1. Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 10 By Herbert I. Gross and Richard A. Medeiros next

2. Suppose we know it’s true that… “If Tom gets 100 on the final exam he will get an A in the course.” Does this statement imply that if Tom doesn’t get 100 on the final exam he won't get an A in the course? Explain. Problem 1a © 2007 Herbert I. Gross Answer: No next

3. Answer: No Solution to 1a: It’s fairly obvious for example that if Tom got a 0 on the final exam, then he didn't get an A in the course. However, if his grade on the final exam was 99, then he probably did get an A in the course. next © 2007 Herbert I. Gross The point is: knowing only what will happen if Tom gets 100 on the final exam doesn’t tell us anything for sure about what will happen if he does not get that 100 on the final exam. next

4. Note 1a People often use faulty logical arguments to arrive at false conclusions. For example, they may jump to such false conclusions as… If all A’s are B’s, then all non-A’s are non-B’s. © 2007 Herbert I. Gross If all A’s are B’s then all non-A’s are non-B’s. So, let’s look at a non-math version of such an argument. If someone says, “All voters are 21 or older” can you conclude that “all non-voters are under 21 years old”? next

5. © 2007 Herbert I. Gross Does the conclusion “All non-voters are under 21” follow from the assumption that “All voters are 21 or older”? So while it is true that every person under 21 is a non-voter, it may not be true that all non-voters are under-21 years old. next If all A’s are B’s then all non-A’s are non-B’s. True/False The answer is “No!” because, while all voters do have to be 21 or older, we know that not all who are 21 or older actually vote. next

6. Again, suppose we know it's true that… “If Tom gets 100 on the final exam, he will get an A in the course.” Does this statement imply that if Tom got an A in the course that he got 100 on the final exam? Explain. Problem 1b © 2007 Herbert I. Gross Answer: No next

7. Answer: No Solution to 1b: It’s certainly possible that Tom got 100 on the final exam. However, this does not follow inescapably from the fact that Tom got an A in the course. next © 2007 Herbert I. Gross For example, Tom might still have received an A even if he only got 95 on the final. next

8. Again, suppose we know it’s true that… “If Tom gets 100 on the final exam he will get an A in the course.” Does this statement imply that if Tom did not get an A in the course that he did not get 100 on the final exam? Explain. Problem 1c © 2007 Herbert I. Gross Answer: Yes next

9. Answer: Yes Solution to 1c: The statement implies that if Tom got 100 on the final exam, then he would get an A in the course. next © 2007 Herbert I. Gross Based on that implication… if he did not get an A, it means he did not get 100 on the final exam. next

10. © 2007 Herbert I. Gross As shown in our figure below, if points (such as c, d, and e) are outside of square B, they must also be outside of square A. next BB bb In other words, if “all A’s are B’s”, then it is also true that “all non-B’s are non-A’s”. cc dd ee A a Review

11. next Is the following argument valid? If all Parisians are Europeans and if all Frenchmen are European, then all Parisians are Frenchmen. Explain. Problem 2 © 2007 Herbert I. Gross Answer: No next

12. Notes… Truth vs. Validity © 2007 Herbert I. Gross next This exercise illustrates the difference between a statement being true and an argument being valid. Namely: truth involves a judgment about a given statement while validity involves knowing whether the conclusion in an argument follows inescapably from the given assumptions.

13. Notes… Truth vs. Validity © 2007 Herbert I. Gross next For example, while the conclusion, “All Parisians are Frenchmen” is true; its truth does not follow inescapably from the given assumptions. Namely: if we let A denote the set of all Parisians; B, the set of all Frenchmen; and C, the set of all Europeans, we see that the argument has the form: “If all A’s are C’s and if all B’s are C’s, then all A’s are B’s.”

14. © 2007 Herbert I. Gross To show that the argument isn’t valid we may represent the given information as follows… C A B and all C’s are B’s, Looks Can Be Deceiving All A’s are C’s and all B’s are C’s, next In this diagram we see that no A’s are B’s. Hence, the conclusion (All A’s are B’s) does not follow inescapably from the given assumptions.

