths. This postulate also applies to any number of angles or their measures. A B C D • If B is between A and C, then A, B, and C are collinear. AB BC AC AB 1 BC 5 AC • If C is between B and D, then B, C, and D are collinear. BC CD BD BC 1 CD 5 BD • If A, B, and C are collinear and B, C, and D are collinear, then A, B, C, and D are collinear, that is, ABCD may conclude:. We AB BC CD AD AB 1 BC 1 CD 5 AD A C D E B • If is a ray in the interior of h BC ABD: mABC mCBD mABD ABC CBD ABD h BD CBE: is a ray in the interior of • If mCBD mDBE mCBE CBD DBE CBE • We can also conclude that: mABC mCBD mDBE mABE ABC CBD DBE ABE 14365C03.pgs 7/12/07 2:52 PM Page 119 The Addition and Subtraction Postulates 119 Note: We write x y to represent the sum of the angles, x and y, only if x and y are adjacent angles. Since AB DE indicates that and mABC mDEF indicates that ABC DEF, that is, since equality implies congruency, we can restate the partition postulate in terms of congruent segments and angles. AB > DE Postulate 3.5.1 A segment is congruent to the sum of all its parts. Postulate 3.5.2 An angle is congruent to the sum of all its parts. The Addition Postulate The addition postulate may be |
stated in symbols or in words as follows: Postulate 3.6 If a b and c d, then a c b d. If equal quantities are added to equal quantities, the sums are equal. The following example proof uses the addition postulate. EXAMPLE 1 Given: g ABC and g DEF with AB DE and BC EF. Prove: AC DF A D B C E F Proof Statements Reasons 1. AB DE 2. BC EF 3. AB BC DE EF 4. AB BC AC DE EF DF 5. AC DF 1. Given. 2. Given. 3. Addition postulate. (Or: If equal quantities are added to equal quantities, the sums are equal.) 4. Partition postulate. (Or: A whole is equal to the sum of all its parts.) 5. Substitution postulate. 14365C03.pgs 7/12/07 2:52 PM Page 120 120 Proving Statements in Geometry Just as the partition postulate was restated for congruent segments and congruent angles, so too can the addition postulate be restated in terms of congruent segments and congruent angles. Recall that we can add line segments AB and if and only if B is between A and C. BC Postulate 3.6.1 If congruent segments are added to congruent segments, the sums are congruent. The example proof just demonstrated can be rewritten in terms of the con- gruence of segments. EXAMPLE 2 Given: g ABC g DEF and with AB > DE and BC > EF. Prove: AC > DF Proof Statements Reasons A D B E C F 1. AB > DE 2. BC > EF 3. AB 1 BC > DE 1 EF 4. AB 1 BC 5 AC 5. DE 1 EF 5 DF 6. AC > DF 1. Given. 2. Given. 3. Addition postulate. (Or: If congruent segments are added to congruent segments, the sums are congruent segments.) 4. Partition postulate. 5. Partition postulate. 6. Substitution postulate (steps 3, 4, 5). When the addition postulate is stated for congruent angles, it is called the angle addition postulate: Postulate 3.6.2 If congruent angles are added to congruent angles, the sums are congruent. Recall that to add angles, the angles must have a common endpoint, a com- |
mon side between them, and no common interior points. In the diagram, ABC CBD ABD. A B C D 14365C03.pgs 7/12/07 2:52 PM Page 121 The Addition and Subtraction Postulates 121 The Subtraction Postulate The subtraction postulate may also be stated in symbols or in words. Postulate 3.7 If a b, and c d, then a c b d. If equal quantities are subtracted from equal quantities, the differences are equal. Just as the addition postulate was restated for congruent segments and congruent angles, so too may we restate the subtraction postulate in terms of congruent segments and congruent angles. Postulate 3.7.1 If congruent segments are subtracted from congruent segments, the differences are congruent. Postulate 3.7.2 If congruent angles are subtracted from congruent angles, the differences are congruent. In Example 3, equal numbers are subtracted. In Example 4, congruent lengths are subtracted. EXAMPLE 3 Given: x 6 14 Prove: x 8 Proof Statements Reasons 1. x 6 14 2. 6 6 3. x 8 1. Given. 2. Reflexive property. 3. Subtraction postulate. 14541C03.pgs 1/25/08 3:51 PM Page 122 122 Proving Statements in Geometry EXAMPLE 4 Given: DEF, E is between D and F D E F Prove: DE > DF 2 EF Proof Statements Reasons 1. E is between D and F. 1. Given. 2. DE 1 EF > DF 3. EF > EF 2. Partition postulate. 3. Reflexive property. 4. DE > DF 2 EF 4. Subtraction postulate. Exercises Writing About Mathematics 1. Cassie said that we do not need the subtraction postulate because subtraction can be expressed in terms of addition. Do you agree with Cassie? Explain why or why not. 2. In the diagram, mABC 30, mCBD 45, and mDBE 15. a. Does mCBD mABC mDBE? Justify your E D answer. b. Is /CBD > /ABC 1 /DBE? Justify your answer. C Developing Skills In 3 and 4, in each case fill in the missing statement or reason in the proof to show |
that the conclusion is valid. B A 3. Given: and AED Prove: AD BC BFC, AE BF, and ED FC Statements Reasons BFC AED 1. and 2. AE ED AD BF FC BC 1. Given. 2. 3. AE BF and ED FC 3. Given. 4. 5. AD BC 4. Substitution postulate. 5. Transitive property (steps 2, 4). D E A C F B 14365C03.pgs 7/12/07 2:52 PM Page 123 The Addition and Subtraction Postulates 123 4. Given: SPR QRP and RPQ PRS Prove: SPQ QRS R Q Statements Reasons 1. 2. 1. Given. 2. If congruent angles are added to congruent angles, the sums are congruent. S P 3. SPR RPQ SPQ 3. 4. 5. SPQ QRS 4. Partition postulate. 5. In 5–8, in each case write a proof to show that the conclusion is valid. 5. Given: AC > BC and MC > NC 6. Given: ABCD with AB > CD Prove: AM > BN Prove: AC > BD. Given: LMN PMQ Prove: LMQ NMP 8. Given: mAEB 180 and mCED 180 Prove: mAEC mBED 14365C03.pgs 7/12/07 2:52 PM Page 124 124 Proving Statements in Geometry 3-8 THE MULTIPLICATION AND DIVISION POSTULATES The postulates of multiplication and division are similar to the postulates of addition and subtraction. The postulates in this section are stated in symbols and in words. The Multiplication Postulate Postulate 3.8 If a b, and c d, then ac bd. If equals are multiplied by equals, the products are equal. When each of two equal quantities is multiplied by 2, we have a special case of this postulate, which is stated as follows: Postulate 3.9 Doubles of equal quantities are equal. The Division Postulate Postulate 3.10 a If a b, and c d, then c b d (c 0 and d 0). If equals are divided by nonzero equals, the quotients are equal. When each of two equal quantities is divided by 2, we have a special case of this postulate, which is stated as |
follows: Postulate 3.11 Halves of equal quantities are equal. Note: Doubles and halves of congruent segments and angles will be handled in Exercise 10. Powers Postulate Postulate 3.12 If a b, then a2 b2. The squares of equal quantities are equal. If AB 7, then (AB)2 (7)2, or (AB)2 49. 14365C03.pgs 7/12/07 2:52 PM Page 125 The Multiplication and Division Postulates 125 Roots Postulate Postulate 3.13 If a b and a 0, then! a b.! Recall that a and b are the positive square roots of a and of b, and so! the postulate can be rewritten as:! Postulate 3.13 Positive square roots of positive equal quantities are equal. If (AB)2 49, then (AB)2 5 " EXAMPLE 1 49, or AB 7. " Given: AB CD, RS 3AB, LM 3CD Prove: RS LM Proof Statements Reasons 1. AB CD 2. 3AB 3CD 3. RS 3AB 4. RS 3CD 5. LM 3CD 6. RS LM EXAMPLE 2 Given: 5x 3 38 Prove: x 7 1. Given. 2. Multiplication postulate. 3. Given. 4. Transitive property of equality (steps 2 and 3). 5. Given. 6. Substitution postulate (steps 4 and 5). Proof Statements Reasons 1. 5x 3 38 2. 3 3 3. 5x 35 4. x 7 1. Given. 2. Reflexive property of equality. 3. Subtraction postulate. 4. Division postulate. 14365C03.pgs 7/12/07 2:52 PM Page 126 126 Proving Statements in Geometry EXAMPLE 3 Given: mABM 1 2m/ABC, mABC 2mMBC Prove: h BM bisects ABC. C M B A Proof Statements Reasons 1 1. mABM 2m/ABC 2. mABC 2mMBC 3. mABC mMBC 1 2 4. mABM mMBC 5. ABM MBC h BM 6. bisects ABC 1. Given. 2. Given. 3. Division postulate (or halves of equal quantities are equal). 4. Transitive property of equality. 5. Congruent angles are angles that have the same measure. 6. |
