Math test with solutions. Introduction to Proofs | MATH 3034, Exams of Mathematics

Download Math test with solutions. Introduction to Proofs | MATH 3034 and more Exams Mathematics in PDF only on Docsity! TEST 1 SOLUTIONS MATH 3034 Spring 2007 1. ( 8 pts) You are given that the statements represented symbolically as follows are all true. • P∨ ∼ Q • ∼ [Q → (R ∧ S)] • Q → (∼ P ∨ S) On your paper list the symbols P , Q, R, and S and label each as true or false. (No justification required.) Solution: P - True Q - True R - False S - True Justification (not required): ∼ [Q → (R ∧ S)] simplifies to Q ∧ (∼ R∨ ∼ S). Therefore Q is true and ∼ R∨ ∼ S is true. Since we are given that P∨ ∼ Q is true and we now know that ∼ Q is false, it follows that P is true. Also, since Q is true and we are given that Q → (∼ P ∨ S) is true, we can conclude that (∼ P ∨ S) is true. But ∼ P is false, so S must be true. We concluded earlier that ∼ R∨ ∼ S is true. But ∼ S is false, so ∼ R must be true; that is, R is false. 2. (8 pts) Each of the statements given in (a) – (d) below is an implication with hypothesis P or Q and with conclusion P or Q, where P and Q are given by: P : 3 > ln 9 Q: 11 is a prime integer. On your paper create a table such as the one indicated below and in the table give the hypothesis (P or Q) and conclusion (P or Q) for each of the statements (a) – (d). (a) For 11 to be a prime integer it is necessary that 3 > ln 9. (b) For 11 to be a prime integer it is sufficient that 3 > ln 9. (c) 3 > ln 9 only if 11 is a prime integer. (d) It must be true that 3 > ln 9 if 11 is a prime integer. Solution: Statement (a) (b) (c) (d) Hypothesis Q P P Q Conclusion P Q Q P 3, (8 pts) Use basic logical equivalences (i.e., theorems) to verify that the statement forms given below are logically equivalent: (P → Q) ∨ (R → S) and (P∧ ∼ Q ∧R) → S 1 Solution: (P → Q) ∨ (R → S) ⇔ (∼ P ∨Q) ∨ (∼ R ∨ S) ⇔ (∼ P ∨Q∨ ∼ R) ∨ S ⇔ ∼ (P∧ ∼ Q ∧R) ∨ S ⇔ (P∧ ∼ Q ∧R) → S 4. (8 pts each) Consider the statement: Statement: For every 2× 2 real matrix A, if there exists a 2× 2 real matrix B such that AB = [ 2 5 1 3 ] then there exists a 2× 2 real matrix C such that AC = [ 1 0 0 1 ] and for every 2× 2 nonzero real matrix D, AD 6= [ 0 0 0 0 ] . (a) Give the (written) contrapositive of the statement in simplified form. (b) Give the (written) negation of the statement in simplified form. Solutions: (a) The contrapositive of the given statement is: For every 2× 2 real matrix A, if either for every 2× 2 real matrix C we haveAC 6= [ 1 0 0 1 ] or there exists a 2× 2 nonzero real matrix D such that AD = [ 0 0 0 0 ] , then for every 2× 2 real matrix B we have AB 6= [ 2 5 1 3 ] (b) The negation of the given statement is: There exist 2× 2 real matrices A and B such that AB = [ 2 5 1 3 ] but either for every 2× 2 real matrix C we have AC 6= [ 1 0 0 1 ] or there exists a 2× 2 nonzero real matrix D such that AD = [ 0 0 0 0 ] . 5. In this problem, consider the following: Given Definition: A function f defined on the real numbers is an increasing function provided for all real numbers a and b, if a ≤ b then f(a) ≤ f(b). (a) (8 pts) Complete the following definition: A function f defined on the real numbers is not an increasing function provided there exist real numbers a and b such that a ≤ b but f(a) > f(b). (b) (11 pts) Let the function f be defined by f(x) = x2 − 4 for all real numbers x. Use the definition from (a) to prove that f is not an increasing function. A Proof: Let a = −2 and b = 0. Then clearly a ≤ b. But f(a) = f(−2) = 0 whereas f(b) = f(0) = −4. Therefore, f(a) > f(b) and f is not increasing. 2