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Exercise - 1.4

Relations and Functions

**Question-1 :-** Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(i) On Z+, define ∗ by a ∗ b = a – b

(ii) On Z+, define ∗ by a ∗ b = ab

(iii) On R, define ∗ by a ∗ b = ab²

(iv) On Z+, define ∗ by a ∗ b = |a – b|

(v) On Z+, define ∗ by a ∗ b = a

(i) On Z+, * is defined by a * b = a − b. e.g., (a, b) = (2, 3) under * is a * b = a − b ⇒ 2 * 3 = 2 − 3 = −1 ∉ Z+. It is not a binary operation.

(ii) On Z+, * is defined by a * b = ab. It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+. This means that * carries each pair (a, b) to a unique element a * b = ab in Z+. e.g., (a, b) = (2, 4) under * is a * b = ab ⇒ 2 * 4 = 8 ∈ Z+. Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab². It is seen that for each a, b ∈ R, there is a unique element ab² in R . This means that * carries each pair (a, b) to a unique element a * b = ab² in R . e.g., (a, b) = (1.2, 3) under * is a * b = ab² ⇒ 1.2 * 3 = 3.6 ∈ R. Therefore, * is a binary operation.

(iv) On Z+, * is defined by a * b = |a − b|. It is seen that for each a, b ∈ Z+, there is a unique element |a − b| in Z+. This means that * carries each pair (a, b) to a unique element a * b = |a − b| in Z+. e.g., (a, b) = (2, 3) under * is a * b = |a − b| ⇒ 2 * 3 = |2 − 3| = 1 ∈ Z+. Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a. It is seen that for each a, b ∈ Z+, there is a unique element a in Z+. This means that * carries each pair (a, b) to a unique element a * b = a in Z+. e.g., (a, b) = (3, 4) under * is a * b = a ⇒ 3 * 4 = 3 ∈ Z+. Therefore, * is a binary operation.

**Question-2 :-** For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(i) On Z, define a ∗ b = a – b

(ii) On Q, define a ∗ b = ab + 1

(iii) On Q, define a ∗ b = ab/2

(iv) On Z+, define a ∗ b = 2^{ab}

(v) On Z+, define a ∗ b = a^{b}

(vi) On R – {– 1}, define a ∗ b = a/(b+1)

(i) On Z, define a ∗ b = a – b For commutative: ∴ [a * b = b * a] a ∗ b = a – b b ∗ a = b – a e.g., (a, b) = (2, 4) ∈ Z 2 * 4 = 2 - 4 = -2 4 * 2 = 4 - 2 = 2 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = (a–b) - c a * (b * c) = a - (b-c) e.g., (a, b, c) = (2, 4, 6) ∈ Z (2 ∗ 4) * 6 = (2–4) - 6 = -2 - 6 = -8 2 * (4 * 6) = 2 - (4-6) = 2 + 2 = 4 Hence, the operation * is not associative. Therefore, it is neither commutative nor associative.

(ii) On Q, define a ∗ b = ab + 1 For commutative: ∴ [a * b = b * a] a ∗ b = ab + 1 b ∗ a = ba + 1 e.g., (a, b) = (1/2, 2/3) ∈ Q 1/2 * 2/3 = 1/2 . 2/3 + 1 = 4/3 2/3 * 1/2 = 2/3 . 1/2 + 1 = 4/3 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = (ab + 1)c + 1 a * (b * c) = a(bc + 1) + 1 e.g., (a, b, c) = (1/2, 2/3, 3/4) ∈ Q (1/2 * 2/3) * 3/4 = (1/2 . 2/3 + 1) x 3/4 + 1 = 4/3 . 3/4 + 1 = 1 + 1 = 2 1/2 * (2/3 * 3/4) = 1/2 x (2/3 . 3/4 + 1) + 1 = 1/2 . 3/2 + 1 = 3/4 + 1 = 7/3 Hence, the operation * is not associative. Therefore, it is commutative but not associative.

