CISP 440 Final Exam
Fill in the blanks below so that argument (b) has the same form as argument (a). Then represent the common form of the arguments using letters to stand for component sentences. ( a ). If Jane is a math major or Jane is a computer science major, then Jane will take Math 150. Jane is a computer science major. Therefore, Jane will take Math 150. ( b ). If logic is easy or ________________ ,then _______________. I will study hard. Therefore, I will get an A in this course.
I (will) study hard. I will get an A in this course. Common form: If p or q, then r. q. Therefore, r.
What is the following statement in the form "If A then B" and the contrapositive of the form? George's attaining age 35 is a necessary condition for his being president of the United States.
If A then B: If George has not attained the age of 35, then he cannot be president of the United States. the contrapositive of the "If A then B" : If George can be president of the United States, then he has attained the age of 35.
p ∨ q p → r q → r ∴ r What is the Rule of Inference?
Proof by Division into Cases
What is the best choice of proving the following statement? The sum of any two rational numbers is rational.
Proof: Suppose r and s are rational numbers. [ We must show that r + s is rational. ] Then, by definition of rational, r = a / b and s = c / d for some integers a, b , c, and d with b ≠≠ 0 and d ≠≠ 0. Thus r + s = abab + cdcd by substitution = ad+bcbdad+bcbd by basic algebra. Let p = a d + b c and q = b d . Then p and q are integer s because products and sums of integers are integers and because a, b, c, and d are all integers. Also q ≠≠ 0 by the zero product property. Thus r + s = pqpq where p and q are integers and q ≠≠ 0 . Therefore, r + s is rational by definition of a rational number. [ This is what was to be shown. ]
Let A = Z+, B = { n ∈ Z | 0 ≤ n ≤ 100}, and C = { 100, 200, 300, 400, 500 }. Evaluate the truth and falsity of the following statement.
True. Every element in C is in C. In general, the definition of subset implies that all sets are subsets of themselves.
De Morgan's laws say (1) that the negation of an and statement is logically equivalent to the ____________ statement in which each component is ______________, and (2) that the negation of an or statement is logically equivalent to the _________________ statement in which each component is _____________ .
or; negated; and; negated
The negation of " if p then q " is ___________________.
p ∧ ∼ q
Use symbols to write the logical form of the following argument. If the argument is valid, identify the rule of inference that guarantees its validity. Otherwise, state whether the converse or the inverse error is made. If I get a Christmas bonus, I'll buy a stereo. If I sell my motorcycle, I'll buy a stereo. ∴ If I get a Christmas bonus or I sell my motorcycle, then I'll buy a stereo.
p ⟶⟶ r q ⟶⟶ r p ∨∨ q ⟶⟶ r valid, proof by division into cases
In each of the following statements determine whether the proposed negation is correct. If it is not, write a correct negation. a. Statement: For all integers n, if n2 is even then n is even. Proposed negation: For all integers n, if n2 is even then n is not even. b. Statement: For all real numbers x1 and x 2 , if x1 = x 2 then x21x12 = x22x22 . Proposed negation: For all real numbers x1 and x 2 , if x21x12 = x22x22 then x1 ≠≠ x 2 .
a. The proposed negation is not correct. Correct negation: There exists an integer n such that n2 is even and n is not even. b. The proposed negation is not correct. Correct negation: ∃ real numbers x1 and x 2 such that x21x12 = x22x22 then x1 ≠≠ x 2 .
Write a negation for the following statements. a. If the square of an integer is odd, then the integer is odd. b. If a function is differentiable then it is continuous
a. There is an integer such that the square of the integer is odd but the integer is not odd. b. There is a function that is differentiable but not continuous.
Write an informal negation for each of the following statements. Be careful to avoid negations that are ambiguous. a. All dogs are friendly. b. All people are happy. c. Some suspicions were substantiated. d. Some estimates are accurate.
a. There is at least one unfriendly dog. b. Some people are unhappy. c. No suspicions were substantiated. d. No estimates are accurate.
"There is an integer that is both prime and even." Let Prime(n) be "n is prime" and Even(n) be "n is even." What would be the best choice of using the notation Prime(n) and Even(n) to rewrite this statement in the following two forms? a. ∃ n such that ___________________ ∧ _______________________________ . b. ∃ _______________________________ n such that __________________________ .
a. ∃ n such that Prime ( n ) ∧ Even ( n ). b. ∃ an even number n such that Prime ( n ).
A negation for "All R have property S" is "There is _______________________ R that _________________________________ ."
some; does not have property S
A universal existential statement is a statement
that is universal because its first part says that a certain property is true for all objects of a given type, and it is existential because its second part asserts the existence of something.
Suppose that r and s are integers. Choose the best answer to prove the following: ∃ an integer k such that 22 r + 18 s = 2 k.
Let k = 11 r + 9 s. Then k is an integer because it is a sum of products of integers; and by substitution, 2 k = 2 ( 11 r + 9 s ), which equals 22 r + 18 s by the distributive law of algebra.
write each of the two statements in symbolic form and determine whether they are logically equivalent. Include a truth table and a few words of explanation. 1. If you paid full price, you didn't buy it at Crown Books. 2. You didn't buy it at Crown Books or you paid full price.
Let p represent "You paid full price" and q represent "You didn't buy it at Crown Books." Thus, "If you paid full price, you didn't buy it at Crown Books" has the form p → q. And "You didn't buy it at Crown Books or you paid full price" has the form q ∨ p.
The rule of universal instantiation says that if some property is true for ______________________________ in a domain, then it is true for __________________________________________.
all elements; any particular element in the domain
Write negation, contrapositive, converse and inverse for the following statement. If P is a square, then P is a rectangle.
Negation: P is a square and P is not a rectangle. Contrapositive : If P is not a rectangle, then P is not a square. Converse : If P is a rectangle, then P is a square. Inverse : If P is not a square, then P is not a rectangle.
If the first two premises of universal modus ponens are written as " If x makes P ( x ) true, then x makes Q ( x ) true " and " For a particular value of a ____________________ ," then the conclusion can be written as "________________________________________________ ."
