\text{Question 1: Construct the composition table for   } \times_4 \text{ on set } S =\{0,1,2, 3\}.

Answer: 

1 \times_4 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 4 = 1

0 \times_4 1 = \text{ Remainder obtained by dividing } 0 \times 1 \text{ by } 4 = 0

2 \times_4 3 = \text{ Remainder obtained by dividing } 2 \times 3 \text{ by } 4 = 2

3 \times_4 3 = \text{ Remainder obtained by dividing } 2 \times 2 \text{ by } 4 = 1

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|}  \hline \ \ \ \times_4 \ \ \ & \ \ \ 0 \ \ \ & \ \ \ 1 \ \ \  & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \   \\  \hline 0 & 0 & 0 & 0 & 0 \\  \hline 1 & 0 & 1 & 2 & 3  \\  \hline 2 & 0 & 2 & 0 & 2   \\  \hline 3 & 0 & 3 & 2 & 1 \\  \hline  \end{array}

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\text{Question 2: Construct the composition table for   } +_5 \text{ on set } S =\{0,1,2, 3, 4\}.

Answer:

1 +_5 1 = \text{ Remainder obtained by dividing } 1 + 1 \text{ by } 5 = 2

3 +_5 4 = \text{ Remainder obtained by dividing } 3 + 4 \text{ by } 5 = 2

4 +_5 4 = \text{ Remainder obtained by dividing } 4 + 4 \text{ by } 5 = 3

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|}  \hline \ \ \ +_5 \ \ \ & \ \ \ 0 \ \ \ & \ \ \ 1 \ \ \  & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  &  \ \ \ 4 \ \ \   \\  \hline 0 & 0 & 1 & 2 & 3 & 4 \\  \hline 1 & 1 & 2 & 3 & 4  & 5 \\  \hline 2 & 2 & 3 & 4 & 0  & 1 \\  \hline 3 & 3 & 4 & 0 & 1 & 2 \\  \hline 4 & 4 & 0 & 1 & 2 & 3 \\  \hline  \end{array}

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\text{Question 3: Construct the composition table for  } \times_6 \text{ on set } S =\{0,1,2, 3, 4, 5 \}.

Answer:

1 \times_6 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 6 = 1

3 \times_6 4 = \text{ Remainder obtained by dividing } 3 \times 4 \text{ by } 6 = 0

4 \times_6 5 = \text{ Remainder obtained by dividing } 4 \times 5 \text{ by } 6 = 2

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|c|}  \hline \ \ \ \times_6 \ \ \ & \ \ \ 0 \ \ \ & \ \ \ 1 \ \ \  & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  &  \ \ \ 4 \ \ \ &  \ \ \ 5 \ \ \   \\  \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\  \hline 1 & 0& 1 & 2 & 3  & 4 & 5 \\  \hline 2 & 0 & 2 & 4 & 0  & 2 & 4 \\  \hline 3 & 0 & 3 & 0 & 3 & 0 & 3 \\  \hline 1 & 0 & 4 & 2 & 0 & 4 & 2 \\  \hline 5 & 0 & 5 & 4 & 3& 2 & 1 \\  \hline  \end{array}

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\text{Question 4: Construct the composition table for   } \times_5 \text{ on } Z_5 = \{ 0, 1, 2, 3, 4 \}.

Answer:

1 \times_5 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 5 = 1

3 \times_5 4 = \text{ Remainder obtained by dividing } 3 \times 4 \text{ by } 5 = 2

4 \times_5 4 = \text{ Remainder obtained by dividing } 4 \times 4 \text{ by } 5 = 1

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|}  \hline \ \ \ \times_5 \ \ \ & \ \ \ 0 \ \ \ & \ \ \ 1 \ \ \  & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  &  \ \ \ 4 \ \ \   \\  \hline 0 & 0 & 0 & 0 & 0 & 0 \\  \hline 1 & 0 & 1 & 2 & 3  & 4 \\  \hline 2 & 0 & 2 & 4 & 1  & 3 \\  \hline 3 & 0 & 3 & 1 & 4 & 2 \\  \hline 4 & 0 & 4 & 3 & 2 & 1 \\  \hline  \end{array}

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\text{Question 5: For the binary operation } \times_{10} \text{ on set } S = \{ 1, 3,7 , 9 \}, \\ \\ \text{ find the inverse of 3.}

Answer:

1 \times_{10} 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 10 = 1

3 \times_{10} 7 = \text{ Remainder obtained by dividing } 3 \times 7 \text{ by } 10 = 1

7 \times_{10} 9 = \text{ Remainder obtained by dividing } 7 \times 9 \text{ by } 10 = 3

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|}  \hline \ \ \ \times_{10} \ \ \ & \ \ \ 1 \ \ \ & \ \ \ 3 \ \ \  & \ \ \ 7 \ \ \  & \ \ \ 9 \ \ \   \\  \hline 1 & 1 & 3 & 7 & 9 \\  \hline 3 & 3 & 9 & 1 & 7  \\  \hline 7 & 7 & 1 & 9 & 3   \\  \hline 9 & 9 & 7 & 3 & 1 \\  \hline  \end{array}

We observe that the elements of the first row are the same as the top most row. So, 1 \in S is the identity element with respect to \times_{10}

Finding inverse of 3:

From the above table we observe, 3 \times_{10} 7 = 1.

