Question 12 Marks
Define a binary operation on a set.
AnswerLet A be a non-empty set. An operation * is called a binary operation on A, if and only if $\text{a}\times\text{b}\in\text{A},\forall\text{a},\text{b}\in\text{A}$
View full question & answer→Question 22 Marks
The binary operation *: R × R → R is defined as a * b = 2a + b. Find (2 * 3) * 4.
AnswerIt is given that, a * b = 2a + b
Now,
(2 * 3) = 2 × 2 + 3
= 4 + 3
= 7
(2 * 3) * 4 = 7 * 4 = 2 × 7 + 4
= 14 + 4
= 18
View full question & answer→Question 32 Marks
Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.
Answer
|
LCM
|
1
|
2
|
3
|
4
|
5
|
|
1
|
1
|
2
|
3
|
4
|
5
|
|
2
|
2
|
2
|
6
|
4
|
10
|
|
3
|
3
|
5
|
3
|
12
|
15
|
|
4
|
4
|
4
|
12
|
4
|
20
|
|
5
|
5
|
10
|
15
|
20
|
5
|
In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.
If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 $\notin$ {1, 2, 3, 4, 5}.
Thus, * is not a binary operation on {1, 2, 3, 4, 5}. View full question & answer→Question 42 Marks
Write the composition table for the binary operation$ \times _5$ $($multiplication modulo $5) $ on the set $S = \{0, 1, 2, 3, 4\}.$
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 $5 = 1$
So, 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 52 Marks
Let * be a binary operation on set of integers I, defined by a * b = 2a + b − 3. Find the value of 3 * 4.
AnswerGiven: a * b = 2a + b -3
Here, 3 * 4 = 2(3) + 4 - 3
= 6 + 4 - 3
= 7
View full question & answer→Question 62 Marks
Define an associative binary operation on a set.
AnswerAn operation * on a set A is called associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:
- $\text{a}\times\text{b}\in\text{A},\forall\text{ a},\text{b}\in\text{A}$ (Binary operation)
- $\text{a}\times\text{b}\times\text{c}=\text{a}\times\text{b}\times\text{c},\forall\text{ a, b, c}\in\text{A}$ (Associative)
View full question & answer→Question 72 Marks
Let $'o'$ be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2} $ for all $\text{a},\text{b}\in\text{Q}_0.$
Find the identity element in $Q_0.$
AnswerWe have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{e}\in\text{Q}_0$ be the identity element with respect to $*.$
By identity property, we have,
$a * e = e * a = a$ for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\Rightarrow\text{e}=2$
Thus the required identity element is $2.$
View full question & answer→Question 82 Marks
Define identity element for a binary operation defined on a set.
AnswerLet S be a non-empty set and * be a binary operation on S.
If there exist an element $\text{e}\in\text{S}$ such that
a * e = e * a = a for all $\text{e}\in\text{S}$
Then e is called the identity element for the binary operation * on S.
'0' is the identity element for '+' on Z
1 is the identity element for '×' on Z.
View full question & answer→Question 92 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'*'$ on N defined by $a * b = a^b$ for all $\text{a, b}\in\text{N.}$
AnswerLet $\text{a, b}\in\text{N}$
Then, $\text{a}^{\text{b}}\in\text{N}$ $\big[$Therefore $\text{a}^{\text{b}}\neq0$ and $a^b$ is positive integer$\big]$
Implies that $\text{a}\ ^*\ \text{b}\in\text{N}$
Therefore, $\text{a}\ ^*\ \text{b}\in\text{N},\ \forall\ \text{a, b}\in\text{N}$
Thus, $*$ is a binary operation on $N.$
View full question & answer→Question 102 Marks
Find the total number of binary operations on {a, b}.
AnswerWe have,
S = {a, b}
The total number of binary operation on S = {a, b} in $2^{2^{2}}= 2^4=16$
View full question & answer→Question 112 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the identity element in Z.
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 - 4 = a and e + a - 4 = a, $\forall\ \text{a}\in\text{Z}$
e = 4, $\forall\ \text{a}\in\text{Z}$
Thus, 4 is the identity element in Z with respect to *.
