MAT 101 Solutions Exam 2 (Logic, Part I) Multiple-Choice Questions 1. D Because this sentence contains exactly ten words, it is stating that it is false. But if it is taken to be false, then it has to be true and if it is assumed to be true then it has to be false based on what it says a contradiction either way! So this sentence is not a statement. Check that all the other sentences are actual statements. The first statement is false 2017 since 2 is an even prime. The second statement is false since 10 is vastly greater 10 2017 10 than 2017 (by comparison: 10 has 2,018 digits while 2017 has 34 digits!) The third statement is false based on the truth table for biconditionals (to be studied later in this chapter) and the fifth statement is false for all integer values of x. 2. B Plug-in the given truth values for the components pqr,, into the compound statement and reduce based on the basic truth tables for negation, conjunction and disjunction: ~ T ~ F F ~ T T F ~ T F ~ T ~ F 3. C First, note that the first part of the compound statement p ~ p is always false. Based on the basic truth table for disjunction, if q is true then p~ p q is true; if q is false then p~ p q is false. Hence, the compound statement is only true when q is true. 4. D. Six components can be either true or false. This implies a total of in the truth table of this compound statement. 6 2 64 rows 5. A Recall that a statement of the form It is not the case that w or h is symbolically written as ~ w h. 6. B The first statement is actually the conjunction ~ w ~ h. The third statement is symbolically written as ~ w h. The fourth statement is an exclusive disjunction logically different from the second statement.
7. B The first and fourth statements are logically equivalent to the conditional r w ~ h. The third statement is written symbolically as the conditional w~ h ~ r. Problem 1 a) Here we identify two component statements: p You may say I m a dreamer. q I am the only one. The statement is then written symbolically as the conjunction p ~ q since the connective but is logically equivalent to and. b) Here we identify two component statements: p You are ready. q Here I come. The statement is then written symbolically as follows: p~ p q. Note that the comma in the quotation acts logically as the connective and. c) Here we identify three component statements: p You lean on me. q You are strong. r I ll be your friend. The statement is then written symbolically as follows: ~ q p ~ q since the connective when is logically equivalent to if.
Problem 2 a) This arithmetical statement is verbalized as 2017 is less than or equal to 2 11, and can thus be written symbolically as the following disjunction: 11 11 2017 2 2017 2. Using De Morgan s Law, we negate this statement as follows: 11 11 11 11 11 11 11 2017 2 2017 2 2017 2. ~ 2017 2 2017 2 ~ 2017 2 ~ 2017 2 This is verbally stated as 2017 is greater than 11 2. b) The statement is negated (straightforwardly) using De Morgan s Law as follows: She was willing to join the party and her car did not break down that day. c) Let p represent the statement Everyone on L.I. was affected by super storm Sandy and q represent the statement Everyone lost power in their homes. ~ q is then the statement Some did not lose power in their homes. The overall statement is therefore written symbolically as the conjunction p ~ q since the word but is logically equivalent to the connective and. Using one of De Morgan s laws, we have the following equivalency: ~ p ~ q ~ p q Since ~ p is the statement Some on L.I. were not affected by super storm Sandy, we can write the negation of the original statement as follows: Some on L.I. were not affected by super storm Sandy or everyone lost power in their homes.
Problem 3 a) Using the alternate method for constructing truth tables, we get the following: p q ~ p ~ p q T T F T F T F F F T F T T T F F F T T F ~ ~ p q p q ~ q p ~ q T T F T T F T T F T F F F F T T Since both statements have different truth tables, they are not logically equivalent. b) By De Morgan s Law, we have: ~ ~ p q ~ ~ p ~ q p ~ q. However, it is clear that p ~ q p ~ q. (e.g. I ordered a burger and fries is not equivalent to I ordered a burger or fries. ) You could also argue with a counterexample: Suppose pq, have the same truth value (i.e. both are true or both are false). Then the statement ~ ~ p q [Check this.] equivalent. has to be false while the statement p ~ q has to be true. This, of course, implies that the two statements are not logically
Problem 4 a) p q r ~ p ~ q ~ p q ~ q r (~ p q) (~ q r ) T T T F F T T T T T F F F T F T T F T F T F T T T F F F T F T T F T T T F T T T F T F T F T F T F F T T T T T T F F F T T T T T The final column for (~ p q) (~ q r) consists only of true values. Therefore, the statement (~ p q) (~ q r) is a tautology. b) The three component statements given here are all clearly false. However, the overall compound statement (~ p q) (~ q r) is still true, even with these false components, by row 8 of the truth table constructed above. Bonus Problem Either p or q is the exclusive disjunction that states p or q, but not both. This is symbolically written as the following conjunction: p q ~ p q. It is false when both components pq, have identical truth values and true when both components pq, have different truth values.