Hence the inverse of 3 is 7.

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\text{Question 6: For the binary operation } \times_7 \text{ on the set } S = \{ 1, 2, 3, 4,5, 6 \}, \\ \\ \text{ compute } 3^{- 1} \times_7 4. 

Answer:

1 \times_7 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 7 = 1

3 \times_7 4 = \text{ Remainder obtained by dividing } 3 \times 4 \text{ by } 7 = 5

4 \times_7 5 = \text{ Remainder obtained by dividing } 4 \times 5 \text{ by } 7 = 6

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|c|}  \hline \ \ \ \times_7 \ \ \ & \ \ \ 1 \ \ \ & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  & \ \ \ 4 \ \ \  &  \ \ \ 5 \ \ \ &  \ \ \ 6 \ \ \   \\  \hline 1 & 1& 2& 3 & 4  & 5 & 6 \\  \hline 2 & 2 & 4 & 6 & 1  & 3 & 5 \\  \hline 3 & 3 & 6 & 2 & 5 & 1 & 4 \\  \hline 4 & 4 & 1 & 5 & 2 & 6 & 3 \\  \hline 5 & 5 & 3 & 1 & 6& 4 & 2 \\  \hline 6 & 6 & 5 & 4 & 3 & 2 & 1 \\  \hline  \end{array}

We observe that the elements of the first row are the same as the top most row. So, the identity element is 1.

3 \times_7 5 = 1 = 5

\text{Hence, } 3^{-1} = 5

\text{Now, } 3^{-1} \times_7 4 = 5 \times_7 4 = 6

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Question 7: Find the inverse of 5 under multiplication modulo 11 on Z_{11}.

Answer:

1 \times_{11} 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 11 = 1

3 \times_{11} 4 = \text{ Remainder obtained by dividing } 3 \times 4 \text{ by } 11 = 1

4 \times_{11} 5 = \text{ Remainder obtained by dividing } 4 \times 5 \text{ by } 11 = 9

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}  \hline \ \ \ \times_{11} \ \ \ & \ \ \ 1 \ \ \ & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  & \ \ \ 4 \ \ \  &  \ \ \ 5 \ \ \ &  \ \ \ 6 \ \ \ &  \ \ \ 7 \ \ \ &  \ \ \ 8 \ \ \ &  \ \ \ 9 \ \ \ &  \ \ \ 10 \ \ \   \\  \hline 1 & 1& 2& 3 & 4  & 5 & 6  &7 & 8& 9&10 \\  \hline 2 & 2 & 4 & 6 & 8  & 10 & 1 &3 &5 &7 &9 \\  \hline 3 & 3 & 6 & 9 & 1 & 4 & 7 &10 &2 &5 &8 \\  \hline 4 & 4 & 8 & 1 & 5 & 9 & 2 &6 &10 & 3&7 \\  \hline 5 & 5 & 10 & 4 & 9& 3 & 8 & 2&7 &1 &6 \\  \hline 6 & 6 & 1 & 7 & 2 & 8 & 3 &9 & 4& 10&5 \\  \hline 7 & 7 & 3 & 10 & 6 & 2 & 9 &5 &1 &8 &4 \\  \hline 8 & 8 & 5 & 2 & 10 & 7 & 4 &1 &9&6 &3 \\  \hline 9 & 9 & 7 & 5 & 3 & 1 & 10 &8 & 6&4 &2 \\  \hline 10 & 10 & 9 & 8 & 7 & 6 &5 & 4&3 &2 & 1\\  \hline  \end{array}

We observe that the elements of the first row are the same as the top most row. So, the identity element is 1.

\text{Also, } 5 \times_{11} 9 = 1.

\text{Hence, } 5^{-1} = 9

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Question 8: Write the multiplication table for the set of integers modulo 5.

Answer:

1 \times_5 1 = \text{ Remainder obtained by dividing } 1 \times 1 \text{ by } 5 = 1

3 \times_5 4 = \text{ Remainder obtained by dividing } 3 \times 4 \text{ by } 5 = 2

4 \times_5 4 = \text{ Remainder obtained by dividing } 4 \times 4 \text{ by } 5 = 1

Hence, the composition table is as follows:

\begin{array}{|c|c|c|c|c|c|}  \hline \ \ \ \times_5 \ \ \ & \ \ \ 0 \ \ \ & \ \ \ 1 \ \ \  & \ \ \ 2 \ \ \  & \ \ \ 3 \ \ \  &  \ \ \ 4 \ \ \   \\  \hline 0 & 0 & 0 & 0 & 0 & 0 \\  \hline 1 & 0 & 1 & 2 & 3  & 4 \\  \hline 2 & 0 & 2 & 4 & 1  & 3 \\  \hline 3 & 0 & 3 & 1 & 4 & 2 \\  \hline 4 & 0 & 4 & 3 & 2 & 1 \\  \hline  \end{array}

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\text{Question 9: Consider the binary operation  } * \text{ and } o \text{ defined by the } \\ \\ \text{ following tables on set } S = \{ a, b, c, d \}.