View full question & answer→Question 122 Marks
Determine whether the following operations define a binary operation on the given set or not:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$ for all $\text{a, b}\in\text{Q.}$
AnswerIf a = 2 and b = -1 in Q,
$\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$
$=\frac{2-1}{-1+1}$
$=\frac{1}{0}$ [which is not defined]
For a = 2 and b = -1,
$\text{a}\ ^*\ \text{b}\notin\text{Q}$
Therefore,
* is a binary operation on Q.
View full question & answer→Question 132 Marks
If the binary operation o is defined by a o b = a + b - ab on the set Q - {-1} of all rational numbers other than 1, shown that o is commutative on Q - [1].
AnswerLet $\text{a, b}\in\text{Q}-1.$ Then,
a o b = a + b - ab
= b + a - ba
= b o a
Therefore,
a o b = b o a, $\forall\ \text{a, b}\in\text{Q}-1$
Thus, o is commutative on Q - {1}.
View full question & answer→Question 142 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Find the identity element in Q − {−1}.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let e be identity element with respect to *.
By identity property,
a * e = a = e * a for all a ∈ Q - {-1}
⇒ a + e + ae = a
⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as }\text{a}\neq-1]$
e = 0 is the identity element with respect to *.
View full question & answer→Question 152 Marks
Determine whether the following operations define a binary operation on the given set or not:
'*' on N defined by a * b = a + b - 2 for all $\text{a, b}\in\text{N.}$
AnswerIf a = 1 and b = 1, a * b = a + b - 2 = 1 + 1 - 2$=0\notin\text{N}$
Thus, there exist a = 1 and b = 1 such that $\text{a}\ ^*\ \text{b}\notin\text{N}$ So, * is not a binary operation on N.
View full question & answer→Question 162 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 $(a, b), (c, d) ∈ A$
Find the identity element in $A$
AnswerLet $(x, y)$ be the identity element in $\text{A}\forall\text{ x, y}\in\text{A}$. Then,
$(a, b) * (x, y) = (a, b) = (x, y) * (a, b)$
Implies that $(a, b) * (x, y) = (a, b)$ and $(x, y) * (a, b) = (a, b)$
Implies that $(ax, by) = (a, b)$ and $(xa, yb) = (a, b)$
Implies that $x = 1$ and $y = 1$
Thus, $(1, 1)$ is the identity element of $A$.
View full question & answer→Question 172 Marks
Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.
AnswerGiven: a * b = 3a + 4b - 2
Here,
4 * 5 = 3(4) + 4(5) - 2
= 12 + 20 - 2
= 30
View full question & answer→Question 182 Marks
Define a commutative binary operation on a set.
AnswerCommutativity: Let S be a non-empty set. A function F: S × S → S is said to be binary operation on S.Mathematically: Let * be a binary operation. It is said to be commutative binary operation if it satisfies commutativity with respect to *.
That is, if $\text{a, b}\in\text{S}$, then
a * b = b * a for all $\text{a, b}\in\text{S}$.
View full question & answer→Question 192 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $R,$ define by $a * b = ab^2.$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{R}$Implies that $\text{a, b}^2\in\text{R}$
Implies that $\text{ab}^2\in\text{R}$
Implies that $\text{a}\ ^*\ \text{b}\in\text{R}$
Thus, $*$ is a binary operation on $R.$
View full question & answer→Question 202 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the invertible elements in Z.
AnswerLet $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
a + b - 4 = 4 and b + a - 4 = 4
$\text{b}=8-\text{a}\in\text{Z}$
Thus, 8 - a is the inverse of $\text{a}\in\text{Z.}$
View full question & answer→Question 212 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+,$ defined * by $a * b = a - b.$
Here, $Z^+$ denotes the set of all non-negative integers.
