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}. 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi) B. False { 5, 2} 1(iii) {x : x an integer and 3 < x < 10}. {4, 5, 6, 7, 8, 9} 3(vii) B. True 4. Find all relations =, or between pairs of 2. Write the following sets in predicate form. 2(i) { a, e, i, o, u }. {x : x is a vowel in English alphabet} 2(ii) { 2, 2 }. {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 {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.
0N1 Mathematics Solutions 2 1. 3(i) A B {b, d} 1(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 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} 3(vi) (A B C) 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}. 2(iii) {x : x R and x 2 < 9}. finite 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. {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 Find 4(i) A B C A = {x U : x is odd }, B = {x U : x 2 > 25}, C = {x U : x is negative }. A B C = {..., 13, 11, 9, 7}. It can also be described as Let Find U = {a, b, c,..., z}. A = {a, b, c, d, e}, B = {b, d, f, h, j}, C = {j, k, l, m, n}. 4(ii) A B A B C = {x Z : x is odd and x < 5}. A B = { 5, 3, 1, 1, 3, 5}
0N1 Mathematics Solutions 2 3 4(iii) A C A C = {1, 3, 5, 7, 9, 11,...} 5(vi) U 5(vii) U A 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 5(i) U A 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. 5(ii) A A A 5(iii) U 5(iv) A A 5(v) A A (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 (iv) A (B C) (A B) C True
0N1 Mathematics Solutions 3 1. 1(i) 1(ii) (A B ) by (8) = (A ) (B ) by (7) = A B so (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. 1(iii) 1(iv) (A B) A by (1) = (B A) A A (A B) by (3) = B (A A ) by (7) = B U by (6) = U by (4) = (A A ) (A B) by (7) = U (A B) by (1) = (A B) U by (6) = 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 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 2 20. Hence x B. So every element of A belongs to B. Therefore A B. 3. Write A = {x R : x 2 3x + 2 < 0} and B = (1, 2). Let x A. Then x 2 3x + 2 < 0, 4. Denote by a, b, c, d, e, f, g, h the number of people in each region of the Venn diagram as shown. A a f d g c e C b B (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 = 5. h
0N1 Mathematics Solutions 3 5 5. (i) (ii) (iii) (iv) (v) (vi) p p p p T F F F T T 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 p q q p q T T F T T F T T F T F F F F T T p q p q (p q) p T T T T T F F F F T F F F F T F 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 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
0N1 Mathematics Solutions 4 1. 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 (a) p q. p q q p q T F T T T (b) ( s q) r. s q r s s q ( s q) r F F T T T T T (c) s (q r). s q r s r q r s (q r) F F T T F F F F 2. (i) 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 Tautology. (ii) 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) 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 Tautology.
0N1 Mathematics Solutions 4 7 (iv) 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 Not a tautology. (v) 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 (vi) 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 Not a tautology. 3. Let p be it is raining ; let q be it is snowing. The given statement is p q. (i) ( p q) p q p q. (ii) (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) p (q p) (p q) (p p) (p q) T p q. (iv) 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. (i) p (p q) ( p p) ( p q) T ( p q) p q.
0N1 Mathematics Solutions 4 8 (ii) p (q p) p (q p) p ( q p) p (p q) ( p p) q T q T (iii) ( 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 r)) ( p p) (q r) F (q r) F. [Using distributive law backwards!]
