1. Write the following sets in list form. 1(i) The set of letters in the word banana. Answer: {a, b, n}.

Transcription

1 Mathematics 0N1 Solutions 1 1. Write the following sets in list form. 1(i) The set of letters in the word banana. {a, b, n}. 1(ii) {x : x 2 + 3x 10 = 0}. { 5, 2} 1(iii) {x : x an integer and 3 < x < 10}. {4, 5, 6, 7, 8, 9} 2. Write the following sets in predicate form. 2(i) { a, e, i, o, u }. 2(ii) { 2, 2 }. {x : x is a vowel in English alphabet} {x R : x 2 2 = 0} 2(iii) { 5, 7, 9, 11, 13, 15, 17 }. {x Z : 5 x 17 and x is odd} 3. Let A = {a, b, c, d, e}, B = {c, e, f } and C = {a, d, e}. Which of the following statements are true? 3(i) c A. True 3(ii) f A. False 3(iii) B A. False 1

2 2 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi) B. False 3(vii) B. True 4. Find all relations =, or between pairs of {r, t, s}, {s, t, r, s}, {t, s, t, r}, {s}, s. s {r, t, s}, s {s, t, r, s}, s {t, s, t, r}, s {s}; {s} {r, t, s}, {s} {s, t, r, s}, {s} {t, s, t, r}, {s} {s}; {r, t, s} = {s, t, r, s} = {t, s, t, r}. Notice that if two sets A and B are equal then A B and B A. This means that {r, t, s} {s, t, r, s}, {s, t, r, s} {r, t, s}, {r, t, s} {t, s, t, r}, {t, s, t, r} {r, t, s}, {s, t, r, s} {t, s, t, r}, {t, s, t, r} {s, t, r, s}. 5. Find all subsets of the set {w, x, y, z}., {w}, {x}, {y}, {z}, {w, x}, {w, y}, {w, z}, {x, y}, {x, z}, {y, z}, {w, x, y}, {w, x, z}, {w, y, z}, {y, x, z}, {w, x, y, z}. This makes 16 = 2 4 subsets altogether.

3 0N1 MATHEMATICS SOLUTIONS 2 16DEC17 3 0N1 Mathematics Solutions (i) If A B is it necessarily true that B A? NO 1(ii) If A B and A C is it necessarily true that B = C? NO 1(iii) If A B and B C is it necessarily true that A C? YES 1(iv) If A B is it necessarily true that A B? YES 1(v) If A B is it necessarily true that A B? NO 1(vi) If A B and B C is it necessarily true that A C? YES 1(vii) If x B and A B is it necessarily true that x A? NO 1(viii) If x A and A B is it necessarily true that x B? YES 2. Which of the following sets are finite, and which infinite? 2(i) The set of even negative integers. INFINITE 2(ii) {x : x Z and x 2 < 9}. FINITE

4 4 2(iii) {x : x R and x 2 < 9}. INFINITE (i(v) [ 1 2, 1]. INFINITE 2(v) { 1 2, 1}. FINITE 3. Let the universal set be the English alphabet (all 26 letters), i.e. U = {a, b, c,..., z}. Let Find A = {a, b, c, d, e}, B = {b, d, f, h, j}, C = {j, k, l, m, n}. 3(i) A B {b, d} 3(ii) A C 3(iii) B C {j} 3(iv) A B {a, b, c, d, e, f, h, j} 3(v) B C {b, d, f, h, j, k, l, m, n}

5 0N1 MATHEMATICS SOLUTIONS 2 16DEC17 5 3(vi) (A B C) {g, i, o, p, q, r, s, t, u, v, w, x, y, z} 4. Let the universal set be the set of all integers, i.e. U = Z. Let A = {x U : x is odd }, B = {x U : x 2 > 25}, C = {x U : x is negative }. Find 4(i) A B C A B C = {..., 13, 11, 9, 7}. It can also be described as A B C = {x Z : x is odd and x < 5}. 4(ii) A B A B = { 5, 3, 1, 1, 3, 5} 4(iii) A C A C = {1, 3, 5, 7, 9, 11,... } 4(iv) B C {..., 7, 6, 5, 4, 3, 2, 1, 6, 7,... } 5. Let A be any subset of a universal set U. Find

6 6 5(i) U A A 5(ii) A A A 5(iii) U 5(iv) A A 5(v) A A 5(vi) U 5(vii) U A U 5(viii) A A U 5(ix) A A A 5(x) A 6. By drawing Venn diagrams, decide whether or not the following statements are always true when A, B and C are subsets of a universal set U.

7 0N1 MATHEMATICS SOLUTIONS 2 16DEC17 7 (i) (A B) = A B FALSE. (ii) (A B C) = A B C TRUE. (iii) (A B) (B C) (C A) = (A B) (B C) (C A) TRUE. See Figure 1 on page 8. (iv) A (B C) (A B) C TRUE. See Figure 2 on page 9. Much harder (non-compulsory) questions 7. If elements of an infinite set A can be numbered by natural numbers 1, 2, 3,..., it is called countable. Examples: The set N of all natural numbers its elements are their own numbers. The set of even natural numbers E = { 2, 4, 6, 8,... }: if e E, its number is e 2. Prove that the following sets are countable: (i) The set of all odd natural numbers. (ii) The set Z of integers. (iii) The set of all odd integers { 3, 1, 1, 3, 5,... }. (iv) The set of all possible sentences in English language. [Answer to this question is likely to involve a very practically useful concept: lexicographical order.]

8 8 FIGURE 1. Venn diagrams for Question 2.6(iii). Top row, from left to right: A B, B C, C A, (A B) (B C) (C A) Bottom row, from left to right: A B, B C, C A, (A B) (B C) (C A) As you can see, the last two diagrams are the same. This proves that is true. (A B) (B C) (C A) = (A B) (B C) (C A)

9 0N1 MATHEMATICS SOLUTIONS 3 16DEC17 9 FIGURE 2. Venn diagrams for Question 2.6(iv). Top row, from left to right: A, B C, A (B C) Bottom row, from left to right: A B, C, (A B) C As you can see, the last set in the top row is part of the last set in the bottom row. This proves that is true. A (B C) (A B) C 0N1 Mathematics Solutions 3 1. Using the laws of Boolean algebra simplify the following expressions. (i) (A B ) (A B ) by (8) = (A ) (B ) by (7) = A B (ii) (A B) A

10 10 (A B) A by (1) = (B A) A by (3) = B (A A ) by (7) = B U by (6) = U (iii) A (A B) A (A B) by (4) = (A A ) (A B) by (7) = U (A B) by (1) = (A B) U by (6) = A B (iv) A (A B) A (A B) by (8) = A ((A ) B ) by (7) = A (A B ) by (3) = (A A) B by (1) = (A A ) B by (7) = B by (1) = B by (6) = 2. Let A = {x R : x 1 > 2 and x < 4} and let Prove that A B. B = {x R : 5 x 2 20}.