The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. Exercises Writing About Mathematics 1. Explain why the word “positive” is needed in the postulate “Positive square roots of posi- tive equal quantities are equal.” 2. Barry said that “c 0” or “d 0” but not both could be eliminated from the division postu- late. Do you agree with Barry? Explain why or why not. 14365C03.pgs 7/12/07 2:52 PM Page 127 The Multiplication and Division Postulates 127 Developing Skills In 3 and 4, in each case fill in the missing statement or reason in the proof to show that the conclusion is valid. 3. Given: AB 1 4BC Prove: CD 4AB and BC CD 4. Given: ma 3mb and mb 20 Prove: ma 60 Statements Reasons Statements Reasons 1 4BC 1. AB 2. 4 4 3. 4AB BC 4. BC CD 5. 4AB CD 6. CD 4AB 1. 2. 3. 4. 5. 6. 1. 2. 3. 3mb 60 1. Given. 2. Given. 3. 4. 4. Transitive property of equality. In 5–7, in each case write a proof to show that the conclusion is valid. 5. Given: LM 2MN and MN 1 2NP Prove: LM > NP 6. Given: 2(3a 4) 16 Prove: a 4 7. Given: PQRS, PQ 3QR, and QR 1 3RS Prove: PQ > RS Applying Skills 8. On Monday, Melanie walked twice as far as on Tuesday. On Wednesday, she walked one- third as far as on Monday and two-thirds as far as on Friday. Prove that Melanie walked as far on Friday as she did on Tuesday. 9. The library, the post office, the grocery store, and the bank are located in that order on the same side of Main Street. The distance from the library to the post office is four times the distance from the post office to the grocery store. The distance from the grocery store to the bank is three times the distance from the post office to the grocery store. Prove that the distance from the library to the post office is equal to the distance from the post office to the bank |
. (Think of Main Street as the line segment LPGB.) 10. Explain why the following versions of Postulates 3.9 and 3.11 are valid: Doubles of congruent segments are congruent. Halves of congruent segments are congruent. Doubles of congruent angles are congruent. Halves of congruent angles are congruent. 14365C03.pgs 8/2/07 5:39 PM Page 128 128 Proving Statements in Geometry CHAPTER SUMMARY Postulates Geometric statements can be proved by using deductive reasoning. Deductive reasoning applies the laws of logic to a series of true statements to arrive at a conclusion. The true statements used in deductive reasoning may be the given, definitions, postulates, or theorems that have been previously proved. Inductive reasoning uses a series of particular examples to lead to a general conclusion. Inductive reasoning is a powerful tool in discovering and making conjectures. However, inductive reasoning does not prove or explain conjectures; generalizations arising from direct measurements of specific cases are only approximate; and care must be taken to ensure that all relevant examples are examined. A proof using deductive reasoning may be either direct or indirect. In direct reasoning, a series of statements that include the given statement lead to the desired conclusion. In indirect reasoning, the negation of the desired conclusion leads to a statement that contradicts a given statement. The Reflexive Property of Equality: a a 3.1 The Symmetric Property of Equality: If a b, then b a. 3.2 The Transitive Property of Equality: If a b and b c, then a c. 3.3 3.4 A quantity may be substituted for its equal in any statement of equality. 3.5 A whole is equal to the sum of all its parts. 3.5.1 A segment is congruent to the sum of all its parts. 3.5.2 An angle is congruent to the sum of all its parts. If equal quantities are added to equal quantities, the sums are equal. 3.6 3.6.1 If congruent segments are added to congruent segments, the sums are congruent. 3.6.2 If congruent angles are added to congruent angles, the sums are 3.7 congruent. If equal quantities are subtracted from equal quantities, the differences are equal. 3.7.1 If |
congruent segments are subtracted from congruent segments, the differences are congruent. 3.7.2 If congruent angles are subtracted from congruent angles, the differences are congruent. If equals are multiplied by equals, the products are equal. 3.8 3.9 Doubles of equal quantities are equal. 3.10 If equals are divided by nonzero equals, the quotients are equal. 3.11 Halves of equal quantities are equal. 3.12 The squares of equal quantities are equal. 3.13 Positive square roots of equal quantities are equal. VOCABULARY 3-1 Generalization • Inductive reasoning • Counterexample • Conjecture 14365C03.pgs 7/12/07 2:52 PM Page 129 Review Exercises 129 3-3 Proof • Deductive reasoning • Given • Prove • Two-column proof • Paragraph proof 3-4 Direct proof • Indirect proof • Proof by contradiction 3-5 Postulate • Axiom • Theorem • Reflexive property of equality • Symmetric property of equality • Transitive property of equality • Equivalence relation 3-6 Substitution postulate 3-7 Partition postulate • Addition postulate • Angle addition postulate • Subtraction postulate 3-8 Multiplication postulate • Division postulate • Powers postulate • Roots postulate REVIEW EXERCISES In 1–3, in each case: a. Write the given definition in a conditional form. b. Write the converse of the statement given as an answer to part a. c. Write the biconditional form of the definition. 1. An obtuse triangle is a triangle that has one obtuse angle. 2. Congruent angles are angles that have the same measure. 3. Perpendicular lines are two lines that intersect to form right angles. 4. Explain the difference between a postulate and a theorem. 5. Name the property illustrated by the following statement: If AB CD, then CD AB. 6. Is the relation “is greater than” an equivalence relation for the set of real numbers? Explain your answer by demonstrating which (if any) of the properties of an equivalence relation are true and which are false. In 7–12, in each case draw a figure that illustrates the given information and write a proof to show that the conclusion is valid. 7. Given: g bisects AB |
Prove: CM MD CD at M. 8. Given: Prove: is a line segment, RMST RM > ST RM > MS, and MS > ST. 9. Given: Prove ABCD AB > CD is a line segment and AC > BD. 10. Given: SQRP is a line segment and SQ RP. Prove: SR QP 14541C03.pgs 1/25/08 3:51 PM Page 130 130 Proving Statements in Geometry 11. Given: h BC bisects ABD and mCBD mPQR. Prove: mABC mPQR 12. Given: CD Prove: CD AB and AB bisect each other at E and CE BE. 13. A student wrote the following proof: Given: Prove: AB > CD CD'BC and AB'BC Statements Reasons 1. 2. 3. AB > CD AB'BC CD'BC 1. Given. 2. Given. 3. Substitution postulate. What is the error in this proof? 14. A palindrome is a sequence of numbers or letters that reads the same from left to right as from right to left. a. Write the definition of a palindrome as a conditional statement. b. Write the converse of the conditional statement in a. c. Write the definition as a biconditional. Exploration The following “proof” leads to the statement that twice a number is equal to the number. This would mean, for example, that if b 1, then 2(1) 1, which is obviously incorrect. What is the error in the proof? Given: a b Prove: b 2b Statements 1. a b 2. a2 ab 3. a2 b2 ab b2 4. (a b)(a b) b(a b) 5. a b b 6. b b b 7. 2b b Reasons 1. Given. 2. Multiplication postulate. 3. Subtraction postulate. 4. Substitution postulate. 5. Division postulate. 6. Substitution postulate. 7. Substitution postulate. 14365C03.pgs 7/12/07 2:52 PM Page 131 CUMULATIVE REVIEW Part I Cumulative Review 131 CHAPTERS 1–3 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. If y 2x 7 and y 3 |
