Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.5

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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