Question 513 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 523 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 533 Marks
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and $\text{h(z)}=\sin\text{z}$ for all $\text{x, y, z}\in\text{N.}$ Show that ho(gof) = (hog)of.
AnswerGiven, f : N → N, g : N → N and h : N → R
⇒ gof : N → N and hog : N → R
⇒ ho(gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4 ....(1)
(hog)(x) = h(g(x)) = h(3x + 4) = sin(3x + 4) ......(2)
Now,
(ho(gof))(x) = h((gof)(x)) = h(6x + 4) = sin(6x + 4) [from (1)]
((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = sin(6x + 4) [from (2)]
So, (ho(gof))(x) = ((hog)of)(x), $\forall\text{ x}\in\text{N}$
Hence, ho(gof) = (hog)of
View full question & answer→Question 543 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 553 Marks
Find $\mathrm{f}^{-1}$ if it exists: $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$, where, $\mathrm{A}=\{1,3,5,7,9\} ; B=\{0,1,9,25,49,81\}$ and $f(x)=x^2$.
Answer$A=\{1,3,5,7,9\} ; B=\{0,1,9,25,49,81\}$
$f: A \rightarrow B$ be a function defined by $f(x)=x^2$
Since different elements of $A$ have different images in $B$.
$\therefore \mathrm{f}$ is one-one.
Again, $0 \in \mathrm{~B}$ does not have a preim-age in $A .$
$\therefore \mathrm{f}$ is not onto.
Hence, $\mathrm{f}^{-1}$ does not exist.
View full question & answer→Question 563 Marks
Three relation $R_1$ is defined in set $A = \{a, b, c\}$ as follows:
$R_1 = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)\}$
Find whether or not the relation $R_1$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
AnswerConsider R, Reflexive: Clearly, $(a, a), (b, b)$ and $(c, c)$ $\in\text{R}_1$Therefore, $R_1$ is reflexive.
Symmetric: We see that the ordered pairs obtained by interchanging the components of $R_1$ are also in $R_1$
Therefore,
$R_1$ is symmetric.
Transitive: Here, $\text{a, b}\in\text{R}_1,\ \text{b, c}\in\text{R}_1$ and also $\text{a, c}\in\text{R}_1$
Therefore, $R_1$ is transitive.
View full question & answer→Question 573 Marks
Each of the following defines a relation on N:
$\text{x}+\text{y}=10,\ \text{x},\ \text{y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
Answer$\therefore\ \text{R}=\{(\text{x},\text{y});\ \text{x}+\text{y}=10,\ \text{x},\text{y}\in\text{N}\}$$\therefore$ R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}
Clearly $(1,1)\notin\text{R}$
So, R is not reflexive.
$(\text{x},\text{y})\in\text{R}\Rightarrow\ (\text{y},\text{x})\in\text{R}$
Thus, R is symmetric.
Now $(1,9)\in\text{R},\ (9,1)\in\text{R},$ but $(1,1)\notin\text{R}$
Hence, R is not transitive.
View full question & answer→Question 583 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 593 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(x, y): x is a person, y is the mother of x}
Answerf = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
Therefore, f is the function.
Injection test: As, y can be mother of two or more persons.
Therefore,
f is not injective.
Surjection test: For every mother y defined by (x, y), there exists a person x for whom y is mother. Therefore, f is surjective.
View full question & answer→Question 603 Marks
If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta\alpha\text{x}+\beta,$ then find the values of $\alpha$ and $\beta.$
AnswerWe have,
A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by $\text{g(x)}=\alpha\text{x}+\beta$
As, g(1) = 1 and g(2) = 3
Therefore, $\alpha(1)+\beta=1$
$=\alpha+\beta=1\ ....(\text{i})$
and $\alpha(2)-\beta=3$
$2\alpha-\beta=3\ ....(\text{ii})$
(ii) - (i) we get
$2\alpha-\alpha=2$
$\alpha=2$
Substituting $\alpha=2$ in (i), we get
$2+\beta=1$
$\beta=1$
View full question & answer→Question 613 Marks
Let $A$ and $B$ be sets. Show that $f: A \times B \rightarrow B \times A$ such that $f (a, b) = (b, a)$ is bijective function.
