Relations and Functions 1.1 NCERT Solutions

1. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A=\{1,2,3,\dots,13,14\} defined as

R=\{(x,y):3x-y=0\}

Solution

We are given

3x-y=0

or

y=3x

Now, x,y\in A=\{1,2,3,\dots,14\} .

Checking possible values of x :

x=1 \Rightarrow y=3;

x=2 \Rightarrow y=6; x=3 \Rightarrow y=9;

and x=4 \Rightarrow y=12

For x\geq5 , we get y>14 , which is not in the set A .

Therefore,

R=\{(1,3),(2,6),(3,9),(4,12)\}

Now, we check the three properties.

Reflexive:

A relation is reflexive if (a,a)\in R for every a\in A .

But ordered pairs like

(1,1),(2,2),(3,3)

are not in R .

Therefore, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Now,

(1,3)\in R

but

(3,1)\notin R

Therefore, the relation is not symmetric.

Transitive:

A relation is transitive if

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

Now,

(1,3)\in R \text{ and } (3,9)\in R

But

(1,9)\notin R

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

Another way to check it (without finding the ordered pairs):

The given relation is

3x-y=0 \Rightarrow y=3x

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (x,x)\in R for every x\in A .

For (x,x)\in R , we must have

3x-x=0

2x=0

This is possible only when x=0 , but 0\notin A .

Therefore, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(x,y)\in R \Rightarrow (y,x)\in R

Suppose (x,y)\in R . Then

y=3x

For (y,x)\in R , we must have

x=3y

But using y=3x , we get

x=3(3x)=9x

which is not true for all x .

Therefore, the relation is not symmetric.

Transitive:

Suppose

(x,y)\in R \text{ and } (y,z)\in R

Then

y=3x \text{ and } z=3y

Substituting y=3x in z=3y , we get

z=3(3x)=9x

For (x,z)\in R , we must have

z=3x

But z=9x , which is not equal to 3x in general.

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

(ii) Relation R in the set N of natural numbers defined as

R=\{(x,y): y=x+5 \text{ and } x<4\}

Solution

Since x<4 and x \in \mathbb{N} , the possible values of x are:

x=1,2,3

Using y=x+5 , we get:

R=\{(1,6),(2,7),(3,8)\}

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (x,x)\in R for every x\in\mathbb{N} .

But no ordered pair of the form (x,x) is present in R .

Therefore, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(x,y)\in R \Rightarrow (y,x)\in R

Here, (1,6)\in R , but (6,1)\notin R .

Therefore, the relation is not symmetric.

Transitive:

A relation is transitive if

(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R

In the relation R=\{(1,6),(2,7),(3,8)\} , there are no two ordered pairs such that the second element of one pair equals the first element of another pair.

Hence, the transitive condition is satisfied automatically.

Therefore, the relation is transitive.

Hence, the relation is transitive but neither reflexive nor symmetric.

(iii) Relation R in the set A=\{1,2,3,4,5,6\} defined as

R=\{(x,y): y \text{ is divisible by } x\}

Solution

Here, (x,y)\in R whenever x divides y .

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (x,x)\in R for every x\in A .

Every natural number divides itself, so

(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\in R

Therefore, the relation is reflexive.

Symmetric:

A relation is symmetric if

(x,y)\in R \Rightarrow (y,x)\in R

Now, (2,4)\in R because 4 is divisible by 2 .

But (4,2)\notin R because 2 is not divisible by 4 .

Therefore, the relation is not symmetric.

Transitive:

A relation is transitive if

(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R

Suppose

(x,y)\in R \text{ and } (y,z)\in R

Then y is divisible by x and z is divisible by y .

This implies that z is also divisible by x .

Hence, (x,z)\in R .

Therefore, the relation is transitive.

Hence, the relation is reflexive and transitive but not symmetric.

(iv) Relation R in the set \mathbb{Z} of all integers defined as

R=\{(x,y):x-y \text{ is an integer}\}

Solution

Since x,y\in\mathbb{Z} , the difference x-y is always an integer.