P ( a ) is true; Q ( a ) is true
Choose the best choice for the following. Evaluate the expressions in the following. I. a. 43 div 9 b. 43 mod 9 II. a. 50 div 7 b. 50 mod 7 III. a. 28 div 5 b. 28 mod 5 IV. a. 30 div 2 b. 30 mod 2
I. a. 4 b. 7 II. a. 7 b. 1 III. a. 5 b. 3 IV a. 15 b. 0
What is the best choice to fill in the following blanks? If a and b are positive integers and a | b, then _____ is less than or equal to _____.
a ; b
What would be the best choice for the following ? Suppose that in standard factored form a = pe11p1e1 pe22p2e2 · · · pekkpkek , where k is a positive integer; p1 , p2 , . . . , pk are prime numbers ; and e1 , e2 ,..., ek are positive integers. a. What is the standard factored form for a2 ? b. Find the least positive integer n such that 25 · 3 · 52 · 73 · n is a perfect square. Write the resulting product as a perfect square. c. Find the least positive integer m such that 22 · 35 · 7 · 11 · m is a perfect square. Write the resulting product as a perfect square.
a. p2e11p12e1 p2e22p22e2 · · · p2ekkpk2ek b. n=42, 25 · 3 · 52 ·73 · n = 58802 c. m = 3 · 7 · 11product = 22 · 35 · 7 · 112 · m = 22 · 36 · 72 · 112 = ( 2 · 33 · 7 · 11 )2 = 41582
What is the best choice for the following statements? a. 32 div 9 b. 32 mod 9
a. 3 b. 5
What is the best choice to answer the following questions? For each of the following values of n and d, find integers q and r such that n = d q + r and 0 ≤ r < d . a. n = 54 , d = 4 b. n = −54 , d = 4 c. n = 54 , d = 70
a. 54 = 4 · 13 + 2 ; hence q = 13 and r = 2. b. -54 = 4 · ( −14 ) + 2 ; hence q = −14 and r = 2 . c. 54 = 70 · 0 + 54 ; hence q = 0 and r = 54.
If n and d are integers and d ≠≠ 0 then n is divisible by d if, and only if, n equals d times some integer. What is the best choice that includes all equivalences of " n is divisible by d," from the following list? a. n is a multiple of d, b. n % d c. d is a factor of n d. n/d e. d is a divisor of n f. d|n
a, c, e and f
Consider the following statement: ∀ basketball players x , x is tall. Which of the following are equivalent ways of expressing this statement? a. Every basketball player is tall. b. Among all the basketball players, some are tall. c. Some of all the tall people are basketball players. d. Anyone who is tall is a basketball player. e. All people who are basketball players are tall. f. Anyone who is a basketball player is a tall person.
a, e and f
Here is a proof that the sum of any two even integers is even. Suppose m and n are particular but arbitrarily chosen even integers. Then m = 2 r for some integer r ( 1 ) , and n = 2 s for some integer s. ( 2 ) Hence m + n = 2 r + 2 s by substitution = 2 ( r + s ) ( 3 ) by factoring out the 2. Now r + s is an integer, (4 ) and so 2 ( r + s ) is even. (5) Thus m + n is even What would be the best choice that above proof steps are valid by universal modus ponens?
( 1 ) If an integer is even, then it equals twice some integer. m is a particular even integer. ∴ m equals twice some integer r . ( 2 ) If an integer is even, then it equals twice some integer. n is a particular even integer. ∴ n equals twice some integer s. ( 3 ) If a quantity is an integer, then it is a real number. r and s are particular integers. ∴ r and s are real numbers. For all a, b, and c, if a, b, and c are real numbers, then ab + ac = a(b + c). 2, r , and s are particular real numbers. ∴ 2 r + 2 s = 2( r + s ). ( 4 ) For all u and v, if u and v are integers, then u + v is an integer. r and s are two particular integers. ∴ r + s is an integer. ( 5 ) If a number equals twice some integer, then that number is even. 2 ( r + s ) equals twice the integer r + s. ∴ 2 ( r + s ) is even
What would be the best choice by using diagrams to test for validity for the following statement? All integers are rational numbers
(rational numbers (integers))
What is the best choice to fill in the following blanks ? According to the quotient-remainder theorem, if an integer n is divided by a positive integer d , the possible remainders are _____________________________________. This implies that n can be written in one of the forms _______________________________________________ for some integer q.
0, 1, 2, ..., ( d−1 ) ; dq , dq + 1 , dq + 2 , ..., dq + ( d − 1 )
Choosing from the following list, what are some of the most common mistakes ,which are provided from out text book, people make when writing mathematical proofs? 1. Arguing from examples. 2. Skipping prof steps 3. Using the same letter to mean two different things 4. Jumping to a conclusion 5. Misusing theorems 6. Assuming what is to be proved 7. Confusion between what is known and what is still to be shown 8. Use of any rather than some 9. Misuse of the word if 10. Misinterpretation of symbols
1, 3, 4, 6, 7, 8, and 9
If the first two premises of universal transitivity are written as " Any x that makes P ( x ) true makes Q ( x ) true" and " Any x that makes Q ( x ) true makes R ( x ) true," then the conclusion can be written as " ___________________________________________________ ."
Any x that makes P ( x ) true makes R ( x ) true.
Let D = { 1, 2, 3, 4, 5 }, and consider the statement ∀ x ∈ D, x2 ≥ x. What would be a proper choice about the above statement?
Check that " x2 ≥ x " is true for each individual x in D. 12 ≥ 1, 22 ≥ 2, 32 ≥ 3, 42 ≥ 4, 52 ≥ 5. Hence "∀ x ∈ D, x2 ≥ x " is true
p ∨ q ∼ p ∴ q What is the Rule of Inference?
Elimination
In the following a single conclusion follows when all the given premises are taken into consideration, but it is difficult to see because the premises are jumbled up. Reorder the premises to make it clear that a conclusion follows logically, and state the valid conclusion that can be drawn. (It may be helpful to rewrite some of the statements in if-then form and to replace some statements by their contrapositives.) I. 1. No birds except ostriches are at least 9 feet tall. 2. There are no birds in this aviary that belong to anyone but me. 3. No ostrich lives on mince pies. 4. I have no birds less than 9 feet high. II. 1. All writers who understand human nature are clever. 2. No one is a true poet unless he can stir the human heart. 3. Shakespeare wrote Hamlet. 4. No writer who does not understand human nature can stir the human heart. 5. None but a true poet could have written Hamlet.
I. 2. If a bird is in this aviary, then it belongs to me. 4. If a bird belongs to me, then it is at least 9 feet high. 1. If a bird is at least 9 feet high, then it is an ostrich. 3. If a bird lives on mince pies, then it is not an ostrich. Contrapositive: If a bird is an ostrich, then it does not live on mince pies. ∴ If a bird is in this aviary, then it does not live on mince pies; that is, no bird in this aviary lives on mince pies II. 3. Shakespeare wrote Hamlet. 5. If a person wrote Hamlet, then that person was a true poet.2. If a person is a true poet, then he can stir the human heart. 4. If a writer can stir the human heart, then that writer understands human nature. 1. If a writer understands human nature, then that writer is ∴ Shakespeare was clever.
Choose the best answer for the following. I. The zero product property, says that if aproduct of two real numbers is 0, then one of the numbers must be 0. a. Write this property formally using quantifiers and variables. b. Write the contrapositive of your answer to part (a). c. Write an informal version (without quantifier symbols or variables) for your answer to part (b). II. Assume that a and b are both integers and that a ≠≠ 0 and b ≠≠ 0. Explain why ( b − a ) / ( a b2 ) must be a rational number.
I. 8. a. real numbers x and y, if xy = 0 then = 0 or y = 0.c. If neither of two real numbers is zero, then their product is nonzero. b. ∀realnumbersxandy,ifx̸=0andy̸=0thenxy̸=0. II. Because a and b are integers, b − a and ab2 are both integers (since differences and products of integers are integers). Also, by the zero product property, ab2 ̸= 0 because neither a nor b is zero. Hence (b − a)/ab2 is a quotient of two integers with nonzero denominator, and so it is rational.