, then x is equal to (1) 1 (2) 1 (3) 5 (4) 5 2. The property illustrated in the equality 2(a 4) 2(4 a) is (1) the distributive property. (2) the associative property. (3) the identity property. (4) the commutative property. 3. In biconditional form, the definition of the midpoint of a line segment can be written as (1) A point on a line segment is the midpoint of that segment if it divides the segment into two congruent segments. (2) A point on a line segment is the midpoint of that segment if it divides the segment into two congruent segments. (3) A point on a line segment is the midpoint of that segment only if it divides the segment into two congruent segments. (4) A point on a line segment is the midpoint of that segment if and only if it divides the segment into two congruent segments. 4. The multiplicative identity element is (2) 1 (3) 0 (1) 1 5. The angle formed by two opposite rays is (4) not a real number (1) an acute angle. (2) a right angle. (3) an obtuse angle. (4) a straight angle. 6. The inverse of the statement “If two angles are right angles, then they are congruent” is (1) If two angles are not congruent, then they are not right angles. (2) If two angles are not right angles, then they are not congruent. (3) If two angles are congruent, then they are right angles. (4) Two angles are not right angles if they are not congruent. 7. The statements “Today is Saturday or I go to school” and “Today is not Saturday” are both true statements. Which of the following statements is also true? (1) Today is Saturday. (2) I do not go to school. (3) I go to school. (4) Today is Saturday and I go to school. 14365C03.pgs 7/12/07 2:52 PM Page 132 132 Proving Statements in Geometry 8. is a line segment and B is the midpoint of AC. Which of the fol- ABCD lowing must be true? (1) C is the |
midpoint of (2) AB BC BD (3) AC CD (4) AC BD AD 9. The statements “AB BC” and “DC BC” are true statements. Which of the following must also be true? (1) AB BC AC (2) A, B, and C are collinear (3) B, C, and D are collinear (4) AB DC 10. Triangle LMN has exactly two congruent sides. Triangle LMN is (3) an isosceles triangle. (4) an equilateral triangle. (1) a right triangle. (2) a scalene triangle. Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Give a reason for each step used in the solution of the equation. 3x 7 13 7 7 6 2 __________ __________ __________ Given 3x x 12. Given: DE'EF Prove: DEF is a right triangle. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Given: ABC with D a point on and AC AD DB. AB Prove: ABC is isosceles. is a line segment. PQ 4a 3, QR 3a 2 and PR 8a 6. Is Q 14. PQR the midpoint of PQR? Justify your answer. 14365C03.pgs 7/12/07 2:52 PM Page 133 Cumulative Review 133 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. h bisects ABC, mABD 3x 18 and mDBC 5x 30. If BD mABC 7x 12, is ABC a straight angle? Justify your answer. 16. For each statement, the hypothesis is true. Write the postulate that justifies |
the conclusion. a. If x 5, then x 7 5 7. b. If 2y 3 represents a real number, then 2y 3 2y 3. RST is a line segment, then c. If d. If y 2x 1 and y 15, then 2x 1 15. e. If a 3, then RS 1 ST 5 RT. 5 5 3 a 5. 14365C04.pgs 7/12/07 3:04 PM Page 134 CHAPTER 4 CHAPTER TABLE OF CONTENTS 4-1 Postulates of Lines, Line Segments, and Angles 4-2 Using Postulates and Definitions in Proofs 4-3 Proving Theorems About Angles 4-4 Congruent Polygons and Corresponding Parts 4-5 Proving Triangles Congruent Using Side, Angle, Side 4-6 Proving Triangles Congruent Using Angle, Side, Angle 4-7 Proving Triangles Congruent Using Side, Side, Side Chapter Summary Vocabulary Review Exercises Cumulative Review 134 CONGRUENCE OF LINE SEGMENTS, ANGLES, AND TRIANGLES One of the common notions stated by Euclid was the following:“Things which coincide with one another are equal to one another.” Euclid used this common notion to prove the congruence of triangles. For example, Euclid’s Proposition 4 states, “If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.” In other words, Euclid showed that the equal sides and angle of the first triangle can be made to coincide with the sides and angle of the second triangle so that the two triangles will coincide.We will expand on Euclid’s approach in our development of congruent triangles presented in this chapter. 14365C04.pgs 7/31/07 1:22 PM Page 135 4-1 POSTULATES OF LINES, LINE SEGMENTS, AND ANGLES Postulates of Lines, Line Segments, and Angles 135 Melissa planted a new azalea bush in the fall and wants to protect it from the cold and snow this winter. She drove four parallel stakes into the ground around the |
bush and covered the structure with burlap fabric. During the first winter storm, this protective barrier was pushed out of shape. Her neighbor suggested that she make a tripod of three stakes fastened together at the top, forming three triangles. Melissa found that this arrangement was able to stand up to the storms. Why was this change an improvement? What geometric figure occurs most frequently in weight-bearing structures? In this chapter we will study the properties of triangles to discover why triangles keep their shape. g AB Recall that a line,, is an infinite set of points that extends endlessly in both directions, but a line g segment, and has a finite length. AB to form a line We can choose some point of segment of any length. When we do this, we say that we are extending the line segment. that is not a point of, is a part of g AB AB AB Postulate 4.1 A line segment can be extended to any length in either direction. D B A Postulate 4.2 When we choose point D on g AB so that B is the midpoint of AD, we say that we have extended we have chosen D so that AB BD and AD 2AB. We will also accept the following postulates: but AD AB is not the original segment, AB. In this case, Through two given points, one and only one line can be drawn. Two points determine a line. Through given points C and D, one and only one line can be drawn. C D Postulate 4.3 Two lines cannot intersect in more than one point. g AEB and g CED any other point. intersect at E and cannot intersect at B D C A E 14365C04.pgs 7/12/07 3:04 PM Page 136 136 Congruence of Line Segments, Angles, and Triangles Postulate 4.4 One and only one circle can be drawn with any given point as center and the length of any given line segment as a radius. Only one circle can be drawn that has point O as its center and a radius r equal in length to segment r. O We make use of this postulate in constructions when we use a compass to locate points at a given distance from a given point. Postulate 4.5 At a given point on a given line, one and only one perpendicular can be drawn to the line. At point P on g APB drawn perpendicular to g APB is perpendicular to, exactly one line, g APB and no other line through |