Answer$f: A \times B \rightarrow B \times A$ is defined as $f(a, b) = (b, a).$
Let $(\text{a}_1,\text{b}_1),(\text{a}_2,\text{b}_2)\in \text{A}\times\text{B}$ such that $f(a_1, b_1) = f(a_2, b_2)$
$\Rightarrow (b_1, a_1) = (b_2, a_2) \Rightarrow b_1 = b_2$ and $a_1 = a_2 \Rightarrow (a_1, b_1) = (a_2, b_2) $
$\therefore f$ is one-one.
Now, let $(\text{b},\text{a})\in\text{B}\times\text{A}$ be any element. Then, there exists $(\text{a},\text{b})\in\text{A}\times\text{B}$ such that $f(a, b) = (b, a). [$By definition of $f]$
$\therefore$ f is onto. Hence, $f$ is bijective.
View full question & answer→Question 623 Marks
Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f ^{–1}$ is $f,$ i.e., $(f ^{–1})^{–1} = f.$
AnswerLet $f: X \rightarrow Y$ be an invertible function.
Then $f$ is one-one and onto $\Rightarrow g: y \rightarrow X$ where $g$ is also one-one and onto such that
g$of(x) = I_x$ and $fog(y) = I_y \Rightarrow g = f^{-1}$
Now, $f^{-1}o(f^{-1})^{-1} = I$ and $fo[f^{-1}o(f^{-1})^{-1}] = foI$
$\Rightarrow [fof^{-1}]0(f^{-1})^{-1} = f \Rightarrow Io(f^{-1})^{-1} = f \Rightarrow (f^{-1})^{-1} = f$
View full question & answer→Question 633 Marks
Let A = {1, 2, 3}. Write all one-one from A to itself.
AnswerWe have,
ho(gof)(x) = h(gof(x)) = h(g(f(x)))
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
((hog)of)(x) = (hog)(f(x)) = (hog)(2x)
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
This shows, ho(gof) = (hog)of
View full question & answer→Question 643 Marks
Let f, g, h be real functions given by f(x) = sinx, g(x) = 2x and h(x) = cosx. Prove that fog = go(fh).
Answerf, g and h are real fuctions given by f(x) = sinx, g(x) = 2x and h(x) = cosx
To prove: fog = go(fh)
L.H.S. fog(x) = f(g(x))
= f(2x) = sin2x
⇒ fog(x) = 2sinx.cosx .....(A)
R.H.S. go(fh)(x) = go(f(x).h(x))
= g(sinx.cosx)
⇒ go(fh)(x) = 2sinx.cosx ......(B)
from (A) & (B)
fog(x) = go(fh)(x)
View full question & answer→Question 653 Marks
The following relation are defined on the set of real numbers.
aRb if $|\text{a}|\leq\text{b}$
Find whether these relation are reflexive, symmetric or transitive.
AnswerWe have aRb if $|\text{a}|\leq\text{b}$ Reflexive: Let $\text{a}\in\text{R}$$\Rightarrow\ |\text{a}|\nleq\text{a}$ $[\therefore|-2|=2>-2|]$
⇒ R is not reflexive. Symmetric: Let aRb $\Rightarrow\ |\text{a}|\leq\text{b}$ $\Rightarrow\ |\text{b}|\leq\text{a}$ $\begin{bmatrix}\therefore\ \ \ \text{Let a}=4, \text{b}=6 \\\ \ \ \ \ \ \ \ \ |4|\leq 8 \text{ but } |8|>4\end{bmatrix}$ ⇒ R is not symmetric. Transitive: Let aRb and bRc$\Rightarrow\ |\text{a}|\leq\text{b}$ and $|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq\text{c}$
$\Rightarrow\ \text{aRc}$
⇒ R is transitive.