Choose the best choice to proof the following. I. Given any integer n , if n > 3, could n , n + 2 , and n + 4 all be prime ? Prove or give a counterexample. II. The fourth power of any integer has the form 8 m or 8 m + 1 for some integer m.
I. Given any integer n , the numbers n, n + 2 , and n + 4 cannot all be prime. Proof : Suppose n is any integer with n > 3 . By the quotient-remainder theorem with d = 3 , we know that n = 3 q, or n = 3 q + 1, or n = 3 q + 2 for some integer q. Note that because n is greater than 3 , either q is greater than 1 or q = 1 and n = 4 = 3 q + 1 or q = 1 and n = 5 = 3 q + 2. Case 1 ( q > 1 and n = 3 q ) : In this case, n is not prime because it is a product of 3 and q and both 3 and q are greater than 1. Case 2 ( q > 1 and n = 3 q + 1 ) : In this case, n + 2 = ( 3 q + 1 ) + 2 by substitution = 3 q + 3 = 3 ( q + 1 ) by algebra. So n + 2 is not prime because it is a product of 3 and q + 1 and both 3 and q + 1 are greater than 1. Case 3 ( q > 1 and n = 3 q + 2 ) : In this case, n + 4 = ( 3 q + 2 ) + 4 by substitution = 3 ( q + 2 ) by algebra.So n + 4 is not prime because it is a product of 3 and q + 2 and both 3 and q + 2 are greater than 1. Conclusion : In all three cases, at least one of n or n + 2 or n + 4 is not prime. II. Proof : Suppose n is any integer. By the quotient-remainder theorem with d =2 , n = 2 q or n = 2 q + 1 for some integer q. Case 1 ( n = 2 q for some integer q ) : In this case, by substitution, n4 = ( 2 q )4 = 16 q4 = 8 ( 2 q4 ). Let m = 2 q4 . Then m is an integer because it is a product of integers. Hence n8 = 8 m where m is an integer. Case 2 ( n = 2 q + 1 for some integer q ) : In this case, by substitution, n4 = ( 2 q + 1 )4 by substitution = ( 2 q + 1 )2 ( 2 q + 1 )2 = ( 4 q2 + 4 q + 1 ) ( 4 q2 + 4 q + 1 ) = 16 q4 + 16 q3 + 4 q2 + 16 q3 + 16 q2 + 4 q + 4 q2 + 4 q + 1 = 16 q4 + 32 q3 + 24 q2 + 8 q + 1 = 8 ( 2 q4 + 4 q3 + 3 q2 + q ) + 1 by algebra. Let m = 2 q4 + 4 q3 + 3 q2 + q . Then m is an integer because products and sums of integers are integers. Hence n 4 = 8 m + 1 where m is an integer. Conclusion : In both cases n4 = 8 m or n4 = 8 m + 1 for some integer m.
What is the best choice for the following? I. Suppose b is an integer. If b mod 12 = 5, what is 8 b mod 12 ? In other words, if division of b by 12 gives a remainder of 5, what is the remainder when 8 b is divided by 12? II. Suppose c is an integer. If c mod 15 = 3 , what is 10 c mod 15 ? In other words, if division of c by 15 gives a remainder of 3, what is the remainder when 10 c is divided by 15? III. Prove that for all integers n, if n mod 5 = 3 then n2 mod 5 = 4.
I. Given that b is an integer and b mod 12 = 5, it follows that 8 b mod 12 = 4.Proof : When b is divided by 12, the remainder is 5. Thus there exists an integer m so that b = 12 m + 5. Multiplying this equation by 8 gives 8 b = 8 ( 12 m + 5 ) by substitution = 96 m + 40 = 96 m + 36 + 4 = 12 ( 8 m + 3 ) + 4 by algebra. Since 8 m + 3 is an integer and since 0 ≤≤ 4 < 12, the uniqueness part of the quotient-remainder theorem guarantees that the remainder obtained when 8 b is divided by 12 is 4. II. Given that c is an integer and c mod 15 = 3, it follows that 10 c mod 15 = 0.Proof : When c is divided by 15, the remainder is 3. Thus there exists an integer k so that c = 15 k + 3. Multiplying this equation by 10 gives10 c = 10 ( 15 k + 3 ) by substitution = 10 ⋅⋅ 15 k + 30= 15 ( 10 k + 2 )= 15 ( 10 k + 2 ) + 0 by algebra.Since 10 k + 2 is an integer and since 0 ≤ 0 < 15, the uniqueness part of the quotient-remainder theorem guarantees that the remainder obtained when 10 c is divided by 15 is 0. III. Proof: Suppose n is any [ particular but arbitrarily chosen ] integer such that n mod 5 = 3 . Then the remainder obtained when n is divided by 5 is 3, and so n = 5 q + 3 for some integer q . By substitution, n2 = ( 5 q + 3 )2 = 25 q2 + 30 q + 9 = 25 q2 + 30 q + 5 + 4 = 5 ( 5 q2 + 6 q + 1 ) + 4. Because products and sums of integers are integers, 5 q2 + 6 q + 1 is an integer, and hence n2 = 5 · ( an integer ) + 4. Thus, since 0 ≤ 4 < 5 , the remainder obtained when n2 is divided by 5 is 4 , and so n2 mod 5 = 4.
What is the best choice for the following ? I. Is it possible to have a combination of nickels, dimes, and quarters that add up to $ 4.72 ? Explain. II. Two athletes run a circular track at a steady pace so that the first completes one round in 8 minutes and the second in 10 minutes. If they both start from the same spot at 4 P.M. , when will be the first time they return to the start together?
I. No. The values of nickels, dimes, and quarters are all multiples of 5. A sum of numbers each of which is divisible by 5 is also divisible by 5. So since $4.72is not a multiple of 5, $4.72 cannot be obtained using only nickels, dimes, and quarters. II. Let n be the number of minutes past 4 p.m. when the athletes first return to the start together. Then n is the smallest multiple of 8 that is also a multiple of 10. This number is 40. Hence the first time the athletes will return to the start together will be 4:40 p.m.
Choose the best proof for the following. I. For all real numbers r and c with c ≥ 0 , if − c ≤ r ≤ c, then | r | ≤ c . II. For all real numbers r and c with c ≥ 0 , if | r | ≤ c, then − c ≤ r ≤ c .
