Question 13 Marks
Check the commutativity and associativity of the following binary operations:
'*' on R defined by a * b = a + b - 7 for all a, b ∈ R.
AnswerCommutativity: Let $\text{a, b}\in\text{R}.$ Then,a * b = a + b - 7
= b + a - 7
= b * a
⇒ a * b = b * a
⇒ * is commutative on R.
Associativity: Let $\text{a, b, c}\in\text{R}.$ Then,
(a * b) * c = (a + b - 7) * c
= a + b - 7 + c - 7
= a + b + c - 14 .......(i)
and a * (b * c) = a * (b + c - 7)
= a + b + c - 7 - 7
= a + b + c - 14 ......(ii)
From (i) and (ii)
(a * b) * c = a *( b * c)
* is associative on R.
Thus, * is associative on R.
View full question & answer→Question 23 Marks
Construct the composition table for $\times _5$ on $Z_5 = \{0, 1, 2, 3, 4\}.$
AnswerHere,
$1\times _51 =$ Remainder obtained by dividing $1 \times 1$ by $5 = 1$
$3\times _54 =$ Remainder obtained by dividing $3 \times 4$ by $5 = 2$
$4\times _54 =$ Remainder obtained by dividing $4 \times 4$ by $5 = 1$
Therefore,
The composition table is as follows:
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 33 Marks
For the binary operation $\times _7$ on the set $S = \{1, 2, 3, 4, 5, 6\},$ compute $3^{−1} \times _7 4.$
AnswerFinding identity element:
Here,
$1\times _71 =$ Remainder obtained by dividing $1 \times 1$ by $7 = 1$
$3\times _74 =$ Remainder obtained by dividing $3 \times 4$ by $7 = 5$
$4\times _75 =$ Remainder obtained by dividing $4 \times 5$ by $7 = 6$
So, the composition table is as follows:
| $\times _7$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $2$ |
$2$ |
$4$ |
$6$ |
$1$ |
$3$ |
$5$ |
| $3$ |
$3$ |
$6$ |
$2$ |
$5$ |
$1$ |
$4$ |
| $4$ |
$4$ |
$1$ |
$5$ |
$2$ |
$6$ |
$3$ |
| $5$ |
$5$ |
$3$ |
$1$ |
$6$ |
$4$ |
$2$ |
| $6$ |
$6$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
We observe that all the elements of the first row of the composition table are same as the top-most row.
So, the identity element is $1.$
Also, $3\times _75 = 1$
So, $3^{-1} = 5$
Now,
$3^{-1}\times _7 4 = 5\times _7 4 = 6$ View full question & answer→Question 43 Marks
The binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ on the set Q of all rational numbers. Show that * is associative.
AnswerThe binary operator * is defined as,
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ for all $\text{a, b}\in\text{Q}$
Now,
Associativity: Let $\text{a, b, c}\in\text{Q},$ then
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{7}\ ^*\ \text{c}=\frac{\text{abc}}{49}\ ....(\text{i})$
and $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{bc}}{7}=\frac{\text{abc}}{49}\ .....(\text{ii})$
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ '*' is associative on Q.
View full question & answer→Question 53 Marks
For the binary operation multiplication modulo $10 (\times _{10})$ defined on the set $S = \{1, 3, 7, 9\},$ write the inverse of $3.$
Answer$1 \times _{10}1 =$ Remainder obtained by dividing $1 \times 1$ by $10 = 1$
$3 \times _{10}1 =$ Remainder obtained by dividing $3 \times 1$ by $10 = 3$
$7 \times _{10}3 =$ Remainder obtained by dividing $7 \times 3$ by $10 = 1$
$3 \times _{10}3 =$ Remainder obtained by dividing $3 \times 3$ by $10 = 9$
So, the composition table is as follows:
| $\times _{10}$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $1$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$9$ |
$1$ |
$7$ |
| $7$ |
$7$ |
$1$ |
$9$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$3$ |
$1$ |
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at $1.$
$\Rightarrow a * 1 = 1 * a = a,$ $\forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$ View full question & answer→Question 63 Marks
Let * be a binary operation on $Q_0$ (set of non-zero rational numbers) defined by a * $b=\frac{a b}{5}$ for $a l l a, b \in Q_0$. Show that * is commutative as well as associative. Also, find its identity element if it exists.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}_0$
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5}$
$=\frac{\text{ba}}{5}$
$=\text{b}\ ^*\ \text{a}$
Therefore, $\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\forall\ \text{a, b}\in\text{Q}_0$
Thus, * is commutative on $Q_0.$
Associativity: Let $\text{a, b, c}\in\text{Q}_0$
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore,
$a^*\left(b^* c\right)=\left(a^* b\right)^* c, \forall a, b, c \in Q_0$
Thus, * is associative on $Q _0$.
