2 Marks Questions

Question 12 Marks

Give an example of a relation which is,
Symmetric and transitive but not reflexive.

Answer

Let R be the relation on A such that,
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.

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Question 22 Marks

If $R$ is a symmetric relation on a set $A,$ then write a relation between $R$ and $R^{-1}.$

Answer

Here, $R$ is symmetric on the set $A.$
Let $(\text{a, b})\in\text{R}$
$\Rightarrow\ (\text{b, a})\in\text{R} [$Since $R$ is symmetric$]$
$\Rightarrow\ (\text{a, b})\in\text{R}^{-1} [$By definition of inverse relation$]$
$\Rightarrow\ \text{R}\subset\text{R}^{-1}$
Let $(\text{x, y})\in\text{R}^{-1}$
$\Rightarrow\ (\text{y, x})\in\text{R} [$By definition of inverse relation$]$
$\Rightarrow\ (\text{x, y})\in\text{R} [$Since $R$ is symmetric$]$
$\Rightarrow\ \text{R}^{-1}\subset\text{R}$
Thus, $\text{R}=\text{R}^{-1}$

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Question 32 Marks

For the set A = {1, 2, 3}, define a relation R on the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Write the ordered pairs to be added to R to make the smallest equivalence relation.

Answer

(3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.

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Question 42 Marks

Write the smallest equivalence relation on the set A = {1, 2, 3}.

Answer

The smallest equivalence relation on the set A = {1, 2, 3} is R = {(1, 1), (2, 2), (3, 3)}.

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Question 52 Marks

The following defines a relation on N:
xy is square of an integer, $\text{x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.

Answer

A relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$xy is square of an integer, $\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), \$8, 2), (16, 1), (1, 16), .....\big\}$ $$
This is reflexive as (1, 1), (2, 2), .... are present.
This is also symmetric because if aRb ⇒ bRa, for all $\text{a, b}\in\text{N.}$
This is transitive also because if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{N.}$

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Question 62 Marks

The following defines a relation on N:
$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$ $$
Determine which of the above relations are reflexive, symmetric and transitive.

Answer

A relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$

$(\text{x, y})\in\big\{(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3),\\ (4, 6), (6, 4), (5, 5)\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This only follows the condition of symmetric set as $(1,9)\in\text{R}$ also $(9,1)\in\text{R}$ Similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9), (9, 1)} $\in\text{R}$ but (1, 1) is absent.

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Question 72 Marks

Let R be the equivalence relation on the set Z of the integers given by R = {(a, b): 2 divides a - b}. Write the equivalence class [0].

Answer

$\text{a, b}\in\text{Z}$ and R is given by R = {(a, b): 2 divides a - b}.The equivalence classes can be taken as [0], [1].
Note that, $\text{for}\ 0\leq\text{i}\leq1,$ [i] = {2n + i: $\text{n}\in\text{Z}$}
So equivalence class [0] = {2n: $\text{n}\in\text{Z}$}
It is clear that all the elements of equivalence class [0] are even.
Hence, equivalence class $[0]=\{0,\pm2,\pm4,\pm6\ ...\}$

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Question 82 Marks

The following defines a relation on N:
$\text{x}>\text{y, x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.

Answer

A relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x}>\text{y, x, y}\in\text{N}$
$(\text{x, y})\in\big\{(2, 1), (3, 1),..., (3, 2), (4, 2),....\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This is not symmetric as (2, 1) is present but (1, 2) is absent.
This is transitive as $(3,2)\in\text{R}$ and $(2,1)\in\text{R}$ also $(3,1)\in\text{R},$ similarly this property satisfies all cases.

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Question 92 Marks

The following defines a relation on N:
x + 4y = 10, $\text{x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.

Answer

A relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x} + 4\text{y} = 10, \ \text{x, y}\in\text{N}$, $$
$(\text{x, y})\in\big\{(6, 1), (2, 2)\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This is also symmetric because $(6,1)\in\text{R}$ but (1, 6) is absent.
This is not transitive as there are only two elements in the set having no element common.

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Question 102 Marks

Define a symmetric relation.

