Class 12 RD Sharma Solutions Chapter 3 Binary Operations Exercise 3.5 (original) (raw)

Last Updated : 20 Aug, 2024

Exercise 3.5 in Chapter 3 of RD Sharma's Class 12 Mathematics textbook focuses on binary operations. This exercise likely covers advanced concepts related to binary operations, including their properties, applications, and problem-solving techniques. Students will be challenged to apply their understanding of binary operations to various mathematical situations.

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 x 4 0 = 0 (0 × 0 = 0 and 0 modulo 4 = 0 % 4 = 0)

_As 1 x 4 1 = 1 (1 × 1 = 1 and 1 modulo 4 = 1 % 4 =1)

_As 1 x 4 2 = 2 (1 × 2 = 2 and 2 modulo 4 = 2 % 4 = 2)

_As 3 x 4 _1 = 3 (3 × 1 = 3 and 3 modulo 4 = 3 % 4 = 3)

_As 4 x 4 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:

**x 4 0 1 2 3

0 0 0 0 0

1 0 1 2 3

2 0 2 0 2

3 0 3 2 1

**x 4	1	2	3
0	0	0	0
1	1	2	3
2	2	0	2
3	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

**+ 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 x 6 0 = 0 (0 × 0 = 0 and 0 modulo 6 = 0 % 6 = 0)

_As 1 x 6 1 = 1 (1 × 1 = 1 and 1 modulo 6 = 1 % 6 =1)

_As 1 x 6 2 = 2 (1 × 2 = 2 and 2 modulo 6 = 2 % 6 = 2)

_As 3 x 6 1 = 3 (3 × 1 = 3 and 3 modulo 6 = 3 % 6 = 3)

_As 2 x 6 2 = 4 (2 × 2 = 4 and 4 modulo 6 = 4 % 6 = 4)

_As 5 x 6 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 :

**x 6 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

**x 6	1	2	3	4	5
0	0	0	0	0	0
1	1	2	3	4	5
2	2	4	0	2	4
3	3	0	3	0	3
4	4	2	0	4	2
5	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 x 5 0 = 0 (0 × 0 = 0 and 0 modulo 5 = 0 % 5 = 0)

_As 1 x 5 1 = 1 (1 × 1 = 1 and 1 modulo 5 = 1 % 5 =1)

_As 1 x 5 2 = 2 (1 × 2 = 2 and 2 modulo 5 = 2 % 5 = 2)

_As 2 x 5 2 = 4 (2 × 2 = 4 and 4 modulo 5 = 4 % 5 = 4)

_As 3 x 5 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:

**x 5 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

**x 5	1	2	3	4
0	0	0	0	0
1	1	2	3	4
2	2	4	1	3
3	3	1	4	2
4	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 x 10 1 = 1 (1 × 1 = 1 and 1 modulo 10 = 1 % 10 = 1)

_As 1 x 10 7 = 7 (1 × 7 = 7 and 7 modulo 10 = 2 % 10 = 7)

_As 3 x 10 1 = 3 (3 × 1 = 3 and 3 modulo 10 = 3 % 10 = 3)

_As 9 x 10 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:

**x 10 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 x 10 7 = 1 (3 × 7 = 21 and 21 modulo 10 = 1) in the composition table.

_So the inverse of 3 is 7

**x 10	1	3	7	9
1	1	3	7	9
3	3	9	1	7
7	7	1	9	3
9	9	7	3	1

Summary

Exercise 3.5 in Chapter 3 of RD Sharma's Class 12 Mathematics textbook delves into advanced concepts of binary operations. This exercise challenges students to apply their understanding of binary operations to various mathematical situations, likely covering properties such as associativity and commutativity, identifying identity and inverse elements, solving equations involving binary operations, and analyzing their composition. It serves as a crucial component for solidifying students' grasp of this fundamental concept in abstract algebra and higher mathematics.