View full question & answer→Question 663 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 673 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 683 Marks
Find fog and gof if: $f(x) = x^2 + 2, \text{g(x)}=1-\frac{1}{1-\text{x}}$
Answer$f(x) = x^2 + 2$ and $\text{g(x)}=1-\frac{1}{1-\text{x}}$
Range of $\text{f}=(2,\infty)\ \subset$ Domain of $g = R \Rightarrow gof$ exist
Range of $g = R - [1] \subset$ Domain of $f = R \Rightarrow fog$ exist
Now,
$fog(x) = f(g(x))$
$=\text{f}\Big(\frac{-\text{x}}{1-\text{x}}\Big)=\frac{\text{x}^2}{(1-\text{x})^2}+2$
And,
$gof(x) = g(f(x))$
$=\text{g}(\text{x}^2+2)=\frac{-(\text{x}^2+2)}{1-(\text{x}^2+2)}$
$\Rightarrow\ \text{gof(x)}=\frac{\text{x}^2+2}{\text{x}^2+1}$
View full question & answer→Question 693 Marks
Let f: R → R be the Signum Function defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$ and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
AnswerIt is given that,
f: R → R is defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let $\text{x}\in(0,\ 1].$
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.
$\therefore fo\text{g(x)}=f(\text{g(x)})=f([\text{x}])=\begin{cases}f(1),&\text{if x}=1\\f(0),&\text{if x}\in(0,\ 1)\end{cases}=\begin{cases}1,&\text{if x}=1\\0,&\text{if x}\in(0,\ 1)\end{cases}$
gof(x) = g(f(x))
= g(1) [x > 0]
=[1] = 1
Thus, when $\text{x}\in(0,\ 1),$ we have fog(x) = 0 and gof(x) = 1.
Hence, fog and gof do not coincide in (0, 1].
View full question & answer→Question 703 Marks
Let $f : R \rightarrow R$ be the function defined by $f(x) = 2x – 3 ∀ x \in R$. write $ f^{-1}.$
AnswerGiven $f(x) = 2x - 3 ∀ x \in R$ Now,
Let $a, b \in R$ such that
$f(a) = f(b) \Rightarrow 2a - 3 = 2b - 3\Rightarrow a = b \Rightarrow f(x)$ is One-One.
Also, If $x, y \in R$ such that $f(x) = y \Rightarrow 2x - 3 = y$
$\Rightarrow\ \text{x}=\frac{\text{y}+3}{2}=\text{g}(\text{y})\ \forall\ \text{y}\in\text{R}$
$\Rightarrow f(x)$ is Onto and therefore is bijective implies $f(x)$ has an inverse
Let $f^{-1}$ denote the inverse of $f(x)$
then,$\text{f}^{-1}\text{x}=\text{g}(\text{x})=\frac{\text{x}+3}{2}\ \forall\ \text{x}\in\text{R}$
View full question & answer→Question 713 Marks
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
AnswerA = {1, 2, 3}
Possible onto function from A to A can be the following:
- {(1, 1), (2, 2), (3, 3)}
- {(1, 1), (2, 3), (3, 2)}
- {(1, 2), (2, 2), (3, 3)}
- {(1, 2), (2, 1), (3, 3)}
- {(1, 3), (2, 2), (3, 1)}
- {(1, 3), (2, 1), (3, 2)}
Here, in each function, different elements of the domain have different images.
Therefore, all the function are one-one.
View full question & answer→Question 723 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 733 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 $(a, b), (c, d) \in A$
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 743 Marks
Let f be a function from R to R, such that f(x) = cos(x + 2). Is f invertible? Justify your answer.
AnswerGiven: A and B are two sets with finite elements.
f : A → B and g : B → A are injective map.
To prove: f is bijective.
Proof: Since, f : A → B is injective we need to show f in surjective only.
Now,
g : B → A is injective.
⇒ Each element of B has image in A.
View full question & answer→Question 753 Marks
Give an example of a relation which is,
Reflexive and symmetric but not transitive.
AnswerLet A = {4, 6, 8} Define a relation R on A as: A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}Relation R is reflexive since for every $\text{a}\in\text{A},\ (\text{a, a})\in\text{R}$ i.e., (4, 4), (6, 6), (8, 8) $\in\text{R}$
Relation R is symmetric since $(\text{a, b})\in\text{R}\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{R.}$
Relation R is not transitive since (4, 6), (6, 8) $\in\text{R,}$ but $(4,8)\notin\text{R.}$
Hence, relation R is reflexive and symmetric but not transitive.
