Class 12: Binary Operations – Exercise 3.5 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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:

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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]