Here we provide RD Sharma Class 12 Ex 3.5 Solutions Chapter 3 Binary Operations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 3.5 Solutions Chapter 3 Binary Operations book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 3 |
| Exercise | 3.5 |
| Category | RD Sharma Solutions |
Question 1. Construct the composition table for ×4 on set S = {0,1,2,3}.
Solution:
Given that ×4 on set S = {0, 1, 2, 3}
We have to find the composition table for ×4 by using multiplication and modulo operations.
As 0 x4 0 = 0 (0 × 0 = 0 and 0 modulo 4 = 0 % 4 = 0)
As 1 x4 1 = 1 (1 × 1 = 1 and 1 modulo 4 = 1 % 4 =1)
As 1 x4 2 = 2 (1 × 2 = 2 and 2 modulo 4 = 2 % 4 = 2)
As 3 x4 1 = 3 (3 × 1 = 3 and 3 modulo 4 = 3 % 4 = 3)
As 4 x4 4 = 0 (2 × 2 = 4 and 4 modulo 4 = 4 % 4 = 0)
.
So on. . .
Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 4)
Therefore, the composition table is:
| x4 | 0 | 1 | 2 | 3 |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 |
| 2 | 0 | 2 | 0 | 2 |
| 3 | 0 | 3 | 2 | 1 |
Question 2. Construct the composition table for +5 on set S = {0, 1, 2, 3, 4}.
Solution:
Given that +5 on set S = {0,1,2,3,4}
We have to find the composition table for +5 by using addition and modulo operations.
As 0 +5 0 = 0 (0 + 0 = 0 and 0 modulo 5 = 0 % 5 = 0)
As 1 +5 0 = 1 (1 + 0 = 1 and 1 modulo 5 = 1 % 5 = 1)
As 1 +5 1 = 2 (1 + 1 = 2 and 2 modulo 5 = 2 % 5 =2)
As 1 +5 2 = 3 (1 + 2 = 3 and 3 modulo 5 = 3 % 5 = 3)
As 3 +5 1 = 4 (3 + 1 = 4 and 4 modulo 5 = 4 % 5 = 4)
As 2 +5 2 = 4 (2 + 2 = 4 and 4 modulo 5 = 4 % 5 = 4)
.
So on . . .
Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i,e., 5)
Therefore, the composition table is:
| +5 | 0 | 1 | 2 | 3 | 4 |
| 0 | 0 | 1 | 2 | 3 | 4 |
| 1 | 1 | 2 | 3 | 4 | 0 |
| 2 | 2 | 3 | 4 | 0 | 1 |
| 3 | 3 | 4 | 0 | 1 | 2 |
| 4 | 4 | 0 | 1 | 2 | 3 |
Question 3. Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.
Solution:
Given that ×6 on set S = {0, 1, 2, 3, 4, 5}
We have to find the composition table for ×6 by using multiplication and modulo operations.
As 0 x6 0 = 0 (0 × 0 = 0 and 0 modulo 6 = 0 % 6 = 0)
As 1 x6 1 = 1 (1 × 1 = 1 and 1 modulo 6 = 1 % 6 =1)
As 1 x6 2 = 2 (1 × 2 = 2 and 2 modulo 6 = 2 % 6 = 2)
As 3 x6 1 = 3 (3 × 1 = 3 and 3 modulo 6 = 3 % 6 = 3)
As 2 x6 2 = 4 (2 × 2 = 4 and 4 modulo 6 = 4 % 6 = 4)
As 5 x6 1 = 5 (5 × 1 = 5 and 5 modulo 6 = 5 % 6 = 5)
.
so on . . .
Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 6)
Therefore, the composition table is :
| x6 | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 0 | 2 | 4 |
| 3 | 0 | 3 | 0 | 3 | 0 | 3 |
| 4 | 0 | 4 | 2 | 0 | 4 | 2 |
| 5 | 0 | 5 | 4 | 3 | 2 | 1 |
Question 4. Construct the composition table for ×5 on set S = {0, 1, 2, 3, 4}.
Solution:
Given that ×5 on set S = {0,1,2,3,4}
We have to find the composition table for ×5 by using multiplication and modulo operations.
As 0 x5 0 = 0 (0 × 0 = 0 and 0 modulo 5 = 0 % 5 = 0)
As 1 x5 1 = 1 (1 × 1 = 1 and 1 modulo 5 = 1 % 5 =1)
As 1 x5 2 = 2 (1 × 2 = 2 and 2 modulo 5 = 2 % 5 = 2)
As 2 x5 2 = 4 (2 × 2 = 4 and 4 modulo 5 = 4 % 5 = 4)
As 3 x5 3 = 2 (3 × 4 = 12 and 12 modulo 5 = 12 % 5 = 2)
.
So on . . .
Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 5)
Therefore, the composition table is:
| x5 | 0 | 1 | 2 | 3 | 4 |
| 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 |
| 2 | 0 | 2 | 4 | 1 | 3 |
| 3 | 0 | 3 | 1 | 4 | 2 |
| 4 | 0 | 4 | 3 | 2 | 1 |
Question 5. Construct the composition table for ×10 on set S = {1, 3, 7, 9}.
Solution:
Given that ×10 on set S = {1,3,7,9}
We have to find the composition table for ×10 by using multiplication and modulo operations.
As 1 x10 1 = 1 (1 × 1 = 1 and 1 modulo 10 = 1 % 10 = 1)
As 1 x10 7 = 7 (1 × 7 = 7 and 7 modulo 10 = 2 % 10 = 7)
As 3 x10 1 = 3 (3 × 1 = 3 and 3 modulo 10 = 3 % 10 = 3)
As 9 x10 7 = 3 (9 × 7 = 63 and 63 modulo 10 = 63 % 10 = 3)
.
So on . . .
Here % – (modulo operator) means it gives the remainder of the number when divided by the other number (i.e., 10)
Therefore, the composition table is:
| x10 | 1 | 3 | 7 | 9 |
| 1 | 1 | 3 | 7 | 9 |
| 3 | 3 | 9 | 1 | 7 |
| 7 | 7 | 1 | 9 | 3 |
| 9 | 9 | 7 | 3 | 1 |
From the composition table, we can observe that elements 1 multiples (i.e., 1st row) are same as top most row.
Therefore, 1 ∈ S is the identity element for x10.
We have to find the inverse of 3:
As 3 x10 7 = 1 (3 × 7 = 21 and 21 modulo 10 = 1) in the composition table.
So the inverse of 3 is 7
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