View full question & answer→Question 763 Marks
Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined by $f(x) = x + 1$ and $g(x) = x - 1$. Show that $fog = gof = I_R.$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R\ fog : R \rightarrow R$ and $gof : R \rightarrow R$
$($Also, we know that $I_R : R \rightarrow R)$
Therefore, the domains of all fog, and $I_R $ are the same.
$(fog)(x) = f(g(x)) = f(x - 1) = x - 1 + 1 = x = I_R(x) ..... (1)(gof)(x) = g(f(x)) = g(x + 1) = x + 1 -1 = x = I_R(x) ..... (2)$
From $(1)$ and $(2), (fog)(x) = (gof)(x) = I_R(x),$ $\forall\text{ x}\in\text{R}$
Hence, $fog = gof = I_R$_
View full question & answer→Question 773 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions from A to itself are one-one, onto or bijective:$h(x) = x^2$
Answer$h(x) = x^2$
Injection test: Let $x$ and $y$ be any two elements in the domain $(A),$
such that $f(x) = f(y). f(x) = f(y) x^2 = y^2$
$\text{x}=\pm\text{y}$
So, $f$ is not one-one.
Surjection test: For $y = -1,$ there is no value of $x$ in $A$.
So, $f$ is not onto. So, $f$ is not bijective.
View full question & answer→Question 783 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions defined on $A$ are one-one, onto or bijective:
$\text{f}(\text{x})=\frac{\text{x}}{2}$
AnswerLet $f(x_1) = f(x_2)$
$\Rightarrow\ \frac{\text{x}_1}{2}=\frac{\text{x}_2}{2}\Rightarrow\ \text{x}_1=\text{x}_2$
So, $f(x)$ is one-one.
Now, let $\text{y}=\frac{\text{x}}{2}$
$\Rightarrow\ \text{x}=2\text{y}\notin\text{A},\ \forall\ \text{y}\in\text{A}$
As for $\text{y}=1\in\text{A},\ \text{x}=2\notin\text{A}$
So, $f(x)$ is not onto.
Also, $f(x)$ is not bijective as it is not onto.
View full question & answer→Question 793 Marks
Let f be an invertible real function. Write $\mathrm{(f^{-1}of)(1) + (f^{-1}of)(2) + ..... + (f^{-1}of)(100).}$
AnswerGiven that f is an invertible real function. $\mathrm{f^{-1}of = I,}$
where I is an identity function.
Therefore, $\mathrm{(f^{-1}of)(1) + (f^{-1}of)(2) + ....... + (f^{-1}of)}$
$(100) = I(1) + I(2) + .... + I(100) = 1 + 2 + .... + 100$
$(\text{As Ix}=\text{x},\ \forall\ \text{x}\in\text{R})$
$=\frac{1001(1001+1)}{2}$
$\Big[$ Sum of first n natural numbers
$=\frac{\text{n}(\text{n}+1)}{2}\Big]= 5050$
View full question & answer→Question 803 Marks
If R and S are transitive relations on a set A, then prove that $\text{R}\cup\text{S}$ may not be a transitive relation on A.
AnswerLet A = {a, b, c} and R and S be two relations on a, given by R = {(a, a), (a, b), (b, a), (b, b)}
And S = {(b, b), (b, c), (c, b), (c, c)}
Here, the relations R and S are transitive on A.
$\text{a, b}\in\text{R}\cup\text{S}$ and $\text{b, c}\in\text{R}\cup\text{S}$ But $\text{a, c}\notin\text{R}\cup\text{S}$
Hence, $\text{R}\cup\text{S}$ is not a transitive relation on A.
View full question & answer→Question 813 Marks
Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.
AnswerLet us consider a function f : N → N given by f(x) = x + 1, which is not onto.
[This not onto because if we take 0 in N (co-domain), then, 0 = x + 1, x = -1 $\notin\text{N}$]
Let us consider,
$\text{g(x)}:\begin{Bmatrix}\text{x}=\text{x}-1&\text{if x}>1\\1,&\text{if x}=1\end{Bmatrix}$
Now, let us find (gof)(x)
Case 1: x > 1
(gof)(x) = g(f(x)) = g(x + 1) = x + 1 - 1 = x
Case 2: x = 1
(gof)(x) = g(f(x)) = g(x + 1) = 1
From case 1 and case 2, x = x, $\forall\ \text{x}\in\text{N,}$
which is an identity function and, hence, it is onto.