11 0N1 MATHEMATICS SOLUTIONS 3 16DEC17 11 Let x A. Then x 1 > 2 and x < 4. Hence x > 3 and x < 4. Hence x 2 > 9 and x 2 < 16. Hence x 2 5 and x Hence x B. So every element of A belongs to B. Therefore A B. 3. Prove that {x R : x 2 3x + 2 < 0} = (1, 2). (Here (1, 2) is the open interval from 1 to 2.) Write A = {x R : x 2 3x + 2 < 0} and B =]1, 2[. Let x A. Then so x 2 3x + 2 < 0, (x 2)(x 1) < 0. Hence EITHER x 2 > 0 and x 1 < 0 OR x 2 < 0 and x 1 > 0. The first is impossible. Therefore x 2 < 0 and x 1 > 0. hence 1 < x < 2. Hence x B. Therefore A B. 4. A survey of 25 people was made to ask if they liked brands A, B and C. (1) 6 people liked A and B. (2) 7 people liked B but not A. (3) 2 people liked all three brands. (4) 7 people liked B and C. (5) 3 people liked A and C but not B. (6) 5 people liked A only. (7) 15 people disliked C. How many people liked B only? How many people liked C but not A? Denote by a, b, c, d, e, f, g, h the number of people in each region of the Venn diagram as shown.

12 12 A a f d g e b B c h C (0) a + b + c + d + e + f + g + h = 25 (1) d + g = 6 (2) b + e = 7 (3) g = 2 (4) e + g = 7 (5) f = 3 (6) a = 5 (7) a + b + d + h = 15. From (3), (5), (6), g = 2, f = 3, a = 5. Therefore, from (1), d = 4. From (4), e = 5. From (3), b = 2. From (7), h = 4. From (0), c = 0. Hence number of people who like B only is b = 2. Number of people who like C but not A is c + e = Construct truth tables for each of the following expressions. (i) p p p p p p T F F F T T

13 0N1 MATHEMATICS SOLUTIONS 3 16DEC17 13 (ii) ( p q) p q p p q ( p q) T T F F T T F F F T F T T T F F F T F T (iii) p q p q q p q T T F T T F T T F T F F F F T T (iv) (p q) p p q p q (p q) p T T T T T F F F F T F F F F T F (v) (p (q r)) (p (q r)) p q r r q r p (q r) (p (q r)) (p (q r)) T T T F T T T T T F T T T T T F T F F F F T F F T T T T F T T F T T T F T F T T T T F F T F F T T F F F T T T T (vi) ((p q) p) q p q p q (p q) p ((p q) p) q T T T T T T F F T F F T T F T F F T F T

14 14 Much harder (non-compulsory) questions 5. It frequently happens that connectives can be expressed in terms of other connectives we already know, for example, that p q p q, that is, conditional can be expressed in terms of disjunction and negation. (i) Express conjunction in terms of disjunction and negation. (ii Express disjunction in terms of conditional and negation. (iii) Hence express conjunction in terms of conditional and negation. (iv) Try to rewrite Boolean Laws in terms of conditional and negation. You will see that this is possible, but the logical identities that you get that you get are ugly and not very intuitive. The Boolean Laws have been chosen in the way they are for good reason: they are simple and transparent. 6. Recall that a statement is called a tautology if takes truth value T regardless of truth values of simple component statement involved in it. For example, p p, p p, and p (q p) are tautologies, but p (p q) is not. For statements built from component statements using only conditional one example is (p q) ((q r) (o (r q))), there is quick way of easily checking whether it is a tautology or not. Find it. Hint: p q is F if and only if p T and q F. Given a statement built conditional only, try to analyse how it could happen to be F. Start with a relatively short statement with just two variables, that is, simple statements involved. 7. Try to figure out how to produce a statement with a given truth table. For example, can you find a formula of Boolean Algebra with the following truth table? p q? T T F T F F F T T F F F

15 0N1 MATHEMATICS SOLUTIONS 3 16DEC17 15 (This problem lies at the heart of the digital circuits theory in electronic engineering.)

16 16 0N1 Mathematics Solutions 4 1. Determine the truth value of each of the compound statements (a), (b) and (c) given the following information: (i) The statement Mr Black is taller than Mr Blue is true. (ii) The statement Mr Green is shorter than Mr White is false. (iii) The statement Mr Blue is of average height is true. (iv) The statement Mr Brown has the same height as Mr Black is false. (a) If Mr Black is taller than Mr Blue, then Mr Green is not shorter than Mr White. T. Let p be Mr Black is taller than Mr Blue T Let q be Mr Green is shorter than Mr White F Let r be Mr Blue is of average height T Let s be Mr Brown has same height as Mr Black F p q. p q q p q T F T T (b) If either Mr Brown does not have the same height as Mr Black or Mr Green is shorter than Mr White, then Mr Blue is of average height T. ( s q) r. s q r s s q ( s q) r F F T T T T (c) If Mr Brown does not have the same height as Mr Black, then either Mr Green is shorter than Mr White or Mr Blue is not of average height. F. s (q r). s q r s r q r s (q r) F F T T F F F

17 0N1 MATHEMATICS SOLUTIONS 4 16DEC Which of the following statements are tautologies? (i) (p q) (q p) Tautology. p q p q q p (p q) (q p) T T T T T T F T T T F T T T T F F F F T (ii) p ((p q) r) Tautology. Always T when p is F. So we only need to consider the case where p is T. But then (p q) r is T. Hence p ((p q) r) is T. (iii) p (q (q p)) Tautology. p q q p q (q p) p (q (q p)) T T T T T T F T T T F T F F T F F T T T (iv) ((p q) q) p Not a tautology. p q p q (p q) q ((p q) q) p T T T T T T F F T T F T T T F F F T F T (v) (p q) (p r) Tautology. If p is F then p q is F so (p q) (p r) is T. So we only need to consider the case where p is T. p q r p q p r (p q) (p r) T T T T T T T T F T T T T F T F T T T F F F T T