0N1 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. (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 2. 4. 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). 5. 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 1 2 3 4
0N1 Mathematics Solutions 5 Predicate Logic 10 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
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 = 6. 2. Let U be a universal set. 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. the statement is In plain English, 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 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 (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. (i) (( x)( y)(p(x, y) ( y) q(y))
0N1 Mathematics Solutions 6 Predicate Logic 12 (( x)( y)(p(x, y) ( y) q(y)) ( x) ( y)(p(x, y) ( y) q(y)) ( x)( y) (p(x, y) ( y) q(y)) ( x)( y)( p(x, y) ( y) q(y)) ( x)( y)( p(x, y) ( y) q(y)) ( x)( y)( p(x, y) ( y)q(y)) ( x)( y)(p(x, y) ( y)q(y)) (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)
0N1 Mathematics Solutions 7 1. Express in interval / segment notation 1. A B (i) { x : 1 x 4 }. A B =]0, 4] [1, 4] (ii) { x : 1 < x 4 } 2. A B ]1, 4] A B =]2, 3[ (iii) { x : x > 4 } 3. A C ]4, + [ A C = [ 1, 3[ 2. Given sets 4. A C A = { 2, 0, 1, 2, 3, 4}, B = [ 1, 3], A C =]0, 1] C = ] 2, 1[, 5. B C find 1. A B B C = [ 1, 1] ]2, 4] 6. B C A B = [ 1, 3] { 2, 4} B C = 2. A B A B = {0, 1, 2, 3} 3. A C 4. Find (i) [0, 1] ]1, 2] [0, 2] A C = [ 2, 1] {2, 3, 4} (ii) ], 1] [0, + [. 4. A C [0, 1] A C = {0} 5. B C B C = ] 2, 3] 6. B C 5. We take R for the universal set U. Compute (i) [1, 2] ( ], 1] [2, + [ ). B C = [ 1, 1[ 3. Given intervals / segments A = ]0, 3[, B = ]2, 4], and C = [ 1, 1], { 1, 2 } just two points! (ii) ( ], 1] [2, + [ ). find The interval ]1, 2[.
0N1 Mathematics Solutions 7 14 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. and then as (x 1) 2 < 0. But squares cannot be negative a contradiction. Hence our assumption that x + 1 x < 2 Since x is positive, we can multiply the both sides of this inequality by x and get x 2 + 1 < 2x, which can be rearranged as x 2 2x + 1 < 0 was false, which means that x + 1 x 2 for all positive real numbers x.
0N1 Mathematics Solutions 8 1. Solve the inequalities expressing the answers as segments, intervals, rays, halflines, the whole line or as the empty set. (i) 2x + 1 x + 2 x 1, or x [1, + [. (vi) 2x < x + 1 1 < 3x or x ] 1 3, + [. (vii) x > x (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. 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.
0N1 Mathematics Solutions 8 16 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 0 3 = 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 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 ], 0[ [ 1, 1] =] 1, 0[. Case 2. x > 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 as x x 3 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. 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:
0N1 Mathematics Solutions 8 17 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 x 3 x = (x + 1)x(x 1) 0 is the union (iii) x x 2 { 1 } ] 1, 0[ { 0 } { 1 } ]1, + [ = [ 1, 0] [1, + [. x [ 1, 0]. (i) 2x + 1 x + 2 The negation is 2x + 1 < x + 2, it is equivalent to 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: 3. Write the negations of the following inequalities. (i) 2x + 1 x + 2 (i) x + 1 0 x 1 0 2x + 1 < x + 2. (ii) 2x + 1 > x + 2 2x + 1 x + 2 (iii) x > x + 1 x x + 1 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]. 4. Find solution sets for the negations of the following inequalities.
0N1 Mathematics Solutions 8 18 (ii) (iii) x + 1 0 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 and 1 x 1 1 1, an obvious absurdity. 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 (iv) (iv) matters. Hence the solution is x 0, and the solution set is [0, + [. 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. 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 shoe-strings. 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.
0N1 Mathematics Solutions 8 19 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. 8. Two persons 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?
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 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 2 at page 22. 4. 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 (c) x 1, y 1, x + y 2 (d) x 1, y 1, x + y < 2 See Figure 3 at page 23. See Figure 1 at page 21. 5. Solve inequalities involving the absolute 2. Given a line 2x + 3y = 5, separate the value { points x if x 0 x = x if x < 0 A( 2, 1), B(1, 2), C(1, 2), D(1, 0), E(100, 1001) 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 ]
0N1 Mathematics Solutions 9 21 Figure 1: Graphical solutions to Question 9.1.
0N1 Mathematics Solutions 9 22 Figure 2: Graphical solutions to Question 9.3.
0N1 Mathematics Solutions 9 23 Figure 3: Graphical solutions to Question 9.4.