View full question & answer→Question 823 Marks
The cartesian equation of a line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$ Write its vector form.
AnswerGiven: The Cartesian equation of the line is
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}=\lambda\ (\text{say})$
$\Rightarrow\ \ \text{x}-5=3\lambda,\ \text{y}+4=7\lambda,\ \text{z}-6=2\lambda$
$\Rightarrow\ \ \text{x}=5+3\lambda,\ \text{y}=-4+7\lambda,\ \text{z}=6+2\lambda$
General equation for the required line is
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Putting the values of x, y, z in this equation,
$\vec{\text{r}}=(5+3\lambda)\hat{\text{i}}+(-4+7\lambda)\hat{\text{j}}+(6+2\lambda)\hat{\text{k}}$
$=5\hat{\text{i}}+3\lambda\hat{\text{i}}-4\hat{\text{j}}+7\lambda\hat{\text{j}}+6\hat{\text{k}}+2\lambda\hat{\text{k}}$
$\Rightarrow\ \ \ \ \vec{\text{r}}=\Big(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}\Big)+\lambda\Big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\Big)\ \ \ \ \Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
View full question & answer→Question 833 Marks
Show that the Modulus Function f: R → R, given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is –x, if x is negative.
Answerf: R → R is given by, $\text{f(x)}=|\text{x}|=\begin{cases}\text{x},&\text{if }\text{x }\geq0\\-\text{x},&\text{if }\text{x }<0\end{cases}$ It is seen that f(-1) = |-1| = 1, f(1) = |1| = 1 $\therefore$ f(-1) = f(1), but $-1\neq1.$ $\therefore$ f is not one-one. Now, consider $-1\in\text{R}.$It is known that f(x) = | Hence, the greatest integer functiox| is always non-negative integer. Thus, there does not exist any element x in domain R such that f(x) = |x| = -1.
$\therefore$ f is not onto. n is neither one-one nor onto.
View full question & answer→Question 843 Marks
Let f be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.
AnswerGiven, f : R → R
Since g(x) = 2x is a polynomial, g : R → R
Clearly, gof : R → R and f + f : R → R
So, domain of gof and f + f are the same.
(gof)(x) = g(f(x)) = 2f(x)
(f + f)(x) = f(x) + f(x) = 2f(x)
⇒ (gof)(x) = (f + f)(x), $\forall\ \text{x}\in\text{R}$
Hence, gof = f + f
View full question & answer→Question 853 Marks
Find $f^{-1}$ if it exists: $f : A \rightarrow B,$ where, $A = \{0, -1, -3, 2\}; B = \{-9, -3, 0, 6\}$ and $f(x) = 3x.$
Answer$A = \{0, -1, -3, 2\}; B = \{-9, -3, 0, 6\}$
$f : A \rightarrow B$ is defined by $f(x) = 3x$
Since different elements of $A$ have different images in $B.$
$\therefore f$ is one-one.
Again, each element in $B$ has a preim-age in $A.$
$\because f$ in one-one bijective.
$⇒ f^{-1} : B \rightarrow A$ exists and is given by
$\text{f}^{-1}(\text{x})=\frac{\text{x}}{3}$
View full question & answer→Question 863 Marks
Let $f : R \rightarrow R$ be defined as $\text{f(x)}=\frac{2\text{x}-3}{4}.$ Write $fof^{-1}(1).$
AnswerLet $f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{2\text{x}-3}{4}$
$\Rightarrow\ \text{f}^{-1}\frac{(2\text{x}-3)}{4}=\text{x}$
$\Rightarrow\ \text{f}^{-1}(2\text{x})=4\text{x}+3$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{4\text{x}+3}{2}$
Now, $fof^{-1}(x) = f(f^{-1}(x))$
$=\text{f}\Big(\frac{4\text{x}+3}{2}\Big)$
$=\frac{2\big(\frac{4\text{x}+3}{2}\big)-3}{4}$
$\Rightarrow fof^{-1}(x) = x$
$\therefore fof^{-1}(1) = 1$
View full question & answer→Question 873 Marks
Each of the following defines a relation on N:
x is greater than $\text{y},\ \text{x},\ \text{y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
Answerx is greater than $\text{y},\ \text{x},\ \text{y}\in\text{N}$ If $(\text{x},\text{x})\in\text{R},$ then x > x, which is not true for any $\text{x}\in\text{N}$ Therefore, R is not reflexive. Let $(\text{x},\text{y})\in\text{R}$ ⇒ xRy ⇒ x > y ⇒ y > x, which is not true for any $\text{x},\text{y}\in\text{N}$Thus, R is not symmetric.