18 18 (vi) (p q) (q p) Not a tautology. p q p q q p (p q) (q p) T T T T T T F F T F F T T F F F F T T T 3. Which of the following statements are logically equivalent to the statement Either it is raining or it is snowing. [USE TRUTH TABLES OR THE FUNDAMENTAL LOGICAL EQUIVALENCES, WHICHEVER IS EASIER.] (i) It is not the case that it is not raining and not snowing. Let p be it is raining ; let q be it is snowing. The given statement is p q. ( p q) p q p q. (ii) Either it is raining and snowing or it is snowing. (p q) q. p q p q (p q) q p q T T T T T T F F F T F T F T T F F F F F Last 2 columns are different. So (p q) q p q. (iii) Either it is raining or it is snowing and not raining. p (q p) (p q) (p p) (p q) T p q.

19 0N1 MATHEMATICS SOLUTIONS 4 16DEC17 19 (iv) It is not raining and not snowing. p q (p q). This has opposite truth values to p q. So p q p q. (v) p q. p q q p q p q T T F F T T F T T T F T F F T F F T F F Last 2 columns are different. p q p q. (vi) p q p q p q. 4. Using Boolean Laws, simplify the following statements. (i) p (p q). p (p q) ( p p) ( p q) T ( p q) p q. (ii) p (q p) p (q p) p (q p) p ( q p) p (p q) ( p p) q T q T

20 20 (iii) ( p q) (p q) ( p q) (p q) ( p q) ( p q) p (q q) p T p. [Using distributive law backwards!] (iv) p ((p q) (p r)) p ((p q) (p r)) p (p (q r)) ( p p) (q r) F (q r) F. [Using distributive law backwards!]

21 0N1 MATHEMATICS SOLUTIONS 5 PREDICATE LOGIC 16DEC N1 Mathematics Solutions 5 1. In each of the following find values of x and y which make p(x, y) true and find values which make p(x, y) false. (Only one example of each is required.) (i) p(x, y) denotes x 2 + y 2 = 2. p(1, 1) is true and p(0, 0) is false. (ii) p(x, y) denotes x > y 3. p(1, 0) is true and p(0, 5) is false. (iii) p(x, y) denotes x + y xy. p(1, 0) is true and p(0, 0) is false. 2. Let p(x), q(x, y), r(x, y) and s(x, y) denote the predicates x > 0, x > y, x = y and x + y = 2, respectively. Find whether the following statements are true or false. (i) p(2) (q(1, 1) r(2, 0)). True, because p(2) is true and q(1, 1) is false. (ii) p( 1) p(1). False, because p( 1) is false and p(1) is true. (iii) (q(2, 1) s(1, 1)). False, because q(2, 1) and s(1, 1) are both true. 3. Write the following statements using quantifiers (e.g. ( x)x 2 0): (i) For all A and B, A B = B A. ( A)( B)A B = B A.

22 22 (ii) x + y = y + x for all x and y. ( x)( y)x + y = y + x. (iii) 2x > 50 for some x. ( x)2x > 50. (iv) There exist A and B such that A B. ( A)( B)A B. (v) Given any x there exists y such that y < x. ( x)( y)y < x. (vi) There exists x such that x y 2 for all y. ( x)( y)x y Let p(x, y) denote the predicate Person x answered question y. Write the following statements using predicate notation (e.g. ( x)( y)p(x, y)): (i) There was one question which was answered by everyone. ( y)( x)p(x, y). (ii) Everyone answered at least one question. ( x)( y)p(x, y). (iii) All questions were answered by everyone. ( y)( x)p(x, y). (iv) There was somebody there who answered all the questions. ( x)( y)p(x, y).

23 0N1 MATHEMATICS SOLUTIONS 5 PREDICATE LOGIC 16DEC Let p(x, y) be as in Exercise 4. Suppose Ann, Bill, Carol and Dick answered from a set of questions numbered 1, 2, 3, 4 as shown. Ann Bill Carol Dick Which of the following statements are true? (i) ( x)( y)p(x, y) False (ii) ( x)( y)p(x, y) False; Carol answered no questions. (iii) ( y)( x)p(x, y) False (iv) ( x)( y)p(x, y) True; this person is Dick. (v) ( y)( x)p(x, y) True; every question is answered by someone. (vi) ( x)( y)p(x, y) True

24 24 0N1 Mathematics Solutions 6 1. Find whether the following statements are true or false where the universal set is Z = {..., 3, 2, 1, 0, 1, 2, 3,... }. (i) ( x)( y)( z)z < x + y True (ii) ( z)( x)z < x 4 True; for example, z = 1. (iii) ( x)( y)y 2 = x False; x = 2 is a counterexample. (iv) ( y)( x)y 2 = x True (v) ( x)( y)x + y = 2 False (vi) ( z)( y)yz = 6. True; for example, z = 1 and y = Let U be a universal set which contains at least 2 different elements. Let x stands for any elements of U and let A and B stand for any subsets of U. Which of the following are true? (i) ( A)( x)(x A x A ) True. In plain English, the statement is For all sets A and all elements x, x belongs to A or x does not belong to A. This is obviously always true because P P is a tautology, every statement of that form is always true. For example, the statement

25 0N1 MATHEMATICS SOLUTIONS 6 16DEC17 25 I live on the Moon or I do not live on the Moon is true. Our particular statement ( A)( x)(x A x A ) is logically equivalent to ( A)( x)((x A) (x A)) (because x A is the same as (x A)). But, as we just discussed, (x A) (x A) is always true, and therefore ( A)( x)((x A) (x A)) is true. (ii) ( A)( x)x A False; A = is a counterexample. (iii) ( A)( B)(A B B A) False. However, the assumption that U contains at least 2 elements is essential: the statement is T if U is empty or contains only one element. (iv) ( A)( B)A B. True; take B = U (or B = A). (v) ( A)( B)(A B ( x)(x B (x A))) True 3. Simplify the following statement.

