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(3 there are infinitely many rational numbers of

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Unformatted text preview: (3) There are infinitely many rational numbers of whose denominators are powers of 3 and which are between 0 and 1. (a) (1) (2) (3) T T T (b) (1) (2) (3) T F T (c) (1) (2) (3) F T F (d) (1) (2) (3) F F F (e) (1) (2) (3) T F F (f) (1) (2) (3) T T F Solution Key: 2.9 Solution: 3.9 Question 10: Consider the following sets of real numbers : • The set A of all natural numbers; • The set B of all natural numbers numbers that can be written only with the digit 5; • The set C of all real numbers between 0 and 1 having only 3’s and 5’s after the decimal point. Which of the following statements is correct? 6 (a) A and B have differ- ent cardinalities (b) B and C have the same cardinality (c) A has a smaller cardinality than C (d) C has a smaller cardinality than B (e) A and C have the same cardinality (f) B has a smaller cardinality than C and C has a smaller cardinality than A Solution Key: 2.10 Solution: 3.10 7 2 Solution key (1) (a) (2) (f) (3) (b) (4) (c) (5) (d) (6) (e) (7) (b) (8) (e) (9) (b) (10) (c) 8 3 Solutions Solution of problem 1.1: Since the largest number of wrong answers was 4 it follows that each student in the class got 0, 1, 2, 3, or 4 problems wrong. Since we have 5 possibilities for the number of mistakes and 35 students, the pigeonhole principle implies that at least 7 students made the same number of mistakes. Indeed if we have 5 rooms labeled by the numbers 0, 1, 2, 3, 4, and we ask each student to go to the room labeled with the number of mistakes he or she made on the test, then at least one room will end up with 7 people at the end. The correct answer is (a) . Solution of problem 1.2: Let us ask a more general question: in how many ways can John climb a stairway of n stairs? Denote the number of ways in which John could climb a stairway of n stairs by J n . When the stairway has a relatively few stairs it is easy to find the number of ways. For instance: n = 1 If John has to climb only one stair, then he can do that in only one way. So J 1 = 1. n = 2 If John has to climb a stairway of two stairs, tehn he can do it either one at a time, or the two at once. So he can climb the stairway at J 2 = 2 ways. Suppose now John is climbing a stairway with n > 2 stairs. There are precisely two different ways in which he can begin the climb: either climb one stair first, or climb two stairs first. If with his first step he climbs one stair, then he is left with climbing a stairway of n- 1 stairs, which he can climb in J n- 1 ways. If with his first step he climbs two stairs, then he is left with climbing a stairway of n- 2 stairs, which he can climb in J n- 2 ways. Therefore John can climb a stairway of n stairs in exactly J n- 1 + J n- 2 ways....
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(3 There are infinitely many rational numbers of whose...

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