Therefore, every ordered pair of integers belongs to the relation R .

So, R is a universal relation on \mathbb{Z} .

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (x,x)\in R for every x\in\mathbb{Z} .

Now,

x-x=0

and 0 is an integer.

Therefore, (x,x)\in R for every x\in\mathbb{Z} .

Hence, the relation is reflexive.

Symmetric:

A relation is symmetric if

(x,y)\in R \Rightarrow (y,x)\in R

Suppose (x,y)\in R . Then

x-y is an integer.

Now,

y-x=-(x-y)

Since negative of an integer is also an integer, y-x is an integer.

Therefore, (y,x)\in R .

Hence, the relation is symmetric.

Transitive:

A relation is transitive if

(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R

Suppose

(x,y)\in R \text{ and } (y,z)\in R

Then

x-y and y-z are integers.

Adding these, we get

(x-y)+(y-z)=x-z

which is also an integer.

Therefore, (x,z)\in R .

Hence, the relation is transitive.

Hence, the relation is reflexive, symmetric and transitive.

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R=\{(x,y): x \text{ and } y \text{ work at the same place}\}

Solution

Reflexive:

Every person works at the same place as himself/herself.

Therefore, (x,x)\in R for every x\in A .

Hence, the relation is reflexive.

Symmetric:

If x works at the same place as y , then y also works at the same place as x .

Therefore, the relation is symmetric.

Transitive:

If x and y work at the same place, and y and z also work at the same place, then x and z work at the same place.

Therefore, the relation is transitive.

Hence, the relation is reflexive, symmetric and transitive.

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(b) R=\{(x,y): x \text{ and } y \text{ live in the same locality}\}

Solution

Reflexive:

Every person lives in the same locality as himself/herself.

Therefore, the relation is reflexive.

Symmetric:

If x lives in the same locality as y , then y also lives in the same locality as x .

Therefore, the relation is symmetric.

Transitive:

If x and y live in the same locality, and y and z also live in the same locality, then x and z live in the same locality.

Therefore, the relation is transitive.

Hence, the relation is reflexive, symmetric and transitive.

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(c) R=\{(x,y): x \text{ is exactly 7 cm taller than } y\}

Solution

Reflexive:

No person can be exactly 7 cm taller than himself/herself.

Therefore, the relation is not reflexive.

Symmetric:

If x is 7 cm taller than y , then y cannot be 7 cm taller than x .

Therefore, the relation is not symmetric.

Transitive:

If x is 7 cm taller than y , and y is 7 cm taller than z , then x is 14 cm taller than z , not 7 cm taller.

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

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(d) R=\{(x,y): x \text{ is wife of } y\}

Solution

Reflexive:

No person can be wife of himself/herself.

Therefore, the relation is not reflexive.

Symmetric:

If x is wife of y , then y is husband of x , not wife.

Therefore, the relation is not symmetric.

Transitive:

If x is wife of y , and y is wife of z , such a situation is not possible.

Hence, the transitive condition is not satisfied.

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

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(e) R=\{(x,y): x \text{ is father of } y\}

Solution

Reflexive:

No person can be father of himself.

Therefore, the relation is not reflexive.

Symmetric:

If x is father of y , then y cannot be father of x .

Therefore, the relation is not symmetric.

Transitive:

If x is father of y , and y is father of z , then x is grandfather of z , not father.

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

2. Show that the relation R in the set \mathbb{R} of real numbers, defined as

R=\{(a,b): a\leq b^2\}

is neither reflexive nor symmetric nor transitive.

Solution

We are given that

(a,b)\in R \iff a\leq b^2

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every a\in\mathbb{R} .

For (a,a)\in R , we must have

a\leq a^2

But this is not true for all real numbers.

For example, let a=\dfrac{1}{2} . Then

\dfrac{1}{2}\nleq \left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}

Therefore, (a,a)\notin R for all a\in\mathbb{R} .

Hence, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Let us take a=1 and b=2 .

