Question 1

Which of the following are sets? Justify your answer. (i) The collection of all the months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. (v) The collection of all natural numbers less than 100. (vi) A collection of novels written by the writer Munshi Prem Chand. (vii) The collection of all even integers. (viii) The collection of questions in this Chapter. (ix) A collection of most dangerous animals of the world.

Accepted Answer

(i), (iv), (v), (vi), (vii) and (viii) are sets.. Well-defined test. A collection is a set only when, for every object, it can be decided beyond dispute whether the object belongs to it. Words such as best, most talented and most dangerous record an opinion, so collections described by them fail the test. (i) The months are January, June and July — fixed by their names. Set. (ii) "Ten most talented writers" varies from person to person. Not a set. (iii) "Eleven best batsmen" varies from person to person. Not a set. (iv) The boys in your class can be named exactly. Set. (v) For any number we can check whether it is a natural number below 100. Set. (vi) The novels written by Munshi Prem Chand form a definite, recorded list. Set. (vii) An integer either is even or is not; there is no ambiguity. Set. (viii) The questions printed in this chapter are fixed. Set. (ix) "Most dangerous animals" is a matter of opinion. Not a set. Sets: (i), (iv), (v), (vi), (vii), (viii). Not sets: (ii), (iii), (ix).

Question 2

Let A = 1,2,3,4,5,6. Insert the appropriate symbol ∈ or in the blank spaces: (i) 5...A (ii) 8... A (iii) 0...A (iv) 4...A (v) 2...A (vi) 10...A

Accepted Answer

Membership test. x ∈ A means x is one of the elements listed in A; x A means it is not. Here A = 1,2,3,4,5,6. (i) 5 appears in the list, so 5 ∈ A. (ii) 8 does not appear, so 8 A. (iii) 0 does not appear (the list starts at 1 ), so 0 A. (iv) 4 appears in the list, so 4 ∈ A. (v) 2 appears in the list, so 2 ∈ A. (vi) 10 does not appear, so 10 A. (i) ∈ (ii) ∉ (iii) ∉ (iv) ∈ (v) ∈ (vi) ∉

Question 3

Write the following sets in roster form: (i) A = x: x is an integer and -3 ≤ x<7 (ii) B = x: x is a natural number less than 6 (iii) C = x: x is a two-digit natural number such that the sum of its digits is 8 (iv) D = x: x is a prime number which is divisor of 60\} (v) E = The set of all letters in the word TRIGONOMETRY (vi) F = The set of all letters in the word BETTER

Accepted Answer

(i) A = -3,-2,-1,0,1,2,3,4,5,6 (ii) B = 1,2,3,4,5 (iii) C = 17,26,35,44,53,62,71,80 (iv) D = 2,3,5 (v) E = T, R, I, G, O, N, M, E, Y (vi) F = B, E, T, R. Roster form. List every element once, separated by commas, inside braces. A repeated letter is written only once, because a set records which objects belong, not how many times they occur. (i) The integers from -3 up to 6 (since x<7 ): A = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 (ii) The natural numbers below 6 are 1,2,3,4,5: B = 1,2,3,4,5 (iii) A two-digit number ab needs a+b=8 with a ≥ 1. Taking a=1,2,…,8 gives b=7,6,5,4,3,2,1,0: C = 17,26,35,44,53,62,71,80 (iv) Factorise: 60 = 2^2 × 3 × 5, so the prime divisors are 2, 3, 5: D = 2,3,5 (v) TRIGONOMETRY has letters T, R, I, G, O, N, O, M, E, T, R, Y; keeping each distinct letter once: E = T, R, I, G, O, N, M, E, Y (vi) BETTER has letters B, E, T, T, E, R; keeping each distinct letter once: F = B, E, T, R A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}; B = {1, 2, 3, 4, 5}; C = {17, 26, 35, 44, 53, 62, 71, 80}; D = {2, 3, 5}; E = {T, R, I, G, O, N, M, E, Y}; F = {B, E, T, R}

Question 4

Write the following sets in the set-builder form: (i) (3, 6, 9, 12\} (ii) 2,4,8,16,32 (iii) 5,25,125,625 (iv) 2,4,6, … (v) 1,4,9, …, 100

Accepted Answer

(i) x: x=3 n, n ∈ N and 1 ≤ n ≤ 4 (ii) x: x=2^n, n ∈ N. and.1 ≤ n ≤ 5 (iii) x: x=5^n, n ∈ N. and.1 ≤ n ≤ 4 (iv) x: x is an even natural number (v) x: x=n^2, n ∈ N. and.1 ≤ n ≤ 10. Set-builder form. Spot the rule that generates the listed numbers, then write x: x satisfies the rule. (A set can be described in more than one correct way; any equivalent rule is acceptable.) (i) 3,6,9,12 are the first four multiples of 3: x: x=3n, n ∈ N, 1 ≤ n ≤ 4 (ii) 2,4,8,16,32 are 2^1,2^2,2^3,2^4,2^5: x: x=2^n, n ∈ N, 1 ≤ n ≤ 5 (iii) 5,25,125,625 are 5^1,5^2,5^3,5^4: x: x=5^n, n ∈ N, 1 ≤ n ≤ 4 (iv) The dots show the list never ends; these are all the even natural numbers: x: x is an even natural number (v) 1,4,9,…,100 are 1^2,2^2,3^2,…,10^2: x: x=n^2, n ∈ N, 1 ≤ n ≤ 10 (i) {x: x = 3n, n ∈ N, 1 ≤ n ≤ 4} (ii) {x: x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 5} (iii) {x: x = 5ⁿ, n ∈ N, 1 ≤ n ≤ 4} (iv) {x: x is an even natural number} (v) {x: x = n², n ∈ N, 1 ≤ n ≤ 10}

Question 5

List all the elements of the following sets: (i) A = x: x is an odd natural number (ii) B = x: x. is an integer,.-1/2<x<9/2 (iii) C = x: x. is an integer,.x^2 ≤ 4 (iv) D = x: x is a letter in the word "LOYAL" (v) E = x: x is a month of a year not having 31 days (vi) F = x: x is a consonant in the English alphabet which precedes k.