P g PD, can be D. P A B Postulate 4.6 From a given point not on a given line, one and only one perpendicular can be drawn to the line. From point P not on drawn perpendicular to perpendicular to g CD. g CD g CD, exactly one line, g PE, can be and no other line from P is P E C D Postulate 4.7 For any two distinct points, there is only one positive real number that is the length of the line segment joining the two points. A B Postulate 4.8 E C B A D resented by AB, which is the length of For the distinct points A and B, there is only one positive real number, repAB Since AB is also called the distance from A to B, we refer to Postulate 4.7. as the distance postulate. The shortest distance between two points is the length of the line segment joining these two points. The figure shows three paths that can be taken in going from A to B. 14365C04.pgs 7/12/07 3:04 PM Page 137 Postulates of Lines, Line Segments, and Angles 137 The length of (the path through C, a point collinear with A and B) is less than the length of the path through D or the path through E. The measure of the shortest path from A to B is the distance AB. AB Postulate 4.9 A line segment has one and only one midpoint. has a midpoint, point M, and no other point is a AB midpoint of AB. Postulate 4.10 An angle has one and only one bisector. Angle ABC has one bisector, h BD, and no other ray bisects ABC. A M B C D A B EXAMPLE 1 Use the figure to answer the following questions: a. What is the intersection of m and n? Answers point B b. Do points A and B determine line m line m, n, or l? EXAMPLE 2 Lines p and n are two distinct lines that intersect line m at A. If line n is perpendicular to line m, can line p be perpendicular to line m? Explain. Solution No. Only one perpendicular can be drawn to a line at a given point on the line. Since line n is perpendicular to m and lines n and p are distinct, line p cannot be perpendicular to m. Answer 14365C04.pgs 7/12/07 3:04 PM |
Page 138 138 Congruence of Line Segments, Angles, and Triangles EXAMPLE 3 h BD bisects ABC and point E is not a point on h BD, can If h BE be the bisector of ABC? Solution No. An angle has one and only bisector. Since point E is not a point on h BE Therefore, h BD h, is not the same ray as BD cannot be the bisector of ABC. Answer h BE. A D B E C Conditional Statements as They Relate to Proof To prove a statement in geometry, we start with what is known to be true and use definitions, postulates, and previously proven theorems to arrive at the truth of what is to be proved. As we have seen in the text so far, the information that is known to be true is often stated as given and what is to be proved as prove. When the information needed for a proof is presented in a conditional statement, we use the information in the hypothesis to form a given statement, and the information in the conclusion to form a prove statement. Numerical and Algebraic Applications EXAMPLE 4 Rewrite the conditional statement in the given and prove format: If a ray bisects a straight angle, it is perpendicular to the line determined by the straight angle. Solution Draw and label a diagram. D Use the hypothesis, “a ray bisects a straight angle,” as the given. Name a straight angle using the three letters from the diagram and state in the given that this angle is a straight angle. Name the ray that bisects the angle, using the vertex of the angle as the endpoint of the ray that is the bisector. State in the given that the ray bisects the angle: A B C Given: ABC is a straight angle and h BD bisects ABC. Use the conclusion, “if (the bisector) is perpendicular to the line determined by the straight angle,” to write the prove. We have called the bisector the line determined by the straight angle is h Prove: BD g'AC g ABC. h BD, and 14365C04.pgs 7/12/07 3:04 PM Page 139 Postulates of Lines, Line Segments, and Angles 139 Answer Given: ABC is a straight angle and g'AC Prove: h BD h BD bisects ABC. In geometry, we are interested in proving that statements are true. These true statements can |
then be used to help solve numerical and algebraic problems, as in Example 5. EXAMPLE 5 h bisects RST, mRSQ 4x, and SQ mQST 3x 20. Find the measures of RSQ and QST. R Q Solution The bisector of an angle separates the angle into two congruent angles. Therefore, RSQ QST. Then since congruent angles have equal measures, we may write an equation that states that mRSQ mQST. 4x S 3x 20 T 4x 3x 20 x 20 mRSQ 4x mQST 3x 20 4(20) 80 3(20) 20 60 20 80 Answer mRSQ mQST 80 Exercises Writing About Mathematics 1. If two distinct lines g AEB and g CED intersect at a point F, what must be true about points E and F? Use a postulate to justify your answer. 2. If LM 10, can LM be extended so that LM 15? Explain why or why not. 14365C04.pgs 7/12/07 3:04 PM Page 140 140 Congruence of Line Segments, Angles, and Triangles Developing Skills In 3–12, in each case: a. Rewrite the conditional statement in the Given/Prove format. b. Write a formal proof. 3. If AB AD and DC AD, then AB DC. 5. If m1 m2 90 and mA m2, then m1 mA 90. 4. If then BD > CD AD > CD AD > BD and. 6. If mA mB, m1 mB, and m2 mA, then m1 m2., 7. If AB > CD EF > CD, then AB > EF, and. 9. If CE CF, CD 2CE, and CB 2CF, then CD CB. 11. If AD BE and BC CD, then AC CE. E D C B A In 13–15, Given: g RST, h SQ, and h SP. 8. If 2EF DB and GH 1 2DB, then EF GH. 10. If RT RS, RD then RD RE. 1 2RT, and RE 1 2RS, 12. If CDB CBD and ADB ABD, then CDA CBA. D B C A 13. If QSR PST and mQSP 96, find mQSR. h 14. If mQ |
SR 40 and SQ 15. If mPSQ is twice mQSR and, find mPST. h'SP h'SP, find h SQ Q R S P T mPST. Applying Skills In 16–19, use the given conditional to a. draw a diagram with geometry software or pencil and paper, b. write a given and a prove, c. write a proof. 16. If a triangle is equilateral, then the measures of the sides are equal. 17. If E and F are distinct points and two lines intersect at E, then they do not intersect at F. 18. If a line through a vertex of a triangle is perpendicular to the opposite side, then it separates the triangle into two right triangles. 19. If two points on a circle are the endpoints of a line segment, then the length of the line segment is less than the length of the portion of the circle (the arc) with the same endpoints. 20. Points F and G are both on line l and on line m. If F and G are distinct points, do l and m name the same line? Justify your answer. 14365C04.pgs 7/12/07 3:04 PM Page 141 4-2 USING POSTULATES AND DEFINITIONS IN PROOFS Using Postulates and Definitions in Proofs 141 A theorem was defined in Chapter 3 as a statement proved by deductive reasoning. We use the laws of logic to combine definitions and postulates to prove a theorem. A carefully drawn figure is helpful in deciding upon the steps to use in the proof. However, recall from Chapter 1 that we cannot use statements that appear to be true in the figure drawn for a particular problem. For example, we may not assume that two line segments are congruent or are perpendicular because they appear to be so in the figure. On the other hand, unless otherwise stated, we will assume that lines that appear to be straight lines in a figure actually are straight lines and that points that appear to be on a given line actually are on that line in the order shown. EXAMPLE 1 Given: g ABCD with AB > CD Prove: AC > BD A B C D Proof Statements Reasons 1. A, B, C, and D are collinear 1. Given. with B between A and C and C between B and D. 2. AB 1 BC BC 1 CD AC BD 3. AB > CD 4. BC > BC 5. AB 1 |