AnswerOn $Z^+, *$ is defined by $a * b = a - b$
It is not a binary operation as the image of $(1, 2)$ under $*$ is $1 * 2 = 1 - 2$
$=-1\notin\text{Z}^{+}$
View full question & answer→Question 222 Marks
If the binary operation * on the set Z is defined by a * b = a + b - 5, the find the identity element with respect to *.
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 - 5 = a and e + a - 5 = a, $\forall\ \text{a}\in\text{Z}$
e = 5, $\forall\ \text{a}\in\text{Z}$
Thus, 5 is the identity element in Z with respect to *.
View full question & answer→Question 232 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 $(a, b), (c, d) \in A$
Find the invertible element in $A.$
AnswerLet $(m, n)$ be the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$. Then,
$(a, b) * (m, n) = (1, 1)$
Implies that $(am, bn) = (1, 1)$
Implies that $am = 1\ \&\ bn = 1$
Implies that $\text{m}=\frac{1}{\text{a}}\text{ and }\text{n}=\frac{1}{\text{b}}$
Thus, $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}}\Big)$ is the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$.
View full question & answer→Question 242 Marks
Determine whether the following operations define a binary operation on the given set or not:$'\odot'$ on N defined by $\text{a}\odot\text{b}=\text{a}^{\text{b}}+\text{b}^{\text{a}}$ for all $\text{a, b}\in\text{N.}$
AnswerLet $\text{a, b}\in\text{N.}$ Then,
$\text{a}^{\text{b}},\text{b}^{\text{a}}\in\text{N}$
$\Rightarrow\ \text{a}^{\text{b}}+\text{b}^{\text{a}}\in\text{N}$ $\big[\because$ Addition is binary operation on N$\big]$
$\Rightarrow\ \text{a}\odot\text{b}\in\text{N}$
Thus, $\odot$ is a binary operation on N.
View full question & answer→Question 252 Marks
Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}$
Find 2 * 4, 3 * 5, 1 * 6.
Answera * b = 1.c.m. (a, b)
2 * 4 = 1.c.m. (2, 4)
= 4
3 * 5 = 1.c.m. (3, 5)
= 15
1 * 6 = 1.c.m. (1, 6)
= 6
View full question & answer→Question 262 Marks
Write the identity element for the binary operation * on the set $R_0$ of all non-zero real numbers by the rule $\text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$.
Answer$\because\ \text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$ Let e be the identity element, then
a * e = a
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\ \Rightarrow\text{e}=2$
Thus, e = 2 is the identity element with respect to *.
View full question & answer→Question 272 Marks
Write the identity element for the binary operation * defined on the set R of all real numbers by the rule $\text{a}\times\text{b}=\frac{3\text{ab}}{7}\ \forall\text{ a, b}\in\text{R}$.
AnswerWe have,
$\text{a}\times\text{b}=\frac{3\text{ab}}{7}$
Let e be the identity element with respect to *. Then
a * e = a
$\Rightarrow\frac{3\text{ae}}{7}=\text{a}\ \Rightarrow\text{e}=\frac{7}{3}$
View full question & answer→Question 282 Marks
Let $S=\{a, b, c\}$. Find the total number of binary operations on $S$.
AnswerNumber of binary operations on a set with n elements is $n^2$.
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is $3^{3^{2}}=3^9$
View full question & answer→Question 292 Marks
Let 'o' be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Find the invertible elements of $Q_0$.
AnswerWe have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{b}\in\text{Q}_0$ be the inverse of $\text{a}\in\text{Q}_0$ with respect to *, then,
a * b = b * a = e for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ab}}{2}=\text{e}\Rightarrow\frac{\text{ab}}{2}=2$
$\Rightarrow\text{b}=\frac{4}{\text{a}}$
Thus, $\text{b}=\frac{4}{\text{a}}$ is the inverse of a with respect to *.
