cisp 440

Choose the best answer to fill in the blank. Zero is a rational number because ___________________________________.

0 = 0/1

What is the best choice for the following statements? a. 32 div 9 b. 32 mod 9

a. 3 b. 5

Choose the best answer for the following questions. The following questions determine whether the property is true for all integers, true for no integers, or true for some integers and false for other integers. Justify your answers. 1. (a+b)2 = a2 + b2 2. −an = ( − a )n 3. The average of any two odd integers is odd.

1. This property is true for some integers and false for other integers. For instance, if a = 0 and b = 1, the property is true because ( 0 + 1 )2 = 02 + 12, but if a = 1 and b = 1, the property is false because ( 1 + 1 ) 2 = 4 and 12 + 12 =2 and 4 ≠ 2. 2. According to the order of operations for real numbers, —an = - ( an) . The following table shows that the property is true for some values of a and n and false for other values. 3. This property is true for some integers and false for other integers. For example, the average of 1 and 5 is 1+52 = 62 =3 , which is odd. However, the average of 1 and 3 is 1+32 = 42 = 2 , which is not odd.

What is the best choice that answers the following questions? Give a reason for your answer in each of the following. Assume that all variables represent integers. 1. Is 4 a factor of 2 a · 34 b ? 2. Does 7 | 34 ? 3.. Does 13 | 73 ? 4. If n = 4 k+1, does 8 divide n2 − 1 ? 5. If n = 4 k + 3, does 8 divide n2 −1 ?

1. Yes : 2 a • 34 b = 4 ( 17 a b ) and 17 a b is an integer because a and b are integers and products of integers are integers. 2. No, 34 / 7 ≅ 4.86, which is not an integer. 3. No, 73 / 13 ≅ 5.6, which is not an integer. 4. Yes, n2 − 1 = ( 4 k + 1 )2 −1 = ( 16 k2 + 8 k + 1 ) −1 = 16 k2 + 8 k = 8 ( 2 k2 + k ) , and 2 k2 + k is an integer because k is an integer and sums and products of integers are integers. 5. Yes, n2 − 1 = ( 4 k + 3 )2 −1 = ( 16 k2 + 24 k + 9 ) −1 = 16 k2 + 24 k + 8 = 8 ( 2 k2 + 3 k +1) , and 2 k2 + 3 k + 1 is an integer because k is an integer and sums and products of integers are integers.

Choose the best answer for the following. Consider the statement : The negative of any rational number is rational. 1. Write the statement formally using a quantifier and a variable. 2. Determine whether the statement is true or false and justify your answer.

1. ∀ real numbers r , if r is rational then − r is rational. 2. The statement is true. Proof : Suppose r is a [ particular but arbitrarily chosen ] rational number. [ We must show that − r is rational. ] By definition of rational, r = a / b for some integers a and b with b ≠ 0. Then − r = - ab by substitution = −ab by algebra. But since a is an integer, so is −a ( being the product of −1 and a ). Hence − r is a quotient of integers with a nonzero denominator, and so −r is rational [ as was to be shown ].

Choose the best choice for the following. I. On a Monday a friend says he will meet you again in 30 days. What day of the week will that be? II. If today is Tuesday, what day of the week will it be 1,000 days from today? III. January 1, 2000, was a Saturday, and 2000 was a leap year. What day of the week will January 1, 2050, be?

I. By the formula( DayN = ( DayT + N ) mod 7 ), the answer is (1 + 30) mod 7 = 31 mod 7 = 3, which is Wednesday. II. Use the formula DayN = ( DayT + N ) mod 7 , letting DayT = 2 ( Tuesday ) and N = 1000. ThenDayN = ( 2 + 1000) mod 7 = 1002 mod 7 = 1,which is a Monday. III. There are 13 leap year days between January 1, 2000 and January 1, 2050 ( once every four years in 2000, 2004, 2008, 2012, . . . , 2048 ). So 13 of the years have 366 days and the remaining 37 years have 365 days. This gives a total of13 · 366 + 37 · 365 = 18, 263days between the two dates. Using the formula DayN = (DayT + N) mod 7 , and letting DayT = 6 ( Saturday ) and N = 18, 263 givesDayN = ( 6 + 18263 ) mod 7 = 18269 mod 7 = 6,which is also a Saturday.

Choose the best choice to proof the following. I. For all integers m , m2 = 5 k ,or m2 = 5 k + 1 ,or m2 = 5 k + 4 for some integer k. II. Every prime number except 2 and 3 has the form 6 q + 1 or 6 q + 5 for some integer q.