\text{(i)    }\begin{tabular}{ |c |c | c| c|c| }  \hline  * & a & b & c & d  \\   \hline  a & a & b & c & d  \\  \hline  b & b & a & d & c  \\  \hline  c & c & d & a & b  \\  \hline  d & d &c & b & a  \\  \hline  \end{tabular}

\text{(ii)    }\begin{tabular}{ |c |c | c| c|c| }  \hline  o & a & b & c & d  \\   \hline  a & a & a & a & a  \\  \hline  b & a & b & c & d  \\  \hline  c & a & c & d & b  \\  \hline  d & a &d & b & c  \\  \hline  \end{tabular}

Show that both the binary operations are commutative and associative. Write down the identities and list the inverse of elements.

Answer:

(i) We observe the following: 

a \ast b = b \ast a = b

c \ast a = a \ast c = c

a \ast d = d \ast a = d

b \ast c = c \ast b = d

b \ast d = d \ast b = c

c \ast d = d \ast c = b

There ' \ast ' is commutative. 

\text{ Also, } a \ast (b \ast c) = a \ast (d) = d (\text{ From above })

(a \ast b) \ast c = (b) \ast c = d (\text{ Also from above })

Hence, '\ast ' is associative too. 

Therefore, to find the identity element, e \text{ for } e \text{ belong to } S, we need: 

a \ast e = e \ast a = a, a \text{ belongs to } S.

\text{ Therefore, } a \ast e = a

e = a (\text{ since, } a \ast a = a, \text{ from the given table })

\text{ To find out the inverse, } a \ast x = e = b \ast x, x \text{ belongs to } S

a \ast x = e

x = a (\text{ From the given table })

\text{ Therefore, the inverse of } a \text{ is } a, b \text{ is } b, c \text{ is } c \text{ and } d \text{ is } d.

(ii) We observe the following: 

a \ast b = b \ast a = a a \ast a = a 

c \ast a = a \ast c = a b \ast b = b

a \ast d = d \ast a = a c \ast c = d

b \ast c = c \ast b = c d \ast d = c

b \ast d = d \ast b = d

c \ast d = d \ast c = b

There ' \ast ' is commutative. 

Also, a \ast (b \ast c) = a \ast (c) = a (\text{ From above })

(a \ast b) \ast c = (a) \ast c = a (\text{ Also from above })

Hence, ' \ast ' is associative too. 

Therefore, to find the identity element, e \text{ for } e \text{ belong to } S, we need: 

a \ast e = e \ast a = a, a \text{ belong to } S.

\text{ Therefore,  } a \ast e = a

We find that there is no unique element e which satisfies the condition. 

e = a or b or c or d for a. 

Since the identity is not unique, the inverse will also be not unique.

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\text{Question 10: Define a binary operation} \ast \text{ on the set } \{ 0, 1, 2,3,4,5 \} \text{ as }

a \ast b = \Bigg\{ \begin{array}{lll} a+b & , & \text{ if } a+b < 6 \\  a+b-6 & , & \text{ if } a+b \geq 6  \end{array}

Show that 0 is the identity for this operation and each element a \neq 0 of the set is invertible with 6 -a being the inverse of a.              [CBSE 2011]

Answer:

\text{Let } X = \{ 0, 1, 2,3,4,5 \}

The operation \ast \text{ on } X is defined as

a \ast b = \Bigg\{ \begin{array}{lll} a+b & , & \text{ if } a+b < 6 \\  a+b-6 & , & \text{ if } a+b \geq 6  \end{array}

\text{ An element } e \in X \text{ is the identity element for the operation } \ast, \text{ if } a \ast e = a = e \ast a \text{ for all } a \in X

\text{ For } a \in X, \text{ we have }

a \ast  0 = a + 0 = a \hspace{1.0cm} [ a \in X \Rightarrow a + 0 < 6 ]

0 \ast  a = 0 + a = a \hspace{1.0cm} [ a \in X \Rightarrow 0 + a < 6 ]

\text{ Therefore } a \ast 0=a=0 \ast a \text{ for all  } a \in X

Thus, 0 is the identity element for the given operation \ast  .

\text{ An element } a \in X \text{ is invertible if there exists } b \in X \text{ such that } a \ast  b = 0 = b \ast  a.

\text{i.e. } \Bigg\{ \begin{array}{lll} a+b = 0 = b+a & , & \text{ if } a+b < 6 \\  a+b-6=0= b+a-6 & , & \text{ if } a+b \geq 6  \end{array}

\Rightarrow a = - b \text{ or } b = 6 - a

\text{ But, } X = \{ 0, 1, 2,3,4,5 \} \text{ and } a, b \in X. \text{ Then } a \neq -b

\text{ Therefore } b = 6 - a \text{ is the inverse of } a \text{ for all } a \in X

\text{ Hence, the inverse of an element } a \in X, a \neq 0 \text{ is } 6-a \text{ i.e., } a^{-1} = 6 - a