Maths M.C.Q (1 Marks)

2. Reflexive, transitive but not symmetric.

Solution:

The relation S is reflexive, since for any $(\text{a, a})\in\text{S}$ the condition a2b holds,

The relation S is not symmetric since, for any ​​​​​​​$(\text{a, b}]\in\text{S}$ but $(\text{b, a})\notin\text{S}$

The relation S is transitive since, for any $(\text{a, b}]\in\text{S}$ and $(\text{b, c})\in\text{S}$

Therefore, $(\text{a, c})\notin\text{S}$

4. An equivalence relation.

Solution:

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

Then,

aa = a² > 0 $\Rightarrow\ \text{a, }\forall$

So, S is reflexive on R.

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

Then,

$\text{a, b}\in\text{S}\Rightarrow\ \text{ab}\geq0$ $\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{b, a}\in\text{S}\ \forall\ \text{a, b}\in\text{R}$

So, S is symmetric on R.

Transitivity: If $\text{a, b, b, c}\in\text{S}\Rightarrow\ \text{ab}\geq0$ and $\text{bc}\geq0\Rightarrow\ \text{ab}\times\text{bc}\geq0$ $\Rightarrow\ \text{ac}\geq0$

$\text{b}^2\geq0\Rightarrow\ \text{a, c}\in\text{S}$ for all $\text{a, b, c}\in\text{set R}$

Hence, S is an equivalence relation on R.

3. An equivalence relation.

Solution:

R = {(a, b): a = b and a, b $\in\text{A}$}

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

Here,

a = a

$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Let $\text{a, b}\in\text{A}$ such that $ (\text{a, b})\in\text{R}.$ Then,

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

$\Rightarrow\ \text{a}=\text{b}$

$\Rightarrow\ \text{b}=\text{a}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

Transitive: Let $\text{a, b, c}\in\text{A}$ such that $ (\text{a, b})\in\text{R}$ and $ (\text{b, c})\in\text{R}.$ Then,

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

and $ (\text{b, c})\in\text{R}\Rightarrow\ \text{b}=\text{c}$

$\Rightarrow\ \text{a}=\text{c}$

$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is transitive on A.

Hence, R is an equivalence relation on A.

2. Symmetric.

​​​​​​​Solution:

Given that L denote the set of all straight lines in a plane.

A relation R be defined by lRm if is perpendicular to m for all l, m ∈ L.

R is not reflexive. R is symmetric as we can say $\text{l}\bot\text{m}$ or $\text{m}\bot\text{l}.$

1. xRy : if $\text{x}\leq\text{y}$

​​​​​​​Solution:

In the set of Z of all integers xRy : if $\text{x}\leq\text{y}$ is not an equivalence relation.

For the relation $\text{x}\leq\text{y}(\text{x, y})\in\text{R}$ but (y, x) not belongs to y as $\text{y}\geq\text{x}$ given.

Hence, it is not an equivalence relation.

4. All the three options.

Solution:

R = a, b : a = b and $\text{a, b}\in\text{A}$

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

Then, a = a

$\Rightarrow\ \text{a, a}\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Let $\text{a, b}\in\text{A}$ such that $\text{a, b}\in\text{R.}$

Then, $\text{a, b}\in\text{R}$

⇒ a = b ⇒ b = a ⇒ b, $\text{a}\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

1. Symmetric and transitive only.

Solution:

Given that A = {a, b, c, d} then a relation R = {(a, b), (b, a), (a, a)} on A.

(a, b), (b, a) $\in\text{R}$

⇒ R is symmetric.

Also for (a, a) R is symmetric.

2. $\text{S}\subset\text{R}$

​​​​​​​Solution:

Given that R is the largest relation on A and S is any relation on A.

We know that R is always subset of A × A.

Hence, $\text{S}\subset\text{R}.$

3. $\{0,\pm3,\pm4,\pm5\}$

Solution:

As aRb ⇔ a < b

does not satisfy reflexive and symmetric relation.

3. {1}

Solution:

Here, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$ ⇒ R = 2, 1, 3, 1

Thus, Range of R = {1}

1. Symmetric.

Solution:

Given R is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$

It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$

2. R is reflexive and transitive but not symmetric.

Solution:

Reflexivity: Clearly, ​​​​​​​$(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,}$ R is not symmetric on A.

Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R},$ R is transitive on A.

4. Transitive but not symmetric.

Solution:

We have,

R = {(a, b): a is brother of b}

Let $(\text{a, b})\in\text{R}.$ Then,

a is brother of b.

but b is not necessary brother of a (As, b can be sister of a)

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

So, R is not symmetric.

