Question
1
:
Construct the composition table for ×4 onset S = {0, 1, 2, 3}.
Answer
1 :
Given that ×4 onset S = {0, 1, 2, 3}
Here,
1 ×4 1= remainder obtained by dividing 1 × 1 by 4
= 1
0 ×4 1= remainder obtained by dividing 0 × 1 by 4
= 0
2 ×4 3= remainder obtained by dividing 2 × 3 by 4
= 2
3 ×4 3= remainder obtained by dividing 3 × 3 by 4
= 1
So, the compositiontable is as follows:
| ×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 |
Question
2
:
Construct the composition table for +5 on set S = {0, 1,2, 3, 4}
Answer
2 :
1 +5 1= remainder obtained by dividing 1 + 1 by 5
= 2
3 +5 1= remainder obtained by dividing 3 + 1 by 5
= 2
4 +5 1= remainder obtained by dividing 4 + 1 by 5
= 3
So, the compositiontable is as follows:
| +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 onset S = {0, 1, 2, 3, 4, 5}.
Answer
3 :
1 ×6 1= remainder obtained by dividing 1 × 1 by 6
= 1
3 ×6 4= remainder obtained by dividing 3 × 4 by 6
= 0
4 ×6 5= remainder obtained by dividing 4 × 5 by 6
= 2
So, the compositiontable is as follows:
| ×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 |
Question
4
:
Construct the composition table for ×5 on set Z5 ={0, 1, 2, 3, 4}
Answer
4 :
Here,
1 ×5 1= remainder obtained by dividing 1 × 1 by 5
= 1
3 ×5 4= remainder obtained by dividing 3 × 4 by 5
= 2
4 ×5 4= remainder obtained by dividing 4 × 4 by 5
= 1
So, the compositiontable is as follows:
| ×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 |
Question
5
:
For the binary operation ×10 set S = {1, 3, 7, 9}, findthe inverse of 3.
Answer
5 :
Here,
1 ×10 1= remainder obtained by dividing 1 × 1 by 10
= 1
3 ×10 7= remainder obtained by dividing 3 × 7 by 10
= 1
7 ×10 9= remainder obtained by dividing 7 × 9 by 10
= 3
So, the compositiontable is as follows:
| ×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 table we canobserve that elements of first row as same as the top-most row.
So, 1 ∈ S is the identity element with respect to ×10
Now we have to findinverse of 3
3 ×10 7= 1
So the inverse of 3 is7.
Question
6
:
Answer
6 :
Question
7
:
Answer
7 :
Question
8
:
Answer
8 :
Question
9
:
Answer
9 :
Question
10
:
Answer
10 :