BC CD 1 BC 6. AC > BD 2. Partition postulate. 3. Given. 4. Reflexive property. 5. Addition postulate. 6. Substitution postulate. EXAMPLE 2 Given: M is the midpoint of AB. Prove: AM 1 2AB and MB 1 2AB A M B 14365C04.pgs 7/12/07 3:04 PM Page 142 142 Congruence of Line Segments, Angles, and Triangles M A Proof B Statements Reasons 1. M is the midpoint of AB. 1. Given. AM > MB 2. 3. AM MB 4. AM MB AB 5. AM AM AB or 2AM AB 6. AM 1 2AB 7. MB MB AB or 2MB AB 8. MB 1 2AB 2. Definition of midpoint. 3. Definition of congruent segments. 4. Partition postulate. 5. Substitution postulate. 6. Halves of equal quantities are equal. 7. Substitution postulate. 8. Halves of equal quantities are equal. Note: We used definitions and postulates to prove statements about length. We could not use information from the diagram that appears to be true. Exercises Writing About Mathematics 1. Explain the difference between the symbols ABC and ACB. Could both of these describe segments of the same line? g ABC g DBE CBE, and EBA also 90? Justify your answer. and 2. Two lines,, intersect and mABD is 90. Are the measures of DBC, Developing Skills In 3–12, in each case: a. Rewrite the conditional statement in the Given/Prove format. b. Write a proof that demonstrates that the conclusion is valid. 3. If AB, CB bisects CB FD bisects AD AB, and. CE, then FE 4. If h CA bisects DCB, DAB, and DCB DAB, then CAB DCA. bisects h AC 14365C04.pgs 7/12/07 3:04 PM Page 143 5. If AD BE, then AE. BD 6. If ABCD is a segment and AB CD, then AC BD. Using Postulates and Definitions in Proofs 143 C A D E B ABCD AC 7. If of AB BC CD. is a segment, B is the midpoint, then, and C is the midpoint of BD A B C D A B |
C D 8. If P and T are distinct points and P is, then T is not the RS the midpoint of. midpoint of RS R T P S 9. If DF BE, then DE. BF 10. If AD BC, E is the midpoint of, then BC, AD and F is the midpoint of AE. FC D C E F B A 11. If AC DB AC each other at E, then and and AE DB. EB bisect D A C B E Applying Skills D E A B C F 12. If h DR bisects CDA, 3 1, and 4 2, then 3 4 13. The rays h DF and h DG separate CDE into three congruent angles, CDF, FDG, and GDE. If mCDF 7a 10 and mGDE 10a 2, find: a. mCDG d. Is CDE acute, right, or obtuse? c. mCDE b. mFDE 14365C04.pgs 7/12/07 3:04 PM Page 144 144 Congruence of Line Segments, Angles, and Triangles 14. Segment is a bisector of BD BC x 10. a. Draw a diagram that shows ABC and BD. ABC and is perpendicular to ABC. AB 2x 30 and b. Find AB and BC. c. Find the distance from A to BD. Justify your answer. 15. Two line segments, and and RN LN. RS 3x 9, and LM 5x 17. a. Draw a diagram that shows and LM LM RS RS., bisect each other and are perpendicular to each other at N, b. Write the given using the information in the first sentence. c. Prove that RS LM. d. Find RS and LM. e. Find the distance from L to RS. Justify your answer. 4-3 PROVING THEOREMS ABOUT ANGLES In this section we will use definitions and postulates to prove some simple theorems about angles. Once a theorem is proved, we can use it as a reason in later proofs. Like the postulates, we will number the theorems for easy reference. Theorem 4.1 If two angles are right angles, then they are congruent. Given ABC and DEF are right angles. C Prove ABC DEF Proof Statements 1. ABC and DEF are right angles. 2. mABC 90 and mDEF 90 3. |
mABC mDEF 4. ABC DEF F B A E D Reasons 1. Given. 2. Definition of right angle. 3. Transitive property of equality. 4. Definition of congruent angles. We can write this proof in paragraph form as follows: Proof: A right angle is an angle whose degree measure is 90. Therefore, mABC is 90 and mDEF is 90. Since mABC and mDEF are both equal to the same quantity, they are equal to each other. Since ABC and DEF have equal measures, they are congruent. 14365C04.pgs 7/12/07 3:04 PM Page 145 Theorem 4.2 If two angles are straight angles, then they are congruent. Proving Theorems About Angles 145 Given ABC and DEF are straight angles. A B C Prove ABC DEF D E F The proof of this theorem, which is similar to the proof of Theorem 4.1, is left to the student. (See exercise 17.) Definitions Involving Pairs of Angles DEFINITION Adjacent angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common Angle ABC and CBD are adjacent angles because as their common they have B as their common vertex, h BC side, and no interior points in common. However, XWY and XWZ are not adjacent angles. B D C A Although XWY and XWZ have W as their common vertex and as their common side, they have interior points in common. For example, point P is in the interior of both XWY and XWZ. h WX DEFINITION Complementary angles are two angles, the sum of whose degree measures is 90. Each angle is called the complement of the other. If mc 40 and md 50, then c and d are complementary angles. If ma 35 and mb 55, then a and b are complementary angles. Complementary angles may be adjacent, as in the case of c and d, or they may be nonadjacent, as in the case of a and b. Note that if the two complementary angles are adjacent, their sum is a right angle: c d WZY, a right angle. Since mc md 90, we say that c is the complement of d, and that d is the complement of c. When the degree measure of an angle is k, the degree measure of the complement of the angle is |
(90 k) because k (90 k) 90. DEFINITION Supplementary angles are two angles, the sum of whose degree measures is 180. 14365C04.pgs 7/12/07 3:04 PM Page 146 146 Congruence of Line Segments, Angles, and Triangles When two angles are supplementary, each angle is called the supplement of the other. If mc 40 and md 140, then c and d are supplementary angles. If ma 35 and mb 145, then a and b are supplementary angles Supplementary angles may be adjacent, as in the case of c and d, or they may be nonadjacent, as in the case of a and b. Note that if the two supplementary angles are adjacent, their sum is a straight angle. Here c d TQR, a straight angle. Since mc md 180, we say that c is the supplement of d and that d is the supplement of c. When the degree measure of an angle is k, the degree measure of the supplement of the angle is (180 k) because k (180 k) 180. EXAMPLE 1 Find the measure of an angle if its measure is 24 degrees more than the measure of its complement. Solution Let x measure of complement of angle. Then x 24 measure of angle. The sum of the degree measures of an angle and its complement is 90. x x 24 90 2x 24 90 2x 66 x 33 x 24 57 Answer The measure of the angle is 57 degrees. Theorems Involving Pairs of Angles Theorem 4.3 If two angles are complements of the same angle, then they are congruent. Given 1 is the complement of 2 and 3 is the comple- ment of 2. Prove 1 3 C D E 23 1 B A 14365C04.pgs 7/12/07 3:04 PM Page 147 Proof Statements 1. 1 is the complement of 2. 2. m1 m2 90 3. 3 is the complement of 2. 4. m3 m2 90 Proving Theorems About Angles 147 Reasons 1. Given. 2. Complementary angles are two angles the sum of whose degree measures is 90. 3. Given. 4. Definition of complementary angles. 5. m1 m2 m3 m2 5. Transitive property of equality 6. m2 m2 7. m1 m3 8. 1 3 (steps 2 and 4). 6. Reflexive property of equality. 7. Sub |
traction postulate. 8. Congruent angles are angles that have the same measure. Note: In a proof, there are two acceptable ways to indicate a definition as a reason. In reason 2 of the proof above, the definition of complementary angles is stated in its complete form. It is also acceptable to indicate this reason by the phrase “Definition of complementary angles,” as in reason 4. We can also give an algebraic proof for the theorem just proved. Proof: In the figure, mCBD x. Both ABD and CBE are complements to CBD. Thus, mABD 90 x and mCBE 90 x, and we conclude ABD and CBE have the same measure. Since angles that have the same measure are congruent, ABD CBE. C D x E (90 x) B ( 9 0 x ) A Theorem 4.4 If two angles are congruent, then their complements are congruent. Given ABD EFH CBD is the complement of ABD. GFH is the complement of EFH. C G D H Prove CBD GHF B A F E This theorem can be proved in a manner similar to Theorem 4.3 but with the use of the substitution postulate. We can also use an algebraic proof. 14365C04.pgs 7/12/07 3:05 PM Page 148 148 Congruence of Line Segments, Angles, and Triangles C D B A F G Proof Congruent angles have the same measure. If ABD EFH, we can represent the measure of each angle by the same variable: mABD mEFH x. Since CBD is the complement of ABD, and GFH is the complement of EFH, then mCBD 90 x and mGFH 90 x. Therefore, mCBD mGFH and CBD GHF. H E Theorem 4.5 If two angles are supplements of the same angle, then they are congruent. Given ABD is the supplement of DBC, and EBC is the supplement of DBC. Prove ABD EBC B C E Theorem 4.6 If two angles are congruent, then their supplements are congruent. Given ABD EFH, CBD is the supplement of ABD, and GFH is the supplement of EFH. Prove CBD GFH The proofs of Theorems 4.5 and 4.6 are similar to the |