Now,

1\leq 2^2=4

So, (1,2)\in R .

But for (2,1)\in R , we must have

2\leq 1^2=1

which is false.

Therefore, (2,1)\notin R .

Hence, the relation is not symmetric.

Transitive:

A relation is transitive if

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

Let us take

a=3,\quad b=2,\quad c=1.5

Then

3\leq 2^2=4

So, (3,2)\in R .

Also,

2\leq (1.5)^2=2.25

So, (2,1.5)\in R .

But for (3,1.5)\in R , we need

3\leq (1.5)^2=2.25

which is false.

Therefore, (3,1.5)\notin R .

Hence, the relation is not transitive.

Therefore, the relation is neither reflexive nor symmetric nor transitive.

3. Check whether the relation R defined in the set \{1,2,3,4,5,6\} as

R=\{(a,b): b=a+1\}

is reflexive, symmetric or transitive.

Solution

Since b=a+1 , the ordered pairs belonging to the relation are:

R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every element of the set.

But no ordered pair of the form (a,a) is present in R .

Therefore, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Now, (1,2)\in R , because

2=1+1

But (2,1)\notin R , since

1\neq2+1

Therefore, the relation is not symmetric.

Transitive:

A relation is transitive if

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

Now,

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

But (1,3)\notin R , because

3\neq1+1

Therefore, the relation is not transitive.

Hence, the relation is neither reflexive, nor symmetric, nor transitive.

4. Show that the relation R in \mathbb{R} defined as

R=\{(a,b): a\leq b\}

is reflexive and transitive but not symmetric.

Solution

We are given that

(a,b)\in R \iff a\leq b

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every a\in\mathbb{R} .

Now, for every real number a , we have

a\leq a

Therefore, (a,a)\in R for every a\in\mathbb{R} .

Hence, the relation is reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Let us take a=2 and b=5 .

Since

2\leq5

we have (2,5)\in R .

But

5\nleq2

Therefore, (5,2)\notin R .

Hence, the relation is not symmetric.

Transitive:

A relation is transitive if

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

Suppose

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

Then

a\leq b \text{ and } b\leq c

Using the transitive property of inequalities, we get

a\leq c

Therefore, (a,c)\in R .

Hence, the relation is transitive.

Therefore, the relation is reflexive and transitive but not symmetric.

5. Check whether the relation R in \mathbb{R} defined by

R=\{(a,b): a\leq b^3\}

is reflexive, symmetric or transitive.

Solution

We are given that

(a,b)\in R \iff a\leq b^3

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every a\in\mathbb{R} .

For (a,a)\in R , we must have

a\leq a^3

But this is not true for all real numbers.

For example, let a=\dfrac{1}{2} . Then

\dfrac{1}{2}\nleq\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}

Therefore, (a,a)\notin R for all a\in\mathbb{R} .

Hence, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Let us take a=1 and b=2 .

Now,

1\leq2^3=8

So, (1,2)\in R .

But for (2,1)\in R , we must have

2\leq1^3=1

which is false.

Therefore, (2,1)\notin R .

Hence, the relation is not symmetric.

Transitive:

A relation is transitive if

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

Suppose

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

Then

a\leq b^3 \text{ and } b\leq c^3

Since b\leq c^3 , cubing both sides gives

b^3\leq c^9

Thus, from a\leq b^3 , we get

a\leq c^9

But this does not necessarily imply that

a\leq c^3

Let us take

a=16,\quad b=4,\quad c=2

Then

16\leq4^3=64

So, (16,4)\in R .

Also,

4\leq2^3=8

So, (4,2)\in R .

But for (16,2)\in R , we need

16\leq2^3=8

which is false.

Therefore, (16,2)\notin R .

Hence, the relation is not transitive.

Therefore, the relation is neither reflexive, nor symmetric, nor transitive.

6. Show that the relation R in the set \{1,2,3\} given by

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

is symmetric but neither reflexive nor transitive.

Solution

We are given

A=\{1,2,3\}

and

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

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every element a of the set.

