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Question
For the binary operation multiplication modulo 10 (×10) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.
Solution
Here,
1 \[\times_{10}\] 1 = Remainder obtained by dividing 1 \[\times\] 1 by 10
= 1
3 \[\times_{10}\] 1 = Remainder obtained by dividing 3 \[\times\] 1 by 10
= 3
7 \[\times_{10}\] 3 = Remainder obtained by dividing 7 \[\times\] 3 by 10
= 1
3 \[\times_{10}\] 3 = Remainder obtained by dividing 3 \[\times\] 3 by 10
= 9
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 first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at 1.
\[\Rightarrow a * 1 = 1 * a = a, \forall a \in S\]
So, the identity element is 1.
Also,
3 \[\times_{10}\] 7 = 1
3-1 = 7
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