SOLUTIONS FOR HOMEWORK 1 Bonus problem: 1.25. Answer: 9, 2, 2. To see this, denote the ages of the daughters, in the non-increasing order, by a1 , a2 , a3 . We know that: (1) a1 > a2 a3 (see the reference to the eldest daughter). (2) a1 a2 a3 = 36. (3) Mo
SECOND TAKE-HOME MIDTERM, FALL 2006 DUE FRIDAY 11/17 (1) [Hint: I recommend that you start by working out the example in part (g).] (a) (4 points): Write down the denitions of injective and well-dened. Denition: An equivalence relation is called trivial,
MATH 347: SOLUTIONS FOR MIDTERM 1
NOTE: 100% = 40 points. 1 (10 points): Suppose C and D are subsets of the domain of a function f. By the denition of a subset, we have to show that any element of f (C D ) is also an element of f (C ) f (D ). To this end,
MATH 347 MIDTERM 1: PRACTICE PROBLEMS The test will be given on Wednesday, October 7. It will be based on Homeworks 1-5 (part of Chapters 1-3). In preparing for the test, you can practice solving the problems from the list below. In addition, take a look
Math 347 Practice Exam #2 Name All problems except # 2 require proofs and/or explanations. 1. Let f : Z Z be dened by f (n) = 3n + 1. (a) Is f an injection ? (b) Is f a surjection ? Explain: Explain:
2. Let f, g : R R. Answer each as True or False (no exp
MATH 247 FALL 2000 FINAL EXAM BRIEF SOLUTIONS
NAME:
Total: 200 points. Do 8 out of 12 questions. You MUST indicate which 8 questions are to be graded; otherwise, just the rst 8 problems will be graded. EXPLAIN every answer. No books, notes, calculators or
Math 347 C1
HOUR EXAM III
5 August 2009 SOLUTIONS 1. Let a, b, c be integers, relatively prime in pairs, that is, gcd(a, b) = gcd(a, c) = gcd(b, c) = 1. Suppose that a2 + b2 = c2 . Show that a and b must have different parities. SOLUTION Assume that they
Math 347 C1
HOUR EXAM II
21 July 2009 SOLUTIONS 1. Consider the equation: 5x + 2y = 37. a) Find integers x, y, which satisfy the equation. b) Find integers x, y, with y positive, which satisfy the equation. (If your answer to part a) has y > 0, then you a
Math 347 C1
HOUR EXAM I
30 June 2009 SOLUTIONS 1. Let a be a non-zero real number. Show that a2 + SOLUTION Assume the opposite: a2 + 1 < 2. a2 1 2. a2
Then, since a2 > 0 we can multiply by a2 without changing the direction of the inequality. Thus a4 + 1 <
Math 347 Honors, Spring 2009, Problem Solutions HW# 1 1.4 The square has the largest area among all rectangles with a given perimeter. The problem may be rewritten as follows: Given p > 0, prove that among all rectangles with perimeter p, the square of si
Math 347 Exam #1 Name
Spring 2005
All problems except # 1 require proofs and/or explanations. 1. (16 points) Write the negation of each statement. (a) x R, if x2 > x, then x < 0 or x > 1. Negation: (b) > 0, N N such that for each integer n N , |an L| < .
347-G1, Fall 2009
Practice Exam 1 Solutions
1
1. (15 points) For each of the claims below, rewrite them in logical notation, negate them and then write out an English sentence which expresses the negated claim (without using words of negation). (a) For ev
Math 347-G1. (Milin)
Solutions to Exam 1
1
1. (20 points) For both claims below, rewrite them in logical notation, negate the logical expressions and then write out an English sentence which expresses the negated claim without using words of negation. (a)
SOLUTIONS FOR HOMEWORK 6
4.13. This is a bonus problem. Here we consider the dierence x between a number n = and its opposite n(o) = . When computing this dierence, we subtract the smaller number from the larger one. For instance, when x = 239, we have n
SOLUTIONS FOR HOMEWORK 5 2.18. This is a bonus problem. Write our polynomial as p(x) = n=0 ak xk . Denote the k set of even and odd integers between 0 and n by E and O, respectively. Then p(x) = k k k O ak . Then p(1) = A + B , while k E ak and B = k E ak
SOLUTIONS FOR HOMEWORK 4 2.23. We say that g S if (a, c) (0, )2 )(x (a, )(|g (x)| c|f (x)|). Negating this statement, we see that g S if and only if / (a, c) (0, )2)(x (a, )(|g (x)| > c|f (x)|), or, in plain English, for any positive real numbers a and c,
SOLUTIONS FOR HOMEWORK 3
2.40. (a) Suppose, for sake of contradiction, that the chessboard with corners removed (CWCR, for short) is covered by dominoes. The number of dominoes required equals 62/2 = 31 (each domino covers 2 squares, out of the 62 availab
SOLUTIONS FOR HOMEWORK 2
1.51. (a) Yes, If (X Y ) = If (X ) If (Y ). To establish this equality, note that a If (X Y ) i f (a) X Y . The last equality holds when either f (a) X , or f (a) Y . By the denition of If , f (a) X is equivalent to a If (X ), and
1
Part I Solutions
SOLUTIONS FOR PART I
1. NUMBERS, SETS, AND FUNCTIONS
1.1. We have at least four times as many chairs as tables. The number of
chairs (c) is at least () four times the number of tables (t ). Hence c 4t .
1.2. Fill in the blanks. The equa