But the ordered pairs

(1,1),(2,2),(3,3)

are not present in R .

Therefore, the relation is not reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Now,

(1,2)\in R

and its reverse ordered pair

(2,1)\in R

Similarly, corresponding to (2,1)\in R , we have

(1,2)\in R .

Thus, whenever (a,b)\in R , we also have (b,a)\in R .

Therefore, the relation is symmetric.

Transitive:

A relation is transitive if

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

Now,

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

For transitivity, this should imply that

(1,1)\in R

But (1,1)\notin R .

Therefore, the relation is not transitive.

Hence, the relation is symmetric but neither reflexive nor transitive.

7. Show that the relation R in the set A of all the books in a library of a college, given by

R=\{(x,y): x \text{ and } y \text{ have same number of pages}\}

is an equivalence relation.

Solution

We are given that two books x and y are related if they have the same number of pages.

To prove that R is an equivalence relation, we must show that it is reflexive, symmetric and transitive.

Reflexive:

A relation is reflexive if (x,x)\in R for every x\in A .

Every book has the same number of pages as itself.

Therefore, (x,x)\in R for every x\in A .

Hence, the relation is reflexive.

Symmetric:

A relation is symmetric if

(x,y)\in R \Rightarrow (y,x)\in R

Suppose (x,y)\in R .

Then books x and y have the same number of pages.

Therefore, book y also has the same number of pages as book x .

Hence, (y,x)\in R .

Therefore, the relation is symmetric.

Transitive:

A relation is transitive if

(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R

Suppose

(x,y)\in R \text{ and } (y,z)\in R

Then books x and y have the same number of pages.

Also, books y and z have the same number of pages.

Therefore, books x and z also have the same number of pages.

Hence, (x,z)\in R .

Therefore, the relation is transitive.

Since the relation is reflexive, symmetric and transitive, R is an equivalence relation.

8. Show that the relation R in the set A=\{1,2,3,4,5\} given by

R=\{(a,b): |a-b| \text{ 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\} .

Solution

We are given

A=\{1,2,3,4,5\}

and

R=\{(a,b): |a-b| \text{ is even}\}

Recall that an even number is divisible by 2 .

Now, we check the three properties one by one.

Reflexive:

A relation is reflexive if (a,a)\in R for every a\in A .

Now,

|a-a|=0

and 0 is even.

Therefore, (a,a)\in R for every a\in A .

Hence, the relation is reflexive.

Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Suppose (a,b)\in R .

Then

|a-b| is even.

But

|b-a|=|a-b|

So, |b-a| is also even.

Therefore, (b,a)\in R .

Hence, the relation is symmetric.

Transitive:

A relation is transitive if

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

Suppose

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

Then |a-b| and |b-c| are even.

Therefore, there exist integers k and l such that

a-b=2k \quad \text{and} \quad b-c=2l

Adding these equations, we get

(a-b)+(b-c)=2k+2l

a-c=2(k+l)

Since k+l is an integer, a-c is even.

Hence, |a-c| is also even.

Therefore, (a,c)\in R .

So, the relation is transitive.

Since the relation is reflexive, symmetric and transitive, R is an equivalence relation.

Now, let us examine the related elements.

For the set \{1,3,5\} :

|1-3|=2,\quad |1-5|=4,\quad |3-5|=2

All these differences are even.

Therefore, all the elements of \{1,3,5\} are related to each other.

For the set \{2,4\} :

|2-4|=2

which is even.

Therefore, the elements 2 and 4 are related to each other.

Now, consider one element from \{1,3,5\} and one element from \{2,4\} .

For example,

|1-2|=1,\quad |3-4|=1,\quad |5-2|=3

All these differences are odd.

Therefore, no element of \{1,3,5\} is related to any element of \{2,4\} .

9. Show that each of the relations R in the set

A=\{x\in\mathbb{Z}:0\leq x\leq12\}

given by

(i) R=\{(a,b): |a-b| \text{ is a multiple of } 4\}

(ii) R=\{(a,b): a=b\}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution

Here,

A=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}

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(i) R=\{(a,b): |a-b| \text{ is a multiple of } 4\}

We check the three properties one by one.

