Question

Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as

$a*b=\left\{\begin{array}{ll}a+b& ,\mathrm{if}a+b<6\\ a+b-6& ,\mathrm{if}a+b\ge 6\end{array}\right.$

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:

$$\begin{gathered} \text{Let}a\in \left\{0,1,2,3,4,5\right\}\text{and}b\in \left\{0,1,2,3,4,5\right\}\text{such that} \\ a*b=b*a=e \\ a*b=e\text{and}b*a=e \\ \text{Case 1: Let us assume that}a+b<6 \\ \text{Then,} \\ a*b=e\text{and}b*a=e \\ a+b=0\text{and}b+a=0 \\ a=-b,\text{which is not possible because all the elements of the given set are non-negative.} \\ \text{Case 2: Let us assume that}a+b\ge 6 \\ \text{Then,} \\ a*b=e\text{and}b*a=e \\ a+b-6=0\text{and}b+a-6=0 \\ b=6-a\text{(from the table we can observe that this is true for all}a\ne 0\text{)} \\ \text{Thus,}6-a\text{is the inverse of}a. \end{gathered}$$