Accepted Answer

(i) A = 1,3,5, … (ii) B = 0,1,2,3,4 (iii) C = -2,-1,0,1,2 (iv) D = L, O, Y, A (v) E = February, April, June, September, November \} (vi) F = b, c, d, f, g, h, j. Roster form. Read the defining property, then write out every object that satisfies it. (i) The odd natural numbers form an unending list: A = 1,3,5,7,9,… (ii) We need integers strictly between -1/2 and 9/2=4.5. The integers in that range are 0,1,2,3,4: B = 0,1,2,3,4 (iii) x^2 ≤ 4 means -2 ≤ x ≤ 2, and x is an integer: C = -2,-1,0,1,2 (iv) LOYAL uses the letters L, O, Y, A, L; each distinct letter is written once: D = L, O, Y, A (v) The months with 31 days are January, March, May, July, August, October, December; the rest are: E = February, April, June, September, November (vi) The letters before k are a,b,c,d,e,f,g,h,i,j; removing the vowels a, e, i leaves the consonants: F = b,c,d,f,g,h,j A = {1, 3, 5, 7, 9, …}; B = {0, 1, 2, 3, 4}; C = {−2, −1, 0, 1, 2}; D = {L, O, Y, A}; E = {February, April, June, September, November}; F = {b, c, d, f, g, h, j}

Question 6

Match each of the set on the left in the roster form with the same set on the right described in set-builder form: (a) x: x is a prime number and a divisor of 6 (b) x: x is an odd natural number less than 10 (c) x: x is natural number and divisor of 6 (d) x: x is a letter of the word MATHEMATICS. (i) 1,2,3,6 (ii) 2,3 (iii) \{M,A,T,H,E,I,C,S\} (iv) 1,3,5,7,9

Accepted Answer

(i) ↔ (c) (ii) ↔ (a) (iii) ↔ (d) (iv) ↔ (b). Convert each set-builder description to roster form, then match. (a) The divisors of 6 are 1,2,3,6; of these the primes are 2 and 3. So the set is 2,3 — this is (ii). (b) The odd natural numbers less than 10 are 1,3,5,7,9. So the set is 1,3,5,7,9 — this is (iv). (c) The natural numbers dividing 6 are 1,2,3,6. So the set is 1,2,3,6 — this is (i). (d) MATHEMATICS uses the letters M, A, T, H, E, M, A, T, I, C, S; the distinct letters are M, A, T, H, E, I, C, S — this is (iii). (a) ↔ (ii), (b) ↔ (iv), (c) ↔ (i), (d) ↔ (iii)

Question 7

Which of the following are examples of the null set (i) Set of odd natural numbers divisible by 2 (ii) Set of even prime numbers (iii) x: x is a natural numbers, x<5 and x 7 (iv) y: y is a point common to any two parallel lines

Accepted Answer

(i), (iii), (iv). Null-set test. A set is the null (empty) set φ when no object satisfies its defining property. So look for a description that is impossible to fulfil. (i) An odd number leaves remainder 1 on division by 2, so it can never be divisible by 2. No such number exists — null set. (ii) 2 is even and prime, so the set is 2; it has one element — not a null set. (iii) No number can satisfy x<5 and x 7 at the same time — null set. (iv) Parallel lines, by definition, never meet, so they have no point in common — null set. Null sets: (i), (iii) and (iv). (ii) is not a null set, since it equals {2}.

Question 8

Which of the following sets are finite or infinite (i) The set of months of a year (ii) 1,2,3, … (iii) 1,2,3, … 99,100 (iv) The set of positive integers greater than 100 (v) The set of prime numbers less than 99

Accepted Answer

Finite or infinite. A set is finite if its elements can be counted and the counting comes to an end; otherwise it is infinite. (i) A year has exactly 12 months — finite. (ii) 1,2,3,… is the set of all natural numbers, and the list never ends — infinite. (iii) 1,2,3,…,99,100 has exactly 100 elements — finite. (iv) The integers 101,102,103,… go on for ever — infinite. (v) The primes below 99 are a part of the 98 numbers 1,2,…,98, so the counting stops — finite. Finite: (i), (iii), (v). Infinite: (ii), (iv).

Question 9

State whether each of the following set is finite or infinite: (i) The set of lines which are parallel to the x -axis (ii) The set of letters in the English alphabet (iii) The set of numbers which are multiple of 5 (iv) The set of animals living on the earth (v) The set of circles passing through the origin (0,0)

Accepted Answer

Finite or infinite. Ask whether the counting of elements comes to an end. (i) One line parallel to the x -axis passes through each point of the y -axis, and there are endlessly many such points — infinite. (ii) The English alphabet has exactly 26 letters — finite. (iii) The multiples 5,10,15,… never stop — infinite. (iv) The animals alive on the earth are enormously many, but they can in principle be counted — finite. (v) A circle through the origin is fixed by choosing its centre, and the centre may be any point of the plane, so there are endlessly many — infinite. Finite: (ii), (iv). Infinite: (i), (iii), (v).

Question 10

In the following, state whether A = B or not: (i) A = a, b, c, d, B = d, c, b, a (ii) A = 4,8,12,16 B = 8,4,16,18 (iii) A = 2,4,6,8,10 B = x: x is positive even integer and x ≤ 10 (iv) A = x: x is a multiple of 10, B = 10,15,20,25,30, …

Accepted Answer

(i) Yes (ii) No (iii) Yes (iv) No. Equality of sets. A = B when every element of A lies in B and every element of B lies in A. The order in which elements are written, and any repetition, make no difference. (i) A = a,b,c,d, B = d,c,b,a: the same four letters, only reordered. A = B. (ii) A = 4,8,12,16, B = 8,4,16,18: 12 ∈ A but 12 B. A ≠ B. (iii) The positive even integers with x ≤ 10 are 2,4,6,8,10, so B = 2,4,6,8,10 = A. A = B. (iv) A = 10,20,30,40,…, while 15 ∈ B and 15 is not a multiple of 10. A ≠ B. A = B in (i) and (iii); A ≠ B in (ii) and (iv).

Question 11

Are the following pair of sets equal? Give reasons. (i) A = 2,3, B = x: x. is solution of.x^2+5 x+6=0 (ii) A = x: x is a letter in the word FOLLOW B = y: y is a letter in the word WOLF

Accepted Answer

(i) No (ii) Yes. Equality of sets. Two sets are equal only if they contain exactly the same elements, so first put each set into roster form. (i) Solve x^2+5x+6=0: x^2+5x+6=(x+2)(x+3)=0 ⇒ x=-2 or x=-3 So B = -2,-3, while A = 2,3. Since 2 B, the sets are not equal. (ii) FOLLOW uses the letters F, O, L, L, O, W, so A = F, O, L, W. WOLF gives B = W, O, L, F. These are the same four letters, so the sets are equal. (i) A ≠ B, because B = {−2, −3}. (ii) A = B = {F, O, L, W}.

