3 Marks

Question 13 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, transitive but not symmetric.

Answer

The relation on A having properties of being reflexive, transitive, but not symmetric is,

R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}

Relation R satisfies reflexivity and transitivity.

$ \Rightarrow(1, 1), (2, 2), (3, 3) \in\text{R}$$$

and $(1, 1), (2, 1) \in \text{R}\Rightarrow(1, 1)\in \text{R}$

However, $(2,1)\in\text{R},$ but $(1,2)\notin\text{R}$

View full question & answer→

Question 23 Marks

Let A = {1, 2, 3}, and let R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}. Find whether or not the relations R₁on A is:

Reflexive.
Symmetric.
Transitive.

Answer

We have A = {1, 2, 3}, and R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}
$\therefore$ (1, 1), (2, 2) and (3, 3) $\in\text{R}_1$
$\therefore$ R₁ is not reflexive.
Now,
$\therefore\ (2,1)\in\text{R}_1$ but $(1,2)\notin\text{R}_1$
$\therefore$ R₁ is not symmetric.
Again,
$\therefore\ (2,1)\in\text{R}_1$ and $(1,3)\in\text{R}_1$ but $(2,3)\notin\text{R}_1$
$\therefore$ R₁ is not transitive.

View full question & answer→

Question 33 Marks

Let A = {1, 2, 3}, and let R₃ = {(1, 3), (3, 3)}. Find whether or not the relations R₃on A is:

Reflexive.
Symmetric.
Transitive.

Answer

R₃ = {(1, 3), (3, 3)}
$\therefore\ (1,1)\notin\text{R}_3$
⇒ R₃ is not reflexive.
Now, $(1,3)\in\text{R}_3$ but $(3,1)\in\text{R}_3$
$\therefore$ R₃ is not symmetric.
Again, it is clear that R₃ is transitive.

View full question & answer→

Question 43 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.

Answer

We 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 53 Marks

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

Answer

Let 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 63 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.

Answer

Let 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 73 Marks

Three relation R₂is defined in set A = {a, b, c} as follows:
R₂ = {(a, a)}
Find whether or not the relation R₂on A is:

Reflexive.
Symmetric.
Transitive.

Answer

R₂ is Reflexive: Clearly $\text{a, a}\in\text{R}_2$
Therefore, R₂ is reflexive.
Symmetric: Clearly, $\text{a, a}\in\text{R}\Rightarrow\ \text{a, a}\in\text{R}.$
Therefore, R₂ is symmetric.
Transitive: R₂ is clearly a transitive relation, since there is only one element in it.

View full question & answer→

Question 83 Marks

Show that the relation $''\geq''$ on the set R of all real numbers is reflexive and transitive but not symmetric.

Answer

We have,
relation $\text{R}=\ ''\geq''$ on the set R of all real numbers
Reflexivity: Let $\text{a}\in\text{R}$
$\Rightarrow\ \text{a}\geq\text{a}$
$\Rightarrow\ ''\geq''$ is reflexive.
Symmetric: Let $\text{a, b}\in\text{R}$
Such that $\text{a}\geq\text{b}\Rightarrow\ \text{b}\geq\text{a}$
$\therefore\ ''\geq''$ not symmetric.
Transitivity: Let $\text{a, b, c}\in\text{R}$
and $\text{a}\geq\text{b}\ \&\ \text{b}\geq\text{c}$
$\Rightarrow\ \text{a}\geq\text{c}$
$\Rightarrow\ ''\geq''$ is transitive.

View full question & answer→

Question 93 Marks

Let A = {1, 2, 3}, and let R₂ = {(2, 2), (3, 1), (1, 3)}. Find whether or not the relations R₂on A is:

Reflexive.
Symmetric.
Transitive.

Answer

R₂ = {(2, 2), (3, 1), (1, 3)}
$\therefore\ (1,1)\notin\text{R}_2$
⇒ R₁ is not reflexive.
Now, $(1,3)\in\text{R}_2$
$\Rightarrow\ (3,1)\in\text{R}_2$
$\therefore$ R₂ is symmetric.
Again,
$(3,1)\in\text{R}_2$ and $(1,3)\in\text{R}_2$ but $(3,3)\notin\text{R}_2$
$\therefore$ R₂ is not transitive.

View full question & answer→

Question 103 Marks

Give an example of a relation which is,

Symmetric but neither reflexive nor transitive.