(iii) On Q, define a ∗ b = ab/2 For commutative: ∴ [a * b = b * a] a ∗ b = ab/2 b ∗ a = ba/2 e.g., (a, b) = (2/3, 4/3) ∈ Q 2/3 * 4/3 = (2/3 x 4/3)/2 = 8/18 4/3 * 2/3 = (2/3 x 4/3)/2 = 8/18 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = ((ab/2) x c)/2 = abc/4 a * (b * c) = (a x (bc/2))/2 = abc/4 e.g., (a, b, c) = (2/3, 4/3, 1/3) ∈ Q (2/3 ∗ 4/3) * 1/3 = (2/3 x 4/3 x 1/3)/4 = 8/108 2/3 * (4/3 * 1/3) = (2/3 x 4/3 x 1/3)/4 = 8/108 Hence, the operation * is associative. Therefore, it is commutative and associative.

(iv) On Z+, define a ∗ b = 2^{ab}For commutative: ∴ [a * b = b * a] a ∗ b = 2^{ab}b ∗ a = 2^{ba}e.g., (a, b) = (2, 1) ∈ Z+ 2 * 1 = 2^{2}= 4 1 * 2 = 2^{2}= 4 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = 2^{ab}* c a * (b * c) = a * 2^{bc}e.g., (a, b, c) = (2, 1, 3) ∈ Z+ (2 ∗ 1) * 3 = 2^{2}* 3 = 4 * 3 = 2^{12}2 * (1 * 3) = 2 * 2^{3}= 2 * 8 = 2^{16}Hence, the operation * is not associative. Therefore, it is commutative but not associative.

(v) On Z+, define a ∗ b = a^{b}For commutative: ∴ [a * b = b * a] a ∗ b = a^{b}b ∗ a = b^{a}e.g., (a, b) = (2, 1) ∈ Z+ 2 * 1 = 2¹ = 2 1 * 2 = 1² = 1 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = a^{b}* c a * (b * c) = a * b^{a}e.g., (a, b, c) = (2, 1, 3) ∈ Z+ (2 ∗ 1) * 3 = 2¹ * 3 = 2 * 3 = 2³ = 8 2 * (1 * 3) = 2 * 1³ = 2 * 1 = 2¹ = 2 Hence, the operation * is not associative. Therefore, it is neither commutative nor associative.

(vi) On R – {– 1}, define a ∗ b = a/(b+1) For commutative: ∴ [a * b = b * a] a ∗ b = a/(b+1) b ∗ a = b/(a+1) e.g., (a, b) = (3, -2) ∈ R – {– 1} 3 * -2 = 3/(-2+1) = -3 -2 * 3 = -2/(3+1) = -2/4 = -1/2 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a ∗ b) * c = a/(b+1) * c a * (b * c) = a * b/(c+1) e.g., (a, b, c) = (2, 1, 3) ∈ R – {– 1} (2 ∗ 1) * 3 = 2/(1+1) * 3 = 1 * 3 = 1/(3+1) = 1/4 2 * (1 * 3) = 2 * 1/(3+1) = 2 * 1/4 = 2/(1/4 + 1) = 8/5 Hence, the operation * is not associative. Therefore, it is neither commutative nor associative.

**Question-3 :-** Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}. Write the operation table of the operation ∧ .

Let A = {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.

**Question-4 :-** Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).

(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

(ii) Is ∗ commutative?

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

(Hint: use the following table)

(i) (2 * 3) * 4 = 1 * 4 = 1 2 * (3 * 4) = 2 * 1 = 1 (ii) For every a, b ∈{1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative. (iii) (2 * 3) = 1 and (4 * 5) = 1 ∴(2 * 3) * (4 * 5) = 1 * 1 = 1

**Question-5 :-** Let ∗′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a ∗′ b = H.C.F. of a and b. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer.

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b. The operation table for the operation *′ can be given as:

**Question-6 :-** Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find

(i) 5 * 7, 20 * 16

(ii) Is * commutative?

(iii) Is * associative?

(iv) Find the identity of * in

(v) Which elements of N are invertible for the operation *?