Finding identity element:
Let e be the identity element in Z with respect to * such that, $a^* e = a = e ^* a , \forall a \in Q _0$
$a^* e=a$ and $e^* a=a, \forall a \in Q_0$
Implies that $\frac{ ae }{5}= a$ and $\frac{ ea }{5}= a , \forall a \in Q _0$
Implies that $e =5, \forall a \in Q _0[\because a \neq 0]$
Thus, 5 is the identity element in with respect to *.
View full question & answer→Question 73 Marks
Find the identity element in the set of all rational numbers except -1 with respect to * defined by a * b = a+ b + ab.
AnswerLet R - {-1} be the set and * be a binary operator, given by a * b = a + b + ab for all $\text{a, b}\in\text{R}-\{-1\}$ Now, Let $\text{a}\in\text{R}-\{-1\}$ and $\text{e}\in\text{R}-\{-1\}$ be the identity element with respect to *. by identity property, we have, a * e = e * a = a ⇒ a + e + ae = a ⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as a }\neq-1]$$\therefore$ The required identity element is 0.
View full question & answer→Question 83 Marks
On the set Z of integers, if the binary operation * is defined by a * b = a + b + 2, then find the identity element.
AnswerLet e be the identity element in Z with respect to * such that a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$ a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$ a + e + 2 = a and e + a + 2 = a, $\forall\ \text{a}\in\text{Z}$e = -2, $\forall\ \text{a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
View full question & answer→Question 93 Marks
Let $A=R_0 \times R$, where $R_0$ denote the set of all non-zero real numbers. $A$ binary operation ' $\odot$ ' is defined on $A$ as follows: $(a, b) \odot(c, d)=(a c, b c+d)$ for all $(a, b),(c, d) \in R_0 \times R$.
Find the identity element in A .
AnswerLet$ E = (x, y)$ be the identity element in A with respect to ⊙,$\forall\ \text{x}\in\text{R}_0\ \&\text{ y}\in\text{R}$ such that
$x ⊙ E = X = E ⊙ X,$ $\forall\text{ x}\in\text{A}$
$\Rightarrow X ⊙ E = X$ and $E ⊙ X = X$
$\Rightarrow (ax, bx + y) = (a, b) and (xa, ya + b) = (a, b)$
Considering $(ax, bx + y) = (a, b)$
$\Rightarrow ax = a$
$\Rightarrow x = 1$
$ bx + y = b$
$\Rightarrow y = 0$ $[\because\text{ x}=1]$
Considering $(xa, ya + b) = (a, b)$
$\Rightarrow xa = a$
$\Rightarrow x = 1$
$ ya + b = b$
$\Rightarrow y = 0$ $[\because\text{ x}=1]$
$[\because\ 1,0]$ is the identity element in A with respect to ⊙.