Reflexive:

For every a\in A ,

|a-a|=0

and 0 is a multiple of 4 .

Therefore, (a,a)\in R for every a\in A .

Hence, the relation is reflexive.

Symmetric:

Suppose (a,b)\in R .

Then

|a-b| is a multiple of 4 .

But

|b-a|=|a-b|

Therefore, |b-a| is also a multiple of 4 .

Hence, (b,a)\in R .

Therefore, the relation is symmetric.

Transitive:

Suppose

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

Then |a-b| and |b-c| are multiples of 4 .

Therefore, there exist integers k and l such that

a-b=4k \quad \text{and} \quad b-c=4l

Adding these equations, we get

(a-b)+(b-c)=4k+4l

a-c=4(k+l)

Since k+l is an integer, a-c is a multiple of 4 .

Hence, |a-c| is also a multiple of 4 .

Therefore, (a,c)\in R .

So, the relation is transitive.

Since the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Elements related to 1 :

We need all elements x\in A such that

|x-1| is a multiple of 4 .

Checking the elements of A , we get

|1-1|=0,\quad |5-1|=4,\quad |9-1|=8

All are multiples of 4 .

Therefore, the set of all elements related to 1 is

\{1,5,9\}

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(ii) R=\{(a,b): a=b\}

This relation is the equality relation.

We again check the three properties.

Reflexive:

For every a\in A ,

a=a

Therefore, (a,a)\in R for every a\in A .

Hence, the relation is reflexive.

Symmetric:

If (a,b)\in R , then

a=b

Therefore,

b=a

Hence, (b,a)\in R .

Therefore, the relation is symmetric.

Transitive:

If

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

then

a=b \text{ and } b=c

Therefore,

a=c

Hence, (a,c)\in R .

Therefore, the relation is transitive.

Since the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Elements related to 1 :

An element x is related to 1 only if

x=1

Therefore, the set of all elements related to 1 is

\{1\}

10. Give an example of a relation which is

• (i) symmetric but neither reflexive nor transitive.
• (ii) transitive but neither reflexive nor symmetric.
• (iii) reflexive and symmetric but not transitive.
• (iv) reflexive and transitive but not symmetric.
• (v) symmetric and transitive but not reflexive.

Solution

(i) Symmetric but neither reflexive nor transitive

Let

A=\{1,2,3\}

and

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

Symmetric:

Since (1,2)\in R , and (2,1)\in R , the relation is symmetric.

Not reflexive:

The ordered pairs (1,1),(2,2),(3,3) are not in R .

Therefore, the relation is not reflexive.

Not transitive:

Since (1,2)\in R and (2,1)\in R , transitivity would require (1,1)\in R .

But (1,1)\notin R .

Therefore, the relation is not transitive.

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(ii) Transitive but neither reflexive nor symmetric

Let

A=\{1,2,3\}

and

R=\{(1,2)\}

Transitive:

There are no ordered pairs of the form (2,a) , so the transitive condition is automatically satisfied.

Therefore, the relation is transitive.

Not reflexive:

The ordered pairs (1,1),(2,2),(3,3) are not in R .

Hence, the relation is not reflexive.

Not symmetric:

Since (1,2)\in R but (2,1)\notin R , the relation is not symmetric.

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(iii) Reflexive and symmetric but not transitive

Let

A=\{1,2,3\}

and

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

Reflexive:

All ordered pairs of the form (a,a) are present.

Therefore, the relation is reflexive.

Symmetric:

Whenever (a,b)\in R , the reverse ordered pair (b,a) is also in R .

Therefore, the relation is symmetric.

Not transitive:

Since (1,2)\in R and (2,3)\in R , transitivity would require (1,3)\in R .

But (1,3)\notin R .

Therefore, the relation is not transitive.

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(iv) Reflexive and transitive but not symmetric

Let

A=\{1,2,3\}

and

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

Reflexive:

All ordered pairs of the form (a,a) are present.