The binary operation * on N is defined as a * b = L.C.M. of a and b. (i) 5 * 7 = L.C.M. of 5 and 7 = 35 20 * 16 = L.C.M of 20 and 16 = 80 (ii) It is known that: L.C.M of a and b = L.C.M of b and a where a, b ∈ N. ∴a * b = b * a Thus, the operation * is commutative. (iii) For a, b, c ∈ N, we have: (a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c ∴(a * b) * c = a * (b * c) Thus, the operation * is associative. (iv) It is known that: L.C.M. of a and 1 = a = L.C.M. 1 and a where a ∈ N ⇒ a * 1 = a = 1 * a, where a ∈ N Thus, 1 is the identity of * in N. (v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a. Here, e = 1 This means that: L.C.M of a and b = 1 = L.C.M of b and a This case is possible only when a and b are equal to 1. Thus, 1 is the only invertible element of N with respect to the operation *.

**Question-7 :-** Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.

The operation * on the set A = {1, 2, 3, 4, 5} is defined as a * b = L.C.M. of a and b. Then, the operation table for the given operation * can be given as: It can be observed from the obtained table that: 3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A 3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A Hence, the given operation * is not a binary operation.

**Question-8 :-** Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N ?

The binary operation * on N is defined as: a * b = H.C.F. of a and b It is known that: H.C.F. of a and b = H.C.F. of b and a where a, b ∈ N. ∴a * b = b * a Thus, the operation * is commutative. For a, b, c ∈ N, we have: (a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c a *(b * c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c ∴(a * b) * c = a * (b * c) Thus, the operation * is associative. Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a where a ∈ N. But this relation is not true for any a ∈ N. Thus, the operation * does not have any identity in N.

**Question-9 :-** Let * be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a − b, (ii) a * b = a² + b², (iii) a * b = a + ab, (iv) a * b = (a − b)², (v) a * b = ab/4, (vi) a * b = ab²

Find which of the binary operations are commutative and which are associative.

(i) a * b = a − b For Commutative : ∴ [a * b = b * a] a * b = a − b b * a = b - a e.g., (a, b) = (1/2, 1/3) ∈ Q 1/2 * 1/3 = 1/2 - 1/3 = 1/6 1/3 * 1/2 = 1/3 - 1/2 = -1/6 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (a − b) - c a * (b * c) = a - (b - c) e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q (1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4 = 1/6 - 1/4 = -1/12 1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4) = 1/2 - 1/12 = 5/12 Hence, the operation * is not associative.

(ii) a * b = a² + b² For Commutative : ∴ [a * b = b * a] a * b = a² + b² b * a = b² + a² e.g., (a, b) = (1/2, 1/3) ∈ Q 1/2 * 1/3 = (1/2)² + (1/3)² = 1/4 + 1/9 = 13/36 1/3 * 1/2 = (1/3)² + (1/2)² = 1/9 + 1/4 = 13/36 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (a² + b²) + c² a * (b * c) = a² + (b² + c²) e.g., (a, b, c) = (1, 2, 3) ∈ Q (1 * 2) * 3 = (1² + 2²) * 3 = (1 + 4) * 3 = 5 * 3 = 5² + 3² = 25 + 9 = 34 1 * (2 * 3) = 1 * (2² + 3²) = 1 * (4 + 9) = 1 * 13 = 1² + 13² = 1 + 169 = 170 Hence, the operation * is associative.

(iii) a * b = a + ab = a(1 + b) For Commutative : ∴ [a * b = b * a] a * b = a(1 + b) b * a = b(1 + a) e.g., (a, b) = (1/2, 1/3) ∈ Q 1/2 * 1/3 = 1/2 x (1 + 1/3) = 1/2 x 4/3 = 2/3 1/3 * 1/2 = 1/3 x (1 + 1/2) = 1/3 x 3/2 = 1/2 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (a − b) - c a * (b * c) = a - (b - c) e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q (1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4 = 1/6 - 1/4 = -1/12 1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4) = 1/2 - 1/12 = 5/12 Hence, the operation * is not associative.

(iv) a * b = (a − b)² For Commutative : ∴ [a * b = b * a] a * b = (a − b)² b * a = (b − a)² e.g., (a, b) = (3, 2) ∈ Q 3 * 2 = (3 - 2)² = 1 2 * 3 = (2 - 3)² = 1 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (a − b)² * c a * (b * c) = a * (b - c)² e.g., (a, b, c) = (3, 2, 1) ∈ Q (3 * 2) * 1 = (3 - 2)² * 1 = 1 * 1 = (1 - 1)² = 0 3 * (2 * 1) = 3 * (2 - 1)² = 3 * 1 = (3 - 1)² = 4 Hence, the operation * is not associative.

