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,
$a ^* b=\frac{ ab }{2}$ for all $a , b \in Q _0$
Let $e \in Q _0$ be the identity element with respect to *.
By identity property, we have,
$a^* e=e^* a=a \text { for all } a \in Q_0$
$\Rightarrow \frac{ae}{2}=a \Rightarrow e=2$
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:
${ }^2{ }^{\prime}$ ' on $N$ defined by $a^* b=a^b$ for all $a, b \in N$.
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 ' $k$ ' is a binary operation on adefined by, $(a, b)^*(c, d)=(a c, b d)$ for all $(a, b),(c, d) \in A$
Find the identity element in A
AnswerLet $(x, y)$ be the identity element in $A \forall x , y \in A$. 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 $(a x, b y)=(a, b)$ and $(x a, y b)=(a, b)$
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 $a^* b=a b^2$.
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 $a ^{\text {* }} b = a - b$
It is not a binary operation as the image of $(1,2)$ under * is 1 * $2=1-2$
$=-1 \notin Z^{+}$
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 ' $'k'$ is a binary operation on adefined by, $(a, b)^*(c, d)=(a c, b d)$ for all $(a, b),(c, d) \in A$
Find the invertible element in $A$ .
AnswerLet $( m , n )$ be the inverse of $( a , b ) \forall( a , b ) \in A$. Then,
$(a, b)^*(m, n)=(1,1)$
Implies that $( am , bn )=(1,1)$
Implies that $am =1 \& bn =1$
Implies that $m =\frac{1}{ a }$ and $n =\frac{1}{b}$
Thus, $\left(\frac{1}{a}, \frac{1}{b}\right)$ is the inverse of $( a , b ) \forall( a , b ) \in A$.
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 $a \times b=\frac{a b}{2}$ for all $a, b \in R_0$.
Answer$\because a \times b =\frac{ ab }{2}$ for $all a , b \in R _0$ Let e be the identity element, then
$a^* e=a$
$\Rightarrow \frac{ ae }{2}= a \Rightarrow e =2$
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 $a l l a, b \in I ^{+}$.
AnswerLet e be the identity element in $I^{+}$with respect to * such that
$a^* e=a=e^* a, \forall a \in I^{+}$
$a^* e=a \text { and } e^* a=a, \forall a \in I^{+}$
$a+e=a \text { and } e+a=a, \forall a \in I^{+}$
$e=0, \forall a \in I^{+}$
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}^{+}$$\Rightarrow\ \text{ab}\in\text{Z}^+$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\in\text{Z}^+$
Therefore,
$\text{a}\ ^*\ \text{b}\in\text{Z}^+,\ \forall\ \text{a, b}\in\text{Z}^+$
Thus, $*$ is a binary operation on $Z^+.$
View full question & answer→Question 342 Marks
Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
On $\mathrm{Z}^{+}$, define * by a * $\mathrm{b}=\mathrm{a}$
Here, $\mathrm{Z}^{+}$denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{Z}^{+}$$\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 $a^* b=a+3 b^2$, find the value of $2 * 4$.
Answer
Given: $a$ * $b=a+3 b^2$
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 $\text{A}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)$ 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]$$\therefore\ \text{a}+_6\text{b}\in\text{S}$ for $\text{a, b}\in\text{S}$
$\therefore +_6 $ defined a binary operation on S.
View full question & answer→Question 412 Marks
Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
On $R$, define * by $a$ * $b=a+4 b^2$
Here, $Z ^{+}$denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{R}$$\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 $Ob = a ^{ b }$ for $all a , b \in Z$.
AnswerWe have,
$a$ O $b=a^b$ for all $a, b \in Z$
Let $a \in Z$ and $b \in Z$
$\Rightarrow a^{b} \notin Z \Rightarrow aOb \notin Z$
For example, if $a=2, b=-2$
$\Rightarrow a^{b}=2^{-2}=\frac{1}{4} \notin Z$
$\therefore$ 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 $b = 3,$
$\text{a}\times_6\text{b}= 2\times_63$ = 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 $a , b \in Z ^{+}$,
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.
View full question & answer→