MCQ 11 Mark
On the power set $P$ of a non$-$empty set $A,$ we define an operation $\triangle \text{ by }\text{X}\triangle\text{Y}=(\text{X}\cap\text{Y})∪(\text{X}∩\text{Y})\text{X}\triangle\text{Y}=\text{X}∩\text{Y}∪\text{X}∩\text{Y}$
Then which are of the following statements is true about $\triangle$
- A
Commutative and associative without an identity.
- B
Commutative but not associative with an identity.
- C
Associative but not commutative without an identity.
- ✓
Associative and commutative with an identity.
AnswerCorrect option: D. Associative and commutative with an identity.
Commutativity:
$\text{X}\triangle\text{Y}=(\overline{\text{X}}\cap\text{Y})\cup(\text{X}\cap\overline{\text{Y}})$
$=(\overline{\text{Y}}\cap\text{X})\cup(\text{Y}\cap\overline{\text{X}})$
$=\text{Y}\triangle\text{X}$
Thus,
$\text{X}\triangle\text{Y}=\text{Y}\triangle\text{X}$
Hence, $\triangle$ is commutative on $A.$
Let $\phi$ be the identity element for $\triangle$ on $P.$
$\text{A}\triangle\phi=\big(\overline{\text{A}}\cap\phi\big)\cup\big(\text{A}\cap\overline{\phi}\big)$
$=\phi\cup\text{A}$
$=\text{A}$
and,
$\phi\triangle\text{A}=\big(\overline{\phi}\cap\text{A}\big)\cup\big(\phi\cap\overline{\text{A}}\big)$
$=\text{A}\cup\phi$
$=\text{A}$
View full question & answer→MCQ 21 Mark
Let $*$ be a binary operation defined on $Q^+$ by the rule $\text{a}*\text{b}=\frac{\text{ab}}3\forall\text{ a, b}\in \text{Q}^+$. The inverse of $4 * 6$ is:
- ✓
$\frac{9}{8}$
- B
$\frac{2}3$
- C
$\frac{3}2$
- D
AnswerCorrect option: A. $\frac{9}{8}$
Let e be the identity element in $Q^+$ with respect to $*$ such that
$a * e = a = e * a$, $\forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a$, $\forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 3$, $\forall\text{ a}\in\text{Q}^+$
Thus, $3$ is the identity element in $Q^+$ with respect to $*$.
Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of a. Then,
$a * b = e = b * a$
$a * b = e$ and $b * a = e$
$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$
$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$
Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.
Given: $\text{a}*\text{b}=\frac{\text{ab}}3$
$4*6=\frac{4\times6}3=8$
Now,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4*6)^{-1}=8^{-1}$
$=\frac{9}8$
View full question & answer→MCQ 31 Mark
Let $*$ be a binary operation defined on set $Q − \{1\}$ by the rule $a * b = a + b − ab.$ Then, the identify element for $*$ is:
AnswerLet $e$ be the identity element in $Q - \{1\}$ with respect to $*$ such that
$a * e = a = e * a, \forall\text{ a}\in\text{Q}-\{-1\}$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}-\{-1\}$
$a + e - ae = a$ and $e + a - ea = a, \forall\text{ a}\in\text{Q}-\{-1\}$
$e(1 - a) = 0, \forall\text{ a}\in\text{Q}-\{-1\}$ $[\because \text{a}\neq1]$
Thus, $0$ is the identity element in $Q - \{1\}$ with respect to $*.$
View full question & answer→MCQ 41 Mark
The number of commutative binary operation that can be defined on a set of $2$ elements is:
AnswerThe number of commutative binary operations on a set of $n$ elements is $\text{n}\frac{\text{n}(\text{n}-1)}{2}$.