Question 12

From the sets given below, select equal sets:

Accepted Answer

Equality of sets. Compare the sets element by element; order and repetition are irrelevant. B = 1,2,3,4 and D = 3,1,4,2 contain exactly the numbers 1,2,3,4, so B = D. E = -1,1 and G = 1,-1 contain exactly -1 and 1, so E = G. No other pair matches: - A = 2,4,8,12 and C = 4,8,12,14 differ, since 2 ∈ A but 2 C. - F = 0,a and H = 0,1 differ, since a is a letter and not the number 1. B = D and E = G

Question 13

Make correct statements by filling in the symbols or in the blank spaces: (i) 2,3,4 … 1,2,3,4,5 (ii) a, b, c … b, c, d (iii) x: x is a student of Class XI of your school … x: x student of your school (iv) x: x is a circle in the plane … x: x is a circle in the same plane with radius 1 unit\} (v) x: x is a triangle in a plane … x: x is a rectangle in the plane (vi) x: x is an equilateral triangle in a plane … x: x is a triangle in the same plane (vii) x: x is an even natural number … x: x is an integer

Accepted Answer

(i) (ii) (iii) ᄃ (iv) (v) (vi) ᄃ (vii). Subset test. P Q means every element of P is also an element of Q. To show P Q it is enough to exhibit one element of P that is missing from Q. (i) 2,3,4 all appear in 1,2,3,4,5: 2,3,4 1,2,3,4,5. (ii) a b,c,d: a,b,c b,c,d. (iii) Every Class XI student of your school is a student of your school:. (iv) A circle of radius 2 units lies in the plane but is not of radius 1 unit:. (v) A triangle is never a rectangle:. (vi) An equilateral triangle is a triangle:. (vii) Every even natural number is an integer:. (i) ⊂ (ii) ⊄ (iii) ⊂ (iv) ⊄ (v) ⊄ (vi) ⊂ (vii) ⊂

Question 14

Examine whether the following statements are true or false: (i) a, b b, c, a (ii) a, e x: x is a vowel in the English alphabet (iii) 1,2,3 1,3,5 (iv) a a, b, c (v) a ∈ a, b, c (vi) x: x is an even natural number less than 6 x: x is a natural number which divides 36\}

Accepted Answer

(i) False (ii) True (iii) False (iv) True (v) False (vi) True. Distinguish ⊂ from ∈. P Q compares two sets and asks whether every element of P lies in Q; x ∈ Q asks whether the single object x is one of the listed elements of Q. (i) Both a and b belong to b,c,a, so a,b is a subset. The statement a,b b,c,a is therefore false. (ii) The vowels are a,e,i,o,u, and both a and e are among them — true. (iii) 2 ∈ 1,2,3 but 2 1,3,5 — false. (iv) The only element of a is a, and a ∈ a,b,c — true. (v) The elements of a,b,c are the letters a, b, c, not the set a — false. (vi) The left set is 2,4; the divisors of 36 are 1,2,3,4,6,9,12,18,36, which contains both 2 and 4 — true. True: (ii), (iv), (vi). False: (i), (iii), (v).

Question 15

Let A = 1,2, 3,4, 5. Which of the following statements are incorrect and why? (i) 3,4 A (ii) 3,4 ∈ A (iii) 3,4 A (iv) 1 ∈ A (v) 1 A (vi) 1,2,5 A (vii) 1,2,5 ∈ A (viii) 1,2,3 A (ix) φ ∈ A (x) φ A (xi) φ A

Accepted Answer

(i) as 3,4 ∈ A, (v) as 1 ∈ A, (vii) as 1,2,5 A, (viii) as 3 A, (ix) as φ A, (xi) as φ A,. Read the elements of A first. A = 1,2, 3,4,5 has exactly four elements: the numbers 1, 2, 5, and the set 3,4. Note that 3 and 4 themselves are not elements of A. Throughout, ∈ links an object to a set, while links a set to a set. (i) 3,4 A — incorrect. This would need 3 ∈ A and 4 ∈ A, but neither is an element of A. (ii) 3,4 ∈ A — correct. The set 3,4 is itself one of the four elements listed. (iii) 3,4 A — correct. Its only element is 3,4, and 3,4 ∈ A. (iv) 1 ∈ A — correct; 1 is listed. (v) 1 A — incorrect. 1 is a number, not a set, so cannot be used with it. ( 1 A would be correct.) (vi) 1,2,5 A — correct; each of 1,2,5 is an element of A. (vii) 1,2,5 ∈ A — incorrect; the set 1,2,5 is not one of the four elements of A. (viii) 1,2,3 A — incorrect; 3 A. (ix) φ ∈ A — incorrect; φ is not one of the four elements listed. (x) φ A — correct; the empty set is a subset of every set. (xi) φ A — incorrect. This would need φ ∈ A, which is false by (ix). Incorrect: (i), (v), (vii), (viii), (ix), (xi). Correct: (ii), (iii), (iv), (vi), (x).

Question 16

Write down all the subsets of the following sets (i) a (ii) a, b (iii) \{1, 2, 3\} (iv) φ

Accepted Answer

(i) φ, a (ii) φ, a, b, a, b (iii) φ, 1, 2, 3, 1,2, 1,3, 2,3, 1,2,3 (iv) φ. Listing subsets. A subset is formed by choosing, for each element, whether to include it or leave it out, so a set with n elements has 2^n subsets. Always remember φ and the set itself. (i) a has 2^1=2 subsets: φ, a. (ii) a,b has 2^2=4 subsets: φ, a, b, a,b. (iii) 1,2,3 has 2^3=8 subsets: φ, 1, 2, 3, 1,2, 1,3, 2,3, 1,2,3. (iv) φ has 2^0=1 subset: φ itself. (i) φ, {a} (ii) φ, {a}, {b}, {a, b} (iii) φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} (iv) φ

Question 17

Write the following as intervals: (i) x: x ∈ R,-4<x ≤ 6 (ii) x: x ∈ R,-12<x<-10 (iii) x: x ∈ R, 0 ≤ x<7 (iv) x: x ∈ R, 3 ≤ x ≤ 4

Accepted Answer

(i) (- 4, 6] (ii) (-12, -10) (iii) [0,7) (iv) [ 3, 4 ]. Interval notation. For real numbers, a round bracket excludes the end point (it goes with < ) and a square bracket includes it (it goes with ≤ ). (i) -4<x ≤ 6: -4 excluded, 6 included — (-4, 6]. (ii) -12<x<-10: both excluded — (-12, -10). (iii) 0 ≤ x<7: 0 included, 7 excluded — [0, 7). (iv) 3 ≤ x ≤ 4: both included — [3, 4]. (i) (−4, 6] (ii) (−12, −10) (iii) [0, 7) (iv) [3, 4]

