Question

A binary operation * on the set \left\\{ {0,1,2,3,4,5} \right\\} defined as,
a*b = \left[ {\begin{array}{*{20}{c}} {a + b;}&{it\;a + b < 6} \\\ {a + b - 6;}&{it\;a + b > 6} \end{array}} \right]
Show that zero is the identity element of this operational each element ‘a’ of the set is invertible with 6 – 0 being the inverse of ‘a’

Answer 1

Hint : -At first we have to define the a*b for the different elements of the set containing \left\\{ {0,1,2,3,4,5} \right\\} with satisfying this condition
a*b = \left[ {\begin{array}{*{20}{c}} {a + b;}&{it\;a + b < 6} \\\ {a + b - 6;}&{it\;a + b > 6} \end{array}} \right]

Complete step-by-step answer :

| ${C_1}$ | ${C_2}$ | ${C _3}$ | ${C_4}$ | ${C_5}$ | ${C_6}$
---|---|---|---|---|---|---
| 0| 1| 2| 3| 4| 5
${R_1} \to 0$ | 0| 1| 2| 3| 4| 5
${R_2} \to 1$ | 1| 2| 3| 4| 5| 0
${R_3} \to 2$ | 2| 3| 4| 5| 0| 1
${R_4} \to 3$ | 3| 4| 5| 0| 1| 2
${R_5} \to 4$ | 4| 5| 0| 1| 2| 3
${R_6} \to 5$ | 5| 0| 1| 2| 3| 4

R= Row
C= Coloumn
$\begin{gathered} 0 + 0 = 0 < 6 \\\ 0 + 1 = 1 < 6 \\\ 0 + 2 = 2 < 6 \\\ 0 + 3 = 3 < 6 \\\ 0 + 4 = 4 < 6 \\\ 0 + 5 = 5 < 6 \\\ 0 + 6 = 6 = 6 \\\ \end{gathered}$
So it is 6 -6 =0 according to the condition $\therefore a + b = a + b - 6 \geqslant 6.$ we have to do for the rest of the operations in the same manner.
Identity element
Properties, it is identical with $\ell$
Then $a*\ell = a,$
So for finding the identity element we have to apply this property
For, 0, $\begin{gathered} 0 + \ell = 0 \\\ \;\;\;\;\;\ell = 0, \\\ \end{gathered}$
For 1, $\begin{gathered} 1 + \ell = 1 \\\ \;\;\;\ell = 1 - 1 = 0 \\\ \end{gathered}$
For 2, $\begin{gathered} 2 + \ell = 2 \\\ \;\;\;\;\;\ell = 0 \\\ \end{gathered}$
For 3, $\begin{gathered} 3 + \ell = 3 \\\ \;\;\;\;\ell = 0 \\\ \end{gathered}$
For 4, $\begin{gathered} 4 + \ell = 4 \\\ \;\;\;\;\;\ell = 0 \\\ \end{gathered}$
For 5, $\begin{gathered} 5 + \ell = 5 \\\ \;\;\;\;\ell = 0 \\\ \end{gathered}$
So for all the element Identity elements is same and it is = 0.
Inverse element
Properties, If is inverse of a,
Then, $a\;*\;I = \ell$ $\left[ {\ell = Identity\;element} \right]$
For 0, $\begin{gathered} 0 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 0, \\\ \end{gathered}$
For 1, $\begin{gathered} 1 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 1, \\\ \end{gathered}$
For2, $\begin{gathered} 2 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 2, \\\ \end{gathered}$
For 3, $\begin{gathered} 3 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 3, \\\ \end{gathered}$
For 4, $\begin{gathered} 4 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 4, \\\ \end{gathered}$
For 5, $\begin{gathered} 5 + I - 6 = 0 \\\ \;\;\;\;\;I = 6 - 5, \\\ \end{gathered}$
So for each element, the inverse element exists and it is in the form of $6 - a$
And for the inverse of each element exists that’s why the operation is invertible with 6 –a being the inverse of ‘a’.

Note : We have to care about the finding of inverse elements, if for any element inverse does not exist operation can’t be Invertible at all. Therefore the inverse must exist for all the elements else the function cannot be invertible.