Therefore,
Number of commutative binary operations an a set of $2$ elements $=2\frac{2(2-1)}{2}=2^1$
$=2$
View full question & answer→MCQ 51 Mark
On the set $Q^+$ of all positive rational numbers a binary operation $*$ is defined by $\text{a}*\text{b}=\frac{\text{ab}}2\forall\text{ a, b}\in \text{Q}^+$. The inverse of $8$ is:
- A
$\frac{1}{8}$
- ✓
$\frac{1}2$
- C
$2$
- D
$4$
AnswerCorrect option: B. $\frac{1}2$
Let e be the identity element in $Q^+$ with respect to $*$ such that
$a * e = a = e * a$, $\forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a$, $\forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{2}=\text{a}\text{ and }\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 2$, $\forall\text{ a}\in\text{Q}^+$
Thus, 2 is the identity element in $Q^+$ with respect to $*$.
Let $\text{b}\in\text{Q}^+$ be the inverse of $8$. Then,
$8 * b = e = b * 8$
$8 * b = e$ and $b * 8 = e$
$\frac{(8)\text{b}}2=2\text{ and }\frac{\text{b}(8)}2=2$ $[\because\ \text{e}=2]$
$b = 12$
Thus, $\frac{1}2$ is the inverse of $8$.
View full question & answer→MCQ 61 Mark
For the multiplication of matrices as a binary operation on the set of all matrices of the form $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix},\text{a, b}\in\text{R}$ the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is:
- A
$\begin{bmatrix}-2&3\\-3&-2\end{bmatrix}$
- B
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}$
- ✓
$\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
- D
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
Let the identity of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ be $\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}$, then
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}=\begin{bmatrix}2&3\\-3&3\end{bmatrix}$
$\Rightarrow 2e - 3f = 2 \rightarrow (1)$
$2f + 3e = 3 \rightarrow (2)$
Solving $(1)$ and $(2)$ we get $e = 1$ and $f = 0$
So, the identity is $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.
Let the inverse be $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}$, then
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow 2a - 3b = 1 \rightarrow (1)$
$2b + 3a = 0 \rightarrow (2)$
Solving $(1)$ and $(2),$ we get $\text{a}=\frac{2}{13}$ and $\text{b}=\frac{-3}{13}$
So, the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
View full question & answer→MCQ 71 Mark
If $G$ is the set of all matrices of the form $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$, where $\text{x}\in\text{R}-\{0\}$, then the identity element with respect to the multiplication of matrices as binary operation, is:
- A
$\begin{bmatrix}1&1\\1&1\end{bmatrix}$
- B
$\begin{bmatrix}-\frac{1}2&-\frac{1}2\\-\frac{1}2&-\frac{1}2\end{bmatrix}$
- ✓
$\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
- D
$\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
Let $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\in\text{G}$ and $\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}\in\text{G}$ such that
$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$\begin{bmatrix}2\text{ex}&2\text{ex}\\2\text{ex}&2\text{ex}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$2\text{ex}=\text{x}$
$\text{e}=\frac{1}2\in\text{R}-\{0\}$
Thus, $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}\in\text{G}$, is the identity element in $G.$
View full question & answer→MCQ 81 Mark
For the binary operation $*$ defined on $R − \{1\}$ by the rule $a * b = a + b + ab$ for all $a, b \in R − \{1\}$, the inverse of $a$ is:
AnswerCorrect option: B. $-\frac{\text{a}}{\text{a}-1}$
Let e be the identity element in $R - \{1\}$ with respect to $*$ such that
$a * e = a = e * a, \forall\text{ a}\in\text{R}-\{1\}$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{R}-\{1\}$
Then,
$a + e + ae = a$ and $e + a + ea = a, \forall\text{ a}\in\text{R}-\{1\}$
$e(1 + a) = 0, \forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\text{R}-\{1\}$
Thus, 0 is the identity element in $R - \{1\}$ with respect to $*.$
Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of $a.$ Then,
$a * b = e = b * a$
$a * b = e$ and $b * a = e$
$\Rightarrow a + b + ab = 0$ and $b + a + ba = 0$
$\Rightarrow\text{b}(1+\text{a})=-\text{a}\in\text{R}-\{1\}$
$\Rightarrow\text{b}=\frac{-\text{a}}{\text{a}-1}\in\text{R}-\{1\}$
Thus, $\frac{-\text{a}}{\text{a}-1}$ is the inverse of $\text{a}\in\text{R}-\{1\}$.
