Question

For the binary operation ×₁₀ on set S = {1, 3, 7, 9}, find the inverse of 3.

Answer 1

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.