I. Proof : Let c be any positive real number and let r be any real number. Suppose that - c ≤≤ r < c. ( * ) By the trichotomy law ( see Appendix A, T 17 ), either r ≥≥ 0 or r < 0. Case 1 ( r ≥≥ 0 ) : In this case | r | = r, and so by substitution into ( * ) , - c ≤≤ | r | ≤≤ c. In particular, | r | ≤≤ c . Case 2 ( r < 0 ): In this case | r | = - r, and so r = - | r | . Hence by substitution into ( * ) , - c ≤≤ - | r | ≤≤ c. In particular, - c ≤≤ - | r | . Multiplying both sides by -1 gives c ≥≥ | r | , or equivalently, | r | ≤≤ c . Therefore, regardless of whether r ≥≥ 0 or r < 0, | r | ≤≤ c [ as was to be shown ]. II. Proof : Suppose that | r | ≤≤ c. ( ** ) By the trichotomy law, either r ≥≥ 0 or r < 0. Case 1 ( r ≥≥ 0 ): In this case | r | = r, and so by substitution into ( ** ) , r ≤≤ c. Since r ≥≥ 0 and c ≥≥ r , then c ≥≥ 0 by transitivity of order ( Appendix A, T 18 ) . Then, by property T 24 of Appendix A, 0 ≥≥ - c, and, again by transitivity of order, r ≥≥ - c . Hence - c ≤≤ r ≤≤ c. Case 2 ( r < 0 ) : In this case | r | = - r, and so by substitution into ( ** ) , - r ≤≤ c. Multiplying both sides of this inequality by —1 gives r ≥≥ - c . Also since r < 0 and c ≥≥ 0 , then c ≥≥ r . Thus - c ≤≤ r ≤≤ c. Conclusion : Regardless of whether r ≥≥ 0 or r < 0, we have that - c ≤≤ r ≤≤ c [ as was to be shown ].
Choose the best choice to proof the following. I. For any integer n, n2 + 5 is not divisible by 4. II. The sum of any four consecutive integers has the form 4 k + 2 for some integer k.
I. Proof : Let n be any integer. [ We must show that n2 + 5 is not divisible by 4 for any integer n . ] By the quotient-remainder theorem, n = 2 q or n = 2 q + 1 for some integer q. Case 1 ( n is even ) : In this case, n = 2 q for some integer q, and so n2 + 5 = ( 2 q )2 + 5 by substitution = 4 q2 + 4 + 1 = 4 ( q2 + 1 ) + 1 by algebra.Let k = q2 + 1. Then k is an integer because it is a product of integers, and thus n2 + 5 has the form 4 k + 1 for some integer k. So, by the uniqueness part of the quotient-remainder theorem, the remainder obtained when n2 + 5 is divided by 4 is 1, not 0, and so n2 + 5 is not divisible by 4. Case 2 ( n is odd ): In this case, n = 2 q + 1 for some integer q, and so n2 + 5 = ( 2 q + 1 )2 + 5 by substitution = ( 4 q2 + 4 q +1 ) + 5 = ( 4 q2 + 4 q + 4 ) + 2 = 4 ( q2 + q + 1 ) + 2 by algebra. Let k = q2 + q + 1 . Then k is an integer because it is a product of integers, and thus n2 + 5 has the form 4 k + 2 for some integer k. So, by the uniqueness part of the quotient-remainder theorem, the remainder obtained when n2 + 5 is divided by 4 is 2 , not 0 , and so n2 + 5 is not divisible by 4. Conclusion : In both possible cases n2 + 5 is not divisible by 4. II. Proof : Consider any four consecutive integers. Call the smallest n. Then the sum of the four integers is n + ( n + 1 ) + ( n + 2 ) + ( n + 3 ) = 4 n + 6 = 4 ( n + 1 ) + 2. Let k = n + 1 . Then k is an integer because it is a sum of integers. Hence n can be written in 4 k + 2 form for some integer k .
Define a relation from R to R as follows: For all ( x, y ) ∈ R × R, ( x, y ) ∈ L means that y=x−1. Is L a function? If it is, find L( 0 ) and L( 1 )
L is a function. For each real number x, y = x−1 is a real number, and so there is a real number y with ( x, y ) ∈ L. Also if ( x, y ) ∈ L and ( x, z ) ∈ L, then y = x − 1 and z = x − 1, and so y = z. So, L(0) = 0 − 1 = −1 and L(1) = 1 − 1 = 0. You can also check these results by inspecting the graph of L, shown below. Note that for every real number x, the vertical line through (x, 0) passes through the graph of L exactly once. This indicates both that every real number x is the first element of an ordered pair in L and also that no two distinct ordered pairs in L have the same first element.
Determine the following statements truth or falsity Prove each true statement directly from the definitions, and give a counterexample for each false statement. In case the statement is false, determine whether a small change would make it true. If so, make the change and prove the new statement. Follow the directions for writing proofs on page 118. I. For all real numbers a and b, if a < b then a < a+b2a+b2 < b. ( You may use the properties of inequalities in T17-T27 of Appendix A. ) II. Given any two rational numbers r and s with r < s, there is another rational number between r and s.
I. The statement is true.Proof : Suppose a and b are any real numbers with a < b. By properties T19 and T20 in Appendix A, we may add b to both sides to obtain ( a + b ) < 2 b,and we may divide both sides by 2 to obtain a+b2a+b2 < b. Similarly, since a < b, we may add a to both sides, which gives2 a < ( a + b ) ,and we may divide both sides by 2, which givesa < a+b2a+b2By combining the inequalities, we havea < a+b2a+b2 < b. II. The statement is true.Proof:Suppose r and s are any two distinct rational numbers with r < s. Let x = r+s2r+s2 [ First to proof r+s2r+s2 is rational ] By definition of rational, r = abab and s = cdcd for some integers a, b, c, and d with b ≠≠ 0 and d ≠≠ 0 .By substitution and the laws of algebra, r+s2r+s2 = ab+cd2ab+cd2 = ad+bcbd2ad+bcbd2 = ad+bc2bdad+bc2bd . Now a d + b c and 2 b d are integers because a, b, c, and d are integers and products and sums of integers are integers. And 2 b d ≠≠ 0 by the zero product property.Hence r+s2r+s2 is a quotient of integers with a nonzero denominator, and so r+s2r+s2 is rational. [ Second to proof r < r+s2r+s2 < s ] By properties T19 and T20 in Appendix A, we may add s to both sides to obtain ( r + s ) < 2 s,and we may divide both sides by 2 to obtain r+s2r+s2 < s. Similarly, since r < s , we may add a to both sides, which gives2 s < ( r + s ) ,and we may divide both sides by 2, which givess < r+s2r+s2By combining the inequalities, we haves < r+s2r+s2 < r. So there exists another rational number between r and s.
Choose the best answer for the following. Find the mistakes in the "proofs" shown in below. I. Theorem: The product of an even integer and an odd integer is even. " Proof : Suppose m is an even integer and n is an odd integer. If m · n is even, then by definition of even there exists an integer r such that m · n = 2 r . Also since m is even, there exists an integer p such that m = 2 p , and since n is odd there exists an integer q such that n = 2 q + 1. Thus m n = ( 2 p ) ( 2 q + 1 ) = 2 r, where r is an integer. By definition of even, then, m · n is even, as was to be shown." II. Theorem: The sum of any two even integers equals 4 k for some integer k. " Proof: Suppose m and n are any two even integers. By definition of even, m = 2 k for some integer k and n = 2 k for some integer k. By substitution, m + n = 2 k + 2 k = 4 k. This is what was to be shown."