View full question & answer→MCQ 91 Mark
Let $*$ be a binary operation on $R$ defined by $a * b = ab + 1.$ Then, $*$ is:
- ✓
Commutative but not associative.
- B
Associative but not commutative.
- C
Neither commutative nor associative.
- D
Both commutative and associative.
AnswerCorrect option: A. Commutative but not associative.
Commutativity:
Let $\text{a, b}\in\text{R}$
$a * b = ab + 1$
$= ba + 1$
$= b * a$
Therefore,
$a * b = b * a, \forall\text{ a, b}\in\text{R}$
Therefore, $*$ is commutative on $R.$
Associativity:
Let $\text{ a, b, c}\in\text{R}$
$a * (b * c) = a * (bc + 1)$
$= a(bc + 1) + 1$
$= abc + a + 1$
$(a * b) * c = (ab + 1) * c$
$= (ab + 1)c + 1$
$= abc + c + 1$
$\therefore a * (b * c) \neq (a * b) * c$
For example: $a = 1, b = 2$ and $c = 3 [$which belong to $R]$
Now,
$1 * (2 * 3) = 1 * (6 + 1)$
$= 1 * 7$
$= 7 + 1$
$= 8$
$(1 * 2) * 3 = (2 + 1) * 3$
$= 3 * 3$
$= 9 + 1$
$= 10$
$\Rightarrow 1 * (2 * 3) \neq (1 * 2) * 3$
Therefore, $\exists a = 1, b = 2$ and $c = 3$ which belong to $R$ such that
$a * (b * c) \neq (a * b) * c$
Hence, $*$ is not associative on $R.$
View full question & answer→MCQ 101 Mark
If the binary operation $^*$ on $Z$ is defined by $a ^* b = a^2 − b^2 + ab + 4$, then value of $(2 ^* 3) ^* 4$ is:
AnswerGiven that $a ^* b = a^2 - b^2 + ab + 4$
So,
$2 ^* 3$
$= 2^2 - 3^2 + 2.3 + 4$
$= 4 - 9 + 6 + 4$
$= 5$
Now,
$(2 ^* 3) ^* 4$
$= 5 ^* 4$
$= 5^2 - 4^2 + 5.4 + 4$
$= 25- 16 + 20 + 4$
$= 33$
View full question & answer→MCQ 111 Mark
$Q^+$ is the set of all positive rational numbers with the binary operation $*$ defined by $\text{a}*\text{b}=\frac{\text{ab}}2\ \forall\text{ a, b}\in\text{Q}^+$. The inverse of an element $\text{a}\in\text{Q}^+$ is:
- A
$\text{a}$
- B
$\frac{1}{\text{a}}$
- C
$\frac{2}{\text{a}}$
- ✓
$\frac{4}{\text{a}}$
AnswerCorrect option: D. $\frac{4}{\text{a}}$
Let e be the identity element in $Q^+$ with respect to $*$ such that
$a * e = a = e * a$, $\forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a$, $\forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$, $\forall\text{ a}\in\text{Q}^+$
$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$
Thus, $2$ is the identity element in $Q^+$ with respect to $*$.
Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of a.
Then,
$a * e = a = e * a$
$a * b = e$ and $b * a = e$
$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$
$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$
Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.