Let xRy and yRz
⇒ x > y and y > z
⇒ x > z ⇒ xRz So, R is transitive.
View full question & answer→Question 883 Marks
Three relation $R_2$ is defined in set $A = \{a, b, c\}$ as follows:
$R_2 = \{(a, a)\}$
Find whether or not the relation $R_2$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_2$ is Reflexive: Clearly $\text{a, a}\in\text{R}_2$
Therefore, $R_2$ is reflexive.
Symmetric: Clearly, $\text{a, a}\in\text{R}\Rightarrow\ \text{a, a}\in\text{R}.$
Therefore, $R_2$ is symmetric.
Transitive: $R_2$ is clearly a transitive relation, since there is only one element in it.
View full question & answer→Question 893 Marks
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
- R = {(a, b) : |a – b| is a multiple of 4}
- R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer
- It is given that the relation R in the set $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\},$ given by
R = {(a, b) : |a - b| is a multiple of 4}
For any element $\text{a}\in\text{A},$ we have $(\text{a},\text{a})\in\text{R}$ as |a - a| = 0 is a multiple of 4.
Therefore, R is reflexive.
Now, Let $(\text{a},\text{a})\in\text{R}$
⇒ |a - b| is a multiple of 4
⇒ |b - a| = |a - b| is a multiple of 4
$\Rightarrow(\text{b},\text{a})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{a},\text{b}),(\text{b},\text{c})\in\text{R}$
⇒ |a - b| is a multiple of 4 and |b - c| is a multiple of 4
⇒ |a - c| = |(a - b) + (b - c)| is a multiple of 4
$\Rightarrow(\text{a},\text{c})\in\text{R}$
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9}
|1 - 1| = 0 is multiple of 4
|5 - 1| = 4 is multiple of 4
|9 - 1| = 8 is multiple of 4.
- It is given that the relation R in the set $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\},$ given by
R = {(a, b) : a = b}
For any element $\text{a}\in\text{A},$ we have $(\text{a},\text{a})\in\text{R}$ as a = a
Therefore, R is reflexive.
Now, Let $(\text{a},\text{a})\in\text{R}$
⇒ a = b
⇒ b = a
$\Rightarrow(\text{b},\text{a})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{a},\text{b}),(\text{b},\text{c})\in\text{R}$
⇒ a = b and b = c
⇒ a = c
$\Rightarrow(\text{a},\text{c})\in\text{R}$
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of elements related to 1 will be those elements from set Awhich are equal to 1.
Therefore, the set of elements related to 1 is {1}. View full question & answer→Question 903 Marks
Show that $f: [–1, 1] \rightarrow R,$ given by $f(\text{x})=\frac{\text{x}}{(\text{x}+2)}$ is one-one. Find the inverse of the function $f:[-1, 1] \rightarrow $ Range $f.$
(Hint: For $\text{y}\in\text{Range f},\text{y}=f(\text{x})=\frac{\text{x}}{\text{x}+2},$ for some $x$ in $[-1, 1],$ i.e., $\text{x}=\frac{2\text{y}}{(1-\text{y})}$)
AnswerPart $I: f: [-1, 1] \rightarrow R$ given by $f(\text{x})=\frac{\text{x}}{(\text{x}+2)},\text{x}\neq-2$
$\text{Let }\text{x}_1,\text{x}_2\in[-1,1],\text{then }f(\text{x}_1)=\frac{\text{x}_1}{\text{x}_1+2}\text{ and }f(\text{x}_2)=\frac{\text{x}_2}{\text{x}_2+2}$
When $f(x_1) = f(x_2),$ then $\frac{\text{x}_1}{\text{x}_1+2}=\frac{\text{x}_2}{\text{x}_2+2}$
$\Rightarrow x_1x_2 + 2x_1 = x_1x_2 + 2x_2 \Rightarrow x_1 = x_2$
$\therefore$ f is one-one.
