Question
For the binary operation multiplication modulo $10 (\times _{10})$ defined on the set $S = \{1, 3, 7, 9\},$ write the inverse of $3.$
✓
Answer
$1 \times _{10}1 =$ 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:
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 $1.$
$\Rightarrow a ^* 1 = 1 ^* a = a, \forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$
$3 \times _{10}1 =$ 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:
| $\times _{10}$ | $1$ | $3$ | $7$ | $9$ |
| $1$ | $1$ | $3$ | $7$ | $9$ |
| $3$ | $3$ | $9$ | $1$ | $7$ |
| $7$ | $7$ | $1$ | $9$ | $3$ |
| $9$ | $9$ | $7$ | $3$ | $1$ |
These two intersect at $1.$
$\Rightarrow a ^* 1 = 1 ^* a = a, \forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$
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