15. © 2007 Herbert I. Gross In this diagram notice that both of the assumptions and the conclusion are obeyed; namely. … (2) All B’s are C’s. Looks Can Be Deceiving However, for an argument to be valid it isn’t enough that the conclusion might follow from the assumptions. Rather it must follow from the assumptions. next (1) All A’s are C’s. (3) All A’s are B’s. CC BB A

16. © 2007 Herbert I. Gross next If all A’s are C’s, and all B’s are C’s, then all A’s are B’s. A represents Germans (instead of Parisians), C represents Europeans, and B represents Frenchmen, then the statements “all A’s are C’s” (that is, all Germans are Europeans), and “all B’s are C’s” (that is, all Frenchmen are Europeans) are true, but the conclusion all A’s are B’s (all Germans are Frenchmen) is false. As a specific “real life” example that shows the above argument is invalid, suppose..

17. next Important Point © 2007 Herbert I. Gross If the assumptions are true and the argument is valid, then the conclusion will also be true. In terms of our course… the connection between truth and validity is that… next

18. © 2007 Herbert I. Gross The statement… “If A is true, then B is true”. does not mean the same thing as the statement… “If A is not true, then B is not true”. next Summary On the other hand, the statement… “If A is true, then B is true” does mean the same thing as the statement “If B is not true, then A is not true”

19. © 2007 Herbert I. Gross In terms of how our game is played, until we can show that a mathematical statement follows inescapably from the rules we’ve accepted; we cannot use it as a “fact. Summary

20. next © 2007 Herbert I. Gross Application to Algebra For example, when we find that the value of x for which x + 3 = 7 is x = 4 what we have proven is that if x + 3 = 7, then x = 4. In terms of our present discussion, this can be paraphrased into the equivalent form … if x ≠ 4, then x + 3 ≠ 7.

21. next © 2007 Herbert I. Gross next Application to Algebra In other words, checking to see whether your answer is correct is actually part of the proof. Let’s go over this more slowly. When we solve the equation x + 3 = 7 to obtain x = 4, we have shown that if there is an answer then the answer must be x = 4. We then have to replace x by 4 to see whether x + 3 = 7 is a true statement. If it is true, then x = 4 is the only value of x for which this is true; but if it is false, then there is no value of x for which x + 3 = 7.

22. The Power of Deductive Reasoning © 2007 Herbert I. Gross We could have solved x + 3 = 7 by inductive reasoning (trial and error). For example… next x x + 3 14 25 36 47 4 7 next

23. The Power of Deductive Reasoning © 2007 Herbert I. Gross However, while our inductive procedure showed that x = 4 was a solution of the equation x + 3 = 7; in itself it did not prove it was the only solution. next On the other hand, our algebraic solution (i.e. deductive reasoning) proved that if x was not equal to 4, then x + 3 was not equal to 7.

24. Preface to Problem 3 © 2007 Herbert I. Gross Now that we’ve explained how the game is played, we will become more informal and accept without proof such statements as 9 + 6 = 15 and that equals subtracted from equals are equal, etc. next

25. For what value of m does… (9 + 6) + m = 35? Problem 3a © 2007 Herbert I. Gross Answer: 20 next

26. © 2007 Herbert I. Gross (9 + 6) + m = 35 15 Answer: 20 Solution to 3a: (9 + 6) + m = 35 We may replace 9 + 6 by 15 in the equation… And we may then subtract 15 from both sides of our equation to obtain m = 20. 15 + m = 35 – 15 – 15 m = 20 next

27. Note © 2007 Herbert I. Gross (9 + 6) + m ≠ 35 In terms of our previous discussion, what we have shown so far is that if m ≠ 20 then… (9 + 6) + = 35 20 So to complete our proof we must replace m by 20 in our equation (9 + 6) + m = 35 and see if this results in a true statement. Doing this we obtain… which is indeed a true statement. next