Question 18

Write the following intervals in set-builder form: (i) (-3, 0) (ii) [6, 12] (iii) (6,12] (iv) [-23, 5)

Accepted Answer

Set-builder form of an interval. An interval is a set of real numbers, so write x: x ∈ R, …. A round bracket becomes a strict inequality <; a square bracket becomes ≤. (i) (-3,0): both ends excluded — x: x ∈ R, -3<x<0 (ii) [6,12]: both ends included — x: x ∈ R, 6 ≤ x ≤ 12 (iii) (6,12]: 6 excluded, 12 included — x: x ∈ R, 6<x ≤ 12 (iv) [-23,5): -23 included, 5 excluded — x: x ∈ R, -23 ≤ x<5 (i) {x: x ∈ R, −3 < x < 0} (ii) {x: x ∈ R, 6 ≤ x ≤ 12} (iii) {x: x ∈ R, 6 < x ≤ 12} (iv) {x: x ∈ R, −23 ≤ x < 5}

Question 19

What universal set(s) would you propose for each of the following: (i) The set of right triangles. (ii) The set of isosceles triangles.

Accepted Answer

Universal set. A set U can serve as a universal set for a given set A only if A U, i.e. U must contain every object being discussed. Here both sets consist of triangles, so the natural choice is a set that holds all triangles. (i) Every right triangle is a triangle, so take U = x: x is a triangle in a plane. Any larger collection also works — the set of all polygons, or the set of all plane figures. (ii) Every isosceles triangle is likewise a triangle, so the same choice serves: U = x: x is a triangle in a plane. The set of all triangles is a suitable universal set in both (i) and (ii).

Question 20

Given the sets A = 1,3,5, B = 2,4,6 and C = 0,2,4,6,8, which of the following may be considered as universal set (s) for all the three sets A, B and C (i) 0,1,2,3,4,5,6 (ii) φ (iii) \{0,1,2,3,4,5,6,7,8,9,10\} (iv) 1,2,3,4,5,6,7,8

Accepted Answer

Test each candidate for containment. A set U is a universal set for A, B, C only if A U, B U and C U — equivalently, only if U contains A B C = 0,1,2,3,4,5,6,8. (i) 0,1,2,3,4,5,6: does not contain 8 ∈ C, so C U. Not a universal set. (ii) φ: the empty set contains no element at all, so it cannot contain A, B or C. Not a universal set. (iii) 0,1,2,3,4,5,6,7,8,9,10: it contains 0,1,2,3,4,5,6,8, hence A U, B U and C U. This one works. (iv) 1,2,3,4,5,6,7,8: does not contain 0 ∈ C. Not a universal set. Only (iii) 0,1,2,3,4,5,6,7,8,9,10 can be taken as a universal set for A, B and C.

Question 21

Find the union of each of the following pairs of sets: (i) X = 1,3,5 Y = 1,2,3 (ii) A =[a, e, i, o, u B = a, b, c (iii) A = x: x is a natural number and multiple of 3 B = x: x is a natural number less than 6 (iv) A = x: x is a natural number and 1<x ≤ 6 B = x: x is a natural number and 6<x<10 (v) A = 1,2,3, B =φ

Accepted Answer

(i) X Y = 1,2,3,5 (ii) A B = a, b, c, e, i, o, u (iii) A B = x: x=1,2,4,5 or a multiple of 3 (iv) A B = x: 1<x<10, x ∈ N (v) A B = 1,2,3. Union of two sets. A B is the set of elements belonging to A, or to B, or to both; an element common to the two is listed only once. (i) X = 1,3,5, Y = 1,2,3. Collecting all elements and keeping 1 and 3 once each: X Y = 1,2,3,5 (ii) A = a,e,i,o,u, B = a,b,c. The letter a is common, listed once: A B = a,b,c,e,i,o,u (iii) A = x: x is a natural number and a multiple of 3 = 3,6,9,12,… and B = x: x is a natural number less than 6 = 1,2,3,4,5. Their union adds 1,2,4,5 to the multiples of 3: A B = 1,2,3,4,5,6,9,12,15,… = x: x = 1, 2, 4, 5 or x is a multiple of 3 (iv) A = x: x ∈ N, 1 < x ≤ 6 = 2,3,4,5,6 and B = x: x ∈ N, 6 < x < 10 = 7,8,9: A B = 2,3,4,5,6,7,8,9 = x: x ∈ N, 1 < x < 10 (v) A = 1,2,3, B = φ. The empty set contributes nothing, so A B = 1,2,3 (i) {1, 2, 3, 5} (ii) {a, b, c, e, i, o, u} (iii) {1, 2, 3, 4, 5, 6, 9, 12, 15, …} (iv) {2, 3, 4, 5, 6, 7, 8, 9} (v) {1, 2, 3}

Question 22

Let A = a, b, B = a, b, c. Is A B? What is A B?

Accepted Answer

Yes, A B = a, b, c. Subset test, then union. A B means every element of A is also an element of B. Here A = a,b and B = a,b,c. Checking each element of A: a ∈ B and b ∈ B. So every element of A lies in B, i.e. A B. For the union, collect the elements of both sets, writing repeated ones once: A B = a,b a,b,c = a,b,c = B Yes, A ⊂ B; and A ∪ B = {a, b, c} = B

Question 23

If A and B are two sets such that A C B, then what is A B?

