Binary Composition Tables Class 12 OP Malhotra Exe-3C ISC Maths Solutions Ch-3 Binary Operations. Properties and example with solutions of Binary Operation composition questions and answers. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.
Binary Composition Tables Class 12 OP Malhotra Exe-3C ISC Maths Solutions Ch-3 Binary Operations
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-3 | Binary Operations |
| Writer | OP Malhotra |
| Exe-3(C) | Binary Composition Tables |
Binary Composition Tables Questions and Answer
Class 12 OP Malhotra Exe-3C ISC Maths Solutions Ch-3 Binary Operations
Que-1: 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 *.
Sol: Let A = {1, 2, 3, 4, 5} and binary operation * on A defined by a* b = min {a, b}
The operation table is given as under :
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 3 | 3 | 3 |
| 4 | 1 | 2 | 3 | 4 | 4 |
| 5 | 1 | 2 | 3 | 4 | 5 |
Que-2: Construct the composition table for the operation * defined on A = {1, 2,3} by a * b = a + b². Is * a binary operation ?
Sol: The composition table is given below :
| * | 1 | 2 | 3 |
| 1 | 2 | 5 | 10 |
| 2 | 3 | 6 | 11 |
| 3 | 4 | 7 | 12 |
Clearly all elements of table are not belongs to A.
∴ * is not a binary operation.
Que-3: Consider a binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = HCF {a, b}. Write the operation table.
(i) Is it a binary operation
(ii) Is it commutative ?
(iii) Compute (2 * 3) * 4 and 2 * (3 * 4)
(iv) Compute (2 * 3) * (4 * 5)
Sol: Given binary operation * on set
A = {1, 2, 3, 4, 5} defined by a * b = HCF (a, b)
The operation table as given below :
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
(i) Since all entries of the table are belongs to set A
∴ * is a binary operation on A.
(ii) Clearly all entries are symmetrical about the principal diagonal.
∴ * is commutative on A.
(iii) (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(iv) (2 * 3) * (4 * 5) = 1 * 1 = 1
Que-4: Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
(i) Compute (a) (2 * 3) * 4 and
(b) 2 * (3 * 4).
(ii) Is * commutative ?
(iii) Compute (2 * 3) * (4 * 5).
Sol: (i) (a) (2 * 3) * 4 = 1 * 4 = 1
(b) 2 * (3 * 4) = 2 * 1 = 1
(ii) Clearly the composition table is symmetrical about main diagonal. Hence the operation x is commutative.
| * | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 1 | 2 | 1 |
| 3 | 1 | 1 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 | 4 | 1 |
| 5 | 1 | 1 | 1 | 1 | 5 |
(iii) (2 * 3) * (4 * 5) = 1 * 1 = 1
Que-5: Let * be defined on the set A ={ 1, 2, 3, 4, 5, 6} by a* b = max. {a, b}.
(i) Is * a binary operation ?
(ii) Is * a commutative ?
(iii) Find the identity element.
(iv) Find invertible elements with respect to * and find their inverses.
Sol: The operation table for * defined on A is given as under:
| * | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 2 | 3 | 4 | 5 | 6 |
| 3 | 3 | 3 | 3 | 4 | 5 | 6 |
| 4 | 4 | 4 | 4 | 4 | 5 | 6 |
| 5 | 5 | 5 | 5 | 5 | 5 | 5 |
| 6 | 6 | 6 | 6 | 6 | 6 | 6 |
(i) Since all entries of composition table belongs to set A.
Thus * is binary operation on A.
(ii) Since the composition table is symmetrical about principal diagonal.
Thus * is commutative on A.
(iii) Since the row headed by 1 is coincide with top most row and column headed by 1 is same as left most column. Thus 1 is the identity element.
(iv) Since all rows and columns except first row and column do not contains identity element
1. So all elements are not invertible.
Clearly 1 be the only invertible element as 1 * 1 = 1.
Que-6: A binary operation * is defined on the set {0, 1, 2, 3, 4, 5, 6} as a * b = {a+b if a+b<7 , a+b−7 if a+b≥7 . Write the composition table of operation. Is it a binary operation? Is it commutative? Prove that zero is the identity for this operation and each element a* 0 of the set is invertible with 7 – a being inverse of a.
Sol: The binary operation table is as under :
| * | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 0 |
| 2 | 2 | 3 | 4 | 5 | 6 | 0 | 1 |
| 3 | 3 | 4 | 5 | 6 | 0 | 1 | 2 |
| 4 | 4 | 5 | 6 | 0 | 1 | 2 | 3 |
| 5 | 5 | 6 | 0 | 1 | 2 | 3 | 4 |
| 6 | 6 | 0 | 1 | 2 | 3 | 4 | 5 |
(i) Since all elements of binary operation table are belonging to set A. Thus * is a binary operation on A.
(ii) Since composition table is symmetrical about principal diagonal. Thus * is commutative on A.
(iii) Clearly the row headed by 0 is same as the top most row and column headed by 0 is identical with left most column. Thus 0 be the identity element.
(iv) Since every row and column contains identity element 0 thus every element is invertible. Let b be the inverse element of a.
Then a * b = 0 = b * a
Now a * (7 – a) = a + (7 – a) – 7 = 0
(7 – a) * a = 1 – a + a – 7 = 0
Thus (7 – a) be the inverse of a.
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