Answer

A relation R on a set A is said to be symmetric if $\text{a, b}\in\text{R}$
Implies that, $\text{b, a}\in\text{R}$ for all $\text{a, b}\in\text{A}$
That is, aRb implies that bRa for all $\text{a, b}\in\text{A}$

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Question 112 Marks

Write the domain of the relation $R$ defined on the set $Z$ of integers as follows$: (a, b) \in R ⇔ a^2 + b^2 = 25$

Answer

We have,
$R = {(a, b) \in R ⇔ a^2 + b^2 = 25}$ be a relation on $Z.$
The domain of $R$ is the value of $'a\ ' \in Z,$ that satisfies $a^2 + b^2 = 25$
$a^2 + b^2 = 25$
$\Rightarrow\ \text{a}=\pm\sqrt{25-\text{b}^2}$
$\therefore$ Domain of $\text{R}=\{0,\pm3,\pm4,\pm5\}$

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Question 122 Marks

If R = {(x, y): x + 2y = 8} is a relation on N by, then write the range of R.

Answer

R = {(x, y): x + 2y = 8, $\text{x, y}\in\text{N}$}
Then, the values of y can be 1, 2, 3 only.
Also, y = 4 cannot result in x = 0 because x is a natural number.
Therefore, range of R is {1, 2, 3}.

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Question 132 Marks

State the reason for the relation R on the set {1, 2, 3} given by R = {(1, 2), (2, 1)} to be transitive.

Answer

Since $1,2\in\text{R,}$ $2,1\in\text{R}$ but $1,1\notin\text{R,}$ R is not transitive on the set 1, 2, 3. For R to be in a transitive relation, there must have $1,1\in\text{R.}$

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Question 142 Marks

Define an equivalence relation.

Answer

A relation R on a set A is said to be equivalence relation on a if R is:
Reflexive, Symmetric and Transitive.
R = {(x, y): x = y} on the set of real numbers is an equivalence relation.

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Question 152 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Symmetric but neither reflexive nor transitive.

Answer

The relation on A having properties of being symmetric, but neither reflexive nor transitive is,
R = {(1, 2), (2, 1)}
The relation R on A is neither reflexive nor transitive, but symmetric.

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Question 162 Marks

Let the relation R be defined on N by aRb if 2a + 3b = 30. Then write R as a set of ordered pairs.

Answer

If $\text{a, b}\in\text{N}$ then b must be an even integer so that $\text{a}\in\text{N}$
Hence only possible values for b are 2, 4, 6, 8.
if b = 2, it gives a = 12
if b = 4, it gives a = 9
if b = 6, it gives a = 6
if b = 8, it gives a = 3
Hence $(\text{a, b})\in\big\{(3, 8), (6, 6), (9, 4), (12, 2)\big\}$ $$

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Question 172 Marks

Let $R = (a, a^3): a$ is a prime number less than $5$ be a relation. Find the range of $R.$

Answer

We have,
$R = (a, a^3): a$ is a prime number less than $5$
Or, $R = \{(2, 8), (3, 27)\}$
So, the range of $R$ is $\{8, 27\}.$

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Question 182 Marks

Let the relation $R$ be defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 - b^2| < 8\}.$ Write $R$ as a set of ordered pairs.

Answer

Given$: A = \{1, 2, 3, 4, 5\} $
$R = \{(a, b): |a^2 - b^2| < 8\}$
$R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\}$

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Question 192 Marks

Write the smallest reflexive relation on set $A = \{1, 2, 3, 4\}.$

Answer

The smallest reflexive relation $R$ on any set $A$ is the identity relation $I_A$ on the set $A.$
We are given$, A = \{1, 2, 3, 4\}$
$\therefore R = \{(1, 1), (2, 2), (3, 3), (4, 4)\}$

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Question 202 Marks

If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by "is less than", write R as a set ordered pairs.

Answer

Since, R = x, y: x, y $\in\text{N}$ and x < y,
Hence, R = {(3, 4), (3, 9), (5, 9), (7, 9)}

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Question 212 Marks

Give an example of a relation which is,Reflexive and transitive but not symmetric.

Answer

Let R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.

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Question 222 Marks

Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.

Answer

We have,
A = {a, b, c} and R = {(a, a), (b, c), (a, b)}.
R can be a reflexive relation only when elements (b, b) and (c, c) are added to it.
R can be a transitive relation only when the element (a, c) is added to it.
So, the minimum number of ordered pairs to be added in R is 3.

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Question 232 Marks

Define a transitive relation.

