Question

For the binary operation multiplication modulo 10 (×₁₀) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.

Answer 1

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