26 26 (i) (( x)( y)(p(x, y) ( z) q(z))) (( x)( y)(p(x, y) ( z) q(z))) ( x) ( y)(p(x, y) ( z) q(z)) ( x)( y) (p(x, y) ( z) q(z)) ( x)( y)( p(x, y) ( z) q(z)) ( x)( y)( p(x, y) ( z) q(z)) ( x)( y)( p(x, y) ( z)q(z)) ( x)( y)(p(x, y) ( z)q(z)) (ii) ( x) ( y)(p(x, y) q(y)) ( x) ( y)(p(x, y) q(y)) ( x)( y) (p(x, y) q(y)) ( x)( y) ( p(x, y) q(y)) ( x)( y)( p(x, y) q(y)) ( x)( y)(p(x, y) q(y)) (iii) ( x)( y)( z)p(x, y, z) ( x)( y)( z) p(x, y, z) (iv) (( x)( y) p(x, y)) ( x)q(x) (( x)( y) p(x, y)) ( x)q(x) (( x)( y) p(x, y)) ( x)q(x) ( ( x)( y) p(x, y)) ( x)q(x) (( x) ( y) p(x, y)) ( x)q(x) (( x)( y) p(x, y)) ( x)q(x) (( x)( y)p(x, y)) ( x)q(x)

27 0N1 MATHEMATICS SOLUTIONS 6 16DEC Find a counterexample to each of the following statements. (i) Every even integer is divisible by 4. For example, 2 is even but is not divisible by 4. (ii) If A and B are sets such that A B then B A. For example, A = {1}, B = {2}. (iii) If x and y are real numbers such that x 2 = y 2 then x 3 = y 3. For example, x = 1 and y = 1. (iv) For all real numbers x, y and z, if x > y then xy > yz. For example, x = 2, y = 1, z = A contrapositive to the statement p q is the statement q p. (i) Use truth tables to prove that the conditional statement p q and its contrapositive q p are logically equivalent. At this stage, it should be a routine truth table. A contrapositive to the statement is the statement It follows from (i) that ( x)(p(x) q(x)) ( x)( q(x) p(x). ( x)(p(x) q(x)) and ( x)( q(x) p(x) are logically equivalent.

28 28 Write the contrapositive form of each of the following statements. The universal set is Z. (ii) If x is odd then x 2 is odd. If x 2 is even then x is even. (iii) If x 2 = 4 then x < 3. If x 3 then x 2 4. (iv) If there exists y such that y 2 = x then x 0. If x < 0 then for y 2 x for every y. (iv) If x 1 and x 2 then x 2 3x If x 2 3x + 2 = 0 then x = 1 or x = 2. Discuss why the contrapositive statements in (ii) (iv) are equivalent to the original statements (that is, the both are true or false simultaneously). 6. A converse of a conditional statement p q is q p. (i) Show by real life examples, and/or by constructing truth tables that p q and q p are not logically equivalent. p: I am at home. q: The light is on. p q: If I am at home then the light is on. q p: If the light is on I am at home then I am at home. p q and q p are not the same: I can leave home but forget to switch off the light.

29 0N1 MATHEMATICS SOLUTIONS 6 16DEC17 29 (ii) Prove that is we start with a statement p q and form its contrapositive, and then the converse of the contrapositive, then the resulting statement is logically equivalent to the contrapositive of the converse. The contrapositive to p q is q p, and the converse of contrapositive is p q. If we now take the converse of p q, we get q p, and then take contrapositive of the converse, we again get the same p q.

30 30 0N1 Mathematics Solutions 7 1. Express in interval / segment notation (i) { x : 1 x 4 }. [1, 4] (ii) { x : 1 < x 4 } ]1, 4] (iii) { x : x > 4 } ]4, + [ 2. Given sets find (1) A B (2) A B A = { 2, 0, 1, 2, 3, 4}, B = [ 1, 3], C = ] 2, 1[, A B = [ 1, 3] { 2, 4} A B = {0, 1, 2, 3} (3) A C A C = [ 2, 1] {2, 3, 4} (4) A C A C = {0}

31 0N1 MATHEMATICS SOLUTIONS 7 16DEC17 31 (5) B C B C = ] 2, 3] (6) B C B C = [ 1, 1[ 3. Given intervals / segments find (1) A B A = ]0, 3[, B = ]2, 4], and C = [ 1, 1], A B =]0, 4] (2) A B A B =]2, 3[ (3) A C A C = [ 1, 3[ (4) A C A C =]0, 1] (5) B C B C = [ 1, 1] ]2, 4] (6) B C 4. Find B C = (i) [0, 1] ]1, 2] [0, 2] (ii) ], 1] [0, + [. [0, 1] 5. We take R for the universal set U. Compute

32 32 (i) [1, 2] ( ], 1] [2, + [ ). { 1, 2 } just two points! (ii) ( ], 1] [2, + [ ). The interval ]1, 2[. 6. [A much harder problem not compulsory!] Suppose x is a positive real number. Prove, by contradiction, that x + 1 x 2. Assume, by the way of contradiction, that x + 1 x < 2. Since x is positive, we can multiply the both sides of this inequality by x and get x < 2x, which can be rearranged as x 2 2x + 1 < 0 and then as (x 1) 2 < 0. But squares cannot be negative a contradiction. Hence our assumption that x + 1 x < 2 was false, which means that x + 1 x 2 for all positive real numbers x.

33 0N1 Mathematics Solutions 8 1. Solve the inequalities expressing the answers as segments, intervals, rays, half-lines, the whole line or as the empty set. (i) 2x + 1 x + 2 x 1, or x [1, + [. (ii) x + 1 < x + 2 x R. Indeed, since 1 < 2, adding an arbitrary x to the both sides, we see that x + 1 < x + 2 is true for all x R. (iii) x + 1 > x + 2 x (that is, no solution). Indeed, if the inequality holds for some real number x then, subtracting x from the both sides of the inequality, we get 1 > 2, an obvious contradiction. (iv) 2x < x x ], 0[. (v) x < 2x x ]0, + [. Indeed, the equation can be rearranged as 0 < 3x, and, after dividing the both sides by 3, as 0 < x. (vi) 2x < x < 3x or x ] 1 3, + [.

