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Home » For the binary operation set , find the inverse of 3 .
For the binary operation \mathrm{x}_{10} set \mathbf{S}=\{1,3,7,9\}, find the inverse of 3 .
Binary Operations | CBSE | Class 12 | Exercise 3.5 | Maths | RD Sharma
For the binary operation \mathrm{x}_{10} set \mathbf{S}=\{1,3,7,9\}, find the inverse of 3 .

Solution:

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
Therefore, the composition table is as follows:

    \[\begin{tabular}{|l|l|l|l|l|} \hline$\times_{10}$ & 1 & 3 & 7 & 9 \\ \hline 1 & 1 & 3 & 7 & 9 \\ \hline 3 & 3 & 9 & 1 & 7 \\ \hline 7 & 7 & 1 & 9 & 3 \\ \hline 9 & 9 & 7 & 3 & 1 \\ \hline \end{tabular}\]

It is observed from the table that elements of first row is same as the top-most row.
Therefore, 1 \in S is the identity element with respect to \times_{10}
We now have to find the inverse of 3
3 \times_{10} 7=1
As a result, the inverse of 3 is 7 .

i

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