Therefore, the relation is reflexive.

Transitive:

The only non-trivial ordered pair is (1,2) .

Since there is no ordered pair of the form (2,a) except (2,2) , the transitive condition is satisfied.

Therefore, the relation is transitive.

Not symmetric:

Since (1,2)\in R but (2,1)\notin R , the relation is not symmetric.

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(v) Symmetric and transitive but not reflexive

Let

A=\{1,2,3\}

and

R=\{(1,1)\}

Symmetric:

Since (1,1) is its own reverse ordered pair, the relation is symmetric.

Transitive:

Since

(1,1)\in R \text{ and } (1,1)\in R

implies (1,1)\in R , the relation is transitive.

Not reflexive:

The ordered pairs (2,2) and (3,3) are not in R .

Therefore, the relation is not reflexive.

There can be many different examples for each type of relation. Try making your own examples and checking their properties — it greatly improves conceptual understanding.

11. Show that the relation R in the set A of points in a plane given by

R=\{(P,Q): OP=OQ\}

(distance of point P from the origin is same as the distance of point Q from the origin), is an equivalence relation. Further, show that the set of all points related to a point P\neq(0,0) is the circle passing through P with origin as centre.

Solution

Let O(0,0) be the origin and OP denote the distance of point P from the origin.

We are given:

(P,Q)\in R \iff OP = OQ

We check reflexive, symmetric and transitive properties.

Reflexive:

For any point P ,

OP = OP

Hence (P,P)\in R . So, R is reflexive.

Symmetric:

If (P,Q)\in R , then OP = OQ .

So, OQ = OP , hence (Q,P)\in R .

Therefore, R is symmetric.

Transitive:

If (P,Q)\in R and (Q,R)\in R , then

OP = OQ \quad \text{and} \quad OQ = OR

So, OP = OR , hence (P,R)\in R .

Therefore, R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

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Set of all points related to a fixed point P \neq (0,0)

Let P(x_1,y_1) . Any point Q(x,y) is related to P if:

OQ = OP

Now,

OQ = \sqrt{x^2+y^2}, \quad OP = \sqrt{x_1^2+y_1^2}

So,

\sqrt{x^2+y^2} = \sqrt{x_1^2+y_1^2}

Squaring both sides, we get:

x^2+y^2=x_1^2+y_1^2

This is of the form

x^2+y^2=r^2

which represents a circle with centre at the origin and radius

r=\sqrt{x_1^2+y_1^2}

Hence, the set of all points related to P is the circle passing through P with origin as centre.

12. Show that the relation R defined in the set A of all triangles as

R=\{(T_1,T_2): T_1 \text{ is similar to } T_2\}

is an equivalence relation. Also consider three right-angled triangles T_1, T_2 and T_3 with sides:

T_1: 3,4,5 \quad T_2: 5,12,13 \quad T_3: 6,8,10

Find which triangles are related.

Solution

We are given:

(T_1,T_2)\in R \iff T_1 \text{ is similar to } T_2

We verify reflexive, symmetric and transitive properties.

Reflexive:

Every triangle is similar to itself.

T_1 \sim T_1

Hence (T_1,T_1)\in R . So, R is reflexive.

Symmetric:

If (T_1,T_2)\in R , then T_1 \sim T_2 .

Similarity of triangles is mutual, hence

T_2 \sim T_1

So, (T_2,T_1)\in R . Therefore, R is symmetric.

Transitive:

If (T_1,T_2)\in R and (T_2,T_3)\in R , then

T_1 \sim T_2 \quad \text{and} \quad T_2 \sim T_3

Hence,

T_1 \sim T_3

So, (T_1,T_3)\in R . Therefore, R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

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Checking which triangles are related

Compare T_1(3,4,5) and T_3(6,8,10) :

\frac{6}{3}=\frac{8}{4}=\frac{10}{5}=2

So, corresponding sides are proportional.