I. This incorrect " proof " assumes what is to be proved. The second sentence states a conclusion that follows from the assumption that m • n is even. The next-to-last sentence states this conclusion as if it were known to be true. But it is not known to be true. In fact, it is the main task of a genuine proof to derive this conclusion, not from the assumption that it is true but from the hypothesis of the theorem. II. The mistake in the " proof " is that the same symbol, k, is used to represent two different quantities. By setting both m and n equal to 2 k, the " proof " specifies that m = n, and, therefore, it only deduces the conclusion in case m = n. If m ≠≠ n, the conclusion is often false.For instance, 6 + 4 = 10 but 10 ≠≠ 4 k for any integer k.
Choose the best choice for the following. I. a. Use the quotient-remainder theorem with d = 3 to prove that the square of any integer has the form 3 k or 3 k + 1 for some integer k. b. Use the mod notation to rewrite the result of part a. II. a. Use the quotient-remainder theorem with d = 3 to prove that the product of any two consecutive integers has the form 3 k or 3 k + 2 for some integer k. b. Use the mod notation to rewrite the result of part a.
I. a. Proof : Suppose n is any integer. By the quotient-remainder theorem with d = 3, we have that n = 3 q , or n = 3 q + 1, or n = 3 q + 2 for some integer q.Case 1 ( n = 3 q for any integer q ) : In this case, n2 = ( 3 q )2 by substitution= 3 ( 3 q2 ) by algebra.Let k = 3 q2. Then k is an integer because it is a product of integers. Hence n2 = 3 k for some integer k.Case 2 ( n = 3 q + 1 for some integer q ): In this case, n2 = ( 3 q + 1 )2 by substitution = 9 q2 + 6 q + 1 = 3 ( 3 q2 + 2 q )+ 1. by algebra. Let k = 3 q2 + 2 q . T hen k is an integer because sums and products of integers are integers. Hence n2 = 3 k + 1 for some integer k.Case 3 ( n = 3q + 2 for some integer q ): In this case, n2 = ( 3 q + 2 )2 by substitution = 9 q2 + 12 q + 4 = 9 q2 + 12 q + 3 +1= 3 ( 3 q2 + 4 q + 1 ) + 1 by algebra. Let k = 3q2 + 4 q + 1. Then k is an integer because sums and products of integers are integers.Hence n2 = 3 k + 1 for some integer k. Conclusion : In all three cases, either n2 = 3 k or n2 = 3 k + 1 for some integer k [ as was to be shown ]. b. Given any integer n, n2 mod 3 ≠≠ 1. II. a. Proof: Suppose n and n + 1 are any two consecutive integers. By the quotient-remaindertheorem with d = 3 , we know that n = 3 q, or n = 3 q + 1 , or n = 3 q + 2 for some integer q.Case 1 ( n = 3 q for some integer q ) : In this case,n ( n + 1 ) = 3 q ( 3 q + 1 ) by substitution = 3 [ q ( 3 q + 1) ] by algebra.Let k = q ( 3 q + 1 ). Then k is an integer because sums and products of integers are integers. Hence n ( n + 1 ) = 3 k for some integer k. Case 2 ( n = 3 q + 1 for some integer q ) : In this case,n ( n + 1 ) = ( 3 q + 1 ) ( 3 q + 2 ) by substitution = 9 q2 + 9 q + 2 = 3 [ ( 3 q2 + 3 q ) + 2 by algebra.Let k = 3 q2 + 3 q. Then k is an integer because sums and products of integers are integers. Hence n ( n + 1 ) = 3 k + 2 for some integer k. Case 3 ( n = 3 q + 2 for some integer q ) : In this case, n ( n + 1 ) = ( 3 q + 2 ) ( 3 q + 3 ) by substitution = 3 [ ( 3 q + 2 ) ( q + 1 ) ] by algebra. Let k = ( 3 q + 2 ) ( q + 1 ) . Then k is an integer because sums and products of integers are integer. Hence n ( n + 1 ) = 3 k for some integer k. Conclusion : In all three cases, the product of the two consecutive integers either equals 3 k or it equals 3 k + 2 for some integer k [as was to be shown]. b. Given any integer n, m n mod 3 ≠≠ 1.
In the following statements, (a) rewrite the statement formally using quantifiers and variables, and (b) write a negation for the statement. I. Every action has an equal and opposite reaction. II. There is a program that gives the correct answer to every question that is posed to it.
I. a. Statement : ∀ actions A, ∃ a reaction R such that R is equal and opposite to A. b. Negation : ∃ an action A such that ∀ reactions R, R is not equal to A or is not opposite to A. II. a. Statement : ∃ a program P such that ∀ questions Q posed to P, P gives the correct answer to Q. b. Negation : ∀ programs P, there is a question Q that can be posed to P such that P does not give the correct answer to Q.
What is the best choice for the following ? I. Consider the following statement : The negative of any multiple of 3 is a multiple of 3. a. Write the statement formally using a quantifier and a variable. b. Determine whether the statement is true or false and justify your answer. II. Show that the following statement is false : For all integers a and b, if 3 | ( a + b ) then 3 | ( a − b ).
I. a. ∀ integers n if n is a multiple of 3 then − n is a multiple of 3. b. The statement is true. Proof : Suppose n is any integer that is a multiple of 3. [ We must show that − n is a multiple of 3. ] By definition of multiple, n = 3 k for some integer k. Then − n = − ( 3 k ) by substitution = 3 ( − k ) by algebra. Hence, by definition of multiple, − n is a multiple of 3 [ as was to be shown ]. II. Counterexample : Let a = 2 and b = 1. Then a + b = 2 + 1 = 3, and so 3 | ( a + b ) because 3 = 3 · 1 . On the other hand, a − b = 2 − 1 = 1, and 3 ∤∤ 1 because 1 / 3 is not an integer. Thus 3 ∤∤ ( a − b ) . [ So the hypothesis of the statement is true but its conclusion is false. ]
What is the best choice for the following statement? Prove: The square of any odd integer has the form 8 m + 1 for some integer m
Proof: Suppose n is a [particular but arbitrarily chosen] odd integer. By the quotient-remainder theorem, n can be written in one of the forms 4 q or 4 q + 1 or 4 q + 2 or 4 q + 3 for some integer q . In fact, since n is odd and 4 q and 4 q + 2 are even , n must have one of the forms 4 q + 1 or 4 q + 3 . Case 1 ( n = 4 q + 1 for some integer q ) : [We must find an integer m such that n 2 = 8 m + 1 . ] Since n = 4 q + 1, n 2 = ( 4 q + 1 ) 2 by substitution = ( 4 q + 1 ) ( 4 q + 1 ) by definition of square = 16 q 2 + 8 q + 1 = 8 ( 2 q 2 + q ) + 1 by the laws of algebra. Let m = 2 q 2 + q . Then m is an integer since 2 and q are integers and sums and products of integers are integers. Thus, substituting, n 2 = 8 m + 1 where m is an integer. Case 2 ( n = 4 q + 3 for some integer q ) : [We must find an integer m such that n 2 = 8 m + 1. ] Since n = 4 q + 3, n 2 = ( 4 q + 3 ) 2 by substitution = ( 4 q + 3 ) ( 4 q + 3 ) by definition of square = 16 q 2 + 24 q + 9 = 16 q 2 + 24 q + ( 8 + 1 ) = 8 ( 2 q 2 + 3 q + 1 ) + 1 by the laws of algebra. [The motivation for the choice of algebra steps was the desire to write the expression in the form 8 · (some integer) + 1.] Let m = 2 q 2 + 3 q + 1 . Then m is an integer since 1, 2, 3, and q are integers and sums and products of integers are integers. Thus, substituting, n 2 = 8 m + 1 where m is an integer. Cases 1 and 2 show that given any odd integer, whether of the form 4 q + 1 or 4 q + 3, n 2 = 8 m + 1 for some integer m. [This is what we needed to show.]