Accepted Answer

B. Union when one set is contained in the other. Suppose A B. Take any x ∈ A B. Then x ∈ A or x ∈ B. If x ∈ A, then x ∈ B as well, because A B. Either way x ∈ B, so A B B. Conversely B A B always holds, since every element of B is in the union. Both inclusions together give equality. If A ⊂ B, then A ∪ B = B

Question 24

If A = 1,2,3,4, B = 3,4,5,6, C = 5,6,7,8 and D = 7,8,9,10; find (i) A B (ii) A C (iii) B C (iv) B D (v) A B C (vi) A B D (vii) B C D

Accepted Answer

Union of sets. List all elements that occur in at least one of the sets, each written once. Here A = 1,2,3,4, B = 3,4,5,6, C = 5,6,7,8, D = 7,8,9,10. (i) A B = 1,2,3,4,5,6 (3 and 4 are common, kept once) (ii) A C = 1,2,3,4,5,6,7,8 (iii) B C = 3,4,5,6,7,8 (5 and 6 are common) (iv) B D = 3,4,5,6,7,8,9,10 (v) A B C = ( A B ) C = 1,2,3,4,5,6 5,6,7,8 = 1,2,3,4,5,6,7,8 (vi) A B D = 1,2,3,4,5,6 7,8,9,10 = 1,2,3,4,5,6,7,8,9,10 (vii) B C D = 3,4,5,6,7,8 7,8,9,10 = 3,4,5,6,7,8,9,10 (i) {1,2,3,4,5,6} (ii) {1,2,3,4,5,6,7,8} (iii) {3,4,5,6,7,8} (iv) {3,4,5,6,7,8,9,10} (v) {1,2,3,4,5,6,7,8} (vi) {1,2,3,4,5,6,7,8,9,10} (vii) {3,4,5,6,7,8,9,10}

Question 25

Find the intersection of each pair of sets of question 1 above.

Accepted Answer

(i) X Y = 1,3 (ii) A B = a (iii) \{ 3 \} (iv) φ (v) φ. Intersection of two sets. A B is the set of elements that belong to A and to B. Taking the same five pairs as in Question 1: (i) X = 1,3,5, Y = 1,2,3. The common elements are 1 and 3: X Y = 1,3 (ii) A = a,e,i,o,u, B = a,b,c. Only a is common: A B = a (iii) A = 3,6,9,12,…, B = 1,2,3,4,5. The only multiple of 3 below 6 is 3: A B = 3 (iv) A = 2,3,4,5,6, B = 7,8,9. No element is common: A B = φ (v) A = 1,2,3, B = φ. The empty set has no element to share: A B = φ (i) {1, 3} (ii) {a} (iii) {3} (iv) φ (v) φ

Question 26

If A = 3,5,7,9,11, B = 7,9,11,13, C = 11,13,15 and D = 15,17; find (i) A B (ii) B C (iii) A C D (iv) A C (v) B D (vi) A ( B C ) (vii) A D (viii) A ( B D ) (ix) ( A B ) ( B C ) (x) ( A D ) ( B C )

Accepted Answer

Intersection, and brackets first. A B keeps only the elements common to both sets; where a bracket appears, evaluate it before intersecting. Here A = 3,5,7,9,11, B = 7,9,11,13, C = 11,13,15, D = 15,17. (i) A B = 7,9,11 (ii) B C = 11,13 (iii) A C D = ( A C ) D = 11 15,17 = φ (iv) A C = 11 (v) B D = φ (B has no element in common with 15,17 ) (vi) B C = 7,9,11,13,15, so A ( B C ) = 7,9,11 (vii) A D = φ (viii) B D = 7,9,11,13,15,17, so A ( B D ) = 7,9,11 (ix) ( A B ) ( B C ) = 7,9,11 7,9,11,13,15 = 7,9,11 (x) A D = 3,5,7,9,11,15,17 and B C = 7,9,11,13,15, so ( A D ) ( B C ) = 7,9,11,15 (i) {7,9,11} (ii) {11,13} (iii) φ (iv) {11} (v) φ (vi) {7,9,11} (vii) φ (viii) {7,9,11} (ix) {7,9,11} (x) {7,9,11,15}

Question 27

If A = x: x is a natural number, B = x: x is an even natural number C = x: x is an odd natural number andD = x: x is a prime number \}, find (i) A B (ii) A C (iii) A D (iv) B C (v) B D (vi) C D

Accepted Answer

Intersection of described sets. Read each intersection as a condition: an element must satisfy the defining property of both sets. Here A = N, B = 2,4,6,…, C = 1,3,5,…, D = 2,3,5,7,11,…. (i) Every even natural number is a natural number, so B A and A B = B = x: x is an even natural number (ii) Similarly C A, so A C = C = x: x is an odd natural number (iii) Every prime number is a natural number, so A D = D = x: x is a prime number (iv) No natural number is both even and odd, so B C = φ (v) A number in B D is an even prime. An even number greater than 2 has 2 as a divisor besides 1 and itself, so it is not prime. Hence B D = 2 (vi) A number in C D is an odd prime, i.e. every prime except 2: C D = x: x is an odd prime number = D - 2 (i) B (ii) C (iii) D (iv) φ (v) {2} (vi) the set of odd prime numbers

Question 28

Which of the following pairs of sets are disjoint (i) 1,2,3,4 and x: x is a natural number and 4 ≤ x ≤ 6 (ii) a, e, i, o, u and c, d, e, f (iii) x: x is an even integer and x: x is an odd integer

Accepted Answer

(iii). Disjoint test. Two sets are disjoint when they have no element in common, i.e. when A B = φ. So form the intersection in each case. (i) 1,2,3,4 and x: x ∈ N, 4 ≤ x ≤ 6 = 4,5,6. The element 4 lies in both, so the intersection is 4 ≠ φ — not disjoint. (ii) a,e,i,o,u and c,d,e,f. The letter e lies in both, so the intersection is e ≠ φ — not disjoint. (iii) x: x is an even integer and x: x is an odd integer. An integer is either even or odd and never both, so the intersection is φ — disjoint. Only the pair in (iii) is disjoint; (i) and (ii) are not.

Question 29

If A = 3,6,9,12,15,18,21, B = 4,8,12,16,20, C = 2,4,6,8,10,12,14,16, D = 5,10,15,20; find (i) A - B (ii) A - C (iii) A - D (iv) B - A (v) C - A (vi) D - A (vii) B - C (viii) B - D (ix) C - B (x) D - B (xi) C - D (xii) D - C

Accepted Answer

Difference of sets. A - B consists of those elements of A that do not belong to B; start from the first set and strike out whatever the second set also holds. Here A = 3,6,9,12,15,18,21, B = 4,8,12,16,20, C = 2,4,6,8,10,12,14,16, D = 5,10,15,20. (i) A - B: strike 12 from A ⇒ 3,6,9,15,18,21 (ii) A - C: strike 6, 12 ⇒ 3,9,15,18,21 (iii) A - D: strike 15 ⇒ 3,6,9,12,18,21 (iv) B - A: strike 12 from B ⇒ 4,8,16,20 (v) C - A: strike 6, 12 from C ⇒ 2,4,8,10,14,16 (vi) D - A: strike 15 from D ⇒ 5,10,20 (vii) B - C: C holds 4,8,12,16 ⇒ 20 (viii) B - D: strike 20 ⇒ 4,8,12,16 (ix) C - B: strike 4,8,12,16 from C ⇒ 2,6,10,14 (x) D - B: strike 20 ⇒ 5,10,15 (xi) C - D: strike 10 ⇒ 2,4,6,8,12,14,16 (xii) D - C: strike 10 from D ⇒ 5,15,20 (i) {3,6,9,15,18,21} (ii) {3,9,15,18,21} (iii) {3,6,9,12,18,21} (iv) {4,8,16,20} (v) {2,4,8,10,14,16} (vi) {5,10,20} (vii) {20} (viii) {4,8,12,16} (ix) {2,6,10,14} (x) {5,10,15} (xi) {2,4,6,8,12,14,16} (xii) {5,15,20}