View full question & answer→Question 103 Marks
Check the commutativity and associativity of the following binary operations:
'*' on N defined by $a * b = 2^{ab}$ for all $a, b \in N.$
AnswerCommutative: Let $\text{a, b}\in\text{N},$ Then $a * b = 2^{ab} = 2^{ba} = b * a$
$\therefore$ a * b = b * a $\therefore$ * is commutative on $N.$
Associative: Let $\text{a, b, c}\in\text{N},$ Then$(\text{a}\ ^*\ \text{b}) *\ \text{c}=2^{\text{ab}}\ ^*\ \text{c}=2^{2{\text{ab}}.\text{c}}\ ...(\text{i})$
and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ 2^{\text{bc}}=2^{\text{a}.2^{\text{bc}}}\ ....(\text{ii})$ From (i) and (ii), we get$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
$\therefore$ * is not associative on $N.$
View full question & answer→Question 113 Marks
Write the composition table for the binary operation multiplication modulo $10 (\times _{10})$ on the set $S = \{2, 4, 6, 8\}.$
Answer$2 \times _{10}4 =$ Remainder obtained by dividing $2 \times 4$ by $10 = 8$
$4 \times _{10}6 =$ Remainder obtained by dividing $4 \times 6$ by $10 = 4$
$2 \times _{10}8 =$ Remainder obtained by dividing $2 \times 8$ by $10 = 6$
$3 \times _{10}4 =$ Remainder obtained by dividing $3 \times 4$ by $10 = 2$
Therefore, the composition table is as follows:
| $\times _{10}$ |
$2$ |
$4$ |
$6$ |
$8$ |
| $2$ |
$4$ |
$8$ |
$2$ |
$6$ |
| $4$ |
$8$ |
$6$ |
$4$ |
$2$ |
| $6$ |
$2$ |
$4$ |
$6$ |
$8$ |
| $8$ |
$6$ |
$2$ |
$8$ |
$4$ |
View full question & answer→Question 123 Marks
Check the commutativity and associativity of the following binary operations:
$'⊙'$ on $Q$ defined by $a ⊙ b = a^2 + b^2$ for all $a, b ∈ Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}$
then, $a ⊙ b = a^2 + b^2 = b^2 + a^2 = b ⊙ a$
Therefore, $a ⊙ b = b ⊙ a, \forall\ \text{a, b}\in\text{Q}$
Thus, $⊙$ is commuatative on $Q.$
Associativity: Let $\text{a, b, c}\in\text{Q.}$
$a ⊙ (b ⊙ c) = a ⊙ (b^2 + c^2) = ab2 + (b^2 + c^2)^2 = ab^2 + b^4 + c^4 + 2b^2c^2 $
$(a ⊙ b) ⊙ c = (a^2 + b^2) ⊙ c = (a^2 + b^2)^2 + c^2 = a^4 + b^4 + 2a^2b^2 + c^2$ Therefore,$\text{a}\odot\text{b}\odot\text{c}\neq\text{a}\odot\text{b}\odot\text{c}$
Thus, $⊙$ is not associative on $Q.$
View full question & answer→Question 133 Marks
Construct the composition table for $+_5$ on set $S = \{0, 1, 2, 3, 4\}.$
Answer$a +_5 b =$ the remainder when $a + b$ is divided by $5.$
eg. $2 + 4 = 6 \Rightarrow 2 +_5 4 = 1 \because [$we get $1$ as remainder when $6$ is divided by $5]$
$2 + 4 = 7 \Rightarrow 3 +_5 4 = 2 \because$ [we get 2 as remainder when $7$ is divided by $5]$
The composition table for $+_5 $ on set $S = \{0, 1, 2, 3, 4\}.$
| $+_5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$0$ |
| $2$ |
$2$ |
$3$ |
$4$ |
$0$ |
$1$ |
| $3$ |
$3$ |
$4$ |
$0$ |
$1$ |
$2$ |
| $4$ |
$4$ |
$0$ |
$1$ |
$2$ |
$3$ |
View full question & answer→Question 143 Marks
Write the multiplication table for the set of integers modulo $5.$
Answer$Z_5 = {0, 1, 2, 3, 4}$
$a\times _5 b$ is the remainder when the product of ab is divided by $5.$
The composition table for $\times _5$ on $Z_5 = \{0, 1, 2, 3, 4\}$
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 153 Marks
On Q, the set of all rational numbers, * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ shown that * is no associative.
AnswerThe binary operator * defined as, $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ for all $\text{a, b}\in\text{Q}.$ Now, Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{a}-\text{b}}{2}\ ^*\ \text{c}=\frac{\frac{\text{a}-\text{b}}{2}-\text{c}}{2}$ $=\frac{\text{a}-\text{b}-2\text{c}}{4}\ ....(\text{i})$ and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{b}-\text{c}}{2}=\frac{\text{a}-\frac{\text{b}-\text{c}}{2}}{2}$ $=\frac{2\text{a}-\text{b}+\text{c}}{4}\ .....(\text{ii})$ From (i) and (ii),$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
Hence, '*' is not associative on Q.
