1986
... 6. A class consists of 21 boys and 9 girls. On an exam the average of the class was 84 and the average of the boys was 80. The integer nearest to the average of the girls is (a) 86 (b) 87 (c) 89 (d) 91 (e) 93 7. The graph of y = f(x) may be obtained from the graph of y = f(2(x)) by (a) shifting it u ...
... 6. A class consists of 21 boys and 9 girls. On an exam the average of the class was 84 and the average of the boys was 80. The integer nearest to the average of the girls is (a) 86 (b) 87 (c) 89 (d) 91 (e) 93 7. The graph of y = f(x) may be obtained from the graph of y = f(2(x)) by (a) shifting it u ...
ProofSpace Problem Set
... a) {x ∈ Z | (3 divides x and 3 divides x2 )}. b) {n ∈ N | (∀m ∈ N)(n + m > 4)} c) {p ∈ Z | (p2 < 0) ⇒ (p = 4)}. 2 For each of the following, write the set in set builder notation. a) The set of integers y such that for all integers x, (x + y) is an integer. b) The set of all real numbers p such that ...
... a) {x ∈ Z | (3 divides x and 3 divides x2 )}. b) {n ∈ N | (∀m ∈ N)(n + m > 4)} c) {p ∈ Z | (p2 < 0) ⇒ (p = 4)}. 2 For each of the following, write the set in set builder notation. a) The set of integers y such that for all integers x, (x + y) is an integer. b) The set of all real numbers p such that ...
Practice questions for Exam 1
... 11. For each of the following, determine if a set is a subset, proper subset, or equal to the other set, or state that none of these properties can be inferred. (a) What can we say for the sets A and B if we know that A ∪ B = A? (b) What can we say for the sets A and B if we know that A − B = A? ...
... 11. For each of the following, determine if a set is a subset, proper subset, or equal to the other set, or state that none of these properties can be inferred. (a) What can we say for the sets A and B if we know that A ∪ B = A? (b) What can we say for the sets A and B if we know that A − B = A? ...
Elementary Number Theory
... Sum of the first n natural numbers Sum of the squares of the first n natural numbers Sum of the cubes of the first n natural numbers’ Problems Divisibility Tests Divisibility tests for 2,3,4,5,8,9,10,11,7,13 Co-primes product rule for divisibility Problems Divisibility, LCM, GCD of naturals and exte ...
... Sum of the first n natural numbers Sum of the squares of the first n natural numbers Sum of the cubes of the first n natural numbers’ Problems Divisibility Tests Divisibility tests for 2,3,4,5,8,9,10,11,7,13 Co-primes product rule for divisibility Problems Divisibility, LCM, GCD of naturals and exte ...
Full text
... data was found to show that, in most cases, approximately 1/10.of the numbers considered end in a particular digit b e {0, 1, ..., 9}, or approximately 10 of them end in a particular sequence of digits b± , 2?2, . .. , 2? e {0, 1, . .. , 9} . In view of the elementary example (2.1) above, surely ver ...
... data was found to show that, in most cases, approximately 1/10.of the numbers considered end in a particular digit b e {0, 1, ..., 9}, or approximately 10 of them end in a particular sequence of digits b± , 2?2, . .. , 2? e {0, 1, . .. , 9} . In view of the elementary example (2.1) above, surely ver ...
The Pigeonhole Principle
... Moreover, a1+14, a2+14, . . ., a30+14 is also an increasing sequence of distinct positive integers with 15 aj + 14 59 . Together the two sequences, each containing 30 integers, contain 60 positive integers, all of which are less than or equal to 59. By the pigeonhole principle, at least two of ...
... Moreover, a1+14, a2+14, . . ., a30+14 is also an increasing sequence of distinct positive integers with 15 aj + 14 59 . Together the two sequences, each containing 30 integers, contain 60 positive integers, all of which are less than or equal to 59. By the pigeonhole principle, at least two of ...
PDF
... Now expand this as a power series. Given a partition of n with ai parts of size i ≥ 1, we get a term xn in this expansion by choosing xa1 from the first term in the product, x2a2 from the second, x3a3 from the third and so on. Clearly any term xn in the expansion arises in this way from a partition ...
... Now expand this as a power series. Given a partition of n with ai parts of size i ≥ 1, we get a term xn in this expansion by choosing xa1 from the first term in the product, x2a2 from the second, x3a3 from the third and so on. Clearly any term xn in the expansion arises in this way from a partition ...
PDF
... A beprisque number n is an integer which is either one more than a prime number and one less than a perfect square, or one more √ than a square √ and one less than a prime. That is, either (n − 1) ∈ P and n + 1 ∈ Z or n − 1 ∈ Z and (n + 1) ∈ P. The beprisque numbers below a thousand are 1, 2, 3, 8, ...
... A beprisque number n is an integer which is either one more than a prime number and one less than a perfect square, or one more √ than a square √ and one less than a prime. That is, either (n − 1) ∈ P and n + 1 ∈ Z or n − 1 ∈ Z and (n + 1) ∈ P. The beprisque numbers below a thousand are 1, 2, 3, 8, ...
Test 2 Solutions
... Either one of the following Proof by Contraposition: Assume that it is not true that m is even or n is even. Then both m and n are odd. Since the product of two odd numbers is odd (see part a), mn is odd. We have shown that if it is not true that m is even or n is even, then it is not true that mn i ...
... Either one of the following Proof by Contraposition: Assume that it is not true that m is even or n is even. Then both m and n are odd. Since the product of two odd numbers is odd (see part a), mn is odd. We have shown that if it is not true that m is even or n is even, then it is not true that mn i ...
