Answer
given below
Work Step by Step
(a)
According to closure property, a set is said to be closed if all elements of the set belong to the table of the clock hour addition operation.
All elements of provided set \[\left\{ \left. 0,1,2,3,4 \right\} \right.\] are present in the table of 5-hour clock addition operation.
Hence, it has been proved that the provided set is closed under 5-hour clock addition operation.
(b)
The symbol \[\oplus \]is the standard notation for clock hour addition.
Applying clock hour addition operation on the provided expression on both sides.
Clock hour addition of \[4\]and \[2\]is\[4\oplus 2=1\].
Clock hour addition of \[2\]and \[3\] is\[2\oplus 3=0\].
Clock hour addition of \[1\] and \[3\]is\[1\oplus 3=1\].
Clock hour addition of \[4\]and \[0\]is\[4\oplus 0=4\].
\[\begin{align}
& \left( 4\oplus 2 \right)\oplus 3=4\oplus \left( 2\oplus 3 \right) \\
& 1\oplus 3=4\oplus 0 \\
& 4=4
\end{align}\]
Thus, the left side of the expression is equal to the right side hence, associative property is applicable in the clock hour addition operation.
Hence, the associative property is verified for the provided operation.
(c)
The 5-hour clock and the table for 5-hour clock addition operation is shown below:
According to identity property value of an element doesn’t change in clock addition.
In the first column of addition table all entries when added with 0 gives same values as shown in the second column.
\[0+0=0\], \[0+1=1\], \[0+2=2\],\[0+3=3\], \[0+4=4\].
Thus,\[0\] is an identity element in clock addition system.
(d)
According to inverse property when any element added to its inverse the resultant is identity element.
So, using the above property:
\[\begin{align}
& \left( \text{inverse}\ \text{of}\ 0 \right)+0=0 \\
& 0+0=0
\end{align}\]
The inverse of\[0\] is\[0\].
\[\begin{align}
& \left( \text{inverse}\ \text{of}\ 1 \right)+1=0 \\
& 4+1=0
\end{align}\]
So, the inverse of \[1\] is\[4\].
\[\begin{align}
& \left( \text{inverse}\ \text{of}\ 2 \right)+2=0 \\
& 3+2=0
\end{align}\]
So, the inverse of 2is3.
\[\begin{align}
& \left( \text{inverse}\ \text{of}\ 3 \right)+3=0 \\
& 2+3=0
\end{align}\]
So, the inverse of \[3\] is\[2\].
\[\begin{align}
& \left( \text{inverse}\ \text{of 4} \right)+4=0 \\
& 1+4=0
\end{align}\]
So, the inverse of \[4\]is\[1\].
Hence, the inverse of \[0\]is0; inverse of \[1\] is4; inverse of \[2\]is3; inverse of \[3\]is2 and
the inverse of \[4\]is1.
(e)
By using the property of clock addition.
\[4\oplus 3=2\] ……(1)
\[3\oplus 4=2\] ……(2)
Comparing the equation (1) and (2),\[4\oplus 3=3\oplus 4\].
By using the property of clock addition.
\[3\oplus 2=0\] ……(3)
\[2\oplus 3=0\] ……(4)
Comparing the equations (3) and (4), \[3\oplus 2=2\oplus 3\].
Hence, the commutative property is verified.