View full question & answer→Question 163 Marks
Construct the composition table for $\times _6$ on set $S =\{0, 1, 2, 3, 4, 5\}.$
AnswerHere,
$1\times _61 =$ Remainder obtained by dividing $1 \times 1$ by $6 = 1$
$3\times _64 =$ Remainder obtained by dividing $3 \times 4$ by $6 = 0$
$4\times _65 =$ Remainder obtained by dividing $4 \times 5$ by $6 = 2$
So, the composition table is as follows:
| $\times _6$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$0$ |
$2$ |
$4$ |
| $3$ |
$0$ |
$3$ |
$0$ |
$3$ |
$0$ |
$3$ |
| $4$ |
$0$ |
$4$ |
$2$ |
$0$ |
$4$ |
$2$ |
| $5$ |
$0$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 173 Marks
Let S be the set of all rational numbers of the for $\frac{\text{m}}{\text{n}},$ where $\text{m}\in\text{Z}$ and n = 1, 2, 3. Prove that * on sdefined by a * b = ab is not a binary operation.
Answer$\text{S}=\Big\{\text{a}=\frac{\text{m}}{\text{n}}:\text{m}\in\text{Z},\text{ n}\in\{1, 2, 3\}\Big\}$Let $\text{a}=\frac{1}{3},\ \text{b}=\frac{5}{3}\in\text{S}$
$\text{a}\ ^*\ \text{b}=\text{ab}$
$=\frac{1}{3}\times\frac{5}{3}$
$=\frac{5}{9}\notin\text{S}\ \big[\because\ 9\notin\{1,2,3\}\big] $
Therefore, $\exists\text{ a, b}\in\text{S},$ such that $\text{a}\ ^*\ \text{b}\notin\text{S}$
Thus, * is not a binary operation.
View full question & answer→Question 183 Marks
Find the inverse of $5$ under multiplication modulo $11$ on $Z_{11}.$
Answer$Z_{11} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
Multiplication modulo $11$ is defined as follows:
For $\text{a},\text{b}\in\text{Z}_{11}$,
$a\times _{11}b$ is the remainder when $a \times b$ is divided by $11.$
Here,
$1\times _{11}1 = $ Remainder obtained by dividing $1 \times 1$ by $11 = 1$
$3\times _{11}4 =$ Remainder obtained by dividing $3 \times 4$ by $11 = 1$
$4\times _{11}5 =$ Remainder obtained by dividing $4 \times 5$ by $11 = 9$
| $\times _{11}$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
| $2$ |
$2$ |
$4$ |
$6$ |
$8$ |
$10$ |
$1$ |
$3$ |
$5$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$6$ |
$9$ |
$1$ |
$4$ |
$7$ |
$10$ |
$2$ |
$5$ |
$8$ |
| $4$ |
$4$ |
$8$ |
$1$ |
$5$ |
$9$ |
$2$ |
$6$ |
$10$ |
$3$ |
$7$ |
| $5$ |
$5$ |
$10$ |
$4$ |
$9$ |
$3$ |
$8$ |
$2$ |
$7$ |
$1$ |
$6$ |
| $6$ |
$6$ |
$1$ |
$7$ |
$2$ |
$8$ |
$3$ |
$9$ |
$4$ |
$10$ |
$5$ |
| $7$ |
$7$ |
$3$ |
$10$ |
$6$ |
$2$ |
$9$ |
$5$ |
$1$ |
$8$ |
$4$ |
| $8$ |
$8$ |
$5$ |
$2$ |
$10$ |
$7$ |
$4$ |
$1$ |
$9$ |
$6$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$5$ |
$3$ |
$1$ |
$10$ |
$8$ |
$6$ |
$4$ |
$2$ |
| $10$ |
$10$ |
$9$ |
$8$ |
$7$ |
$6$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
We observe that the first row of the composition table is same as the top-most row.
Therefore,
The identity element is $1.$
Also,
$5\times _{11}9 = 1$
Hence, $5 - 1 = 9$ View full question & answer→Question 193 Marks
Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}.$
Check the commutativity and associativity of '*' on N.