What would be the following argument using quantifiers, variables, and predicate symbols? Write the major premise in conditional form. Is this argument valid? Why? All human beings are mortal. Zeus is not mortal. ∴ Zeus is not human.
The major premise: ∀ x, if x is human then x is mortal. Let H ( x ) be " x is human, " let M ( x ) be " x is mortal, " and let Z stand for Zeus. The argument becomes ∀ x, if H ( x ) then M ( x ) ∼ M ( Z ) ∴ ∼ H ( Z ) This argument has the form of universal modus tollens and is therefore valid.
Given the following information about a computer program, find the mistake in the program. a. There is an undeclared variable or there is a syntax error in the first five lines. b. If there is a syntax error in the first five lines, then there is a missing semicolon or a variable name is misspelled. c. There is not a missing semicolon. d. There is not a misspelled variable name.
The program contains an undeclared variable.
Indicate whether the arguments in following are valid or invalid. Support your answers by drawing diagrams. a. All teachers occasionally make mistakes. No gods ever make mistakes. ∴ No teachers are gods. b. No vegetarians eat meat. All vegans are vegetarian. ∴ No vegans eat meat.
Valid. The major and minor premises can be diagrammed as follows: According to the diagram, the set of teachers and the set of gods can have no common elements. Hence, if the premises are true, then the conclusion must also be true, and so the argument is valid. b. Valid. The only drawing representing the truth of the premises also represents the truth of the conclusion.
Diagrams can be helpful in testing an argument for validity. However, if some possible configurations of the premises are not drawn, a person could conclude that an argument ___________________________ was when it was actually ________________________.
valid; invalid
Choose the best answer for the following. Suppose a, b, and c are integers and x, y, and z are nonzero real numbers that satisfy the following equations: xyx+yxyx+y = a , xzx+zxzx+z = b , and yzy+zyzy+z = c. Is x rational ? If so, express it as a ratio of two integers.
Yes.Proof : Suppose a, b, and c are integers and x, y, and z are nonzero real numbers, where xyx+yxyx+y = a ............(1) xzx+zxzx+z = b .............(2) , and yzy+zyzy+z = c ................(3)Note that because a, b, and c are real numbers, none of the denominators x + y, z + x or y + z can equal zero.[ The strategy will be to express x in terms of the integers a, b, and c in hopes of showing that x can be written as a ratio of integers with a nonzero denominator. ]First observe that, by the zero product property,a ≠≠ 0 because from (1) x y = a ( x + y ) and x y ≠≠ 0 b ≠≠ 0 because from (2) z x = b ( z + x ) and z x ≠≠ 0c ≠≠ 0 because from (3) y z = c ( y + z ) and y z ≠≠ 0.i. Solve equation (1) for y in terms of a and x:x y = a ( x + y ) ⇔⇔ x y = a x + a y ⇔⇔ x y - a y = a x ⇔⇔ ( x - a ) y = a x. Now because a x is a product of nonzero real numbers, a x ≠≠ 0, and so ( x - a ) ≠≠ 0 . Thus we may divide by x - a to obtain y = axx−aaxx−a ii. Similarly, solve equation (2) for z in terms of b and x:z x = b ( z + x ) ⇔⇔ z x = b z + b x ⇔⇔ z x - b z = b x ⇔⇔ ( x - b ) z = b x.And because b x is a product of nonzero real numbers, b x ≠≠ 0. Thus ( x -b ) ≠≠ 0, and we may divide by x — b to obtain z = bxx−bbxx−b iii. Substitute the results of i and ii into equation (3) ( axx−aaxx−a ) ( bxx−bbxx−b ) = c ( axx−aaxx−a + bxx−bbxx−b ) and solve for x in terms of a, b, and c by first multiplying both sides ( x - a) ( y — b ) to obtain ( a x ) ( b x ) = c a x ( x - b) + c b x ( x - a ). Because x ≠≠ 0, both sides may be divided by x to yielda b x = c a x — c a b + c b x — c b a,or, equivalently (by putting all the terms involving x on the right-hand side and all the other terms on the left-hand side), 2 a b c = a c x + b c x — a b x = x ( a c + b c — a b).Because 2 a b c ≠≠ 0, by the zero product property, a c + b c — a b cannot be zero either. Thus we may divide both sides by a c + b c — a b to obtainx = 2abcac+bc−ab2abcac+bc−ab Finally, because products and sums of integers are integers, we see that x has been expressed s a ratio of integers with a nonzero denominator.
For each of the following equations, determine which of the following statements are true: (1) For all real numbers x, there exists a real number y such that the equation is true. (2) There exists a real number x, such that for all real numbers y, the equation is true. Note that it is possible for both statements to be true or for both to be false. a. 2 x + y = 7 b. y + x = x + y c. x2 − 2 x y + y2 =0 d. ( x − 5 ) ( y − 1 ) = 0 e. x2 + y2 = −1
a. (1) The statement " ∀ real numbers x, ∃ a real number y such that 2 x + y = 7 " is true. (2) The statement " ∃ a real number x such that ∀ real numbers y, 2 x + y = 7 " is false. b. Both statements (1) " ∀ real numbers x, ∃ a real number y such that x + y = y + x " and (2) " ∃ a real number x such that ∀ real numbers y, x + y = y + x " are true. c. Statement (1) is true because x2 — 2 x y + y2 = ( x — y )2. Thus given any real number x, take y = x, then x — y = 0, and so x2 — 2 x y + y2 = 0. Statement (2) is false. Given any real number x, choose a real number y with y ≠≠ x. Then x2 — 2 x y + y2 = ( x — y )2 ≠≠ 0. d. Statement (1) is true because no matter what real number x might be chosen, y can be taken to be 1 so that ( x — 5 ) ( y — 1 ) = ( x — 5 ) ⋅⋅ 0 = 0.Statement (2) is also true. Take x = 5. Then for all real numbers y, ( x — 5 ) ( y — 1 ) = 0 ⋅⋅ ( y — 1 ) = 0. e. Statements (1) and (2) are both false because all real numbers have nonnegative squares and the sum of any two nonnegative real numbers is nonnegative. Hence for all real numbers x and y, x2 + y2 ≠≠ -1.