Question 30

If X = a, b, c, d and Y = f, b, d, g, find (i) X - Y (ii) Y - X (iii) X Y

Accepted Answer

Difference and intersection. X - Y keeps the elements of X that are not in Y; X Y keeps the ones that are in both. With X = a,b,c,d and Y = f,b,d,g, the common letters are b and d. (i) Remove b, d from X: X - Y = a,c (ii) Remove b, d from Y: Y - X = f,g (iii) The letters lying in both sets: X Y = b,d (i) X − Y = {a, c} (ii) Y − X = {f, g} (iii) X ∩ Y = {b, d}

Question 31

If R is the set of real numbers and Q is the set of rational numbers, then what is R - Q?

Accepted Answer

Difference of sets. R - Q is the set of real numbers that are not rational. By definition a real number is called irrational exactly when it cannot be written as p/q with p, q integers and q ≠ 0 — that is, exactly when it is real but not rational. So removing Q from R leaves precisely the irrational numbers, such as √ 2, √ 3, π. R − Q = the set of all irrational numbers

Question 32

State whether each of the following statement is true or false. Justify your answer. (i) 2,3,4,5 and 3,6 are disjoint sets. (ii) a, e, i, o, u and a, b, c, d are disjoint sets. (iii) 2,6,10,14 and 3,7,11,15 are disjoint sets. (iv) 2,6,10 and 3,7,11 are disjoint sets.

Accepted Answer

Disjoint test. Two sets are disjoint when they share no element, i.e. when their intersection is φ. Test each statement by forming the intersection. (i) 2,3,4,5 3,6 = 3 ≠ φ, since 3 belongs to both. False. (ii) a,e,i,o,u a,b,c,d = a ≠ φ, since a belongs to both. False. (iii) 2,6,10,14 3,7,11,15 = φ — the first set has only even numbers, the second only odd ones, so nothing is common. True. (iv) 2,6,10 3,7,11 = φ, again even against odd. True. (i) False (ii) False (iii) True (iv) True

Question 33

Let U = 1,2,3,4,5,6,7,8,9, A = 1,2,3,4, B = 2,4,6,8 and C = 3,4,5,6. Find (i) A^′ (ii) B^′ (iii) ( A C )^′ (iv) ( A B )^′ (v) ( A^′)^′ (vi) (B - C)'

Accepted Answer

Complement. A ' = U - A is the set of elements of the universal set that are not in A. Here U = 1,2,3,4,5,6,7,8,9, A = 1,2,3,4, B = 2,4,6,8, C = 3,4,5,6. (i) Delete 1,2,3,4 from U: A ' = 5,6,7,8,9 (ii) Delete 2,4,6,8 from U: B ' = 1,3,5,7,9 (iii) First A C = 1,2,3,4,5,6, so ( A C )' = 7,8,9 (iv) First A B = 1,2,3,4,6,8, so ( A B )' = 5,7,9 (v) A ' = 5,6,7,8,9, and taking its complement returns the original set (the law ( A ')' = A ): ( A ')' = 1,2,3,4 = A (vi) B - C keeps the elements of B not in C: B - C = 2,8. Hence ( B - C )' = 1,3,4,5,6,7,9 (i) {5,6,7,8,9} (ii) {1,3,5,7,9} (iii) {7,8,9} (iv) {5,7,9} (v) {1,2,3,4} (vi) {1,3,4,5,6,7,9}

Question 34

If U = a, b, c, d, e, f, g, h, find the complements of the following sets: (i) A = a, b, c (ii) B = d, e, f, g (iii) C = a, c, e, g (iv) D = f, g, h, a

Accepted Answer

Complement. With U = a,b,c,d,e,f,g,h, the complement of a set is what is left of U after deleting that set's elements. (i) A = a,b,c: delete a,b,c from U A ' = d,e,f,g,h (ii) B = d,e,f,g: delete d,e,f,g B ' = a,b,c,h (iii) C = a,c,e,g: delete a,c,e,g C ' = b,d,f,h (iv) D = f,g,h,a: delete a,f,g,h D ' = b,c,d,e (i) {d, e, f, g, h} (ii) {a, b, c, h} (iii) {b, d, f, h} (iv) {b, c, d, e}

Question 35

Taking the set of natural numbers as the universal set, write down the complements of the following sets: (i) x: x is an even natural number (ii) x: x is an odd natural number (iii) x: x is a positive multiple of 3 (iv) x: x is a prime number (v) x: x is a natural number divisible by 3 and 5 (vi) x: x is a perfect square (vii) x: x is a perfect cube (viii) x: x+5=8 (ix) x: 2 x+5=9 (x) x: x ≥ 7 (xi) x: x ∈ N and 2 x+1 10