I. Proof : Let m be any integer. [ We must show that m2 = 5 k or m2 = 5 k + 1 , or m2 = 5 k + 4 for some integer k. ] By the quotient-remainder theorem, there is an integer q so that m = 5 q or m = 5 q + 1 or m = 5 q + 2 or m = 5 q + 3 or m = 5 q + 4. Case 1 ( m = 5q for some integer q ) : In this case, m2 = ( 5 q )2 by substitution = 5 ( 5 q2 ) by algebra.Let k = 5 q2 . Then k is an integer because it is a product of integers, and hence m2 = 5 k for some integer k. Case 2 ( m = 5 q + 1 for some integer q) : In this case, m2 = ( 5 q + 1 )2 by substitution = 25 q2 + 10 q + 1 = 5 ( 5 q2 + 2 q ) + 1 by algebra. Let k = 5 q2 + 2 q . Then k is an integer because it is a sum of products of integers, and hence m2 = 5 k + 1 for some integer k. Case 3 ( m = 5 q + 2 for some integer q ) : In this case, m2 = ( 5 q + 2 )2 by substitution = 25 q2 + 20 q + 4 = 5 ( 5 q2 + 4 q ) + 4 by algebra. Let k = 5 q2 + 4 q . Then k is an integer because it is a sum of products of integers, and hence m2 = 5 k + 4 for some integer k. Case 4 ( m = 5 q + 3 for some integer q ): In this case, m2 = ( 5 q + 3 )2 by substitution = 25 q2 + 30 q + 9 = 25 q2 + 30 q + 5 + 4 = 5 ( 5 q2 + 6 q + 1 ) + 4 by algebra. Let k = 5 q2 + 6 q + 1 . Then k is an integer because it is a sum of products of integers, and hence m2 = 5 k + 4 for some integer k. Case 5 ( m = 5 q + 4 for some integer q): In this case, m2 = ( 5 q + 4 )2 by substitution = 25 q2 + 40 q + 16 = 25 q2 + 40 q + 15 + 1 = 5 ( 5 q2 + 8 q + 3 ) + 1 by algebra. Let k = 5 q2 + 8 q + 3. Then k is an integer because it is a sum of products of integers, and hence m2 = 5 k + 1 for some integer k. Conclusion : In all cases m has one of the required forms. II. Proof: Let p be any prime number except 2 or 3. By the quotient-remainder theorem, there is an integer k so that p can be written as p = 6 k or p = 6 k+1 or p = 6 k + 2 or p = 6 k + 3 or p = 6 k + 4 or p = 6 k + 5 . Since p is prime and p ≠ 2, p is not divisible by 2 . Consequently, p ≠ 6 k , p ≠ 6 k + 2 , and p ≠ 6 k + 4 for any integer k [ because all of these numbers are divisible by 2 ]. Furthermore, since p is prime and p ≠ 3, p is not divisible by 3. Thus p ≠ 6 k + 3 [ because this number is divisible by 3 ]. Therefore, p = 6 k + 1 or p = 6 k + 5 for some integer k.

Choose the best answer for the following. In the following determine whether the statement is true or false. Justify your answer with a proof or a counterexample, as appropriate. In each case use only the definitions of the terms and the Assumptions listed on page 110 not any previously established properties. I. Every positive integer can be expressed as a sum of three or fewer perfect squares. II. (Two integers are consecutive if, and only if, one is one more than the other.) Any product of four consecutive integers is one less than a perfect square. III. If m and n are positive integers and m n is a perfect square, then m and n are perfect squares.

I. The statement is false.Counterexample: The number 28 cannot be expressed as a sum of three or fewer perfect squares. The only perfect squares that could be used to add up to 28 are those that are smaller than 28 : 1, 4, 9, 16, and 25. The method of exhaustion can be used to show that no combination of these numbers add up to 28. (In fact, there are just three ways to express 28 as a sum of four or fewer of these numbers : 28 = 25 + 1 + 1 + 1 = 16 + 4 + 4 + 4 = 9 + 9 + 9 + 1,and in none of these ways are only three perfect squares used.) II. The statement is true.Proof:Consider any product of four consecutive integers. Call the second smallest of the four n. Then the product is ( n - 1 ) n ( n + 1 ) ( n + 2 ).Let m = n2 + n —1.Note that m is an integer because sums, products, and differences of integers are integers. Alsom2 - 1 = ( n2 + n —1 )2 -1 = ( n4 + 2 n3 — n2 — 2 n + 1 ) - 1 = n4 + 2 n3 — n2 — 2 n = ( n - 1 ) n ( n + 1 ) ( n + 2 ) Hence the given product of four consecutive integers is one less than a perfect square. III. The statement is false.Counterexample: Let m = n = 3. Then m n = 3 ⋅ 3 = 9 which is a perfect square, but neither m nor n is a perfect square.

