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Question
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as \[a * b = \begin{cases}a + b & ,\text{ if a + b} < 6 \\ a + b - 6 & , \text{if a + b} \geq 6\end{cases}\]
Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.
Solution
Here,
1 * 1 =1+1 \[\because\] 1+1 \[<\] 6 )
= 2
3 * 4 = 3 + 4 \[-\] 6 ( \[\because\] 3 + 4 \[>\] 6 )
= 7 \[-\] 6
= 1
4 * 5 = 4 + 5 \[-\] 6 (\[\because\] 4 + 5 \[>\] 6 )
= 9 \[-\]6
= 3 etc.
So, the composition table is as follows:
| * | 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 top-most row and the first column coincides with the left-most column.
These two intersect at 0.
So, 0 is the identity element .
\[\Rightarrow a * 0 = 0 * a = a, \forall a \in \left\{ 0, 1, 2, 3, 4, 5 \right\}\]
Finding inverse :-
\[\text{Leta} \in \left\{ 0, 1, 2, 3, 4, 5 \right\} \text{ and }b \in \left\{ 0, 1, 2, 3, 4, 5 \right\} \text{ such that}\]
\[a * b = b * a = e\]
\[a * b = e \text{ and }b * a = e\]
Case 1 :- Let us assume that a + b < 6
Then,
\[a * b = e \text { and }b * a = e\]
\[a + b = 0 \text{ and } b + a = 0\]
a = - b, which is not possible because all the elements of the given set are non-negative.
Case 2 :- Let us assume that a + b ≥ 6
Then,
\[a * b = e \text{ and } b * a = e\]
\[a + b - 6 = 0 \text{ and }b + a - 6 = 0\]
b = 6 - a (from the table we can observe that this is true for all a ≠ 0)
Thus, 6 - a is the inverse of a.
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