View full question & answer→MCQ 121 Mark
An operation $*$ is defined on the set $Z$ of non$-$zero integers by $a * b = ab$ for all $a, b \in Z.$ Then the property satisfied is:
Answer$*$ is not clouser because when $a = 1$ and $b = 2,$
$\text{a}*\text{b}=\frac{\text{a}}{\text{b}}=\frac{1}{2}\in\text{Z}$
$*$ is not commutative because when $a = 1, b = 2$ and $c = 3,$
$1*(2*3)=1*\Big(\frac{2}3\Big)$
$=\frac{1}{\big(\frac{2}{3}\big)}$
$=\frac{3}2$
$(1*2)*3=\frac{1}2*3$
$=\frac{\big(\frac{1}2\big)}{3}$
$=\frac{1}6$
Thus,
$1*(2*3)\neq(1*2)*3$
View full question & answer→MCQ 131 Mark
The law $a + b = b + a$ is called:
AnswerThe law $a + b = b + a$ is commutative.
View full question & answer→MCQ 141 Mark
A binary operation $*$ on $Z$ defined by $a * b = 3a + b$ for all $a, b \in Z,$ is:
- A
- B
- ✓
- D
Commutative and associative.
AnswerLet $\text{a, b}\in\text{Z}$
$a * b = 3a + b$
$b * a = 3b + a$
Thus, $a * b \neq b * a$
If $a = 1$ and $b = 2,$
$1 * 2 = 3(1) + 2$
$= 5$
$2 * 1 = 3(2) + 1$
$= 7$
$1 * 2 \neq 2 * 1$
Thus, $*$ is not commutative on $Z.$
View full question & answer→MCQ 151 Mark
If the binary operation $\odot$ is defined on the set $Q^+$ of all positive rational numbers by $\text{a}\odot\text{b}=\frac{\text{ab}}4$. Then, $3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$ is equal to:
- ✓
$\frac{3}{160}$
- B
$\frac{5}{160}$
- C
$\frac{3}{10}$
- D
$\frac{3}{40}$
AnswerCorrect option: A. $\frac{3}{160}$
Given $\text{a}\odot\text{b}=\frac{\text{ab}}4$
$\Rightarrow\Big(\frac{1}5\odot\frac{1}2\Big)$
$=\frac{\frac{1}5.\frac{1}2}{4}$
$=\frac{1}{40}$
$3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$
$=3\odot\frac{1}{40}$
$=\frac{\frac{1}{40}.3}{4}$
$=\frac{3}{160}$
View full question & answer→MCQ 161 Mark
The number of binary operation that can be defined on a set of $2$ elements is:
AnswerTotal number of binary operations on a set containing $n$ elements is $\text{(n)}^{\text{n}^2}$
so for $n = 2$ we have $(2)^{2^2}=2^4=16$
View full question & answer→MCQ 171 Mark
Subtraction of integers is:
- ✓
Commutative but no associative.
- B
Commutative and associative.
- C
Associative but not commutative.
- D
Neither commutative nor associative.
AnswerCorrect option: A. Commutative but no associative.
Let $\text{a, b}\in\text{Z}$, then
$a * b = a - b$
$b * a = b - a$
$\Rightarrow a * b \neq b * a$
Substraction is not commutative.
$(a * b) * c$
$= (a - b) * c$
$= a - b - c$
$a * (b * c)$
$= a * (b - c)$
$= a - b + c$
$\Rightarrow (a * b) * c \neq a * (b * c)$
Substraction is not associative.
View full question & answer→MCQ 181 Mark
Let $*$ be a binary operation on $N$ defined by $a * b = a + b + 10$ for all $a, b \in N.$ The identity element for $*$ in $N$ is:
- A
$−10$
- B
$0$
- C
$10$
- ✓
Non$-$existent.
AnswerCorrect option: D. Non$-$existent.
Given $a * b = a + b + 10$
Let the identity element be $e$, then
$a * e = a$
$\Rightarrow a + e + 10 = a$
$\Rightarrow e = -10$
But the operation is defined on the set of natural numbers.
So, the identity element doesn't exist.
