Answer
shown below
Work Step by Step
(a)
The set\[\left\{ 0,1,2,3,4,5,6,7 \right\}\]is closed under the operation of clock addition, because the entries of the table are all the elements of the set.
Hence, the set \[\left\{ 0,1,2,3,4,5,6,7 \right\}\]is closed under the operation of clock addition.
(b)
Consider the expression of LHS:
\[\begin{align}
& \left( 4\oplus 6 \right)\oplus 7=2\oplus 7 \\
& =1
\end{align}\]
Now, find RHS as,
\[\begin{align}
& 4\oplus \left( 6\oplus 7 \right)=4\oplus 5 \\
& =1
\end{align}\]
Hence, \[\left( 4\oplus 6 \right)\oplus 7=4\oplus \left( 6\oplus 7 \right)\]. Associative property has been verified.
(c)
Identity element is that element, which does not change anything when clock addition is used.
Now, as we can see that,
\[0\oplus 0=0,\ 0\oplus 1=1,\ 0\oplus 2=2,\ 0\oplus 3=3,\ 0\oplus 4=4,\ 0\oplus 5=5,\ 0\oplus 6=6,\ 0\oplus 7=7\]and,
\[0\oplus 0=0,\ 1\oplus 0=1,\ 2\oplus 0=2,\ 3\oplus 0=3,\ 4\oplus 0=4,\ 5\oplus 0=5,\ 6\oplus 0=6,\ 7\oplus 0=7\]
Thus, 0 is the identity element.
Hence, 0 is the identity element in the 8-hour clock addition.
(d)
When an element is added to its inverse, the result is the identity element. And, as the identity element is 0. Find inverse of each element as,
Let \[x\]be the inverse of any element, such that
\[x\oplus 0=0\], and from the table \[0\oplus 0=0\]
Thus, 0 is the inverse of element 0.
\[x\oplus 1=0\], and from the table \[7\oplus 1=0\]
Thus, 7 is the inverse of element 1.
\[x\oplus 2=0\], and from the table \[6\oplus 2=0\]
Thus, 6 is the inverse of element 2.
\[x\oplus 3=0\], and from the table \[5\oplus 3=0\]
Thus, 5 is the inverse of element 3.
\[x\oplus 4=0\], and from the table \[4\oplus 4=0\]
Thus, 4 is the inverse of element 4.
\[x\oplus 5=0\], and from the table \[3\oplus 5=0\]
Thus, 3 is the inverse of element 5.
\[x\oplus 6=0\], and from the table \[2\oplus 6=0\]
Thus, 2 is the inverse of element 6.
\[x\oplus 7=0\], and from the table \[1\oplus 7=0\]
Thus, 1 is the inverse of element 7.
Hence, inverse of elements 0,1,2,3,4,5,6 and 7 are 0,7,6,5,4,3,2 and 1 respectively.
(e)
\[5\oplus 6=3\]
Now, find RHS as,
\[6\oplus 5=3\]
And,
\[4\oplus 7=3\]
Now, find RHS as,
\[7\oplus 4=3\]
Both the equations follow commutative property.
Hence,\[5\oplus 6=6\oplus 5\]and\[4\oplus 7=7\oplus 4\]follows commutative property and has been verified.