Answer

Let A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as $(5, 5), (6, 6), (7, 7)\notin\text{R.}$ $$
Now, as $(5, 6)\in\text{R}$ and also $(6,5)\in\text{R,}$ R is symmetric.
$\Rightarrow(5, 6), (6, 5)\in\text{R,}\text{but}(5, 5)\notin\text{R}$ $$ $$ $$
Therefore, R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.

View full question & answer→

Question 113 Marks

Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Answer

Let A be a set.

Then $\text{I}_\text{A}=\{(\text{a, a});\text{ a}\in\text{A}\}$ is the identity relation on A.

Hence, every identity relation on a set is reflexive by defination.

Converse:

Let A = {(a, b, c)} be a set.

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation defined on A.

Clearly R is reflexive on set A, but it is not identity relation on set A as $(\text{a, b})\in\text{R}$

Hence, a reflexive relation need not be identity relation.

View full question & answer→

Question 123 Marks

The following relation are defined on the set of real numbers.
aRb if a - b > 0
Find whether these relations are reflexive, symmetric or transitive.

Answer

aRb if a - b > 0

Let R be the set of real numbers.

Reflexivity: Let $\text{a}\in\text{R}$

$\Rightarrow\ \text{a}-\text{a}=0$

$\Rightarrow\ (\text{a, a})\notin\text{R}$

$\therefore$ R is not reflexive.

Symmetric: Let aRb

⇒ a - a > 0

⇒ b - a < 0

$\therefore$ R is not symmetric.

Transitive: Let aRb and bRc

⇒ a - a > and b - c > 0

⇒ a - c > 0

⇒ aRc

$\therefore$ R is transitive.

View full question & answer→

Question 133 Marks

An integer m is said to be related to another integer n if m is a multiple of n.Check if the relation is symmetric, reflexive and transitive.

Answer

$\text{R} = \big\{{(\text{m, n})}(1, 1), (2, 1)\Rightarrow(1, 1): \text{m, n} \in\text{Z,}\text{m} \\=\text{kn},\text{where}\text{k}\in\text{N}\big\}$$$$$$$ $$ $$
$$Reflexivity:

Let m be an arbitrary element of R. Then,

m = km is true for k = 1

$\Rightarrow\ (\text{m, m})\in\text{R}$

Thus, R is reflexive.

Symmetry: Let $(\text{m, n})\in\text{R}$

⇒ m = kn for some $\text{k}\in\text{N}$

$\Rightarrow\ \text{n}=\frac{1}{\text{k}}\text{m}$

$\Rightarrow\ (\text{n, m})\notin\text{R}$

Thus, R is not symmetric.

Transitivity: Let (m, n) and (n, o) $\in\text{R}$

⇒ m = kn and n = lo for some $\text{k, l}\in\text{N}$

⇒ m = (kl)o

Here, $\text{kl}\in\text{N}$

$\Rightarrow\ (\text{m, o})\in\text{R}$

Thus, R is transitive.

View full question & answer→

Question 143 Marks

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmeteric, transitive and reflexive.

Answer

We have,
R = {(1, 2), (2, 3)}
R can be a transitive only when the elements (1, 3) is added
R can be a reflexive only when the elements (1, 1), (2, 2), (3, 3) are added
R can be a symmetric only when the elements (2, 1), (3, 1) and (3, 2) are added
So, the required enlarged relation, R' = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} = A × A

View full question & answer→

Question 153 Marks

Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x is father of and y}

Answer

A be the set of human beings
R = {(x, y): x is father of and y}
Reflexive: Since x can not be father of x
$\therefore\ (\text{x, x})\notin\text{R}$
⇒ R is not reflexive.
Symmetric: Let $(\text{x, y})\in\text{R}$
⇒ x is father of y
⇒ y can not be father of x
$\Rightarrow\ (\text{y, x})\notin\text{R}$
⇒ R is not Symmetric.
Transitive: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}$
⇒ x is father of y and y is father of z
⇒ x is grandfather of z
$\Rightarrow\ (\text{x, z})\notin\text{R}$
⇒ R is not transitive.

View full question & answer→

Question 163 Marks

Three relation R₄is defined in set A = {a, b, c} as follows:
R₄ = {(a, b), (b, c), (c, a)}
Find whether or not the relation R₄on A is:

Reflexive.
Symmetric.
Transitive.