Accepted Answer

(i) x: x is an odd natural number (ii) x: x is an even natural number (iii) x: x ∈ N and x is not a multiple of 3 (iv) x: x is a positive composite number or x=1 (v) x: x is a positive integer which is not divisible by 3 or not divisible by 5 (vi) x: x ∈ N and x is not a perfect square (vii) x: x ∈ N and x is not a perfect cube (viii) x: x ∈ N and x ≠ 3 (ix) x: x ∈ N and x ≠ 2 (x) x: x ∈ N and x<7 (xi) x: x ∈ N. and.x ≤ 9/2. Complement in N. With U = N, the complement of A is A ' = x ∈ N: x A — negate the defining property, keeping x a natural number. (i) A natural number that is not even is odd: x: x is an odd natural number (ii) Likewise, not odd means even: x: x is an even natural number (iii) In N every multiple of 3 is positive, so the complement is x: x ∈ N and x is not a multiple of 3 (iv) The naturals that are not prime are 1 together with the composite numbers: x: x ∈ N, x is not a prime = 1 x: x is a composite number (v) A number divisible by both 3 and 5 is divisible by 15, so the set is 15,30,45,… and x: x ∈ N and x is not divisible by 15 (vi) x: x ∈ N and x is not a perfect square (vii) x: x ∈ N and x is not a perfect cube (viii) x + 5 = 8 ⇒ x = 3, so the set is 3 and x: x ∈ N and x ≠ 3 (ix) 2x + 5 = 9 ⇒ 2x = 4 ⇒ x = 2, so the set is 2 and x: x ∈ N and x ≠ 2 (x) The set is 7,8,9,…, so its complement is x: x ∈ N and x < 7 = 1,2,3,4,5,6 (xi) 2x + 1 10 ⇒ 2x 9 ⇒ x 4.5, so the set is 5,6,7,… and its complement is x: x ∈ N and x ≤ 4 = 1,2,3,4 (i) odd naturals (ii) even naturals (iii) naturals not multiples of 3 (iv) 1 and the composite numbers (v) naturals not divisible by 15 (vi) naturals that are not perfect squares (vii) naturals that are not perfect cubes (viii) {x ∈ N: x ≠ 3} (ix) {x ∈ N: x ≠ 2} (x) {1, 2, 3, 4, 5, 6} (xi) {1, 2, 3, 4}

Question 36

If U = 1,2,3,4,5,6,7,8,9, A = 2,4,6,8 and B = 2,3,5,7. Verify that (i) ( A B )^′= A^′ B^′ (ii) ( A B )^′= A^′ B^′

Accepted Answer

De Morgan's laws, verified by listing. With U = 1,2,3,4,5,6,7,8,9, A = 2,4,6,8 and B = 2,3,5,7, first record the two complements: A ' = 1,3,5,7,9, B ' = 1,4,6,8,9 (i) Left side: A B = 2,3,4,5,6,7,8, so ( A B )' = 1,9 Right side: the elements common to A ' and B ' are 1 and 9, so A ' B ' = 1,9 Both sides are 1,9, so ( A B )' = A ' B '. Verified. (ii) Left side: the only element in both A and B is 2, so A B = 2 and ( A B )' = 1,3,4,5,6,7,8,9 Right side: A ' B ' = 1,3,5,7,9 1,4,6,8,9 = 1,3,4,5,6,7,8,9 Both sides agree, so ( A B )' = A ' B '. Verified. (i) (A ∪ B)′ = A′ ∩ B′ = {1, 9} (ii) (A ∩ B)′ = A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9}

Question 37

Draw appropriate Venn diagram for each of the following: (i) ( A B )^′, (ii) A^′ B^′, (iii) ( A B )^′, (iv) A^′ B^′

Accepted Answer

Shading a Venn diagram. For each part draw the same picture: a rectangle for the universal set U with two overlapping circles A and B inside it. Those circles cut the rectangle into exactly four regions — only A, only B, the lens-shaped overlap A B, and the part of the rectangle lying outside both circles. Deciding which of these four regions to shade is the whole question. (i) ( A B )'. A B is the two circles taken together, so its complement is everything in U that is left over: shade only the part of the rectangle outside both circles. (ii) A ' B '. A point lies here when it is outside A and outside B — that is again the part of the rectangle outside both circles. So the shading is identical to (i); the picture is exactly De Morgan's law ( A B )'= A ' B '. (iii) ( A B )'. A B is only the lens-shaped overlap, so its complement is everything else: shade the whole rectangle except the overlap — that is, only- A, only- B and the region outside both circles. (iv) A ' B '. A point lies here when it is outside A or outside B; the only points excluded are those lying in both circles at once, i.e. the overlap. So the shading is identical to (iii), which is De Morgan's other law ( A B )'= A ' B '. Sketch the rectangle-with-two-circles once and shade a fresh copy of it four times — the four regions named above are what the answer refers to. (i) and (ii) both shade the region of U outside both circles; (iii) and (iv) both shade all of U except the overlap A B.

Question 38

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A'?

Accepted Answer

A' is the set of all equilateral triangles.. Complement inside the stated universal set. By definition A '= x∈ U: x A, so we must describe the triangles that are not in A. A triangle belongs to A when at least one of its angles is different from 60^°. Negating that: a triangle fails to belong to A exactly when no angle of it is different from 60^°, i.e. when all three of its angles equal 60^°. This is consistent, since 60^°+60^°+60^°=180^°, and a triangle is equiangular if and only if it is equilateral. A′ = the set of all equilateral triangles in the plane (equivalently, all triangles each of whose angles is 60^° ).

Question 39

Fill in the blanks to make each of the following a true statement: (i) A A^′=…. (ii) φ^′ A =… (iii) A A^′=…. (iv) U^′ A =….

Accepted Answer

(i) U (ii) A (iii) φ (iv) φ. Complement laws. Recall the four facts being tested: A A '= U, A A '=φ, φ'= U and U '=φ. (i) Every element of U either belongs to A or does not, so A A '= U. (ii) The complement of the empty set is the whole universal set, φ'= U; and U A = A because A U. So φ' A = A. (iii) No element can both belong to A and not belong to A, so A A '=φ. (iv) The complement of the universal set is empty, U '=φ; and φ A =φ. So U ' A =φ. (i) U (ii) A (iii) φ (iv) φ

Question 40

Decide, among the following sets, which sets are subsets of one and another: A = x: x ∈ R. and x satisfy.x^2-8 x+12=0, B = 2,4,6, C = 2,4,6,8, …, D = 6.

Accepted Answer

A B, A C, B C, D A, D B, D C. Roster form first, then compare element by element. X Y means every element of X is an element of Y, so the sets must be listed before they can be compared. Solve the equation defining A: x^2-8x+12=0 ⇒ (x-2)(x-6)=0 ⇒ x=2 or x=6, so A = 2,6. The others are already in roster form: B = 2,4,6, C = 2,4,6,8,… ( all even natural numbers ), D = 6. Now test each pair: - D = 6: 6∈ A, 6∈ B, 6∈ C. Hence D A, D B, D C. - A = 2,6: both 2 and 6 lie in B and in C. Hence A B and A C. - B = 2,4,6: every one of these is an even natural number. Hence B C. No other inclusion holds: 4∈ B but 4 A; 8∈ C but 8 B; 2∈ A but 2 D. D ⊂ A ⊂ B ⊂ C — that is, the subset relations among these four sets are D A, D B, D C, A B, A C and B C.