Use the properties of even and odd integers that are listed in Example 4. 2 . 3 to do the following exercises. Indicate which properties you use to justify your reasoning. a. True or false ? If m is any even integer and n is any odd integer, then m2 + 3 n is odd. Explain. b. True or false ? If a is any odd integer, then a2 + a is even. Explain. c. True or false ? If k is any even integer and m is any odd integer, then ( k + 2 )2 − ( m − 1 )2 is even. Explain.
a. True. Proof : Suppose m is any even integer and n is any odd integer. [ We must show that m2 + 3 n is odd. ] By properties 1 and 3 of Example 4.2.3, m2 is even ( because m2 = m · m ) and 3 n is odd ( because both 3 and n are odd ). It follows from property 5 [ and the commutative law for addition ] that m2 + 3 n is odd [ as was to be shown ]. b True.Proof : Suppose a is any odd integer. Then a2 = a ⋅⋅ ais a product of odd integers and hence is odd by property 3.Therefore, a2 + a is a sum of odd integers and thus even by property 2. c. True.Proof:Suppose k is any even integer and m is any odd integer.By property 1, k + 2 is even because it is a sum of even integers, and thus also by property 1, (k + 2 )2 is even because it is a product of even integers.By property 2, m - 1 is even because it is a difference of odd integers, and thus also by property 1, ( m - 1 ) is even because because it is a product of even integers.Finally, by property 1, ( k + 2 )2 - ( m - 1 )2 is even because it is a difference of even integers.
Refer to the above picture of Tarski's world. Let Above(x, y) mean that x is above y (but possibly in a different column). Determine the truth or falsity of each of the following statements. Give reasons for your answers. a. ∀ u, Circle ( u ) → Gray ( u ). b. ∀ u, Gray ( u ) → Circle ( u ). c. ∃ y such that Square ( y ) ∧ Above ( y, d ). d. ∃ z such that Triangle ( z ) ∧ Above ( f, z ).
a. False. Figure b is a circle that is not gray. b. True. All the gray figures are circles. c. False. The only objects above d are a (a triangle) and b (a circle). d. True. g is a triangle and f is above g.
What would be the best choice to proof the following statements? a. There are real numbers a and b such that a+b−−−−−−√a+b = a−−√a + b√b b. There is an integer n > 5 such that 2n − 1 is prime. c. There is a real number x such that x > 1 and 2x > x10 .
a. For example, let a = 1 and b= 0. Then a+b−−−−−−√a+b = 1-√1 = 1 and also a−−√a + b√b = 1-√1 + 0-√0 = 1 . Hence for these values of a and b, a+b−−−−−−√a+b = a−−√a + b√b In fact, if a is any nonzero integer and b = 0, then a+b−−−−−−√a+b = a+0−−−−−−√a+0 = a−−√a = a−−√a + 0 = a−−√a + 0-√0 = a−−√a + b√b b. For example, let n = 7. Then n is an integer such that n > 5 and 2n − 1 = 127, which is prime. c. For example, let x = 60. Note that to four significant digits 260 ≅≅ 1.153 x 1018 and6010 ≅≅ 6.047 x 1017 and so 2x ≥≥ x10 Examples can also be found in the approximate range 1 < x < 1.077. For instance, 21.07 ≅≅ 2.099 and 1.0710 ≅≅ 1.967,and so 21.07 > 1.0710
What would be the best choice to proof the following statements? a. There are integers m and n such that m > 1 and n > 1 and 1m1m + 1n1n is an integer. b. There are distinct integers m and n such that 1m1m + 1n1n is an integer
a. For example, let m = n = 2. Then m and n are integers such that m > 1 and n > 1 and 1m1m + 1n1n = 1212 + 1212 = 1, which is an integer. b. For example, let m= 1 and n = —1. Then 1m1m + 1n1n = 1111 + 1(−1)1(−1) = 1 + ( - 1 ) = 0 In fact, if k is any nonzero integer, then 1k1k + 1(−k)1(−k) = 1k1k + ( - 1k1k ) = 0.
Let D = E = { −2, −1, 0, 1, 2 }. Write negations for each of the following statements and determine which is true, the given statement or its negation. a. ∀ x in D, ∃ y in E such that x + y = 1. b. ∃ x in D such that ∀ y in E, x + y = − y . c. ∀ x in D, ∃ y in E such that x y ≥ y. d. ∃ x in D such that ∀ y in E, x ≤ y.
a. Negation : ∃ x in D such that ∀ y in E, x + y ≠≠ 1. The negation is true. When x = −2, the only number y with the property that x + y = 1 is y = 3, and 3 is not in E. b. Negation : ∀ x in D, ∃ y in E such that x + y ≠≠ −y. The negation is true and the original statement is false. To see that the original statement is false, take any x in D and choose y to be any number in E with y ≠≠ − x2x2 . Then 2 y ≠≠ −x, and adding x and subtracting y from both sides gives x + y≠≠ −y. c. Negation : ∃ x in D such that ∀ y in E, x y ≱≱ y. (Or: ∃ x in D such that ∀ y in E, x y < y. ) The statement is true. For each number x in D, you can find a y in D so that x y > y. Here is a table showing one way to do this : how all possible choices for x could be matched with a y so that x y > y.: d. Negation : ∀ x in D, ∃ y in E such that x ≰≰ y. (Or: ∀ x in D, ∃ y in E such that x > y.) The statement is true. It says that there is a number in D that is less than or equal to every number in D. In fact, —2 is in D and —2 is less than or equal to every number in D ( —2, —1,0, 1, and 2 ).
Choose the best answer to the following questions . a. Is 1 prime? b. Write the first six prime numbers. c. Write the first six composite numbers.
a. No. A prime number is required to be greater than 1. b. 2, 3, 5, 7, 11, 13 c. 4, 6, 8, 9, 10, 12
a. Is ( 5, −5 ) = ( −5, 5 ) ? b. Is ( −2−4−2−4 , (−2)3 ) = ( 3636 , −8 ) ?
a. No: For two ordered pairs to be equal, the elements in each pair must occur in the same order. In this case the first element of the first pair is 5, whereas the first element of the second pair is —5, and the second element of the first pair is —5 whereas the second element of the second pair is 5. b. Yes: Both pairs have the same elements 1212 and —8.
Write a negation for the following statements. a. ∀ real numbers x, if x2 ≥ 1 then x > 0. b. ∀ integers d, if 6/d is an integer then d = 3.
a. Some real numbers that are less than or equal to zero have squares that are greater than or equal to one. b. ∃ an integer d such that 6/d is an integer and d ≠≠ 3.
