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Question
For the binary operation ×10 on set S = {1, 3, 7, 9}, find the inverse of 3.
Solution
Here,
1 \[\times_{10}\] 1 = Remainder obtained by dividing 1 \[\times\] 1 by 10
=1
3 \[\times_{10}\] 7 = Remainder obtained by dividing 3 \[\times\] 7 by 10
=1
7 \[\times_{10}\] 9 = Remainder obtained by dividing 7 \[\times\]9 by 10
= 3
So, the composition table 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 |
We observe that the elements of the first row are same as the top-most row.
So, \[1 \in S\] is the identity element with respect to \[\times_{10}\] Finding inverse of 3:
From the above table we observe,
3 \[\times_{10}\] 7 = 1
So, the inverse of 3 is 7.
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