Question 41

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If x ∈ A and A ∈ B, then x ∈ B (ii) If A B and B ∈ C, then A ∈ C (iii) If A B and B C, then A C (iv) If A B and B C, then A C (v) If x ∈ A and A B, then x ∈ B (vi) If A B and x B, then x A

Accepted Answer

(i) False (ii) False (iii) True (iv) False (v) False (vi) True. Prove it or break it with one counterexample. Keep the two relations apart while reading each part: ∈ is membership (being one of the listed elements) and is inclusion (every element of the first set being in the second). Most of these parts turn on exactly that difference. (i) If x∈ A and A ∈ B, then x∈ B — False. Take A = 1 and B = 1, 2. Then 1∈ A, and A = 1 is itself one of the elements of B, so A ∈ B. But the elements of B are 1 and 2, and 1 is neither of them, so 1 B. (ii) If A B and B ∈ C, then A ∈ C — False. Take A = 1, B = 1,2, C = 1,2, 3. Then A B and B ∈ C, but the only elements of C are 1,2 and 3, so A = 1 C. (iii) If A B and B C, then A C — True. Let x∈ A. Since A B, x∈ B; since B C, x∈ C. As x was arbitrary, every element of A lies in C, i.e. A C. (iv) If A B and B C, then A C — False. Take A = 1, B = 2, C = 1,3. Here 1 B, so A B; and 2 C, so B C. Yet 1∈ C, so A C — the conclusion fails. (v) If x∈ A and A B, then x∈ B — False. Take A = 3,5,7 and B = 3,4,6. Since 5 B, A B. Now choose x=5: it satisfies x∈ A, but 5 B. ( A B says some element of A escapes B; it says nothing forcing a given x into B.) (vi) If A B and x B, then x A — True. This is just the contrapositive of A B. Suppose, on the contrary, that x∈ A. Then A B would give x∈ B, contradicting x B. Hence x A. (i) False (ii) False (iii) True (iv) False (v) False (vi) True

Question 42

Let A, B, and C be the sets such that A B = A C and A B = A C. Show that B = C.

Accepted Answer

Absorption, then the distributive law. The trick is to start from B, open it up with the absorption law B = B ( A B ), and then feed in the two hypotheses one at a time until C appears. Note that both hypotheses were needed — the union condition alone, or the intersection condition alone, is not enough. B = C

Question 43

Show that the following four conditions are equivalent: (i) A B (ii) A - B =φ (iii) A B = B (iv) A B = A

Accepted Answer

Prove a cycle of implications. To show four statements are equivalent it is enough to prove the single loop ( i )⇒( ii )⇒( iii )⇒( iv )⇒( i ), because then any one of them reaches any other by travelling round the loop. That is four proofs instead of the twelve a pairwise approach would need. (i) ⇒ (ii). Assume A B. If A - B were non-empty there would be an x with x∈ A and x B, contradicting A B. Hence A - B =φ. (ii) ⇒ (iii). Assume A - B =φ. The inclusion B A B always holds. For the other direction let x∈ A B. If x∈ B there is nothing to do; if x∈ A but x B, then x∈ A - B =φ, which is impossible. So A B B, and therefore A B = B. (iii) ⇒ (iv). Assume A B = B. Then A B = A ( A B )= A, the last step being the absorption law A ( A B )= A. (iv) ⇒ (i). Assume A B = A. Let x∈ A. Then x∈ A B, so in particular x∈ B. Hence A B. The loop closes, so each condition implies all the others. (i) A B, (ii) A - B =φ, (iii) A B = B and (iv) A B = A are equivalent.

Question 44

Show that if A B, then C - B C - A.

Accepted Answer

Element chase. To prove one set is contained in another, take an arbitrary element of the first and show it must lie in the second. Let x∈ C - B. By the definition of difference this means x∈ C and x B. Now use the hypothesis A B. If x belonged to A, then A B would force x∈ B, contradicting x B. Hence x A. So x∈ C and x A, which is precisely x∈ C - A. Since x was an arbitrary element of C - B, every element of C - B lies in C - A. If A B, then C - B C - A.

Question 45

Show that for any sets A and B, A=(A B) (A-B) and A (B-A)=(A B)

Accepted Answer

Turn each difference into an intersection with a complement, then distribute. The one rewriting that makes both identities routine is X - Y = X Y '. First identity. In words: each element of A either lies in B — and is then counted by A B — or does not, and is then counted by A - B. Nothing outside A is in either piece. Second identity. A =( A B ) ( A - B ) and A ( B - A )= A B.

Question 46

Using properties of sets, show that (i) A ( A B )= A (ii) A ( A B )= A.

Accepted Answer

Create the pattern the distributive law needs, using the identity elements. The two facts A = A U and A = A φ let us rewrite the lone A so that the distributive law can be applied. These two results are the absorption laws. (i) (ii) A ( A B )= A and A ( A B )= A.

Question 47

Show that A B = A C need not imply B = C.

Accepted Answer

A single counterexample settles a "need not imply". We are not asked to prove anything is always false — only to exhibit one triple of sets for which A B = A C holds while B = C fails. Take A = 1,2, B = 2,3, C = 2,4. Then A B = 2, A C = 2, so A B = A C. But 3∈ B while 3 C, so B ≠ C. The reason is worth naming: intersecting with A only records what B and C do inside A. Anything they do outside A — here the elements 3 and 4 — is invisible to the test, so it can differ freely. A B = A C need not imply B = C; e.g. A = 1,2, B = 2,3, C = 2,4.

Question 48

Let A and B be sets. If A X = B X =φ and A X = B X for some set X, show that A = B. (Hints A = A ( A X ), B = B ( B X ) and use Distributive law)

Accepted Answer

Absorption, then the distributive law (the route the hint points to). Show that each of A and B equals the same set A B. Start with A: Exactly the same computation with the roles of A and B exchanged: Both A and B have been shown equal to A B, hence equal to each other. A = B

Question 49

Find sets A, B and C such that A B, B C and A C are non-empty sets and A B C =φ.

Accepted Answer

We may take A = 1,2, B = 1,3, C = 2,3. Give each pair its own shared element. The three pairwise intersections must be non-empty, yet no element may sit in all three sets at once. So use three different elements and let each one belong to exactly two of the sets — never to the third. Let element 1 be shared by A and C, element 2 by A and B, element 3 by B and C: A = 1,2, B = 2,3, C = 1,3. Check the three pairwise intersections: A B = 2, B C = 3, A C = 1, all non-empty as required. Check the triple intersection: 1 B, 2 C and 3 A, so no element lies in all three sets, and A B C =φ. A = 1,2, B = 2,3, C = 1,3

EXERCISE 1.1 1–6 (part 1 of 6)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6