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.

Answer 1

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.