AnswerCommutativity: Let $\text{a, b}\in\text{N}$a * b = 1.c.m. a, b
= 1.c.m. b, a
= b * a
Therefore,
$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a}\ \forall\ \text{a, b}\in\text{N}$
Thus, * is Commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}$
a * b * c = a * 1.c.m. b, c
= 1.c.m. a, b, c
a * b * c = 1.c.m. a, b * c
= 1.c.m. a, b, c
Therefore,
$\text{a}\ ^*\ \text{b}\ ^*\ \text{c}=\text{a}\ ^*\ \text{b}\ ^*\ \text{c}\ \forall\ \text{a, b, c}\in\text{N}$
Thus, * is associative on N.
View full question & answer→Question 203 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by a * b = ab + 1 for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,a * b = ab + 1
= ba + 1
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Q}$
Thus, * is commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
a * (b * c) = a * (bc + 1) = a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 213 Marks
On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.
AnswerLet $\text{a, b, c}\in\text{Z}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Z.
View full question & answer→Question 223 Marks
Let $R_0$ denote the set of all non-zero real numbers and let $A = R_0 \times R_0.$ If '*' is a binary operation on adefined by,
(a, b) * (c, d) = (ac, bd) for all (a, b), (c, d) ∈ A
Show that '*' is both commutative and associative on A.
AnswerCommutativity:Let $\text{a}, \text{b}\ \&\ \text{c}, \text{d}\in\text{A}\forall\text{ a, b, c, d}\in\text{R}_0$. Then,
$(a, b) * (c, d) = (ac, bd)$
$= (ca, db)$
$= (c, d) * (a, b)$
$\therefore$ (a, b) * (c, d) = (c, d) * (a, b)
Thus, * is commutative on A.
Associativity:
Let $(a, b), (c, d) \& (e, f)$ $\in\text{A}\forall\text{ a, b, c, d, e, f}\in\text{R}_0$. Then,
$(a, b) * ((c, d) * (e, f)) = (a, b) * (ce, df)$
$= (ace, bdf)$
$((a, b) * (c, d)) * (e, f) = (ac, bd) * (e, f)$
= (ace, bdf)
$\therefore$ $(a, b) * ((c, d) * (e, f)) = ((a, b) * (c, d) * (e, f))$
Thus, * is associative on A.
View full question & answer→Question 233 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $a * b = (a - b)^2$ for all $a, b \in Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,$a * b = (a - b)^2$
$= (b - a)^2$
$= b * a$
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Q}$
Thus, * is commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (b - c)^2$
$= a * (b^2 + c^2 - 2bc)$
$= (a - b^2 - c^2 + 2bc)^2$
$(a * b) * c = (a - b)^2 * c$
$= (a^2 + b^2 - 2ab) * c$
$= (a^2 + b^2 - 2ab - c)^2$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 243 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Z defined by a * b = a + b - ab for all a, b ∈ Z.
AnswerCommutativity: Let $\text{a, b}\in\text{Z}.$ Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Z}$
Thus, * is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac + abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c - (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Z}$
Thus, * is associative on Z.
View full question & answer→Question 253 Marks
Let $A = R_0 \times R$, where $R_0$ denote the set of all non-zero real numbers. A binary operation $'⊙'$ is defined on A as follows:
$(a, b) ⊙ (c, d) = (ac, bc + d)$ for all $(a, b), (c, d) \in R_0 \times R.$
Show that $ '⊙'$ is commutative and associative on $A.$
AnswerCommutativity:Let $\text{x}=(\text{a},\text{b})$ and $\text{y}=(\text{c},\text{d})\in\text{A},\forall\text{ a},\text{c}\in\text{R}_0\ \&\text{ b},\text{d}\in\text{R}.$ Then,
$X ⊙ Y = (ac, bc + d )$ & $Y ⊙ X = (ca, da + b)$
Therefore, $x ⊙ Y = Y ⊙ X, \forall\ \text{X},\text{Y}\in\text{A}$
Thus, $⊙$ is commutative on $A.$
Associativity:
Let $\text{X}=(\text{a},\text{b}),\text{Y}=(\text{c},\text{d})$ and $\text{Z}=(\text{e},\text{f}),\forall\ \text{a},\text{c},\text{e}\in\text{R}_0\ \&\text{ b},\text{d},\text{f}\in\text{R}$
$x ⊙ Y ⊙ Z = (a, b) ⊙ (ce, de + f)$
$= (ace, bce + de + f)$
$x ⊙ Y ⊙ Z = (ac, bc + d) ⊙ (e, f)$
$= (ace, (bc + d)e + f)$
$= (ace, bce + de + f)$
$\therefore x ⊙ Y ⊙ Z = x ⊙ Y ⊙ Z, \forall\ \text{X},\text{Y},\text{Z}\in\text{A}$
Thus, $⊙$ is accosiative on $A.$
View full question & answer→Question 263 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Show that '*' is both commutative and associative.