34 34 (vii) x > x x ], 0[. 2. Solve the inequalities expressing the answers as Boolean combinations (that is unions, intersections, etc.) of segments, intervals, rays, half-lines. (i) x < x 2 x ], 0[ ]1, + [. (ii) x x 3 x [ 1, 0] [1, + [. There are at least two possible approaches to this inequality and at least two ways to solve it. Solution 1. The inequality could be much simplified if we divide both its sides by x. This operation has to be done with care. We have to consider as a separate case the possibility when x = 0 and we cannot divide by x. We have treat separately the cases x > 0 and x < 0 when we can divide by x, but remember that division by x > 0 does not change the direction of inequality, but division by x < 0 changes the direction of inequality. So we have to consider three cases: x = 0, x < 0 and x > 0. CASE 0. x = 0 Substituting x = 0 into the inequality x x 3, we see that = 0 is true, hence 0 belongs to the solution set. CASE 1. x < 0 When dividing the both sides of the inequality x x 3 by x < 0 we have to change the direction of the inequality, so we get 1 x 2

35 0N1 MATHEMATICS SOLUTIONS 8 16DEC17 35 which has the solution set [ 1, 1] (or 1 x 1). But we have to remember our assumption x < 0 (or x ], 0[ ), so in this case the solutions form the set CASE 2. x > 0 ], 0[ [ 1, 1] =] 1, 0[. Division by x does not change the direction of inequality, so we get 1 x 2 which has the solution set ], 1] [1, + [. But we have to remember our assumption x > 0 (or x ]0 + [ ), so in this case the solutions form the set (], 1] [1, + [ ) ]0 + [ = [1, + [. Now we have to assemble the three Cases 0, 1, and 2 together: the solution set of is x x 3 { 0 } ] 1, 0[ [1, + [ = ] 1, 0] [1, + [. Solution 2. This approach is more general and works for many inequalities between polynomial expressions. We rearrange x x 3 as 0 x 3 x and factorise the RHS of the inequality: 0 x 3 x = x(x 2 1) = x(x + 1)(x 1). The expression x(x +1)(x 1) is zero, positive, or negative depending of whether the individual factor x, x + 1, and x 1 are zero, positive or negative. They turn into zero when x = 0, x = 1, or x = +1; it is easy to check that, in theses cases, x x 3, so the points x = 0, x = 1, x = +1 belong to the solution set.

36 36 The three points x = 1, 0, 1 divide the real line R into 4 intervals ], 1[, ] 1, 0[, ]0, 1[, ]1, + [ ; at each of these intervals the quantities x + 1, x, and x 1 are either positive or negative but have their signs unchanged. we can easily make a list. If x ], 1[, all three of x + 1, x, and x 1 are negative, so their product is negative: (x + 1)x(x 1) < 0. (just take x = 2). If we move the point x further in the positive direction and go over the boundary point x = 1, x + 1 changes sign, but x and x 1 do not, so the sign of the product (x + 1)x(x 1) changes: if x ] 1, 0[, we have (x + 1)x(x 1) > 0. Similarly, at x = 0 the quantity x changes the sign but x+1 and x 1 do not, and the sign of the product (x +1)x(x 1) changes again: if x ]0, 1[, we have (x + 1)x(x 1) < 0. Similarly, after another change of signs at x = 1 we have if x ]1, + [, we have (x + 1)x(x 1) > 0. Now the solution set of the inequality is the union x 3 x = (x + 1)x(x 1) 0 { 1 } ] 1, 0[ { 0 } { 1 } ]1, + [ = [ 1, 0] [1, + [. (iii) x x 2 x [ 1, 0]. 3. Write the negations of the following inequalities.

37 0N1 MATHEMATICS SOLUTIONS 8 16DEC17 37 (i) 2x + 1 x + 2 2x + 1 < x + 2. (ii) 2x + 1 > x + 2 2x + 1 x + 2 (iii) x > x + 1 x x Find solution sets for the negations of the following inequalities. (i) 2x + 1 x + 2 The negation is it is equivalent to 2x + 1 < x + 2, x < 1 and has the solution set ], 1[. (ii) 2x + 1 > x + 2 The negation is 2x + 1 x + 2, it is equivalent to x 1 and has the solution set ], 1]. (iii) x > x + 1 The negation is x x + 1 and has the solution set [ 1 2, + [. 5. Solve the systems of simultaneous inequalities:

38 38 (i) x x 1 0 x [ 1, 1]. Indeed, the system can be rearranged as x 1 x 1, therefore the solution set is { x : 1 x 1 } = [ 1, 1]. (ii) x x 1 0 The solution set is empty, since the two inequalities can be rearranged as x 1 1 x and contradict each other since they imply 1 x 1 and 1 1, an obvious absurdity. (iii) x + 1 x + 2 x 0 The first inequality, x + 1 x + 2, holds for all x R, therefore it is only the second inequality that matters. Hence the solution is x 0, and the solution set is [0, + [.

39 0N1 MATHEMATICS SOLUTIONS 8 16DEC17 39 (iv) x + 2 x + 1 x 0 The first inequality, x + 2 x + 1, has no solution, therefore the system of simultaneous inequalities which includes it also has no solution. (iv) 2x + 2 x + 1 x 1 The system is equivalent to x 1 x 1 and therefore has solution x = 1. The following three problems are much harder and not compulsory. They are loosely related to the harmonic mean and geometric mean discussed in Lectures 15 16, and well be discussed later. But if the rest of homework is too easy for you, you may wish to have a try now. No solution are provided for time being, 6. In the Manchester Airport, connections between terminals have segments where passengers have to walk on their own, and segments where a travelator, or a moving walkway, is provided. A passenger is in a hurry, but needs to tie his shoestrings. What is speedier: to stop on the usual pedestrian walkway and tie the shoe-strings, or tie the shoe-strings while standing, instead of walking, on a moving walkway? HINT: Buy a stop-watch and take bus 43 to Manchester Airport. In a more serious way, introduce some notation: let T