What is the best choice for the following ? For each of the following statements determines whether the statement is true or false. Prove the statement directly from the definitions if it is true, and give a counterexample if it is false. I. For all integers a and b, if a | b then a2 | b2 . II. For all integers a and n, if a | n2 and a ≤ n then a | n. III. For all integers a and b, if a | 10 b then a | 10 or a | b .

I. The statement is true.Proof:Let a and b be integers such that a | b.By definition of divisibility, b = a k for some integer k. Squaring both sides of this equation givesb2 = ( a k )2 = a2 k2 But k2 is an integer ( being a product of the integer k times itself). Hence by definition of divisibility, a2 | b2. II. The statement is false.Counterexample:Let a = 4 and n = 6.Then a | n2 and a ≤ n because 4 | 36 and 4 ≤ 6, but a ∤ n because 4 ∤ 6 . III. The statement is false.Counterexample:Let a = 25 and b = 5.Then a | 10 b because 25 | 50 but a ∤ 10 because 25 ∤ 10 and a ∤ b because 25 ∤ 5.

Choose the best answer for the following.. In the following find the mistakes in the "proofs" that the sum of any two rational numbers is a rational number. I. " Proof : Any two rational numbers produce a rational number when added together. So if r and s are particular but arbitrarily chosen rational numbers, then r + s is rational." II. " Proof : Let rational numbers r = 14 and s = 12 be given. Then r + s = 14 + 12 = 34, which is a rational number. This is what was to be shown. " III. " Proof : Suppose r and s are rational numbers. By definition of rational, r = a / b for some integers a and b with b ≠ 0 , and s = a / b for some integer s a and b with b ≠ 0 . Then r + s = ab+ ab = 2ab Let p = 2 a. Then p is an integer since it is a product of integers. Hence r + s = p / b , where p and b are integers and b ≠ 0. Thus r + s is a rational number by definition of rational. This is what was to be shown."

I. This "proof" assumes what is to be prove. II. This incorrect proof just shows the theorem to be true in the one case where one of the rational numbers is 1/4 and the other is 1/2. It is an example of the mistake of arguing from examples. A correct proof must show the theorem is true for any two rational numbers. III. By setting both r and s equal to a/b, this incorrect proof violates the requirement that r and s be arbitrarily chosen rational numbers. If both r and s equal a/b, then r = s.

Choose the best answer for the following. In the following determine whether the statement is true or false. Justify your answer with a proof or a counterexample, as appropriate. In each case use only the definitions of the terms and the Assumptions listed on page 110 not any previously established properties. I. The product of any two odd integers is odd. II. The negative of any odd integer is odd. III. The difference of any two odd integers is odd.

I. True. Proof : Suppose m and n are any odd integers. [ We must show that m n is odd. ] By definition of odd, n = 2 r + 1 and m = 2 s+1 for some integers r and s. Then m n = ( 2 r + 1 ) ( 2 s + 1 ) by substitution = 4 r s + 2 r + 2 s + 1 = 2 ( 2 r s + r + s ) + 1 by algebra Now 2 r s + r + s is an integer because products and sums of integers are integers and 2, r, and s are all integers. Hence m n = 2 · ( some integer ) + 1, and so, by definition of odd, m n is odd. II. True. Proof : Suppose n is any odd integer. [ We must show that − n is odd. ] By definition of odd, n = 2 k + 1 for some integer k. By substitution and algebra, − n = − ( 2 k + 1 ) = − 2 k − 1 = 2 ( − k − 1 ) + 1. Let t = − k − 1. Then t is an integer because differences of integers are integers. Thus − n = 2 t + 1, where t is an integer, and so, by definition of odd, −n is odd [ as was to be shown ]. III. False. Counterexample : Both 3 and 1 are odd, but their difference is 3 − 1 = 2, which is even.

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.

What is the best choice for the following question? Prove that for all integers a, b, and c, if a | b and b | c, then a | c .

Proof: Suppose a, b, and c are [ particular but arbitrarily chosen ] integers such that a divides b and b divides c. [ We must show that a divides c . ] By definition of divisibility, b = a r and c = b s for some integers r and s. By substitution c = b s = ( a r ) s = a ( r s ) by basic algebra. Let k = r s . Then k is an integer since it is a product of integers, and therefore c = a k where k is an integer. Thus a divides c by definition of divisibility. [This is what was to be shown.]

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 = ab + cd by substitution = ad+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 = pq 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. ]

What is the best choice of proofing the following statement? For all real numbers r, | − r | = | r |.

Proof: Suppose r is any real number. By Theorem T 23 in Appendix A, if r > 0, then − r < 0 , and if r < 0 , then −r > 0. Thus

What is the best choice to answer the following question? If k is any nonzero integer, does k divide 0 ?

Yes, because 0 = k · 0.