Indicate which of the following statements are true and which are false. Justify your answers as best you can. a. ∀ x ∈ Z+ , ∃ y ∈ Z+ such that x = y + 1. b. ∀ x ∈ Z, ∃ y ∈ Z such that x = y + 1. c. ∃ x ∈ R such that ∀ y ∈ R, x = y + 1. d. ∀ x ∈ R+ , ∃ y ∈ R+ such that x y = 1.
a. This statement says that given any positive integer, there is a positive integer such that the first integer is one more than the second integer. This is false. Given the positive integer x = 1, the only integer with the property that x = y + 1 is y = 0, and 0 is not a positive integer. b. This statement says that given any integer, there is an integer such that the first integer is one more than the second integer. This is true. Given any integer x, take y = x − 1. Then y is an integer, and y + 1 = ( x − 1 ) + 1 = x. c. The statement says that there is a real number that is one greater than every real number. This is false. For instance, if the number is one greater than 3, then it equals 4 and so it is not one greater than 4. d. The statement says that every positive real number has a positive reciprocal. In other words, given any positive real number, we can find a positive real number such that the product of the two equals 1. This is true.
Choose the best choice for the following questions. a. Is 10/3 a rational number? b. Is − 539539 a rational number? c. Is 0.281 a rational number? d. Is 7 a rational number ? e. Is 0 a rational number?
a. Yes b. Yes c. Yes d. Yes e. Yes
Choose the best answer to the following questions by using the definitions of even and odd. a. Is 0 even? b. Is −301 odd? c. If a and b are integers, is 6a2b even? d. If a and b are integers, is 10 a + 8 b + 1 odd?
a. Yes, 0 = 2 · 0. b. Yes, −301 = 2 ( −151 ) + 1. c. Yes, 6a2b = 2 ( 3 a 2 b ), and since a and b are integers, so is 3 a 2 b ( being a product of integers ). d. Yes, 10 a + 8 b + 1 = 2 ( 5 a + 4 b ) + 1, and since a and b are integers, so is 5 a + 4 b (being a sum of products of integers).
What is the best choice that answers the following questions? a. If a and b are integers, is 3 a + 3 b divisible by 3 ? b. If k and m are integers, is 10 k m divisible by 5 ?
a. Yes. By the distributive law of algebra, 3 a + 3 b = 3 ( a + b ) and a + b is an integer because it is a sum of two integers. b. Yes. By the associative law of algebra, 10 k m = 5 · ( 2 k m ) and 2 k m is an integer because it is a product of three integers.
A function F from A to B is a relation from A to B that satisfies the following two properties: a. for every element x of A, there is ________________________. b. for all elements x in A and y and z in B, if __________________________ then __________________________.
a. an element y of B such that ( x, y ) ∈ F (i.e., such that x is related to y by F) b. ( x, y ) ∈ F and ( x, z ) ∈ F; y = z
The following statement is true: " ∀ real numbers x, ∃ an integer n such that n > x. " ∗ For each x given below, find an n to make the predicate " n > x " true. a. x = 15.83 b. x = 108 c. x = 101010101010
a. n = 16. b. n = 108 + 1 c. n = 101010101010 + 1
Use symbols to write the logical form of the following arguments. If the argument is valid, identify the rule of inference that guarantees its validity. Otherwise, state whether the converse or the inverse error is made. a. If I go to the movies, I won't finish my homework. If I don't finish my homework, I won't do well on the exam tomorrow. ∴ If I go to the movies, I won't do well on the exam tomorrow. b. If this number is larger than 2, then its square is larger than 4. This number is not larger than 2. ∴ The square of this number is not larger than 4.
a. p → q q → r ∴ p → r valid: transitivity b. p → q ∼ p ∴ ∼ q invalid: inverse error
Let p be the statement "DATAENDFLAG is off," q the statement "ERROR equals 0," and r the statement "SUM is less than 1,000." Express the following sentences in symbolic notation. a. DATAENDFLAG is off but ERROR is not equal to 0. b. DATAENDFLAG is on and ERROR equals 0 but SUM is greater than or equal to 1,000. c. Either DATAENDFLAG is on or it is the case that both ERROR equals 0 and SUM is less than 1,000.
a. p ∧∧ ~ q b. ( ~ p ∧∧ q ) ∧∧ ~ r c. ~ p ∨∨ ( q ∧∧ r )
Use symbols to write the logical form of the following arguments. If the argument is valid, identify the rule of inference that guarantees its validity. Otherwise, state whether the converse or the inverse error is made. a. If this computer program is correct, then it produces the correct output when run with the test data my teacher gave me. This computer program produces the correct output when run with the test data my teacher gave me. ∴ This computer program is correct. b. Sandra knows Java and Sandra knows C++. ∴ Sandra knows C++.
a. p ⟶⟶ q q ∴ p invalid, converse error b. p ∧∧ q ∴ q valid, generalization
The following statement is true : " ∀ nonzero numbers x, ∃ a real number y such that x y = 1. " For each x given below, find a y to make the predicate " x y = 1 " true. a. x = 2 b. x = − 1 c. x = 3/4
a. y = 1212 b. y = −1 c. Let y = 4343 . Then x y = ( 3434 ) ( 4343 ) = 1.
"If compound X is boiling, then its temperature must be at least 150◦ C." Assuming that this statement is true, which of the following must also be true? a. If the temperature of compound X is at least 150◦ C, then compound X is boiling. b. If the temperature of compound X is less than 150◦ C, then compound X is not boiling. c. Compound X will boil only if its temperature is at least 150◦ C. d. If compound X is not boiling, then its temperature is less than 150◦ C. e. A necessary condition for compound X to boil is that its temperature be at least 150◦ C. f. A sufficient condition for compound X to boil is that its temperature be at least 150◦ C. Correct!
b, c and e.
The following represents the common form of each argument using letters to stand for component sentences, and fill in the blanks so that the argument in part (b) has the same logical form as the argument in part (a). a. If n is divisible by 6, then n is divisible by 3. If n is divisible by 3, then the sum of the digits of n is divisible by 3. Therefore, if n is divisible by 6, then the sum of the dig- its of n is divisible by 3. (Assume that n is a particular, fixed integer.) b. If this function is __________ then this function is differentiable. If this function is ______________ then this function is continuous. Therefore, if this function is a polynomial, then this function _____________________ .
common form: If p then q. If q then r Therefore, if p then r. b. a polynomial; differentiable; is continuous
An argument is called ___________ if, and only if, it is _________ and all its premises are ________. An argument that is not __________ is called ______________. What would be the best choice to fill in blanks for the above question?
sound; valid; true; sound; unsound
What is the best choice to fill in the following blanks? If a and b are integers, the notation a | b denotes _____________________ and the notation a/b denotes ________________________.
the sentence "a divides b"; the number obtained when a is divided by b
Rewrite the statement " No good cars are cheap " in the form " ∀ x, if P ( x ) then ∼ Q ( x ) ." Indicate whether each of the following arguments is valid or invalid, and justify your answers. a. No good car is cheap. A Rimbaud is a good car. ∴ A Rimbaud is not cheap. b. No good car is cheap. A Simbaru is not cheap. ∴ A Simbaru is a good car. c. No good car is cheap. A VX Roadster is cheap. ∴ A VX Roadster is not good. d. No good car is cheap. An Omnex is not a good car. ∴ An Omnex is cheap.
∀ x, if x is a good car, then x is not cheap. a. Valid, universal instantiation b. Invalid, converse error c. Valid, universal modus tollens d. Invalid, inverse error