Also,

Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$

⇒ a is brother of b and b is brother of c

⇒ a is brother of c

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

So, R is transitive.

3. {2, 4, 6}

Solution:

The relation R is defined as R = x, y: $\text{x, y}\in\text{N}$ and x + 2y = 8

⇒ R = x, y: $\text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$

Domain of R is all values of $\text{x}\in\text{N}$ satisfying the relation R.

Also, there are only three values of x that result in y, which is a natural number.

These are {2, 6, 4}.

3. An equivalence relation.

Solution:

We observe the following properties of relation R.

Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$

$\Rightarrow\ \text{a, b}\in\text{N}$

$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$

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

So, R is reflexive on N × N.

Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that (a, b)R(c, d)

$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$

$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$

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

So, R is symmetric on N × N.

Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that (a, b)R(c, d) and (c, d)R(e, f)

⇒ a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e

⇒ a + f = b + e

⇒ (a, b)R(e, f)

So, R is transitive on N × N.

Hence, R is an equivalence relation on N.​​​​​​​

2. Reflexive and symmetric.

Solution:

Reflexivity: Let $\text{x}\in\text{R.}$ Then,

$\text{x}-\text{x}=0<1$

$\Rightarrow\ |\text{x}-\text{x}|\leq1$

$\Rightarrow\ (\text{x, x})\in\text{R}$ for all $\text{x}\in\text{Z}$

So, R is reflexive on Z.

Symmetry: Let $\text{x, y}\in\text{R.}$ Then,

$|\text{x}-\text{y}|\leq0$

$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$

$\Rightarrow\ |(\text{y}-\text{x})|\leq1$ [Since |x - y| = |y - x|]

$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$

So, R is symmetric on Z.

Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,

$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$

⇒ It is not always true that $|\text{x}-\text{y}|\leq1.$

$\Rightarrow\ (\text{x, z})\notin\text{R}$

So, R is not transitive on Z.

1. Reflexive.

Solution:

We have,

$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ $$ is an irrational number, $\text{x, y}\in\text{R}\big\}$

As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$ which is an irrational number

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

So, R is reflexive relation.

Since, $\Big(\sqrt{2},2\Big)\in\text{R}$

i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$ which is an irrational number

but $2-\sqrt{2}+\sqrt{2}=2,$ which is a rational number

$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$

So, R is not symmetric relation.

Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$

$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$

So, R is not transitive relation.

4. $\text{i}\phi1$

Solution:

$\because\ |2+3\text{i}|=\sqrt{13}\neq13$

$|3|\neq-3$

$|1+\text{i}|=\sqrt{2}\neq2$

and $|\text{i}|=1$

So, $(\text{i, }1)\in\phi$

3. Transitive.

Solution:

Reflexivity: Since $(1, 1)\notin\text{B,}$ B is not reflexive on A.

Symmetry: Since $1,2\in\text{B}$ but $2,1\notin\text{B,}$ B is not symmetric on A.

Transitivity: Since $1,2\in\text{B},\ 2,3\in\text{B}$ and $1,3\in\text{B,}$ B is transitive on A.

1. Reflexive but not symmetric.

Solution:

We have,

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

As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive relation.

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

So, R is not symmetric relation.

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

So, R is transitive relation.

4. Reflexive, transitive but not symmetric.

Solution:

We have,

R = {(m, n): n divides m; m, n ∈ N}

As, m divides m

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

So, R is reflexive.

Since, $(2,1)\in\text{R}$ i.e. 1 divides 2

but 2 cannot divide 1 i.e. $(2,1)\notin\text{R}$

So, R is not symmetric.

Let $(\text{m, n})\in\text{R}$ and $(\text{n, p})\in\text{R.}$ Then,

n divides m and p divides n

⇒ p divides m

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

So, R is transitive.​​​​​​​

4. {2, 3, 4, 5}

Solution:

Given that relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. R can be written as,

{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Here we can see that domain means x element which is $2\leq\text{x}\leq5.$

Hence, {2, 3, 4, 5}

2. Reflexive.

Solution:

Reflexivity: Since $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A},$ R is reflexive on A.

Symmetry: Since $(\text{a, b})\in\text{R}$ but $(\text{b, a})\notin\text{R,}$ is not symmetric on A.

⇒ R is not antisymmetric on A.

Also, R is not an identity relation on A.

4. None of these.

Solution:

The relation R is defined as,

$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$

⇒ R = {(1, 3), (2, 6), (3, 9)}