AnswerCommutativity: Let $\text{a, b}\in\text{Z}.$ Then,
a * b = a + b - 4
= b + a - 4
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Z}$
Thus, * is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b + c - 4)
= a + b + c - 4 - 4
= a + b + c - 8
(a * b) * c = (a + b - 4) * c
= a + b - 4 + c - 4
= a + b + c - 8
Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Z}.$
Thus, * is associative on Z.
View full question & answer→Question 273 Marks
A binary operation * is defined on the set R of all real numbers by the rule $\text{a}\times\text{b}=\sqrt{\text{a}^2+\text{b}^2}\ \forall\text{ a, b}\in\text{R}$.
Write the identity element for * on R.
AnswerLet e be the identity element in R with respect to * such thata * e = a = e * a, $\forall\text{ a}\in\text{R}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{R}$
Then,
$\sqrt{\text{a}^2+\text{e}^2}=\text{a}$ and $\sqrt{\text{e}^2+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R}$
Implies that $\sqrt{\text{a}^2+\text{e}}=\text{a}$ and $\sqrt{\text{e}+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R}$ [$\because$ $e^2 = e$]
Implies that $a^2 + e = a^2 and e + a^2 = a^2$, $\forall\text{ a}\in\text{R}$
Implies that $\text{e}=0\in\text{R},\forall\text{ a}\in\text{R}$
Thus, 0 is the identity element in R with respect to *.
View full question & answer→Question 283 Marks
Let $A = R_0 \times R,$ where $R_0$ denote the set of all non-zero real numbers. A binary operation $'⊙'$ is defined on A as follows:
$(a, b) ⊙ (c, d) = (ac, bc + d)$ for all $(a, b), (c, d) \in R_0 \times R.$
Find the invertible elements in A.
AnswerLet $F = (m, n)$ be the inverse in $\text{A }\forall\text{ m}\in\text{R}_0\ \&\text{ n}\in\text{R} x ⊙ F = X = E$ and $F ⊙ X = E$
Implies that $(am, bm + n) = (1, 0)$ and $(ma, na + b) = (1,0)$
Considering $(am, bm + n) = (1, 0)$
Implies that $am = 1$
Implies that $\text{m}=\frac{1}{\text{a}}$
& $bm + n = 0$
Implies that $\text{n}=\frac{-\text{b}}{\text{a}}$ $\Big[\because\ \text{m}=\frac{1}{\text{a}}\Big]$
Considering $(ma, na + b) = (1,0)$
Implies that $ma = 1$
Implies that $\text{m}=\frac{1}{\text{a}}$
& $na + b = 0$
Implies that $\text{n}=\frac{-\text{b}}{\text{a}}$
$\therefore$ The inverse of $(\text{a},\text{b})\in\text{A}$ with respect to $⊙$ is $\Big(\frac{1}{\text{a}},\frac{-\text{b}}{\text{a}}\Big).$
View full question & answer→Question 293 Marks
On the set Z of all integers a binary operation * is defined by a * b = a + b + 2 for all a, b ∈ Z. Write the inverse of 4.
AnswerTo find the identity element, let e be the identity element in Z with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{Z}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{Z}$
Then,
a + e + 2 = a and e + a + 2 = a, $\forall\text{ a}\in\text{Z}$
$\text{e}=-2\in\text{Z},\ \forall\text{ a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
Now,
Let b ∈ Z be the inverse of 4.
Here,
4 * b = e = b * 4
4 * b = e and b * 4 = e
Then,
4 + b + 2 = -2 and b + 4 + 2 = -2
$\text{b}=-8\in\text{Z}$
Thus, -8 is the inverse of 4.
