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 ×101 = Remainder obtained by dividing 1 × 1 by 10 = 1
3 ×101 = Remainder obtained by dividing 3 × 1 by 10 = 3
7 ×103 = Remainder obtained by dividing 7 × 3 by 10 = 1
3 ×103 = Remainder obtained by dividing 3 × 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.
⇒ a * 1 = 1 * a = a, $\forall\text{ a}\in\text{S}$
So, the identity element is 1.
Also,
3 ×107 = 1
3-1 = 7

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free