View full question & answer→Question 302 Marks
Find the identity element in the set $I^+$ of all positive integers defined by $a * b = a + b$ for all $a, b \in I^+.$
AnswerLet e be the identity element in $I^+$ with respect to $*$ such that
$a * e = a = e * a, \forall\ \text{a}\in\text{I}^{+}$
$a * e = a$ and $e * a = a, \forall\ \text{a}\in\text{I}^{+}$
$a + e = a$ and $e + a = a, \forall\ \text{a}\in\text{I}^{+}$
$e = 0,\forall\ \text{a}\in\text{I}^{+}$
Thus, $0$ is the identity element in $I^+$ with respect to $*.$
View full question & answer→Question 312 Marks
Let $'o'$ be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Show that $'o'$ is both commutative and associate.
AnswerWe have, $\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Commutativity:
Let $\text{a},\text{b}\in\text{Q}_0,$ then
$\Rightarrow\text{a }^*\text{ b}=\frac{\text{ab}}{2}=\frac{\text{ba}}{2}=\text{a }^*\text{ b}$
$\Rightarrow\text{a }^*\text{ b}=\text{b }^*\text{ a}$
Thus, $*$ is commutative on $Q_0.$
Associativity:
Let $\text{a},\text{b},\text{c}\in\text{Q}_0,$ then
$\Rightarrow(\text{a }^*\text{ b})\ ^*\ \text{c}=\frac{\text{ab}}{2}\ ....(1)$ and,
$\text{a }^*\ (\text{b }^*\text{ c})=\text{a }^*\ \frac{\text{bc}}{2}=\frac{\text{abc}}{4}\ ....(2)$ From $(1)\ \&\ (2)$
$(\text{a }^*\text{ b})\ ^*\ \text{c}=\text{a }^*\ (\text{b }^*\text{ c})$
$\Rightarrow *$ is accosiative on $Q_0.$
View full question & answer→Question 322 Marks
Write the total number of binary operations on a set consisting of two elements.
AnswerNumber of binary operations on a set with n elements $=\text{n}^{\text{n}^2}$
Here, Number of binary operations on a set with 2 elements $=2^{2^2}$
$= 2^4$
$=16$
View full question & answer→Question 332 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+,$ defined $*$ by $a * b = ab.$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{Z}^{+}$$\Rightarrow\ \text{ab}\in\text{Z}^+$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\in\text{Z}^+$
Therefore,
$\text{a}\ ^*\ \text{b}\in\text{Z}^+,\ \forall\ \text{a, b}\in\text{Z}^+$
Thus, $*$ is a binary operation on $Z^+.$
View full question & answer→Question 342 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+,$ define $*$ by $a * b = a$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{Z}^{+}$$\Rightarrow\ \text{a}\in\text{Z}^{+}$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\in\text{Z}^{+}$
Therefore,
$\text{a}\ ^*\ \text{b}\in\text{Z}^{+},\ \forall\ \text{a, b}\in\text{Z}^{+}$
Thus, $*$ is a binary operation on $Z^+.$
View full question & answer→Question 352 Marks
Write the inverse of $5$ under multiplication modulo $11$ on the set $\{1, 2, ... ,10\}$.
AnswerAs, $e=1: 5 \times 9 \equiv 1(\bmod 11)$
So, the inverse of 5 i.e. $5^{-1}=9$
View full question & answer→Question 362 Marks
If the binary operation $*$ on the set $Z$ of integers is defined by $a * b = a + 3b^2,$ find the value of $2 * 4.$
AnswerGiven: $a * b = a + 3b^2$
Here,
$2 * 4 = 2 + 3(4)^2$
$= 2 + 3(16)$
$= 2 + 48$
$= 50$
View full question & answer→Question 372 Marks
Prove that the operation $^*$ on the set $\text{M}=\Bigg\{\begin{bmatrix}\text{a} & 0 \\0 & \text{b} \end{bmatrix};\text{ a, b}\in\text{R}-\{0\}\Bigg\}$ defined by $A ^* B = AB$ is a binary operation.
AnswerGiven that ^* is an operation that is valid on the set $\text{M}=\Bigg\{\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right):\text{b}\in \text{R}-\big\{0\big\}\Bigg\}$ and it is defined as given: $A ^* B = AB.$
According to the problem it is given that on applying the operation $^*$ fore two given numbers in the set $'M\ '$ it gives a number in the set $'M\ '$ as a result of the operation.