View full question & answer→Question 303 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that '*' is both commutative and associative on Q - {-1}.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Commutativity: Let a, b ∈ Q - {-1}
⇒ a * b = a + b + ab = b + a + ba = b * a
⇒ a * b = b * a
⇒ '*' is commutative on Q - {-1}
Associativity: Let a, b, c ∈ Q - {-1}, then
⇒ (a * b) * c = (a + b + ab) * c
= a + b + ab + c + ac + bc + abc .......(i)
and, a * (b * c) = a * (b + c + bc)
= a + b + c + bc + ab + ac + abc .....(ii)
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ * is associative on Q - {-1}
View full question & answer→Question 313 Marks
Construct the composition table for $\times _4$ on set $S = \{0, 1, 2, 3\}.$
AnswerHere,
$1\times _41 =$ Remainder obtained by dividing $1 \times 1$ by $4 = 1$
$0\times _41 =$ Remainder obtained by dividing $0 \times 1$ by $4 = 0$
$2\times _43 =$ Remainder obtained by dividing $2 \times 3$ by $4 = 2$
$3\times _43 =$ Remainder obtained by dividing $3 \times 3$ by $4 = 1$
Therefore,
The composition table is as follows:
| $x_4$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $2$ |
$0$ |
$2$ |
$0$ |
$2$ |
| $3$ |
$0$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 323 Marks
Check the commutativity and associativity of the following binary operations:
'*' on $N$, de fined by $a * b = a^b$ for all $a, b \in N$.
AnswerCommutativity: Let $\text{a, b}\in\text{N}.$ Then,$\text{a}\ ^*\ \text{b}=\text{a}^{\text{b}}\neq\text{b}^{\text{a}}=\text{b}\ ^*\ \text{a}$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
⇒ '*' is not commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}.$ Then,
$(a * b) * c = a^b * c$
$= (a^b)^c = a^{bc} .....(i)$
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \text{b}^{\text{c}}=(\text{a})^{\text{b}^{\text{c}}}\ ....(\text{ii})$
From (i) and (ii)
$\text{a}^{\text{bc}}\neq(\text{a})^{\text{b}^{\text{c}}}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
⇒ '*' is not associative on N.
View full question & answer→Question 333 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Z defined by a * b = a - b for all a, b ∈ Z.
AnswerCommutativity: Let $\text{a, b}\in\text{Z}.$ Then,a * b = a - b
b * a = b - a
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus '*' is not commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b - c)
= a - (b - c)
= a - b + c
(a * b) * c = (a - b) - c
= a - b - c
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, '*' is not associative on Z.
View full question & answer→Question 343 Marks
On the set Q of all ration numbers if a binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5},$ prove that * is associative on Q.
AnswerLet $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a} \ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Q}$
Thus, * is associative on Q.
View full question & answer→Question 353 Marks
For the binary operation $\times _{10}$ on set $S = \{1, 3, 7, 9\},$ find the inverse of $3.$
Answer$a\times _{10} b =$ the remainder when the product of ab is divided by $10.$ The composition table for $\times _{10}$ on set $S = \{1, 3, 7, 9\}$
| $\times _{10}$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $1$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$9$ |
$1$ |
$7$ |
| $7$ |
$7$ |
$1$ |
$9$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$3$ |
$1$ |
We know that an element $\text{b}\in\text{S}$ will be the inverse of $\text{a}\in\text{S}$
if $a\times _{10} b = 1 [\because 1$ is the identity element with respect to multiplication$.]$
$\Rightarrow 3\times _{10} b = 1$
From the above table $b = 7$
$\therefore$ Inverse of $3$ is $7.$ View full question & answer→Question 363 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let b be the inverse of a ∈ Q - {-1}
Then, a * b = b * a = e [e is the identity element]
⇒ a + b + ab = e
⇒ a + b + ab = 0
⇒ b(1 + a) = -a
$\Rightarrow\text{b}=\frac{-\text{a}}{1+\text{a}}$ $\begin{bmatrix}\because\ \frac{-\text{a}}{1+\text{a}}\neq-1\text{ because if }\frac{-\text{a}}{1+\text{a}}=-1\\\Rightarrow\text{a}=1+\text{a}\Rightarrow1=0\text{ Not possible}\end{bmatrix}$
$\text{b}=\frac{-\text{a}}{1+\text{a}}$ is the inversre of a with respect to *.
