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Question
Let +6 (addition modulo 6) be a binary operation on S = {0, 1, 2, 3, 4, 5}. Write the value of \[2 +_6 4^{- 1} +_6 3^{- 1} .\]
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Solution
Here,
1 \[+_6\] 1 = Remainder obtained by dividing 1 + 1 by 6
= 2
3 \[+_6\] 4 = Remainder obtained by dividing 3 + 4 by 6
= 1
4 \[+_6\] 5 = Remainder obtained by dividing 4 + 5 by 6
= 3
So, the composition table is as follows:
| +6 | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
\[\Rightarrow a +_6 0 = 0 +_6 a = a, \forall a \in S\]
So, 0 is the identity element.
From the table,
\[4 +_6 2 = 0 \]
\[ \Rightarrow 4^{- 1} = 2\]
\[3 +_6 3 = 0 \]
\[ \Rightarrow 3^{- 1} = 3\]
\[\text{Now},\]
\[2 +_6 4^{- 1} +_6 3^{- 1} = 2 +_6 2 +_6 3\]
\[ = 4 +_6 3 \]
\[ = 1\]
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