$\Rightarrow \text{A}^*\text{B}\in \text{M}...(1)$
Let us take $\text{A}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\text{ and }\text{B}=\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$
here $\text{a}\in \text{R},\ \text{c}\in \text{R},\ \text{d}\in \text{R}$ then,
$\Rightarrow \text{AB}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\times\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$
$\Rightarrow \text{AB}=\begin{pmatrix}((\text{a}\times\text{c})+(0\times 0))&((\text{a}\times0)+(0\times \text{d}))(0\times\text{c})+(\text{b}\times 0))&((0\times0)+(\text{b}\times\text{d})) \end{pmatrix}$
$\Rightarrow \text{Ab}=\begin{pmatrix}(\text{ac}+0)&(0+0)0+0)&(0+\text{bd}) \end{pmatrix}$
$\Rightarrow \text{AB}=\begin{pmatrix} \text{ac}&0\\0&\text{bd}\end{pmatrix}$
Since $\text{b}\in \text{R}$ and $\text{c}\in \text{R}$ then $\text{ac}\in \text{R}$
And also $\text{b}\in \text{R}$ and $\text{d}\in \text{R}$ then $\text{bd}\in \text{R}$
$\Rightarrow \text{AB}\in \text{R}$
View full question & answer→Question 382 Marks
Let * be a binary operation, on the set of all non-zero real numbers, given by
$\text{a}\times\text{b}=\frac{\text{ab}}{5}\ \forall\text{ a, b}\in\text{R}-\{0\}$
Write the value of x given by 2 * (x * 5) = 10.
AnswerGiven: 2 * (x * 5) = 10
Here,
$2\times\Big(\frac{5\text{x}}{5}\Big)=10$
Implies that 2 * x = 10
Implies that $\frac{2\text{x}}{5}=10$
Implies that $\text{x}=\frac{10\times5}{2}$
Implies that x = 25
View full question & answer→Question 392 Marks
Let * be a binary operation on N given by a * b = LCM (a, b) for all $\text{a, b}\in\text{N.}$ Find 5 * 7.
AnswerAs, a * b = LCM (a, b)
So, 5 * 7 = LCM (5, 7) = 35
View full question & answer→Question 402 Marks
Determine whether the following operations define a binary operation on the given set or not:
'+6' on $S = \{0, 1, 2, 3, 4, 5\}$ defined by, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$
AnswerWe have, $S = \{0, 1, 2, 3, 4, 5\}$ and, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$
Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that $a + b < 6$
Then $\text{a}+_6\text{b}=\text{a}+\text{b}\in\text{S}$ $\big[\because a + b < 6 = 0, 1, 2, 3, 4, 5 \big]$
Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that $a + b > 6$
Then $\text{a}+_6\text{b}=\text{a}+\text{b}-6\in\text{S}$
$\big[\because\ \text{if a}+\text{b}\geq6$ then $\text{a}+\text{b}-6\geq6 = 0, 1, 2, 3, 4, 5\big]$
$\therefore\ \text{a}+_6\text{b}\in\text{S}$ for $\text{a, b}\in\text{S}$
$\therefore +_6$ defined a binary operation on $S.$
View full question & answer→Question 412 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $R$, define $*$ by $a * b = a + 4b^2$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{R}$$\Rightarrow\ \text{a, 4b}^2\in\text{R}$
$\Rightarrow\ \text{a}+\text{4b}^2\in\text{R}$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\in\text{R} $
Therefore,
$\text{a}\ ^*\ \text{b}\in\text{R},\ \forall\ \text{a, b}\in\text{R}$
Thus, $*$ is a binary operation on $R.$
View full question & answer→Question 422 Marks
Let * be a binary operation on the set I of integers, defined by a * b = 2a + b - 3. Find the value of 3 * 4.