40 40 be time, in seconds, needed for tieing shoe-strings, u speed, in m/sec, of the passenger walking, v speed, in m/sec, of the travelator, and L the length of the travelator. Express the time lost in these two scenarios in terms of T, u, v, and L, and compare the two outcomes. 7. A paddle-steamer takes five days to travel from St Louis to New Orleans, and seven days for the return journey. Assuming that the rate of flow of the current is constant, calculate how long it takes for a raft to drift from St Louis to New Orleans. HINT: The answer involves essentially the same expression as the formula for geometric mean but with a twist. We need to somehow calculate the speed of the steamer downstream and upstream, and for let us introduce some fantasy unit of length, say, lieue, and chose it such a way that the distance between St Louis and New Orleans is 35 lieue; then the speed of the steamer is 35 5 = 7 lieue/day downstream and 35 7 = 5 lieue/day upstream. The difference of upstream and downstream is twice the speed of the current, hence the speed of the current is (7 5) 2 1 lieue/day, and a raft will drift 35 days. 8. Two hikers set out at sunrise and each walked with a constant speed. One went from A to B, and the other went from B to A. They met at noon, and continuing without a stop, they arrived respectively at B at 4pm and at A at 9pm. At what time was sunrise on that day? HINT: Can you imagine that the geometric mean appears in the answer? Assume that the two hikers walked from A to B and from B to A, respectively, and that they met at point M. Then the first hiker covered the distance from A to M in from sunrise to noon and then distance from M to B in 4 hours. Since he walked at constant speed, distance from A to M distance from M to B time from sunrise to noon =. 4 hours

41 0N1 MATHEMATICS SOLUTIONS 8 16DEC17 41 Similarly, for the second hiker distance from M to A distance from B to M = 9 hours time from sunrise to noon. Since it does not matter in which direction we measure distance, from A to M or from M to A, etc., distance from A to M distance from M to A = distance from M to B distance from B to M and consequently we get a proportion time from sunrise to noon = 4 hours Solving it, we have 9 hours time from sunrise to noon. time from sunrise to noon = 4 9 = 36 = 6 hours. Therefore the sunrise was 6 hours before the noon, that it, at 12 6 = 6am hours.

42 0N1 Mathematics Solutions 9 1. (Based on Exercise 4.3 of Schaum s Intermediate Algebra) Graph the following lines using the intercept method. (a) x y = 2 (b) x + y = 2 (c) 3x + 4y = 12 (d) 2x + 4y = 8 (e) 2y 3x = 6 (f) 2x + 3y = 4 See Figure 3 at page Given a line 2x + 3y = 5, separate the points A( 2, 1), B(1, 2), C(1, 2), D(1, 0), E(100, 1001) in two groups that lie on opposite sides of the line. B is one side, A, C, D, E are on another. Relative position of a point x, y with respect to a line 2x + 3y = 5 depends on the sign +, 0 of the value of the linear function 2x + 3y 5 at this point. A simple calculation shows that f (x, y) is positive at B and negative at A, C, D, E. 3. Graph the following linear inequalities in two variables x and y. (a) x 1 3 (b) 2x + y < 2 (c) x 3y > 3 (d) y 2 See Figure 4 at page Graph the following systems of simultaneous linear inequalities in variables x and y. (a) x 1, y 1 (b) x 1, y 1, x + y 3

43 0N1 MATHEMATICS SOLUTIONS 9 16DEC17 43 FIGURE 3. Graphical solutions to Question 9.1.

44 44 (c) x 1, y 1, x + y 2 (d) x 1, y 1, x + y < 2 See Figure 5 at page Solve inequalities involving the absolute value { x if x 0 x = x if x < 0 and express the answers in the interval form. (a) 2x < 8 x ] 4, 4[ (b) 2x < 8 x ] 4, 4[ (c) 2x > 8 x R (d) x + 1 > 1 x ], 2[ ]0, + [ (e) x + 1 > x x R (f) x 1 x x ], 1 2 ] Harder non-compulsory problems: 6. Sketch graphs of the functions: (a) y = x (b) y = x + x (c) y = x x

45 (d) 0N1 MATHEMATICS SOLUTIONS 9 16DEC17 45 y = { x x if x 0 0 if x = 0 7. Prove an inequality for all x, y R: x + y x + y. 8. Sketch graphs of the functions (a) y = x y y = x 0 x (b) y = x 1 y y = x x (c) y = x 1 1 y y = x x

46 46 0N1 Mathematics Solutions Solve the following quadratic inequalities (a) x 2 + 6x + 9 < 0 No solution. (b) x 2 + 4x x = 2. (c) x 2 4x + 3 < 0 x ]1, 3[. (d) x 2 + 3x x ], 2] [ 1, + [. 2. The following inequalities can be rearranged into quadratic or linear inequalities. Solve them. Warning: if you multiply / divide both part of an inequality by a number, take into account the sign of this number! (a) x(x + 2) > 3 x ], 3[ ]1, + [. After opening the brackets and rearrangement, we have an equivalent inequality x 2 + 2x 3 > 0. After completing the square, we have x 2 + 2x 3 = x 2 + 2x = (x + 1) 2 4 = (x + 1) = [(x + 1) + 2] [(x + 1) 2] = (x + 3)(x 1).

47 0N1 MATHEMATICS SOLUTIONS 10 16DEC17 47 Hence our inequality is equivalent to (x + 3)(x 1) > 0. which has solution x ], 3[ ]1, + [. (b) x + 4 x 4 x > 0, or x ]0, + [. We need to multiply the both part of the inequality by x. When x < 0, this will lead to change in the direction of the inequality. But for x < 0 4 < 0, too, so in that case x x + 4 x < 0 and x < cannot be a solution of the original inequality. We also have to exclude the case x = 0 since division by zero is not permitted. So if x is a solution then x > 0 and we can multiply the both parts of the inequality by x without changing the direction of the inequality, obtaining x x 2 4x (x 2) 2 0 But the last inequality holds for all real x. Hence our solutions are bound only by restriction made in the process of rearrangements, x > 0, and therefore the solution set is ]0, + [ (c) 1 x > x Since the expression on LHS involves division by x, the value x = 0 should be excluded from the solution set. CASE 1: x < 0. Multiplying by x, we change the direction of the inequality and get which is equivalent to 1 < x 2 x 2 1 > 0

48 48 or, which is the same (x + 1)(x 1) > 0 Therefore x + 1 and x 1 have to be of the same sign. Since we assume that x < 0, x 1 < 0 and x + 1 < 0. But x 1 < x + 1, therefore x 1 < 0 x + 1 < 0 is the same as x + 1 < 0 which is the same as x < 1. Hence in Case 1 the solution is x ], 1[. CASE 2: x > 0. Multiplying by x, we get which is equivalent to or, which is the same 1 > x 2 x 2 1 < 0 (x + 1)(x 1) < 0 Therefore x + 1 and x 1 have to be of different signs, which is possible only if 1 < x < 1. Since we assume that x > 0, this means that 0 < x < 1 x ]0, 1[. Combining these two cases, we see that the total of the solution set is x ], 1[ ]0, 1[. (d) x 2 x 3 x {0} [1, + [. (e) x 2 x 4 x [ 1, 1]. Indeed, x = 0 belongs to the solution set. Let us look for other solutions: if x 0, we can divide the both parts of the inequality by the positive number x 2 without changing the direction of the inequality, and get x 2 1. We solved this inequality before, its solution set is [ 1, 1]; it contains 0, so we have not lost

