MATH 225: Foundations of Higher Matheamatics. Dr. Morton. Chapter 2: Logic (This is where we begin setting the stage for proofs!)

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1 MATH 225: Foundations of Higher Matheamatics Dr. Morton Chapter 2: Logic (This is where we begin setting the stage for proofs!) New Problem from 2.5 page 3 parts 1,2,4: Suppose that we have the two open sentences P (x) x is a multiple of 3; Q(x) x < 9 over the domain S = N. For which x S are the following true? Make sure to show all work and give all possible scenarios which lead to the sentence being true. Solution: Recall from class today the following: P (x) is true on {3, 6, 9, 12,...} P (x) is false on {1, 2, 4, 5, 7, 8,...} Q(x) is true on {1, 2, 3, 4, 5, 6, 7, 8} Q(x) is false on {9, 10, 11, 12,...} 1. P (x) Q(x) Solution: There are two different possibilities we need to consider for when this is true: First, when both the hypothesis and conclusion are true, and secondly when the hypothesis is false. 1. Hyp/Concl true: Then P (x) is true AND Q(x) is true. Thus x is in both {3, 6, 9, 12,...} and {1, 2, 3, 4, 5, 6, 7, 8}. This is the set {3, 6}. 2. Hyp false: Then Q(x) is false: thus x is in {9, 10, 11, 12,...} For the entire statement to be true, then, it can come from the first type OR the second type, i.e. x can be in {3, 6} or in {9, 10, 11, 12,...}. Thus x is in {3, 6, 9, 10, 11, 12, 13,...} 2. ( P (x)) Q(x) Solution: There is only one possibility we need to consider for when this is true: When both ( P (x) AND Q(x) are true. Thus P (x) is false and Q(x) is true. We did much of this in class in part 3. We found that P (x) was F on {1, 2, 4, 5, 7, 8,...}, while Q(x) is T on {1, 2, 3, 4, 5, 6, 7, 8}. For this entire AND statement to be true, then, x must be in {1, 2, 4, 5, 7, 8,...} and {1, 2, 3, 4, 5, 6, 7, 8}, i.e. x is in {1, 2, 4, 5, 7, 8}. 4. [ P (x)] Q(x) Solution: Again, we did much of this in class today. Recall that in an iff statement, both parts must be true or both parts must be false. 1. Hyp/Concl both true: Then P (x) is true AND Q(x) is true, in other words P (x) is false AND Q(x) is true. We did this in part 2. above; the solution is {1, 2, 4, 5, 7, 8} 1

2 2. Hyp/Concl both false: Then P (x) is false AND Q(x) is false, in other words P (x) is true AND Q(x) is false. We know that P (x) is true on {3, 6, 9, 12,...} and Q(x) is false on {9, 10, 11, 12,...}. Thus for this part to be true, x must be in both {3, 6, 9, 12,...} and {9, 10, 11, 12,...}. Thus this part works when x is in {9, 12, 15, 18,...}. For the entire statement to be true, then, it can come from the first type OR the second type, i.e. x can be in {1, 2, 4, 5, 7, 8} or in {9, 12, 15, 18,...}. Thus x is in {1, 2, 4, 5, 7, 8, 9, 12, 15, 18, 21,...} 2

3 Page 10 Chapter 2 handout problem 4: What is the converse of If 3 is prime then 5 is odd. Is the original true? The converse? Solution: Converse is If 5 is odd then 3 is prime. Here the original is of the form If T then T =T, and the converse is of the form If T then T=T. Page 11 Chapter 2 handout problem 9bc: Are the following true or false? Why? b. Every integer is a real number if and only if every real number is an integer. Solution: This is of the form T if and only if F which is FALSE. c. 16 is prime if and only if π is rational. Solution: This is of the form F if and only if F which is TRUE. Page 11 Chapter 2 handout problem 10: Rephrase Example 9(c) using the equivalent phrasing found in example 8. Solution: is prime if and only if π is rational is prime is equivalent to π being rational is prime iff π is rational is prime is a necessary and sufficient condition for π to be rational. 3

4 Page 11 Chapter 2 handout problem 11: Consider the open sentences where n N. P (n) n is divisible by 6, Q(n) n is divisible by 3, 1. Write the implication P (n) Q(n) in words or symbols. (I.e., form the open sentence P (n) Q(n).) Solution: If n is divisible by 6 then n is divisible by Is part (a) true for each n N? Solution: We need to be careful here. An implication is true when both the hypothesis and conclusion are true OR when the hypothesis is false. The hyp is true on {6, 12, 18, 24,..}. The conclusion is true on {3, 6, 9, 12, 15, 18,...}. For them to both be true, we look at their intersection and find it to be {6, 12, 18,...}. The hyp is false on {1, 2, 3, 4, 5, 7, 8, 9, 10, 11,...} i.e. all things that are not multiples of 6. Thus the entire statement is true when one OR the other of these parts is true. Thus it is true on {6, 12, 18,...} OR {1, 2, 3, 4, 5, 7, 8, 9, 10, 11,...}, which is actually N 3. Write the converse of the implication P (n) Q(n) in words or symbols. (What is the converse?) Solution: If n is divisible by 3 then n is divisible by Is part (c) true for each n N? Solution: We need to be careful here. An implication is true when both the hypothesis and conclusion are true OR when the hypothesis is false. The hyp is true on {3, 6, 9, 12, 15, 18,...}. The conclusion is true on {6, 12, 18, 24,..}. For them to both be true, we look at their intersection and find it to be {6, 12, 18,...}. The hyp is false on {1, 2, 4, 5, 7, 8,,...} i.e. all things that are not multiples of 3. Thus the entire statement is true when one OR the other of these parts is true. Thus it is true on {6, 12, 18,...} OR {1, 2, 4, 5, 7, 8, 10, 11,...}, which is actually {1, 2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 17...} 5. Give a value of n so that P (n) Q(n) is true. It is true when the hypothesis and conclusion are either both true or both false. For example, n = 6 (it is divisible by both 3 and 6). 6. Give a value of n so that P (n) Q(n) is false. It is false when the hypothesis and conclusion do not agree. For example, n = 9 (it is divisible by 3 but not by 6). 4

5 Page 11 Chapter 2 handout problem 12: Consider the open sentences where x, y R. P (x, y) x < y, Q(x, y) x 2 < y 2, 1. Form the open sentence P (x, y) Q(x, y). Solution: x < y iff x 2 < y 2 2. For the open sentence above, give a pair (x, y) so that the statement is true and a pair (x, y) so that the statement is false. Solution: They are both false when (x, y) = (1, 1), making the iff statement true. On ( 1, 1) the hyp is true but the conclusion is false so the iff statement is false. 5

6 Page 13 Chapter 2 handout, Problem 6: Find the truth table for P ( ( P )). P P ( P ) P ( ( P )) T F T T F T F T Page 14 Chapter 2 handout, Problem 7a: ( P ) (Q P ) P Q (Q P ) ( P ) ( P ) (Q P ) T T T F T T F T F T F T F T T F F T T T This is a tautology. 6