View full question & answer→MCQ 191 Mark
The binary operation $^*$ is defined by $a ^* b = a^2 + b^2 + ab + 1$, then $(2 ^* 3) ^* 2$ is equal to:
AnswerGiven: $a ^* b = a^2 + b^2 + ab + 1$
$2 ^* 3 = 2^2 + 3^2 + 2 \times 3 + 1$
$= 4 + 9 + 6 + 1$
$= 20$
$(2 ^* 3) ^* 2 = 20 ^* 2$
$= 20^2 + 2^2 + 20 \times 2 + 1$
$= 400 + 4 + 40 + 1$
$= 445$
View full question & answer→MCQ 201 Mark
Let $^*$ be a binary operation on $Q^+$ defined by $\text{a}^*\text{b}=\frac{\text{ab}}{100}\forall\text{ a, b}\in\text{Q}^+$. The inverse of $0.1$ is:
AnswerCorrect option: A. $10^5$
Let $e$ be the identity element in $Q^+$ with respect to $^*$ such that
$a ^* e = a = e ^* a$, $\forall\text{ a}\in\text{Q}^+$
$a ^* e = a$ and $e ^* a = a$, $\forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}{100}=\text{a}\text{ and }\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$
$\text{e}=100,\forall\text{ a}\in\text{Q}^+$
Thus, 100 is the identity element in $Q^+$ with repect to $^*$.
$0.1 ^* b = e = b ^* 0.1$
$0.1 ^* b = e$ and $b ^* 0.1 = e$
$\frac{(0.1)\text{b}}{100}=100\text{ and }\frac{\text{b}(0.1)}{100}=100$
$\text{b}=\frac{100\times100}{0.1}$
$=10^5\in\text{Q}^+$
Thus, $10^5$ is the inverse of $0.1$.
View full question & answer→MCQ 211 Mark
If $a * b$ denote the bigger among $a$ and $b$ and if $ab = (a * b) + 3,$ then $4.7 =$
Answer$4.7 = (4 * 7) + 3$
$= 7 + 3$
$= 10$
View full question & answer→MCQ 221 Mark
$Q^+$ denote the set of all positive rational numbers. If the binary operation $\text{a }\odot$ on $Q^+$ is defined as:
$\text{a }\odot=\frac{\text{ab}}{2}$, then the inverse of $3$ is:
- ✓
$\frac{4}{3}$
- B
$2$
- C
$\frac{1}3$
- D
$\frac{2}3$
AnswerCorrect option: A. $\frac{4}{3}$
Let us first find the identity element.
We know that if $e$ is the identity element then,
$\text{a}\odot\text{e}=\text{e}$
Given $\text{a}\odot\text{e}=\frac{\text{ae}}2$
$\Rightarrow\text{a}=\frac{\text{ae}}2$
$\Rightarrow\text{e}=2$
Let $b$ be the inverse of $3,$ then
$3\odot\text{b}=\text{e}$
$\Rightarrow\frac{3\text{b}}2=2$
$\Rightarrow\text{b}=\frac{4}3$
View full question & answer→MCQ 231 Mark
Which of the following is true?
- A
$*$ defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on $Z.$
- ✓
$*$ defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on $Q.$
- C
All binary commutative operations are associative.
- D
Subtraction is a binary operation on $N.$
AnswerCorrect option: B. $*$ defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on $Q.$
For option $a$, if we take $3$ and $2$ then
$3*2=\frac{5}2\in\text{Z}$.
So, option $a$ is not true.
For option $b$, if we take any two numbers $a$ and $b$
then $\frac{\text{a + b}}2$ belongs to $Q$ for $\text{a, b}\in\text{Q}$.
So, option $b$ is correct.
For option $d$, if we take $2, 3$ then $2-3=-1\in\text{N}$.
So, option $d$ is not true.
Option $c$ is not true.
View full question & answer→MCQ 241 Mark
The binary operation $*$ defined on $N$ by $a * b = a + b + ab$ for all $a, b \in N$ is:
- A
- B
- ✓
Commutative and associative both.