View full question & answer→Question 373 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $a * b = ab^2$ for all $a, b \in Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,$a * b = ab^2$
$b * a = ba^2$
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (bc^2)$
$= a(bc^2)^2$
$= ab^2c^4$
$(a * b) * c = (ab^2) * c$
$= ab^2c^2$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 383 Marks
Let $+_6 ($addition modulo $6) $ be a binary operation on $S = \{0, 1, 2, 3, 4, 5\}$. Write the value of $2 + _64^{−1}+ _63^{−1}.$
AnswerThe composition table for $+_6$ on the set $S = \{0, 1, 2, 3, 4, 5\}$ is
| $+_6$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $0$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$0$ |
| $2$ |
$2$ |
$3$ |
$4$ |
$5$ |
$0$ |
$1$ |
| $3$ |
$3$ |
$4$ |
$5$ |
$0$ |
$1$ |
$2$ |
| $4$ |
$4$ |
$5$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $5$ |
$5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$'0' $ is the identity element for $+_6$ from the table it is clear that
$4^{-1} = 2$ and $3^{-1} = 3$
Now, $2 + _64^{-1} + _63^{-1} = 2 + _62 + _63$
$= 4 + _63$
$= 1$ View full question & answer→Question 393 Marks
Let A be any set containing more than one element. Let '*' be a binary operation on A defined by a * b = b for all a, b ∈ A. Is '*' commutative or associative on A?
AnswerCommutativity: Let $\text{a, b}\in\text{A.}$ Then, $\text{a}\ ^*\ \text{b}=\text{b}$ $\text{b}\ ^*\ \text{a}=\text{a}$ Therefore, $\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$ Thus, * is not commutative on A. Associativity: Let $\text{a, b, c}\in\text{A.}$ Then, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \text{c}$ $=\text{c}$ $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\text{b}\ ^*\ \text{c}$ $=\text{c}$ Therefore,$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{A}$
Thus, * is associative on A.
View full question & answer→Question 403 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by a * b = a + ab for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,a * b = a + ab
b * a = b + ba
= b + ab
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
a * (b * c) = a * (b + bc)
= a + a(b + bc)
= a + ab + abc
(a * b) * c = (a + ab) * c
= (a + ab) + (a + ab)c
= a + ab + ac + abc
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 413 Marks
Determine which of the following binary operations are associative and which are commutative:
* on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}$ for all $\text{a, b}\in\text{Q}$
Answer$\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}=\frac{\text{b}+\text{a}}{2}=\text{b}\ ^*\ \text{a,}$
Which shows * is commutative.
Further, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{a}+\text{b}}{2}\big)+\text{c}}{2}=\frac{\text{a}+\text{b}+2\text{c}}{4}$
Further, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{b}+\text{c}}{2}\Big)$
$=\frac{\text{a}+\big(\frac{\text{b}+\text{c}}{2}\big)}{2}=\frac{2\text{a}+\text{b}+\text{c}}{2}\neq\frac{\text{a}+\text{b}+2\text{c}}{4}$
Hence, * is not associative.
View full question & answer→Question 423 Marks
Check the commutativity and associativity of the following binary operations:
'o' on Q defined by $\text{a o b}=\frac{\text{ab}}{2}$ for all a, b ∈ Q.
AnswerBinary operation 'o' defined on Q, given by $\text{a o b}=\frac{\text{ab}}{2}$ for all a, b ∈ Q.
Commutative: Let $\text{a, b}\in\text{Q},$ Then
$\text{a o b}=\frac{\text{ab}}{2}=\frac{\text{ba}}{2}=\text{b o a}$
$\Rightarrow\ \text{a o b}=\text{b o a}$
$\therefore$ o is commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q},$ Then,
$(\text{a o b})\text{ o c}=\Big(\frac{\text{ab}}{2}\Big)\text{ o c}=\frac{\text{abc}}{4}\ .....(\text{i})$
$\text{a o }(\text{b o c})=\text{a o }\Big(\frac{\text{bc}}{2}\Big)=\frac{\text{abc}}{4}\ ....(\text{ii})$
From (i) and (ii) we get
(a o b) o c = a o (b o c)
$\therefore$ 'o' is associative on Q.
View full question & answer→