AnswerIt is given that, a * b = 2a + b - 3 Now, 3 * 4 = 2 × 3 + 4 - 3 = 10 - 3= 7
View full question & answer→Question 432 Marks
For the binary operation multiplication modulo $5 (\times _5)$ defined on the set $S = \{1, 2, 3, 4\}.$ Write the value of $(3 \times _5 4^{-1})^{−1}$
AnswerThe composition table for $\times _5$ on the set $S = \{1, 2, 3, 4\}$ is
| $\times _5$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$4$ |
$3$ |
$2$ |
$1$ |
Now,
$(3 \times _5 4^{-1})^{-1} = (3 \times _5 4)^{-1} [\because 4^{-1} = 4]$
$= 2^{-1} [3 \times _5 4 = 2]$
$= 3 [\because 2^{-1} = 3]$ View full question & answer→Question 442 Marks
Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.
AnswerThe binary operator * defined on Z and is given by a * b = 3a + 7b
Commutativity: Let $\text{a, b}\in\text{Z},$ Then,
a * b = 1a + 7b and
b * a = 3b + 7a
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, '*' is not commutative on Z.
View full question & answer→Question 452 Marks
Let * be a binary operation on N given by a * b = HCF (a, b), a, b ∈ N. Write the value of 22 * 4.
AnswerWe have,a * b = HCF (a, b) for all a, b ∈ Z
Now, 22 * 4 = HCF (22, 4) = 2 $\therefore$ 22 * 4 = 2
View full question & answer→Question 462 Marks
Determine which of the following binary operations are associative and which are commutative:
'*' on N defined by a * b = 1 for all $\text{a, b}\in\text{N}.$
AnswerClearly, by defination a * b = 1 = b * a, $\forall\ \text{a, b}\in\text{N}$
Also, (a * b) * c = (1 * c) = 1
and a * (b * c) = (a * 1) = 1 $\forall\ \text{a, b, c}\in\text{N}$
Hence, N is both associative and commutative.
View full question & answer→Question 472 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'O'$ on $Z$ defined by a $O$ $b = a^b$ for all $\text{a, b}\in\text{Z.}$
AnswerWe have,
$a\ O\ b = a^b$ for all $\text{a, b}\in\text{Z}$
Let $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$
$\Rightarrow\ \text{a}^{\text{b}}\notin\text{Z}\ \Rightarrow\ \text{a O b}\notin\text{Z}$
For example, if $a = 2, b = -2$
$\Rightarrow\ \text{a}^{\text{b}}=2^{-2}=\frac{1}{4}\notin\text{Z}$
$\therefore$ The operation $'O'$ does not define a binary operation on $ Z.$
View full question & answer→Question 482 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'\times _6'$ on $S = \{1, 2, 3, 4, 5\}$ defined by, $a \times _6 b =$ Remainder when ab is divided by $6.$
AnswerConsider the composition table,
|
$\times _6$
|
$1$
|
$2$
|
$3$
|
$4$
|
$5$
|
|
$1$
|
$1$
|
$2$
|
$3$
|
$4$
|
$5$
|
|
$2$
|
$2$
|
$4$
|
$0$
|
$2$
|
$4$
|
|
$3$
|
$3$
|
$0$
|
$3$
|
$0$
|
$3$
|
|
$4$
|
$4$
|
$2$
|
$0$
|
$4$
|
$2$
|
|
$5$
|
$5$
|
$4$
|
$3$
|
$2$
|
$1$
|
Here all the elements of the table are not in $S.$
For $a = 2$ and $b = 3$,
$\text{a}\times_6\text{b}= 2 \times_6 3 =$ remainder when $6$ divided by $6= 0 \neq\text{S}$
Thus, $\times _6$ is not a binary operation on $S.$ View full question & answer→Question 492 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+$ define * by $a * b = |a - b|$
Here, $Z^+$ denotes the set of all non-negative integers.
AnswerOn $Z^+, *$ is defined by $a * b = |a - b|.$
It is seen that for each $\text{a, b}\in\text{Z}^{+},$ there is a unique element $|a - b|$ in $Z^+.$
This means that $*$ carries each pair $(a, b)$ to a unique element $a * b = |a - b|$ in $Z^+.$
Therefore, $*$ is a binary operation.
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