Exercise 1.4
The Exercise 1.4 deals with functions, sets and relation and certain other properties of sets .The exercise deals with the integers(Z), real numbers (R) etc.
The exercise 1.4 also include one-one and onto functions. The Exercise 1.4 provides a lot of working problems to make you more familiar and through with the concepts..
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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 |
Solution : (i) On Z+, * is defined by a * b = a − b. It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z+. (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+. 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. 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+. 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+ . Therefore, * is a binary operation |
Question 2: For each binary operation * defined below, determine whether * is 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 Solution : (i) On Z, * is defined by a * b = a − b. It can be observed that 1 * 2 = 1 − 2 = −1 and 2 * 1 = 2 − 1 = 1. ∴1 * 2 ≠ 2 * 1, where 1, 2 ∈ Z Hence, the operation * is not commutative. Also, we have (1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4 1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2 ∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Z Hence, the operation * is not associative. (ii) On Q , * is defined by a * b = ab + 1. It is known that: ab = ba for all a, b ∈ Q ⇒ ab + 1 = ba + 1 for all a, b ∈ Q ⇒ a * b = a * b for all a, b ∈ Q Therefore, the operation * is commutative. It can be observed that (1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10 1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8 ∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q Therefore, the operation * is not associative. (iii) On Q, * is defined by a * b = ab / 2 It is known that: ab = ba for all a, b ∈ Q ⇒ ab / 2 =ab / 2 for all a, b ∈ Q ⇒ a * b = b * a for all a, b ∈ Q Therefore, the operation * is commutative. For all a, b, c ∈ Q, we have (𝑎 ∗ 𝑏) ∗ 𝑐 = (ab / 2)∗c=(ab / 2)c / 2=abc / 4 and 𝑎 ∗ (𝑏 ∗ 𝑐) = a∗(bc / 2)=a(bc / 2) / 2=abc / 4 ∴ (a*b)*c = a*(b*c), where a, b, c ∈ Q Therefore, the operation * is associative. (iv) On Z+ , * is defined by a * b = 2ab . It is known that: ab = ba for all a, b ∈ Z+ ⇒ 2 ^ab = 2 ^ba for all a, b ∈ Z+ ⇒ a * b = b * a for all a, b ∈ Z+ Therefore, the operation * is commutative. It can be observed that (1 ∗ 2) ∗ 3 = 2 ^1×2 ∗ 3 = 4 ∗ 3 = 2 ^ 4×3 = 2 ^12 and 1 ∗ (2 ∗ 3) = 1 ∗ 2 ^2×3 = 1 ∗ 26 = 1 ∗ 64 = 21×64 = 264 ∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Z+ Therefore, the operation * is not associative. (v) On Z+, * is defined by a * b = a^b . It can be observed that 1*2 = 1² = 1 and 2*1 = 2^1 = 2 ∴ 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Z+ Therefore, the operation * is not commutative. It can also be observed that (2 ∗ 3) ∗ 4 = 2^3 ∗ 4 = 8 ∗ 4 = 8^4 = 2^12 and 2 ∗ (3 ∗ 4) = 2 ∗ 3^4 = 2 ∗ 81 = 281 ∴ (2 * 3) * 4 ≠ 2 * (3 * 4), where 2, 3, 4 ∈ Z+ Therefore, the operation * is not associative. (vi) On R, * − {−1} is defined by 𝑎 ∗ 𝑏 = a / b+1 It can be observed that 1 ∗ 2 = 1 / 2+1=1 / 3 and 2 ∗ 1 = 2 / 1+1=2 / 2=1 ∴1 * 2 ≠ 2 * 1, where 1, 2 ∈ R − {−1} Therefore, the operation * is not commutative. It can also be observed that (1 ∗ 2) ∗ 3 = 1 / 2+1∗3=1 / 3 ∗3= 1/ 3/ (3+1)=1 / 12 and 1 ∗ (2 ∗ 3) = 1∗ 2/3+1=1∗2/4=1∗1/2=1/(1/ 2+1)=(1/3)/ 2=2/ 3 ∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ R − {−1} Therefore, the operation * is not 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 ∧. Solution: The binary operation ∧ on the set {1, 2, 3, 4, 5} is defined as a ∧ b = min {a, b} for all a, b ∈ {1, 2, 3, 4, 5}. Thus, the operation table for the given operation ∧ can be given as: |
Question 4: Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. (i) Compute (2 * 3) * 4 and 2 * (3 * 4) (ii) Is * commutative? (iii) Compute (2 * 3) * (4 * 5). (Hint: use the following table) Solution : (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 |
EXERCISE 1.4 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. Solution : 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:We observe that the operation tables for the operations * and *′ are the same. Thus, the operation *′ is same as the operation*. |
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 N (v) Which elements of N are invertible for the operation *? Solution : 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 for all 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 for all a ∈ N ⇒ a * 1 = a = 1 * a for all 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. Solution : 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. |
EXERCISE 1.4 Question 8: |
EXERCISE 1.4 Question 9: |
Question 10: Find which of the operations given above has identity. Solution : An element e ∈ Q will be the identity element for the operation * if a * e = a = e * a, for all a ∈ Q. However, there is no such element e ∈ Q with respect to each of the six operations satisfying the above condition. Thus, none of the six operations has identity. |
EXERCISE 1.4 Question 11: 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. Solution : A = N × N and * is a binary operation on A and is defined by (a, b) * (c, d) = (a + c, b + d) Let (a, b), (c, d) ∈ A Then, a, b, c, d ∈ N We have: (a, b) * (c, d) = (a + c, b + d) (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) [Addition is commutative in the set of natural numbers] ∴ (a, b) * (c, d) = (c, d) * (a, b) Therefore, the operation * is commutative. Now, let (a, b), (c, d), (e, f) ∈ A Then, a, b, c, d, e, f ∈ N We have [(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) ∴ [(𝑎, 𝑏) ∗ (𝑐, 𝑑)] ∗ (𝑒, 𝑓) = (𝑎, 𝑏) ∗ [(𝑐, 𝑑) ∗ (𝑒, 𝑓)] Therefore, the operation * is associative. Let an element e = (e1, e2 ) ∈ A will be an identity element for the operation * if a * e = a = e * a for all a = (a1, a2 ) ∈ A i.e., (a1 + e1, a2 + e2 ) = (a1, a2 ) = (e1 + a1, e2 + a2 ) Which is not true for any element in A. Therefore, the operation * does not have any identity element. |
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 Solution : (I) Define an operation * on N as a * b = a + b ∀ a, b ∈ N Then, in particular, for b = a = 3, we have 3 * 3 = 3 + 3 = 6 ≠ 3 Therefore, statement (i) is false. (ii) R.H.S. = (c * b) * a = (b * c) * a [* is commutative] = a * (b * c) [Again, as * is commutative] = L.H.S. ∴ a * (b * c) = (c * b) * a Therefore, statement (ii) is true. |
EXERCISE 1.4 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? Solution : On N, the operation * is defined as a * b = a³ + b³ . For, a, b, ∈ N, we have a * b = a³ + b³ = b³ + a³ = b * a [Addition is commutative in N] Therefore, the operation * is commutative. It can be observed that (1 * 2) * 3 = (1³ + 2³ ) * 3 = (1 + 8) * 3 = 9 * 3 = 9³ + 3³ = 729 + 27 = 756 and 1* (2 * 3) = 1 * (2³ + 3³ ) = 1* (8 + 27) = 1*35 = 1³ + 35³ = 1 + 42875 = 42876 ∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ N Therefore, the operation * is not associative. Hence, the operation * is commutative, but not associative. Thus, the correct answer is B. |