proofs of Theorems 4.3 and 4.4 and will be left to the student. (See exercises 18 and 19.) More Definitions and Theorems Involving Pairs of Angles DEFINITION A linear pair of angles are two adjacent angles whose sum is a straight angle. C In the figure, ABD is a straight angle and C is not on. Therefore, ABC + CBD ABD. Note that ABC and CBD are adjacent angles and whose remaining sides are opposite rays that whose common side is h BC g ABD D B A together form a straight line, g. AD Theorem 4.7 If two angles form a linear pair, then they are supplementary. Given ABC and CBD form a linear pair. Prove ABC and CBD are supplementary. C D A B 14365C04.pgs 7/12/07 3:05 PM Page 149 Proving Theorems About Angles 149 Proof, and their remaining sides, In the figure, ABC and CBD form a linear pair. They share a common side, h, are opposite rays. The sum of and BC a linear pair of angles is a straight angle, and the degree measure of a straight angle is 180. Therefore, mABC mCBD 180. Then, ABC and CBD are supplementary because supplementary angles are two angles the sum of whose degree measure is 180. h BD h BA Theorem 4.8 If two lines intersect to form congruent adjacent angles, then they are perpendicular. Given Prove g ABC g ABC g DBE and g'DBE with ABD DBC Proof The union of the opposite rays,, and h is the straight angle, ABC. The BC measure of straight angle is 180. By the partition postulate, ABC is the sum of ABD and DBC. Thus, h BA D B E A C mABD mDBC mABC 180 Since ABD DBC, they have equal measures. Therefore, mABD = mDBC 1 2(180) 90 The angles, ABD and DBC, are right angles. Therefore, because perpendicular lines intersect to form right angles. g ABC g'DBE DEFINITION Vertical angles are two angles in which the sides of one angle are opposite rays to the sides of the second angle. A C E D In the figure, AEC and DEB are a pair of vertical angles because h EB are opposite rays and h EA are opposite rays. Also, AED and h are opposite rays and EA h and |
ED CEB are a pair of vertical angles because h ED are opposite rays. In each pair of vertical angles, the opposite rays, h EB h EC h EC and and and B which are the sides of the angles, form straight lines, g AB and g CD. When two straight lines intersect, two pairs of vertical angles are formed. 14365C04.pgs 7/12/07 3:05 PM Page 150 150 Congruence of Line Segments, Angles, and Triangles Theorem 4.9 If two lines intersect, then the vertical angles are congruent. Given g AEB and g CED intersect at E. Prove BEC AED Proof 1. Statements g CED and intersect at E. g AEB h EA h ED 2. and h EB h EC 3. BEC and AED are and are opposite rays. are opposite rays. vertical angles. 4. BEC and AEC are a linear pair. AEC and AED are a linear pair. 5. BEC and AEC are supplementary. AEC and AED are supplementary. 6. BEC AED B D C A E Reasons 1. Given. 2. Definition of opposite rays. 3. Definition of vertical angles. 4. Definition of a linear pair. 5. If two angles form a linear pair, they are supplementary. (Theorem 4.7) 6. If two angles are supplements of the same angle, they are congruent. (Theorem 4.5) In the proof above, reasons 5 and 6 demonstrate how previously proved theorems can be used as reasons in deducing statements in a proof. In this text, we have assigned numbers to theorems that we will use frequently in proving exercises as well as in proving other theorems. You do not need to remember the numbers of the theorems but you should memorize the statements of the theorems in order to use them as reasons when writing a proof. You may find it useful to keep a list of definitions, postulates, and theorems in a special section in your notebook or on index cards for easy reference and as a study aid. In this chapter, we have seen the steps to be taken in presenting a proof in geometry using deductive reasoning: 1. As an aid, draw a figure that pictures the data of the theorem or the prob- lem. Use letters to label points in the figure. 2. State the given, which is the hypothesis of the |
theorem, in terms of the figure. 3. State the prove, which is the conclusion of the theorem, in terms of the figure. 14365C04.pgs 7/12/07 3:05 PM Page 151 Proving Theorems About Angles 151 4. Present the proof, which is a series of logical arguments used in the demonstration. Each step in the proof should consist of a statement about the figure. Each statement should be justified by the given, a definition, a postulate, or a previously proved theorem. The proof may be presented in a two-column format or in paragraph form. Proofs that involve the measures of angles or of line segments can often be presented as an algebraic proof. EXAMPLE 2 If g ABC h BC g DBE and intersect at B bisects EBF, prove that and CBF ABD. A D Solution Given: g and ABC at B and EBF. g hDBE BC bisects intersect B F Prove: CBF ABD Proof Statements E C Reasons h BC bisects EBF. 1. 2. EBC CBF g ABC g intersect at B. 3. DBE 4. EBC and ABD are vertical and angles. 5. EBC ABD 6. CBF ABD 1. Given. 2. Definition of a bisector of an angle. 3. Given. 4. Definition of vertical angles. 5. If two lines intersect, then the vertical angles are congruent. 6. Transitive property of congruence (steps 2 and 5). Alternative Proof An algebraic proof can be given: Let mEBF 2x. It is given that h BC bisects EBF. The bisector of an angle separates the angle into two congruent angles: EBC and CBF. Congruent angles have equal measures. Therefore, mEBC mCBF x. It is also given that intersect at B. If two lines intersect, g DBE g ABC and the vertical angles are congruent and therefore have equal measures: mABD mEBC x. Then since mCBF = x and mABD x, mCBF = mABD and CBF ABD. 14365C04.pgs 7/12/07 3:05 PM Page 152 152 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Josh said that Theorem 4.9 could also have been |
proved by showing that AEC BED. Do you agree with Josh? Explain. 2. The statement of Theorem 4.7 is “If two angles form a linear pair then they are supplemen- tary.” Is the converse of this theorem true? Justify your answer. Developing Skills In 3–11, in each case write a proof, using the hypothesis as the given and the conclusion as the statement to be proved. 3. If mACD mDCB 90, B DCA, and A DCB, then A and B are complements. C A D B h ACFG h BCDE 4. If and intersect at C and ADC BFC, then ADE BFG. 5. If ADB is a right angle and CE'DBE, then ADB CEB. If ABC and BCD are right angles, and EBC ECB, then EBA ECD. 7. If ABC is a right angle and DBF is a right angle, then ABD CBF 14365C04.pgs 7/12/07 3:05 PM Page 153 Proving Theorems About Angles 153 8. If g EF intersects g DC and mBHG mCGH, then BHG DGE. at H and g AB at G, 9. If h ABD and h ACE intersect at A, and ABC ACB, then DBC ECB AEB g CED 10. If and AEC CEB, then intersect at E, and g'CED g AEB. C B A E D 11. If ABC is a right angle, and BAC is complementary to DBA, then BAC CBD. B A D C In 12–15, g AEB and g CED intersect at E. 12. If mBEC 70, find mAED, mDEB, and mAEC. 13. If mDEB 2x 20 and mAEC 3x – 30, find mDEB, mAEC, mAED, and mCEB. 14. If mBEC 5x – 25 and mDEA 7x – 65, find mBEC, mDEA, mDEB, and mAEC. 15. If mBEC y, mDEB 3x, and mDEA 2x – y, find mCEB, mBED, mDEA, and mAEC. D A E B C |