- D
AnswerCorrect option: C. Commutative and associative both.
$a * b = a + b + ab$
$b * a = b + a + ba$
$\Rightarrow a * b = b * a$
So $*$ is commutative.
Now,
$(a * b) * c$
$= (a + b + ab) * c$
$= a + b + ab + c + ca + cb + abc$
$a * (b * c)$
$= a * (b + c + bc)$
$= a + b + c + bc + ab + ac + abc$
$\Rightarrow (a * b) * c = a * (b * c)$
So $*$ is associative.
View full question & answer→MCQ 251 Mark
If $a ^* b = a^2 + b^2$, then the value of $(4 ^* 5) ^* 3$ is:
- A
($4^2 + 5^2) + 3^2$
- B
$(4 + 5)^2 + 3^2$
- ✓
$41^2 + 3^2$
- D
$(4 + 5 + 3)^2$
AnswerCorrect option: C. $41^2 + 3^2$
Given $a ^* b = a^2 + b^2$
So, $4 ^* 5 = 4^2 + 5^2$
Now,
$(4 ^* 5) ^* 3 = (4 ^* 5)^2 + 3^2$
$= (4^2 + 5^2)^2 + 3^2$
$= 41^2 + 3^2$
View full question & answer→MCQ 261 Mark
On $Z$ an operation $*$ is defined by $a * b = a^2 + b^2$ for all $a, b \in Z$. The operation $*$ on $Z$ is:
- A
Commutative and associative.
- B
Associative but not commutative.
- ✓
- D
Answer$a * b = a^2 + b^2$
$b * a = b^2 + a^2$
$\Rightarrow a * b = b * a$
So $*$ is commutative.
Now
$(a * b) * c$
$= (a^2 + b^2) * c$
$= (a^2 + b^2)^2 + c^2$
$a * (b * c)$
$= a * (b^2 + c^2)$
$= a^2 + (b^2 +c^2)^2$
$\Rightarrow (a * b) * c$ $\neq$ $a * (b * c)$
So $*$ is not associative.
View full question & answer→MCQ 271 Mark
If a binary operation $*$ is defined on the set $Z$ of integers as $a * b = 3a − b,$ then the value of $(2 * 3) * 4$ is:
AnswerGiven: $a * b = 3a - b$
$2 * 3 = 3 (2) - 3$
$= 6 - 3$
$= 3$
$(2 * 3) * 4 = 3 * 4$
$= 3(3) - 4$
$= 9 - 4$
$= 5$
View full question & answer→MCQ 281 Mark
Mark the correct alternative in the following question for the binary operation $*$ on $Z$ defined by $a * b = a + b + 1,$ the identity element is:
AnswerWe have,
$a * b = a + b + 1$
Let e be the identity element of $*$.
Then,
$a * e = a = e * a$
$a + e + 1 = a$
$e = a - a - 1$
$e = -1$
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Consider the binary operation $*$ defined on $Q − \{1\}$ by the rule $a * b = a + b − ab$ for all $a, b \in Q − \{1\}.$ The identity element in $Q − \{1\}$ is:
- ✓
$0$
- B
$1$
- C
$\frac{1}2$
- D
$-1$
AnswerLet $e$ be the identity element in $Q - {1}$ with respect to $*$ such that
$a * e = a = e * a, \forall\text{ a}\in\text{Q}-\{-1\}$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}-\{-1\}$
Then,
$a + e - ae = a$ and $e + a - ea = a, \forall\text{ a}\in\text{Q}-\{-1\}$
$e(1 - a) = 0, \forall\text{ a}\in\text{Q}-\{-1\}$
$\text{e}=0\in\text{Q}-\{-1\}$
$[\because\text{ a}\neq1]$
Thus, $0$ is the identity element in $Q - \{1\}$ with respect to *.
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