Answer

A relation R on a set A is said to be transitive if
$(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, b, c}\in\text{R}$
i.e., aRb and bRc
⇒ aRc for all $\text{a, b, c}\in\text{R}$

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Question 242 Marks

If $R = \{(x, y): x^2 + y^2 \leq 4; x, y \in Z\}$ is a relation on $Z,$ write the domain of $R.$

Answer

Domain of $R$ is the set of values of $x$ that satisfies the relation $R.$
Because $x$ must be an integer, the provided values of $x$ are$:$
$0,\pm1,\pm2$
Thus, Domain of $R$ is $0,\pm1,\pm2$

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Question 252 Marks

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.

Answer

We have, A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)}
To make R transitive we shall add (1, 3) only.
$$ $\therefore \text{R}' = \big\{(1, 2), (1, 1), (2, 3), (1, 3)\big\}$

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Question 262 Marks

A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y): y is one half of x; x, y ∈ A} is a relation on A, then write R as a set of ordered pairs.

Answer

Since R = {(x, y): y is one half of x; x, y ∈ A}
So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}

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Question 272 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, symmetric and transitive.

Answer

The relation on A having properties of being symmetric, reflexive and transitive is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.

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Question 282 Marks

Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b if "a is a divisor of b". Write R as a set of ordered pairs.

Answer

We have, A = {2, 3, 4, 5}, B = {1, 3, 4} and relation from A to B is given by aRb if ''is divisor of'' B$\therefore$ R can be written as ordered pair as R = {(2, 4), (3, 3), (4, 4)}

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Question 292 Marks

Let $A = \{3, 5, 7\}, B = \{2, 6, 10\}$ and $R$ be a relation from $A$ to $B$ defined by $R = (x, y): x$ and $y$ are relatively prime. Then, write $R$ and $R^{-1}.$

Answer

$R = (x, y): x$ and $y$ are relatively prime.
Then,
$R = \{(3, 2), (5, 2), (7, 2), (3, 10), (7, 10), (5, 6), (7, 6)\}$
So, $R^{-1} = \{(2, 3), (2, 5), (2, 7), (10, 3), (10, 7), (6, 5), (6, 7)\}$

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Question 302 Marks

Write the identity relation on set A = {a, b, c}.

Answer

Identity set of A is:
I = {(a, a), (b, b), (c, c)}
Every element of this relation is related to itself.

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Question 312 Marks

Define a reflexive relation.

Answer

A relation R on a set A is said to be reflexive if every element of A is related to itself.
Mathematically, reflexive relation is written as R = {(a, a): for all $\text{a}\in\text{A}$}
For example if A = {1, 2, 3}, then a reflexive relation on A will be R = {(1, 1), (2, 2), (3, 3)}

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Question 322 Marks

If $A = \{2, 3, 4\}, B = \{1, 3, 7\}$ and $R = (x, y): x \in A, y \in B$ and $x < y$ is a relation from $A$ to $B,$ then write $R^{-1}.$

Answer

Since $R = (x, y): x \in A, y \in B$ and $x < y$
$R = \{(2, 3), (2, 7), (3, 7), (4, 7)\}$
Hence $, R^{-1} = \{(3, 2), (7, 2), (7, 3), (7, 4)\}$

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Question 332 Marks

Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.

Answer

No, it is not necessary that a relation which is symmetric and transitive is reflexive as well.
For Example, Let $A = \{a, b, c\}$ be a set and
$R_2 = \{(a, a)\}$ is a relation defined on $A.$
Clearly $,R_2$ is symmetric and transitive but not reflexive.

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Question 342 Marks

Give an example of a relation which is,
Transitive but neither reflexive nor symmetric.

Answer

Let R be the relation on A such that
R = {(1, 2), (2, 3), (1, 3)}
The relation R on A is transitive, but neither symmetric nor reflexive.

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Question 352 Marks

Let A = {0, 1, 2, 3} and R be a relation on A defined as R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}. Is R reflexive? symmetric? transitive?

Answer

We have, R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, R is a reflexive relation. Also, $(\text{a, b})\in\text{R}$ and $(\text{b, a})\in\text{R}$ So, R is a symmetric as well And, $(0,1)\in\text{R}$ but $(1,2)\notin\text{R}$ and $(2,3)\notin\text{R}$ So, R is not a transitive relation.

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Question 362 Marks

Let $R = \{(x, y): |x^2 - y^2| < 1\}$ be a relation on set $A = \{1, 2, 3, 4, 5\}.$ Write $R$ as a set of ordered pairs.

Answer

Given: $R = \{(x, y): |x^2 - y^2| < 1\}$ be a relation on $A = \{1, 2, 3, 4, 5\}$
Then$, R = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)\}$

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