(v) a * b = ab/4 For Commutative : ∴ [a * b = b * a] a * b = ab/4 b * a = ba/4 e.g., (a, b) = (4, 2) ∈ Q 4 * 2 = (4 x 2)/4 = 2 2 * 4 = (2 x 4)/4 = 2 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (ab/4) * c a * (b * c) = a * (bc/4) e.g., (a, b, c) = (4, 2, 1) ∈ Q (4 * 2) * 1 = (8/4) * 1 = 2 * 1 = 2/4 = 1/2 4 * (2 * 1) = 4 * (2/4) = 4 * 1/2 = 4/8 = 1/2 Hence, the operation * is associative.

(vi) a * b = ab² For Commutative : ∴ [a * b = b * a] a * b = ab² b * a = ba² e.g., (a, b) = (3, 2) ∈ Q 3 * 2 = 3 x 2² = 12 2 * 3 = 2 x 3² = 18 Hence, the operation * is not commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = ab² * c a * (b * c) = a * bc² e.g., (a, b, c) = (3, 2, 1) ∈ Q (3 * 2) * 4 = (3 x 2²) * 4 = 12 * 4 = 12 x 4² = 192 3 * (2 * 4) = 3 * (2 x 4²) = 3 * 32 = 3 x (32)² = 3072 Hence, the operation * is not associative.

**Question-10 :-** Find which of the operations given above has identity.

Let the identity be I. (i) a * I = a - I (ii) a * I = a² - I² (iii) a * I = a + aI (iv) a * I = (a - I)² (v) a * I = aI/4 (vi) a * I = aI² Therefore, none of the operations given above has identity.

**Question-11 :-** Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d)
Show that * is commutative and associative. Find the identity element for * on A, if any.

A = N x N and * is a binary operation defined on A. (a, b) * (c, d) = (a+c, b+d) = (c+a, d+b) = (c, d) * (a, b) Hence, The operation is commutative. Again, [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f) And (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f) Here, [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)] Hence, The operation is associative. Let identity function be (e, f), then (a, b) * (e, f) = (a+e, b+f) For identity function a = a + e e = 0 and for b + f = b f = 0 As 0 ≠ N, therefore, identity-element does not exist.

**Question-12 :-** State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a ∈ N.

(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a

(i) * being a binary operation on N, is defined as a * a = a∀a ∈ N Hence operation * is not defined, therefore, the given statement is false. (ii) * being a binary operation on N. c * b = b * c and (c * b) * a = (b * c) * a = a * (b * c) Thus, a * (b * c) = (c * b) * a, therefore the given statement is true.

**Question-13 :-** Consider a binary operation ∗ on N defined as a ∗ b = a³ + b³. Choose the correct answer.

(A) Is ∗ both associative and commutative?

(B) Is ∗ commutative but not associative?

(C) Is ∗ associative but not commutative?

(D) Is ∗ neither commutative nor associative?

Binary operation ∗ on N defined as a ∗ b = a³ + b³ For Commutative : ∴ [a * b = b * a] a * b = a³ + b³ b * a = b³ + a³ e.g., (a, b) = (2, 3) ∈ Q 2 * 3 = (2)³ + (3)³ = 8 + 27 = 35 3 * 2 = (3)³ + (2)³ = 27 + 8 = 35 Hence, the operation * is commutative. For associative: ∴ [(a * b) * c = a * (b * c)] (a * b) * c = (a³ + b²) + c³ a * (b * c) = a³ + (b³ + c³) e.g., (a, b, c) = (1, 2, 3) ∈ Q (1 * 2) * 3 = (1³ + 2³) * 3 = (1 + 8) * 3 = 9 * 3 = 9³ + 3³ = 729 + 27 = 756 1 * (2 * 3) = 1 * (2³ + 3³) = 1 * (8 + 27) = 1 * 35 = 1³ + 35³ = 1 + 42875 = 42876 Hence, the operation * is not associative. Therefore, option (B) is correct.

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