Choose the best choice for the following. In the following, use the properties listed in Example 4.2.3. a. Prove that for all integers m and n, m + n and m − n are either both odd or both even. b. Find all solutions to the equation m2 − n2 = 56 for which both m and n are positive integers. c. Find all solutions to the equation m2 − n2 = 88 for which both m and n are positive integers.

a. Proof 1: Suppose m and n are integers. Case 1 (both m and n are even) : In this case both m + n and m — n are even ( by property 1 in Example 4.2.3 ). Case 2 ( one of m and n is even and the other is odd ) : In this case both m + n andm — n are odd ( by properties 5 - 7 in Example 4.2.3 ). Case 3 ( both m and n are odd ) : In this case both m + n and m — n are even (by property 2 in Example 4.2.3 ). Conclusion : In all three possible cases, either both m n and m — n are even or both m + n and m — n are odd [ as was to be shown ]. Proof 2 : Suppose m and n are integers. Case 1 ( m — n is even ) : In this case, m + n = ( m — n ) + 2 n , and so m + n is a sum of two even integers and is therefore even by by property 1 in Example 4.2.3. Case 2 ( m — n is odd ) : In this case also, m + n = ( m — n ) + 2 n . Hence m + n is the sum of an odd integer and an even integer, and is therefore odd by by property 5 in Example 4.2.3. Conclusion : Now either m — n is even or m — n is odd [ by the quotient-remainder theorem ] , and thus either case 1 or case 2 must apply. In each case, m — n and m + n are either both even or both odd [ as was to be shown ]. b. If m2 − n2 = 56 , then 56 = ( m + n ) ( m − n ) . Now 56 = 23 · 7, and by the unique factorization theorem, this factorization is unique. Hence the only representations of 56 as a product of two positive integers are 56 = 7 · 8 = 14 · 4 = 28 · 2 = 56 · 1. By part ( a ), m and n must both be odd or both be even. Thus the only solutions are either m + n = 14 and m − n = 4 or m + n = 28 and m − n = 2. This gives either m = 9 and n = 5 or m = 15 and n = 13 as the only solutions. c. If m2 — n2 = 88 , then 88 = ( m + n ) ( m - n ) . Now 88 = 23 ⋅ 11, and by the unique factorization of integers theorem this factorization is unique up to the order in which the factors are written down. It follows that the only representations of 88 as a product of two positive integers are88 = 1 ⋅ 88 = 2 ⋅ 44 = 4 ⋅ 22 = 8 ⋅ 11.By part (a), m and n must both be odd or both be even. Thus the only solutions are either m + n = 22 and m - n = 4 or m + n = 44 and m - n = 2This gives two solutions: either m = 13 and n = 9 or m = 23 and n = 21

Choose the best choice for the following. If DayT is the day of the week today and DayN is the day of the week in N days, then DayN = ( DayT + N ) mod 7, where Sunday = 0, Monday = 1,..., Saturday = 6. Check the correctness of formula above for the following values of DayT and N . a. DayT = 6 ( Saturday ) and N = 15 b. DayT = 0 (Sunday ) and N =7 c. DayT = 4 ( Thursday ) and N = 12

a. When today is Saturday, 15 days from today is two weeks ( which is Saturday ) plus one day ( which is Sunday ). Hence DayN should be 0 . According to the formula, when today is Saturday, DayT = 6, and so when N = 15, DayN = ( DayT + N ) mod 7 = ( 6 + 15 ) mod 7= 21 mod 7 = 0, which agrees b. When today is Sunday, 7 days from today is Sunday also. Hence DayN should be 0. Substituting DayT = 0 ( Sunday ) and n = 7 into the formula gives DayN = ( DayT + N ) mod 7 = ( 0 + 7 ) mod 7 = 0, which agrees. c. When today is Thursday, twelve days from today is one week ( which is Thursday ) plus five days ( which is Tuesday ) . Hence DayN should be 2. Substituting DayT = 4 ( Thursday ) and N = 12 into the formula gives the correct result:DayN = ( DayT + N ) mod7 = ( 4 + 12 ) mod 7 = 16 mod 7 = 2.

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 1m + 1n is an integer. b. There are distinct integers m and n such that 1m + 1n is an integer

a. For example, let m = n = 2. Then m and n are integers such that m > 1 and n > 1 and 1m + 1n = 12 + 12 = 1, which is an integer. b. For example, let m= 1 and n = —1. Then 1m + 1n = 11 + 1(−1) = 1 + ( - 1 ) = 0 In fact, if k is any nonzero integer, then 1k + 1(−k) = 1k + ( - 1k ) = 0.

What is the best choice to fill in the following blanks? If n and d are integers with d > 0 , n div d is___________________________________________and n mod d is _____________________________________________.

the quotient obtained when n is divided by d ; the nonnegative remainder obtained when n is divided by d