28. For what value of m does… 9 + (6 + m) = 35? Problem 3b © 2007 Herbert I. Gross Answer: 20 next

29. © 2007 Herbert I. Gross Answer: 20 Solution to 3b: We cannot just replace 9 + 6 by 15 in the equation 9 + (6 + m) = 35 because the grouping symbols tell us to add 6 and m first, and only then add that sum to 9. …which is the same equation that we solved in part (a). next (9 + 6) + m = 35 However, by the associative property of addition, we may rewrite our equation, 9 + (6 + m) = 35, in the equivalent form… next

30. © 2007 Herbert I. Gross Solution to 3b: 9 + (6 + m) = 35 We know from problem 3a that m = 20. Thus, to complete our proof we replace m by 20 in our equation, obtaining the true statement… In summary… 9 + ( 6 + m) = 35, if and only if m = 20. next 9 + ( 6 + 20 ) = 35 next

31. Key Point © 2007 Herbert I. Gross The key point here was to illustrate how we use our rules to re-group expressions in ways that make it easier for us to work with them. next For example, even though you might have read Exercise 3b immediately as being equivalent to Exercise 3a, we wanted to emphasize the validity of being able to change the way terms are grouped.

32. For what value of m does… 9 + (m + 6) = 35? Problem 3c © 2007 Herbert I. Gross Answer: 20 next

33. © 2007 Herbert I. Gross Answer: 20 Solution to 3c: 9 + (m + 6) = 35 In 3b, the problem stated was: 9 + (6 + m) = 35. That is, the 9 and the 6 were separated by the placement of the parentheses. In this part of the exercise, they are also separated by m. However, by using the commutative property of addition we may rewrite… into the equivalent form… next 9 + (m + 6) = 35 9 + (6 + m) = 35 next

34. © 2007 Herbert I. Gross Solution to 3c: Thus, an equivalent form of the equation, 9 + (m + 6) = 35 is 9 + (6 + m) = 35… So to complete our proof, we need only replace m by 20 in the original equation and see that we get the true statement… next 9 + ( m + 6 ) = 35. …which is actually the same equation we solved in part (b) to show that m = 20. 9 + (6 + m) = 35 next 20

35. next Comment © 2007 Herbert I. Gross You might have performed all the above steps mentally. However, we did want you to see the subtle ways in which we use our rules, almost without thinking about them.

36. next Evaluate the following expression when m = 20… (9 – 6) – m Problem 4a © 2007 Herbert I. Gross Answer: - 17 next

37. © 2007 Herbert I. Gross Answer: - 17 Solution to 4a: To evaluate the expression (9 – 6) – m, we replace m by 20 to obtain… (9 – 6) – 20 = 3 – 20 = - 17 next 3 + - 20 =

38. next Evaluate the following expression when m = 20… 9 – (6 – m) Problem 4b © 2007 Herbert I. Gross Answer: 23 next

39. © 2007 Herbert I. Gross Answer: 23 Solution: Starting with our expression 9 – (6 – m), next 9 – ( 6 – 20 ). we replace m by 20 to obtain… next and by the “add the opposite” rule... = 9 + + 14 or 23 next 9 – - 14 Thus, 9 – ( 6 – 20 ) = 9 - - 14 We know that 6 – 20 = 6 + - 20 = - 14.

40. next Key Point © 2007 Herbert I. Gross In other words, subtraction does not have the associative property. Notice that without the grouping symbols, both 4a and 4b would look like 9 – 6 – m. But while grouping makes no difference with respect to addition, it does make a difference with respect to subtraction. next

41. Key Point © 2007 Herbert I. Gross Nor does Subtraction have the commutative property. For example: if we replace 6 – m in the expression 9 – (6 – m) by m – 6, we obtain the expression 9 – (m – 6). If we then replace m by 20 in both of 9 – (6 – m) and 9 – (m – 6), we obtain… next 9 – (6 – m) = 9 – (6 – 20) = 9 – - 14 = 23 9 – (m – 6) = 9 – (20 – 6) = 9 – 14 = - 5

42. next PEMDAS Reminder (see Lesson 2) © 2007 Herbert I. Gross Since the expression… 9 – 6 – m does not have grouping symbols, PEMDAS requires that we read it as… (9 – 6) – m.