14365C04.pgs 7/12/07 3:05 PM Page 154 154 Congruence of Line Segments, Angles, and Triangles g RS 16. intersects g LM at P, mRPL x y, mLPS 3x 2y, mMPS 3x – 2y. a. Solve for x and y. b. Find mRPL, mLPS, and mMPS. 17. Prove Theorem 4.2, “If two angles are straight angles, then they are congruent.” 18. Prove Theorem 4.5, “If two angles are supplements of the same angle, then they are congruent.” 19. Prove Theorem 4.6, “If two angles are congruent, then their supplements are congruent.” Applying Skills 20. Two angles form a linear pair. The measure of the smaller angle is one-half the measure of the larger angle. Find the degree measure of the larger angle. 21. The measure of the supplement of an angle is 60 degrees more than twice the measure of the angle. Find the degree measure of the angle. 22. The difference between the degree measures of two supplementary angles is 80. Find the degree measure of the larger angle. 23. Two angles are complementary. The measure of the larger angle is 5 times the measure of the smaller angle. Find the degree measure of the larger angle. 24. Two angles are complementary. The degree measure of the smaller angle is 50 less than the degree measure of the larger. Find the degree measure of the larger angle. 25. The measure of the complement of an angle exceeds the measure of the angle by 24 degrees. Find the degree measure of the angle. 4-4 CONGRUENT POLYGONS AND CORRESPONDING PARTS Fold a rectangular sheet of paper in half by placing the opposite edges together. If you tear the paper along the fold, you will have two rectangles that fit exactly on one another. We call these rectangles congruent polygons. Congruent polygons are polygons that have the same size and shape. Each angle of one polygon is congruent to an angle of the other and each edge of one polygon is congruent to an edge of the other. In the diagram at the top of page 155, polygon ABCD is congruent to polygon EFGH. Note that the congruent polyg |
ons are named in such a way that each vertex of ABCD corresponds to exactly one vertex of EFGH and each vertex of EFGH corresponds to exactly one vertex of ABCD. This relationship is called a one-to-one correspondence. The order in which the vertices are named shows this one-to-one correspondence of points. 14365C04.pgs 7/12/07 3:05 PM Page 155 Congruent Polygons and Corresponding Parts 155 ABCD EFGH indicates that: • A corresponds to E; E corresponds to A. • B corresponds to F; F corresponds to B. • C corresponds to G; G corresponds to C. • D corresponds to H; H corresponds to D. A D B E F H C G Congruent polygons should always be named so as to indicate the corre- spondences between the vertices of the polygons. Corresponding Parts of Congruent Polygons In congruent polygons ABCD and EFGH shown above, vertex A corresponds to vertex E. Angles A and E are called corresponding angles, and A E. In this example, there are four pairs of such corresponding angles: A E B F C G D H In congruent polygons, corresponding angles are congruent. In congruent polygons ABCD and EFGH, since A corresponds to E and B corresponds to F, EF In this example, there are four pairs of such corresponding sides: are corresponding sides, and and AB AB EF. AB EF BC > FG CD > GH DA > HE In congruent polygons, corresponding sides are congruent. The pairs of congruent angles and the pairs of congruent sides are called the corresponding parts of congruent polygons. We can now present the formal definition for congruent polygons. DEFINITION Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent. This definition can be stated more simply as follows: Corresponding parts of congruent polygons are congruent. Congruent Triangles The smallest number of sides that a polygon can have is three. A triangle is a polygon with exactly three sides. In the figure, ABC and DEF are congruent triangles. C F A B D E 14365C04.pgs 7/12/07 3:05 PM |
Page 156 156 Congruence of Line Segments, Angles, and Triangles C F B E A D The correspondence establishes six facts about these triangles: three facts about corresponding sides and three facts about corresponding angles. In the table at the right, these six facts are stated as equalities. Since each congruence statement is equivalent to an equality statement, we will use whichever notation serves our purpose better in a particular situation. Congruences Equalities AB > DE BC > EF AC > DF A D B E C F AB DE BC EF AC DF mA mD mB mE mC mF For example, in one proof, we may and in another AC > DF prefer to write proof to write AC = DF. In the same way, we might write C F or we might write mC = mF. From the definition, we may now say: Corresponding parts of congruent triangles are equal in measure. In two congruent triangles, pairs of corresponding sides are always opposite pairs of corresponding angles. In the preceding figure, ABC DEF. The order in which we write the names of the vertices of the triangles indicates the one-to-one correspondence. 1. A and D are corresponding congruent angles. 2. BC is opposite A, and EF is opposite D. 3. BC and EF are corresponding congruent sides. Equivalence Relation of Congruence In Section 3-5 we saw that the relation “is congruent to” is an equivalence relation for the set of line segments and the set of angles. Therefore, “is congruent to” must be an equivalence relation for the set of triangles or the set of polygons with a given number of sides. 1. Reflexive property: ABC ABC. 2. Symmetric property: If ABC DEF, then DEF ABC. 3. Transitive property: If ABC DEF and DEF RST, then ABC RST. Therefore, we state these properties of congruence as three postulates: Postulate 4.11 Any geometric figure is congruent to itself. (Reflexive Property) Postulate 4.12 A congruence may be expressed in either order. (Symmetric Property) 14365C04.pgs 7/12/07 3:05 PM Page 157 Congruent Polygons and Corresponding Parts 157 Postulate 4.13 Two geometric figures congruent to the same geometric figure are congruent to |
each other. (Transitive Property) Exercises Writing About Mathematics 1. If ABC DEF, then AB > DE. Is the converse of this statement true? Justify your answer. 2. Jesse said that since RST and STR name the same triangle, it is correct to say RST STR. Do you agree with Jesse? Justify your answer. Developing Skills In 3–5, in each case name three pairs of corresponding angles and three pairs of corresponding sides in the given congruent triangles. Use the symbol to indicate that the angles named and also the sides named in your answers are congruent. 3. ABD CBD 4. ADB CBD 5. ABD EBC In 6–10, LMNP is a square and each statement. LSN and PSM bisect each other. Name the property that justifies 6. LSP LSP 7. If LSP NSM, then NSM LSP. 8. If LSP NSM and NSM NSP, then LSP NSP. 9. If LS SN, then SN LS. 10. If PLM PNM, then PNM PLM. P N S L M 14365C04.pgs 7/12/07 3:05 PM Page 158 158 Congruence of Line Segments, Angles, and Triangles 4-5 PROVING TRIANGLES CONGRUENT USING SIDE, ANGLE, SIDE The definition of congruent polygons states that two polygons are congruent if and only if each pair of corresponding sides and each pair of corresponding angles are congruent. However, it is possible to prove two triangles congruent by proving that fewer than three pairs of sides and three pairs of angles are congruent. Hands-On Activity In this activity, we will use a protractor and ruler, or geometry software. Use the procedure below to draw a triangle given the measures of two sides and of the included angle. STEP 1. Use the protractor or geometry software to draw an angle with the given measure. STEP 2. Draw two segments along the rays of the angle with the given lengths. The two segments should share the vertex of the angle as a common endpoint. STEP 3. Join the endpoints of the two segments to form a triangle. STEP 4. Repeat steps 1 through 3 to draw a second triangle using the same angle measure and segment lengths. a. Follow the steps to draw two different triangles with each of the given side- |