0N1 Mathematics Exercises 10 Solutions 24 1. Solve the following quadratic inequalities (a) x 2 + 6x + 9 < 0 No solution. (b) x 2 + 4x + 4 0 x = 2. (c) x 2 4x + 3 < 0 x ]1, 3[. (d) x 2 + 3x + 2 0 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 + 1 1 3 = (x + 1) 2 4 = (x + 1) 2 2 2 = [(x + 1) + 2] [(x + 1) 2] = (x + 3)(x 1). 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 2 + 4 4 x 2 4x + 4 0 (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 1 < x 2 which is equivalent to x 2 1 > 0 or, which is the same (x + 1)(x 1) > 0
0N1 Mathematics Exercises 10 Solutions 25 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 1 > x 2 which is equivalent to x 2 1 < 0 or, which is the same (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[. (e) x 2 x 4 Warning: Previously the solution was incorrect, the solution below is correct now. 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 solutions when divided by x 2. (f) x + 1 x 1 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). 4. Sketch the solution sets of the following systems of simultaneous inequalities in variables x and y. (d) x 2 x 3 x [0] [1, + [. Obviously, x = 0 is a solution. If x 0, we can divide the both parts of the inequality by the positive number x 2 and get 1 < x, that is, x [1, + [. (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 4.
0N1 Mathematics Exercises 10 Solutions 26 Figure 4: Graphical solutions to Question 10.4.
0N1 Mathematics Exercises 11 Solutions 27 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+1. 2. Prove, by induction on n, that 1 + 3 + 5 + + (2n 1) = n 2 for every positive integer n. Was explained in a lecture. 3. Prove, by induction on n, that 1 + 2 + 3 + + n = 1 n(n + 1) 2 for every positive integer n. Solution. Let p n be the statement 1 + 2 + + n = 1 n(n + 1). 2 Then p 1 is the statement This is clearly true. 1 = 1 2 1 2. Suppose p n is true for n = k, i.e. Then 1 + 2 + + k = 1 k(k + 1). 2 1 + 2 + + k + (k + 1) = (1 + 2 + + k) + (k + 1) = 1 k(k + 1) + (k + 1) 2 = 1 2 k(k + 1) + 1 2(k + 1) 2 = 1 (k + 1)(k + 2). 2 Therefore 1 + 2 + + k + (k + 1) = 1 (k + 1)((k + 1) + 1). 2 3. Prove, by induction on n, that if q 1 then 1 + q + q 2 + + q n 1 + q n = 1 qn+1 1 q for every positive integer n. Was explained in a lecture. 4. Let x be any real number 1. Prove by induction that (1 + x) n 1 + nx, for all n 1.
0N1 Mathematics Exercises 11 Solutions 28 Basis of induction. When n = 1, is true. (1 + x) 1 = 1 + x 1 + 1 x Inductive step. Assume that the statement is true for n = k, Then (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 2 1 + (k + 1)x. 5. Recall that, for a positive integer n, n! = 1 2 3 4 (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. Prove by induction that, for all integers n 4, n! > 2 n. Basis of induction. When k = 4, 4! = 1 2 3 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+1. 6. 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 two problems are harder and not compulsory. 7. 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 is a consequence of for every k > 10. p k p k+1, (k + 1) 3 < 2 k+1 k 3 < 2 k, Now that p k is true, that k > 10, and compute: (k + 1) 3 = k 3 (k + 1)3 k ( 3 k + 1 = k 3 k ) 3
0N1 Mathematics Exercises 11 Solutions 29 ( = k 3 1 + 1 ) 3 k ( < k 3 1 + 1 ) 3 10 (we used here that k > 10) = k 3 (1.1) 3 = k 3 1.331 < 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. 8. 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 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 0 + 1 z 0 = 1 + 1 1 = 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 ) k ( x + 1 ) x is an integer. We can rearrange the expression on the RHS further: z k z 1 = (x k + 1x ) k ( x + 1 ) x = x k x + x k 1 x + 1 x x + 1 k x 1 k x = x k+1 + x k 1 + 1 x + 1 k 1 x k+1 = So we get ( x k+1 + 1 x k+1 = z k+1 + z k 1. which means that ) + z k z 1 = z k+1 + z k 1, z k+1 = z k z 1 z k 1. ( x k 1 + 1 x k 1 Since z k 1 is an integer by the inductive assumption, we see that z k+1 is an integer. )