43. next For what value of c is it true that… c ÷ (6 ÷ 2) = 60? Problem 5a © 2007 Herbert I. Gross Answer: 180 next

44. © 2007 Herbert I. Gross Answer: 180 Solution to 5a: c ÷ (6 ÷ 2) = 60 next Since 6 ÷ 2 is 3, we may rewrite the above equation as… c ÷ 3 = 60 Since c is being divided by 3 we “undo” this by multiplying both sides of the equation by 3 to obtain… = 180 next c = 60 × 3

45. next © 2007 Herbert I. Gross Solution to 5a: To complete the proof, we replace c by 180 in our equation c ÷ (6 ÷ 2) = 60 to obtain the true statement… = 180 ÷ 3= 60 next 180 ÷ (6 ÷ 2)

46. next Notes © 2007 Herbert I. Gross Note that … 180 ÷ (6 ÷ 2) = 180 ÷ 3 = 60, while (180 ÷ 6) ÷ 2 = 30 ÷ 2 = 15. This will be discussed again in our solution for problem 5b. next In other words: 180 ÷ (6 ÷ 2) ≠ (180 ÷ 6) ÷ 2. So we see that division does not have the associative property.

47. Notes © 2007 Herbert I. Gross Some people prefer using fraction notation to represent division. For example, the equation c ÷ 3 = 60 can be written in the equivalent form… next c3c3 = 60 and in this form it is relatively easy to see that… c3c3 × 3 = 60 × 3 c = 180 c = 60 × 3 next

48. For what value of c is it true that… (c ÷ 6) ÷ 2 = 60 Problem 5b © 2007 Herbert I. Gross Answer: 720 next

49. © 2007 Herbert I. Gross Answer: 720 Solution to 5b: (c ÷ 6) ÷ 2 = 60 The last step we performed on the left hand side of the above equation was to divide by 2. So to “undo” this, we multiply both sides of the equation by 2, to obtain the equivalent statement… To get c by itself on the left-hand side of the equation, we just “undo” dividing by 6. That is, we multiply both sides of the equation by 6 to obtain… c ÷ 6 = 2 × 60 (or 120). c = 120 × 6 or 720. next

50. © 2007 Herbert I. Gross Solution for 5b: (c ÷ 6) ÷ 2 = 60 To complete the proof, we replace c by 720 in our equation and see that we get the true statement… (720 ÷ 6) ÷ 2= 120 ÷ 2= 60 next

51. Notes © 2007 Herbert I. Gross Notice that problems 5a and 5b, while meaning different things, would look alike without the use of the parentheses. That is, both equations would become… …which again demonstrates that division doesn’t obey the associative rule. next c ÷ 6 ÷ 2 = 60

52. next Comments © 2007 Herbert I. Gross By the PEMDAS agreement, we would read c ÷ 6 ÷ 2 = 60 as (c ÷ 6) ÷ 2 = 60 which is the problem we solved in problem 5b. next Notice that (c ÷ 6) ÷ 2 could also have been rewritten in the fraction form… = c ÷ 12 c6c6 ÷ 2 c6c6 × 1212 c 12 next

53. Equivalence © 2007 Herbert I. Gross Notice that the equation (c ÷ 6) ÷ 2 = 60 is equivalent to the simpler equation c ÷ 12 = 60. Important point: while the expression (c ÷ 6) ÷ 2 is not equivalent to the expression c ÷ (6 ÷ 2), it is equivalent to the expression c ÷ (6 × 2). next

54. Equivalence © 2007 Herbert I. Gross If we wanted the expressions to be equivalent after regrouping (c ÷ 6) ÷ 2, we would have to replace the second division sign by a multiplication sign. That is… next is equivalent to… c ÷ 6 ÷ 2 (c ÷ 6) ÷ 2 × ()

55. next Cornbread Model © 2007 Herbert I. Gross The previous discussion becomes more tangible if we think in terms of dividing a “cornbread” into 6 equally sized pieces and then dividing each of these six pieces into 2 equally sized pieces. next c c ÷ 6 (c ÷ 6) ÷ 2 = c ÷ 12 next