Part II: Let $\text{y}\in\text{Range of f}\Rightarrow\text{y}=f(\text{x})=\frac{\text{x}}{\text{x}+2}$ for some x in [-1, 1]
| As |
$\text{y}=\frac{\text{x}}{\text{x}+2}$ |
$\Rightarrow $ |
$yx + 2y = x$ |
$\Rightarrow $ |
$x(1 - y) = 2y$ |
| $\Rightarrow $ |
$\text{x}=\frac{2\text{y}}{1-\text{y}}$ |
$\Rightarrow $ |
$f^{-1}\text{y}=\frac{2\text{y}}{1-\text{y}}$ |
$\therefore$ |
$f$ is onto. |
Therefore, $f^{-1}\text{x}=\frac{2\text{x}}{1-\text{x}}$ View full question & answer→Question 913 Marks
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
(Hint: Consider f(x) = x + 1 and $\text{g(x)}=\begin{cases}\text{x}-1\ \text{if x}>1\\\ \ \ 1\ \ \ \ \ \text{if x}=1\end{cases}$
AnswerDefine f: N → N by, f(x) = x + 1And, g: N → N by,
$\text{g(x)}=\begin{cases}\text{x}-1\ \text{if x}>1\\\ \ \ 1\ \ \ \ \ \text{if x}=1\end{cases}$ We first show that g is not onto. For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N. $\therefore$ f is not onto. Now, gof: N → N is defined by, gof(x) = g(f(x)) = g(x + 1) = (x + 1) - 1 = x $[\text{x}\in\text{N}\Rightarrow(\text{x}+1)>1]$ Then, it is clear that for $\text{y}\in\text{N},$ there exists $\text{x}=\text{y}\in\text{N}$ such that gof(x) = y. Hence, gof is onto.
View full question & answer→Question 923 Marks
Let $A = \{1, 2, 3\},$ and let $R_2 = \{(2, 2), (3, 1), (1, 3)\}.$ Find whether or not the relations $R_2$ on A is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_2 = \{(2, 2), (3, 1), (1, 3)\}$
$\therefore\ (1,1)\notin\text{R}_2$
$\Rightarrow R_1 $ is not reflexive.
Now, $(1,3)\in\text{R}_2$
$\Rightarrow\ (3,1)\in\text{R}_2$
$\therefore R_2$ is symmetric.
Again,
$(3,1)\in\text{R}_2$ and $(1,3)\in\text{R}_2$ but $(3,3)\notin\text{R}_2$
$\therefore R_2 $ is not transitive.
View full question & answer→Question 933 Marks
If f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, then find fog.
AnswerWe have, f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}
As,
fog(2) = f(g(2)) = f(5) = 2,
fog(3) = f(g(3)) = f(6) = 3,
Therefore, fog : {2, 3} → {2, 3} is defined as fog = {(2, 2), (3, 3)}
View full question & answer→Question 943 Marks
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if $\text{A}\subset\text{B}.$ Is R an equivalence relation on P(X)? Justify your answer.
AnswerSince every set is a subset of itself, ARA for all $\text{A}\in\text{P(X)}.$
$\therefore$ R is reflexive.
Let $\text{ARB}\Rightarrow\text{A}\subset\text{B}.$
This cannot be implied to $\text{B}\subset\text{A}.$
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
$\therefore$ R is not symmetric.
Further, if ARB and BRC, then $\text{A}\subset\text{B}\ \text{and}\ \text{B}\subset\text{C}.$
$\Rightarrow\text{A}\subset\text{C}$
$\Rightarrow\text{ARC}$
$\therefore$ R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.
View full question & answer→Question 953 Marks
Give examples of two one-one functions $f_1$ and $f_2$ from $R$ to $R,$ such that $f_1 + f_2 : R \rightarrow R.$ defined by $(f_1 + f_2)(x) = f_1(x) + f_2(x)$ is not one-one.
AnswerWe know that $f_1: R \rightarrow R$, given by $f_1(x)=x$, and $f_2(x)=-x$ are one-one.