angle-side measures. (1) 3 in., 90°, 4 in. (2) 5 in., 40°, 5 in. (3) 5 cm, 115°, 8 cm (4) 10 cm, 30°, 8 cm b. For each pair of triangles, measure the side and angles that were not given. Do they have equal measures? c. Are the triangles of each pair congruent? Does it appear that when two sides and the included angle of one triangle are congruent to the corresponding sides and angle of another, that the triangles are congruent? This activity leads to the following statement of side-angle-side or SAS triangle congruence, whose truth will be assumed without proof: Postulate 4.14 Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other. (SAS) In ABC and DEF, BA > ED, ABC DEF and. It follows that ABC DEF. The postulate used here is abbreviated SAS. BC > EF B E A C D F 14365C04.pgs 7/12/07 3:05 PM Page 159 Proving Triangles Congruent Using Side, Angle, Side 159 Note: When congruent sides and angles are listed, a correspondence is established. Since the vertices of congruent angles, ABC and DEF, are B and E, B corresponds to E. Since BA ED and B corresponds to E, then A corresponds to D, and since BC EF and B corresponds to E, then C corresponds to F. We can write ABC DEF. But, when naming the triangle, if we change the order of the vertices in one triangle we must change the order in the other. For example, ABC, BAC, and CBA name the same triangle. If ABC DEF, we may write BAC EDF or CBA FED, but we may not write ABC EFD. EXAMPLE 1 Given: ABC, CD CD'AB. is the bisector of AB, and C Prove: ACD BCD Prove the triangles congruent by SAS. Proof Statements 1. CD bisects AB. 1. Given. A D B Reasons 2. D is the midpoint of AB. S 3. AD > DB CD'AB 4. 5. ADC and BDC are right angles. A 6. ADC BDC S 7. CD > CD |
8. ACD BCD 2. The bisector of a line segment intersects the segment at its midpoint. 3. The midpoint of a line segment divides the segment into two congruent segments. 4. Given. 5. Perpendicular lines intersect to form right angles. 6. If two angles are right angles, then they are congruent. 7. Reflexive property of congruence. 8. SAS (steps 3, 6, 7). Note that it is often helpful to mark segments and angles that are congruent with the same number of strokes or arcs. For example, in the diagram for this are marked with a single stroke, ADC and BDC are proof, is marked with an “” to indimarked with the symbol for right angles, and cate a side common to both triangles. and AD DB CD 14365C04.pgs 7/12/07 3:05 PM Page 160 160 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Each of two telephone poles is perpendicular to the ground and braced by a wire that extends from the top of the pole to a point on the level ground 5 feet from the foot of the pole. The wires used to brace the poles are of unequal lengths. Is it possible for the telephone poles to be of equal height? Explain your answer. 2. Is the following statement true? If two triangles are not congruent, then each pair of corre- sponding sides and each pair of corresponding angles are not congruent. Justify your answer. Developing Skills In 3–8, pairs of line segments marked with the same number of strokes are congruent. Pairs of angles marked with the same number of arcs are congruent. A line segment or an angle marked with an “” is congruent to itself by the reflexive property of congruence. In each case, is the given information sufficient to prove congruent triangles using SAS? 3. B F 4. A C D E 6. A C D B B D A 7. C C F A B D E 5. D C A B 8. A B E C D In 9–11, two sides or a side and an angle are marked to indicate that they are congruent. Name the pair of corresponding sides or corresponding angles that would have to be proved congruent in order to prove the triangles congruent by |
SAS. 9. D E A B C 10. C 11. A B A D B E D C 14365C04.pgs 7/12/07 3:05 PM Page 161 Proving Triangles Congruent Using Angle, Side, Angle 161 Applying Skills In 12–14: a. Draw a diagram with geometry software or pencil and paper and write a given state- ment using the information in the first sentence. b. Use the information in the second sentence to write a prove statement. c. Write a proof. and ABC DBE bisect each other. Prove that ABE CBD. 12. 13. ABCD is a quadrilateral; AB = CD; BC = DA; and DAB, ABC, BCD, and CDA are separates the quadrilateral into two congruent tri- AC right angles. Prove that the diagonal angles. 14. PQR and RQS are a linear pair of angles that are congruent and PQ QS. Prove that PQR RQS. 4-6 PROVING TRIANGLES CONGRUENT USING ANGLE, SIDE, ANGLE In the last section we saw that it is possible to prove two triangles congruent by proving that fewer than three pairs of sides and three pairs of angles are congruent, that is, by proving that two sides and the included angle of one triangle are congruent to the corresponding parts of another. There are also other ways of proving two triangles congruent. Hands-On Activity In this activity, we will use a protractor and ruler, or geometry software. Use the procedure below to draw a triangle given the measures of two angles and of the included side. STEP 1. Use the protractor or geometry software to draw an angle with the first given angle measure. Call the vertex of that angle A. STEP 2. Draw a segment with the given length along one of the rays of A. One of the endpoints of the segment should be A. Call the other endpoint B. A A B 14365C04.pgs 7/12/07 3:05 PM Page 162 162 Congruence of Line Segments, Angles, and Triangles STEP 3. Draw a second angle of the triangle using the other given angle measure. Let B be the ver- tex of this second angle and let the sides of this angle. h BA be one of STEP 4. Let C be the intersection of the rays of A |
and B that are not on g AB. STEP 5. Repeat steps 1 through 4. Let D the vertex of the first angle, E the vertex of the second angle, and F the intersection of the rays of D and E. C A B F D E a. Follow the steps to draw two different triangles with each of the given angle-side-angle measures. (1) 65°, 4 in., 35° (2) 60°, 4 in., 60° (3) 120°, 9 cm, 30° (4) 30°, 9 cm, 30° b. For each pair of triangles, measure the angle and sides that were not given. Do they have equal measures? c. For each pair of triangles, you can conclude that the triangles are congruent. The triangles formed, ABC and DEF, can be placed on top of one another so that A and D coincide, B and E coincide, and C and F coincide. Therefore, ABC DEF. This activity leads to the following statement of angle-side-angle or ASA triangle congruence, whose truth will be assumed without proof: Postulate 4.15 Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other. (ASA) Thus, in ABC and DEF, if B E,, and A D, it follows that ABC DEF. The postulate used here is abbreviated ASA. We will now use this postulate to prove two triangles congruent. BA > ED 14365C04.pgs 7/12/07 3:05 PM Page 163 Proving Triangles Congruent Using Angle, Side, Angle 163 EXAMPLE 1 Given: CED and AEB midpoint of BD'BE. intersect at E, E is the AC'AE, and AEB, Prove: AEC BDE Prove the triangles congruent by ASA. D B E A C Proof Statements Reasons g AEB 1. and g CED intersect at E. 1. Given. A 2. AEC BED 2. If two lines intersect, the vertical angles are congruent. 3. E is the midpoint of. AEB 3. Given. S 4. AE > BE AC'AE and 5. 6. A and B are right angles. BD'BE A 7. A B 8. AEC BED Exercises Writing About Mathematics 4. |
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