49 0N1 MATHEMATICS SOLUTIONS 10 16DEC17 49 solutions when divided by x 2. (f) x x x ], 0[. 3. Determine which of the following points A(3, 2), B(1, 2), C(1, 1) lie inside of the triangle formed by the lines 2x + y = 1, 2y x = 2, 3x + y = 6. B(1, 2). First of all, rewrite the equations of the lines as 2x + y 1 = 0 x + 2y 2 = 0 3x + y 6 = 0 and draw the following picture (purely symbolical, without any coordinates or caring about realistic proportions Solving system of linear equations, 2x + y 1 = 0 x + 2y 2 = 0 we find that the point of intersection of these two lines is (0, 1), similarly the point of intersection of the lines 2x + y 1 = 0 3x + y 6 = 0

50 50 is (1, 3), and the point of intersection of 2x + y 1 = 0 x + 2y 2 = 0 is approximately (1.2, 1, 7). We represented this information about the vertexes of the triangle formed by the three lines in the diagram Now, for each vertex we compute the sign taken by the linear function representing the opposite side of the triangle; for example, the function 3x + y 6 = 0 takes at the vertex (0, 1) the value = 2 thus is negative; we assign to the side 3x + y 6 = 0 the sign ; we do the same for the rest of the sides and add this information to the diagram: Observe that the point (x, y) of the plane liez inside of the triangle if and only i if it lies on the half plane with respect to every side as the vertex opposite sides, hence, if and only if the functions 2x + y 1, x + 2y 2, 3x + y 6 takes at the point (x, y) values of signs, +,, respectively. But it is ease to computes, that at the given points these signs are A(3, 2) :,, +, B(1, 2) :, +,, C(1, 1) :,,,.

51 0N1 MATHEMATICS SOLUTIONS 10 16DEC17 51 Hence the point B lies inside of the triangle, while points A and C are outside of the triangle. 4. Sketch the solution sets of the following systems of simultaneous inequalities in variables x and y. (a) 1 x 2, 1 y 1 (b) 1 x + y 2 (c) x y 2x (d) x 2 y 1 (e) 0 y x 2 See Figure [Examination January 2016] Find the maximal values of the functions (i) f (x, y) = x + y (ii) g(x, y) = x + 3y (iii) h(x, y) = x subject to restrictions x 0, y 0, x + 2y 6, 2x + y 6. Coefficients of the linear functions f, g, h and of the restriction functions x + 2y and 2x + y, are non-negative. Therefore we work in the classical set up of linear programming and maxima of f, g, h are reached on the boundary of the polygon defined by the restrictions. Checking the vertices (0, 3), (2, 2), (3, 0) of this polygon, we see that f, g, h reached maxima at (2, 2), (0, 3), and (3, 0), respectively, and the maximal values are f (2, 2) = 4, g(0, 3) = 9, and h(3, 0) = [Examination January 2016] Solve the system of simultaneous inequalities x 2 + y x y 2 0 The system has no solutions. That could be seen from sketching the solution sets of y x 2 1 (the set of points

52 52 in the plane over the parabola) and x y + 2 (the half-plane under the line). Alternatively, substituting x from the second inequality into the first one, one quickly gets the inequality y 2 3y + 3 0, which has no solution because the discriminant of the quadratic function involved is negative. 7. Compute (a) [0, 1] ]2, 3[ [0, 3[ (b) [ 1, 1] [2, 3] [ 3, 3] (c) ([1, 2] [2, 3]) + [3, 4] [5, 10]. 8. [Harder and non-compulsory question.] In this question, we consider only segments and interval with non-negative endpoints. Is multiplication of segments distributive in respect to addition of segments, that is, is true that for non-negative a, b, c, d, e, f [a, b] ([c, d] + [e, f ]) = ([a, b] [c, d]) + ([a, b] [e, f ])? Prove or provide a counterexample.

53 0N1 MATHEMATICS SOLUTIONS N1 Mathematics Solutions 11 Some of these problems were discussed in the lectures, but still worth looking. 1. Prove, by induction on n, that n < 2 n for every positive integer n. BASIS OF INDUCTION. 1 < 2 = 2 1, therefore the basis of induction holds. INDUCTIVE STEP. Assume that the statement is true for n = k, k < 2 k. Adding 1 to the both sides of this inequality, we get k + 1 < 2 k + 1 < 2 k + 2 k = 2 k Prove, by induction on n, that (2n 1) = n 2 for every positive integer n. Was explained in the lecture notes. 3. Prove, by induction on n, that n = 1 n(n + 1) 2 for every positive integer n. Let p n be the statement n = 1 n(n + 1). 2

54 54 Then p 1 is the statement 1 = This is clearly true. Suppose p n is true for n = k, i.e. Then k = 1 k(k + 1) k + (k + 1) = ( k) + (k + 1) Therefore = 1 k(k + 1) + (k + 1) 2 = 1 2 k(k + 1) + 1 2(k + 1) 2 = 1 (k + 1)(k + 2) k + (k + 1) = 1 2 (k + 1)((k + 1) + 1). 4. Prove, by induction on n, that if q 1 then 1 + q + q q n 1 + q n = 1 qn+1 1 q for every positive integer n. BASIS OF INDUCTION. When n = 1, is true. 1 + q = 1 q1+1 1 q = 1 q2 1 q = 1 + q INDUCTIVE STEP. Assume that the statement P(k) 1 + q + q q k 1 + q k = 1 qk+1 1 q is true. We need to prove P(k + 1) 1 + q + q q k 1 + q k + q }{{ k+1 } = 1 q(k+1)+1 1 q }{{}

55 0N1 MATHEMATICS SOLUTIONS Here, the brace }{{} under the line shows parts of the formula P(k +1) different from that of P(k). So let us try to get P(k +1) by adding q k+1 to the both sides of P(k): 1 + q + q q k 1 + q k + q k+1 }{{} = 1 qk+1 1 q + qk+1 }{{} thus proving P(k + 1). = 1 qk+1 + q k+1 (1 q) 1 q = 1 qk+1 + q k+1 q k+2 1 q = 1 q(k+1)+1, 1 q 5. Let x be any real number 1. Prove by induction that (1 + x) n 1 + nx, for all n 1. BASIS OF INDUCTION. When n = 1, (1 + x) 1 = 1 + x x is true. INDUCTIVE STEP. n = k, Then Assume that the statement is true for (1 + x) k 1 + kx. (1 + x) k+1 = (1 + x) k (1 + x) (1 + kx) (1 + x) = 1 + x + kx + kx 2 = 1 = (k + 1)x + kx (k + 1)x. 6. Recall that, for a positive integer n, n! = (n 1) n is the product of all integers from 1 to n, so that 1! = 1, 2! = 1 2 = 2,, 3! = 2 3 = 6, etc.