Proving $f_1$ is one-one: Let $f_1(x)=f_1(y)$
$\Rightarrow x=y$
So, $\mathrm{f}_1$ is one-one.
Proving $\boldsymbol{f}_2$ is one-one: Let $f_2(x)=f_2(y)$
$\Rightarrow-x=-y $
$ \Rightarrow x=y$
So, $\mathrm{f}_2$ is one-one.
Proving $\left(f_1+f_2\right)$ is not one-one:
Given: $\left(f_1+f_2\right)(x)=f_1(x)+f_2(x)=x+(-x)=0$
So, for every real number $x,\left(f_1+f_2\right)(x)=0$
So, the image of ever number in the domain is same as 0 .
Thus, $\left(f_1+f_2\right)$ is not one-one.
View full question & answer→Question 963 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 973 Marks
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
AnswerA = {1, 2, 3, 4, 5} and R = {(a, b) : a − b is even} , then R = {(1, 3), (1, 5), (3, 5), (2, 4)}
-
| For (a, a), |a − a| = 0 which is even. |
$\therefore$ |
R is reflexive |
| If |a − b| is even, then |b − a| is also even. |
$\therefore$ |
R is symmetric |
| Now, if |a − b| and |b − c| is even then |a − b + b − c| ⇒ |a − c| |
|
|
| is also even. |
$\therefore$ |
R is transitive |
Therefore, R is an equivalence relation.
- Elements of {1, 3, 5} are related to each other.
Since |1− 3| = 2, |3− 5| = 2, |1− 5| = 4, all are even numbers
⇒ Elements of {1, 3, 5} are related to each other.
Similarly elements of (2, 4) are related to each other.
Since |2 − 4| = 2 an even number, then no element of the set {1, 3, 5} is related to any element of (2, 4). View full question & answer→Question 983 Marks
Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by $\text{A}*\text{B}=\text{A}\cap\text{B}\ \ \forall\ \text{A},\text{ B}$ in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *.
AnswerIt is given that *: P(X) × P(X) → P(X) is defined as $\text{A}*\text{B}=\text{A}\cap\text{B}\ \ \forall\ \text{A},\text{ B}\in\text{P(X)}$
We know that $\text{A}\cap\text{X}=\text{A}=\text{X}\cap\text{A}\forall\text{A}\in\text{P(X)}$
Thus, X is the identity element for the given binary operation *.
Now, an element $\text{A}\in\text{P(X)}$ is invertibleif there exists $\text{B}\in\text{P(X)}$ such that
A * B = X = B * A. (As X is the identity element)
i.e.,
$\text{A}\cap\text{B}=\text{X}=\text{B}\cap\text{A}$
This case is possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation *.
Hence, the given result is proved.
View full question & answer→Question 993 Marks
Show that the logarithmic function $\text{f}:\text{R}0^+\rightarrow \text{R}$ given by $f(x) = \log_a x, a > 0$ is a bijection.
AnswerWe have, $f : A \rightarrow B$ and $g : B \rightarrow C$ are one-one functions.
Now we have to prove: $gof : A \rightarrow C$ in one-one.
Let $\text{x, y}\in\text{A}$ such that
$gof(x) = gof(y)$
$\Rightarrow g(f(x)) = g(f(y))$
$\Rightarrow f(x) = f(y) [\because g$ in one-one$]$
$\Rightarrow x = y [\because f$ in one-one$]$
$\therefore$ gof is one-one function.
View full question & answer→Question 1003 Marks
Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
AnswerReflexivity: Let a be an arbitrary element of R. Then, a = a + 1 cannot be true for all $\text{a}\in\text{A.}$ $\Rightarrow\ (\text{a, a})\notin\text{R}$ So, R is not reflexive on A. Symmetry: Let $(\text{a, b})\in\text{R}$ ⇒ b = a + 1 ⇒ -a = -b + 1 ⇒ a = b - 1 Thus, $(\text{b, a})\notin\text{R}$ So, R is not symmetric on A.Transitivity: Let (1, 2) and (2, 3) $\in\text{R}$
⇒ 2 = 1 + 1 and 3 = 2 + 1 is true. But $3\neq1+1$ $\Rightarrow\ (1,3)\notin\text{R}$ So, R is not transitive on A.
View full question & answer→