Answer

R₄ is reflexive: Here, $\text{a}\notin\text{R}_4,\ \text{b},\ \text{b}\notin\text{R}_4,\ \text{c},\ \text{c}\notin\text{R}_4$
Therefore, R₄ is not reflexive.
Symmetric: Here, $\text{a, b}\notin\text{R}_4,$ but $\text{b, b}\notin\text{R}_4$
Therefore, R₄ is not symmetric.
Transitive: Here, $\text{a, b}\notin\text{R}_4,\ \text{b, c}\in\text{R}_4$
But $\text{a, c}\notin\text{R}_4$
Therefore, R₄ is not transitive.

View full question & answer→

Question 173 Marks

Check whether the relation R on R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

Answer

The relation R on R is defined by R.

We observe that $(-2)\leq(-2)^3$ is not true.

Therefore, R is not reflexive.

Since $1\leq\Big(3^{\frac{1}{3}}\Big)^3$ but $3^\frac{1}{3}\leq1$ i.e. $\Big(1,3^\frac{1}{3}\Big)\in\text{R.}$

Therefore, R is not symmetric.

Hence, R is not transitive because $(5,2)\in\text{R}$ and $\Big(2,2^\frac{1}{3}\Big)\in\text{R}$ but $\Big(5,2^\frac{1}{3}\Big)\notin\text{R.}$

View full question & answer→

Question 183 Marks

The following relation are defined on the set of real numbers.
aRb if 1 + ab > 0
Find whether these relation are reflexive, symmetric or transitive.

Answer

We have aRb if 1 + ab > 0
Let R be the set of real numbers
Reflexive: Let $\text{a}\in\text{R}$
⇒ 1 + a² > 0
⇒ aRa
⇒ R is reflexive.
Symmetric: Let aRb
⇒ 1 + ab > 0
⇒ 1 + ba > 0
⇒ bRa
⇒ R is symmetric
Transitive: Let aRb and bRc
⇒ 1 + ab > 0 and 1 + bc > 0
⇒ 1 + ac > 0
⇒ R is not transitive.

View full question & answer→

Question 193 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.

Answer

Reflexivity: 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→

Question 203 Marks

Test whether the following relations R₂are:

Reflexive.
Symmetric.
Transitive.

R₂ on Z defined by $(\text{a, b})\in\text{R}_2\Leftrightarrow\ |\text{a}-\text{b}|\leq5$

Answer

Reflexivity: Let a be an arbitrary element of R₂. Then,

$\text{a}\in\text{R}_2$

$\Rightarrow\ |\text{a}-\text{a}|=0\leq5$

So, R₂ is reflexive.

Symmetry: Let $(\text{a, b})\in\text{R}_2$

$\Rightarrow\ |\text{a}-\text{b}|\leq5$

$\Rightarrow\ |\text{b}-\text{a}|\leq5$ [Since, |a - b| = |b - a|]

$\Rightarrow\ (\text{b, a})\in\text{R}_2$

So, R₂ is symmetric.

Transitivity: Let $(1, 3)\in\text{R}_2$ and $(3,7)\in\text{R}_2$

$\Rightarrow\ |1-3|\leq5$ and $|3-7|\leq5$

But $|1-7|\nleq5$

$\Rightarrow\ (1,7)\notin\text{R}_2$

So, R₂ is transitive.

View full question & answer→

Question 213 Marks

Three relation R₁is defined in set A = {a, b, c} as follows:
R₁ = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
Find whether or not the relation R₁on A is:

Reflexive.
Symmetric.
Transitive.

Answer

Consider R, Reflexive: Clearly, (a, a), (b, b) and (c, c) $\in\text{R}_1$

Therefore, R₁ is reflexive.

Symmetric: We see that the ordered pairs obtained by interchanging the components of R₁ are also in R₁

Therefore,

R₁is 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, R₁ is transitive.

View full question & answer→

Question 223 Marks

Three relation R₃is defined in set A = {a, b, c} as follows:
R₃ = {(b, c)}
Find whether or not the relation R₃on A is:

Reflexive.
Symmetric.
Transitive.

Answer

R₃ is Reflexive: Here $\text{b, b}\notin\text{R}_3$ neither $\text{c, c}\notin\text{R}_3$

Therefore, R₃ is not reflexive.

Symmetric: Here, $\text{b, c}\notin\text{R}_3,$ but $\text{c, c}\notin\text{R}_3$

So, R₃ is not symmetric.

Transitive: Here, R₃ has only two elements.

Hence, R₃ is transitive.

View full question & answer→

Maths 3 Marks

Test yourself on this topic