56 56 Prove by induction that, for all integers n 4, n! > 2 n. BASIS OF INDUCTION. When k = 4, 4! = = 24 > 16 = 2 4. INDUCTIVE STEP. If k! > 2 k, then multiplying both sides of inequality by k 1, we get k! (k + 1) > 2 k (k + 1) > 2 k 2 = 2 k Prove by induction on n that for each n 3, the angles of any n-gon in the plane have the sum equal to (n 2)π radians. HINT. Check that for a triangle (n = 3), then cut from an n-gone a triangle. The following problems are harder and not compulsory. 8. Prove that, for all integers n > 10, n 3 < 2 n. Let p k be the statement k 3 < 2 k BASIS OF INDUCTION. Please notice that the inequality is false if n 10, so the induction starts at the first true statement p 11, for k = 11, 11 3 = 1331 < 2048 = 2 11 ; that is true, and therefore p 11, the basis of induction, is true. INDUCTIVE STEP. Assume that the k 11 and that the statement p k, k 3 < 2 k, is true. we want to prove that is, that p k p k+1, (k + 1) 3 < 2 k+1

57 is a consequence of 0N1 MATHEMATICS SOLUTIONS k 3 < 2 k, for every k > 10. Now that p k is true, that k > 10, and compute: (k + 1) 3 = k 3 (k + 1)3 k ( 3 ) 3 k + 1 = k 3 k ( = k ) 3 k ( < k ) 3 10 (we used here that k > 10) = k 3 (1.1) 3 = k < k 3 2 < 2 k 2 (by p k k 3 < 2 k ) = 2 k+1. This proves p k+1, hence proves the inductive step p k p k+1 for k > 10. Hence all statements p k for k > 10 are true. 9. Let x be a real number such that x + 1 x is an integer. Prove by induction that then x n + 1 x n is an integer for all positive integers n. This problem is much harder than the previous problems. Denote z n = x n + 1 x n

58 58 BASIS OF INDUCTION. The case n = 1 is given as the assumption of the problem. It is also useful to check the statement in the case n = 0: z z 0 = = 2 is an integer. INDUCTIVE STEP. Assume that z l = x l + 1 x l is an integer for all l k. We want to prove that z k+1 is also an integer. Multiplying z k by z 1, we see that z k z 1 = (x k + 1x ) ( x + 1 ) k x is an integer. We can rearrange the expression on the RHS further: z k z 1 = (x k + 1x ) ( x + 1 ) k x So we get = x k x + x k 1 x + 1 x x + 1 k x 1 k x = x k+1 + x k x + 1 k 1 x k+1 = which means that ( x k x k+1 = z k+1 + z k 1. ) + z k z 1 = z k+1 + z k 1, z k+1 = z k z 1 z k 1. ( x k x k 1 Since z k 1 is an integer by the inductive assumption, we see that z k+1 is an integer. ) The principle of mathematical induction as we know it now was first published by the British mathematician Augustus De Morgan (the one of the De Morgan laws in Boolean Algebra) in 1838 in The Penny Cyclopedia of the Society for the Diffusion of Useful Knowledge. The

59 0N1 MATHEMATICS SOLUTIONS title of this encyclopedia clearly said that it was produced for the general public. You can find the text of De Morgan s original article at Fill in details in the last paragraph of De Morgan s article: There are cases in which the successive induction only brings any term within the general rule, when two, three, or more of the terms immediately preceding are brought within it. Thus, in the application of this method to the deduction of the well-known consequence of x + 1 x = 2 cos θ, namely, x n + 1 = 2 cos nθ, x n it can only be shown that any one case of this theorem is true, by showing that the preceding two cases are true: thus its truth, when n = 5 and n = 6, makes it necessarily follow when n = 7. In this case the two first instances must be established (when n = 1 by hypothesis, and when n = 2 by independent demonstration), which two establish the third, the second and third establish the fourth, and so on. 11. [The sudden examination paradox] 1 Friday afternoon, just before school lets out, a teacher promises his class that they will have a quiz on one of the five days of the coming week. Moreover, he guarantees the students that the quiz will be a surprise in that they won t be able to predict the night before that it will happen the next day. The class is dismayed until one of the students realizes that something fishy is going on. She reasons: The quiz can t be given on Friday, for sure, because that s the last possible day, so we would be able to predict it Thursday night. So Friday is out. That makes Thursday the last possible day the quiz can be given. But then the quiz also can t be given on Thursday, because Wednesday night we would know it was coming the next day! And in the same way we can eliminate Wednesday, Tuesday, and even Monday from the list of possible days for the quiz. 1 Adam Bjorndahl, Puzzles and Paradoxes in Mathematical Induction,

60 60 This argument is good enough to convince the rest of the class, who gleefully go about their business, content in the certainty that there can be no surprise quiz. Tuesday morning comes, however, and the teacher hands out a quiz sheet to each student. There are, of course, objections: You can t give this quiz! We already figured out that you couldn t make it a surprise no matter what day you gave it on! But the teacher is unperturbed. You figured that out, did you? Well, here s the quiz. Aren t you surprised? The students reluctantly agreed that they were. But where did their logic go awry? Students argument is self-referential, it refers to their way of arguing. [Self-referential statements and paradoxes generated by them had been discussed in the lectures.]

61 0N1 MATHEMATICS SOLUTIONS FIGURE 4. Graphical solutions to Question 9.3.

62 62 FIGURE 5. Graphical solutions to Question 9.4.

63 0N1 MATHEMATICS SOLUTIONS FIGURE 6. Graphical solutions to Question 10.4.