Question

For the binary operation multiplication modulo 5 (×₅) defined on the set S = {1, 2, 3, 4}. Write the value of ${\left(3{\times }_5{4}^{-1}\right)}^{-1}$.

Answer 1

Here,

1 ${\times }_5$1 = Remainder obtained by dividing 1 $\times$ 1 by 5
= 1

3 ${\times }_5$ 4 = Remainder obtained by dividing 3 $\times$ 4 by 5
= 2

4 ${\times }_5$ 4 = Remainder obtained by dividing 4 $\times$ 4 by 5
= 1
So, the composition table is as follows:

${\times }_5$ ×	1	2	3	4
1	1	2	3	4
2	2	4	1	3
3	3	1	4	2
4	4	3	2	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.
$\Rightarrow a{\times }_{5}1=1{\times }_{5}a=a,\forall a\in S$
Thus, 1 is the identity element.

$$\begin{gathered} \text{Now,} \\ {\left(3{\times }_5{4}^{-1}\right)}^{-1} \\ ={\left(3{\times }_{5}4\right)}^{-1}\left[\text{∵ 4}{\times }_5\text{4 = 1}\right] \\ =2^{-1} \\ =3\left[\because \text{2}{\times }_5\text{3 = 1}\right] \end{gathered}$$