Hence, T_1 \sim T_3 and (T_1,T_3)\in R .

Now compare T_1(3,4,5) and T_2(5,12,13) :

\frac{5}{3}\neq\frac{12}{4}\neq\frac{13}{5}

Therefore, T_1 \not\sim T_2 .

Similarly,

\frac{6}{5}\neq\frac{8}{12}\neq\frac{10}{13}

So, T_2 \not\sim T_3 .

Final Answer:

Only T_1 and T_3 are related.

(T_1,T_3)\in R \quad \text{and} \quad (T_3,T_1)\in R

13. Show that the relation R defined in the set A of all polygons as

R=\{(P_1,P_2): P_1 \text{ and } P_2 \text{ have same number of sides}\}

is an equivalence relation. Also find the set of all elements in A related to the right-angled triangle T with sides 3, 4 and 5.

Solution

We are given:

(P_1,P_2)\in R \iff P_1 \text{ and } P_2 \text{ have same number of sides}

We verify reflexive, symmetric and transitive properties.

Reflexive:

Every polygon has the same number of sides as itself.

P_1 \text{ has same number of sides as } P_1

Hence (P_1,P_1)\in R . So, R is reflexive.

Symmetric:

If (P_1,P_2)\in R , then P_1 and P_2 have the same number of sides.

Therefore, P_2 and P_1 also have the same number of sides.

Hence (P_2,P_1)\in R . Therefore, R is symmetric.

Transitive:

If (P_1,P_2)\in R and (P_2,P_3)\in R , then

P_1 and P_2 have same number of sides, and P_2 and P_3 also have same number of sides.

Therefore, P_1 and P_3 have same number of sides.

Hence (P_1,P_3)\in R . Therefore, R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

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Set of all elements related to triangle T(3,4,5)

The triangle T has 3 sides.

Any polygon related to T must also have 3 sides.

Therefore, the set of all elements in A related to T is the set of all triangles.

Final Answer:

The equivalence class of T is the set of all triangles in A .

14. Let L be the set of all lines in XY plane and R be the relation in L defined as

R=\{(L_1,L_2): L_1 \text{ is parallel to } L_2\}

Show that R is an equivalence relation. Also find the set of all lines related to the line y = 2x + 4 .

Solution

We are given:

(L_1,L_2)\in R \iff L_1 \parallel L_2

We verify reflexive, symmetric and transitive properties.

Reflexive:

Every line is parallel to itself.

L_1 \parallel L_1

Hence (L_1,L_1)\in R . So, R is reflexive.

Symmetric:

If (L_1,L_2)\in R , then L_1 \parallel L_2 .

Parallelism is mutual, so L_2 \parallel L_1 .

Hence (L_2,L_1)\in R . Therefore, R is symmetric.

Transitive:

If (L_1,L_2)\in R and (L_2,L_3)\in R , then

L_1 \parallel L_2 \quad \text{and} \quad L_2 \parallel L_3

Hence, L_1 \parallel L_3 .

So, (L_1,L_3)\in R . Therefore, R is transitive.

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

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Set of all lines related to y = 2x + 4

Any line parallel to y = 2x + 4 must have the same slope.

The slope of y = 2x + 4 is 2 .

Therefore, all lines related to it are of the form:

y = 2x + c \quad \text{where } c\in\mathbb{R}

Final Answer:

The equivalence class of the line y = 2x + 4 is

\{\, y = 2x + c : c\in\mathbb{R} \,\}

15. Let R be the relation in the set \{1,2,3,4\} given by

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

• (A) R is reflexive and symmetric but not transitive.
• (B) R is reflexive and transitive but not symmetric.
• (C) R is symmetric and transitive but not reflexive.
• (D) R is an equivalence relation.

Solution

Given set:

A=\{1,2,3,4\}

We check reflexive, symmetric and transitive properties.

1. Reflexive:

For a relation to be reflexive, all ordered pairs

(1,1),(2,2),(3,3),(4,4)

must belong to R .

All these ordered pairs are present in R .

Therefore, R is reflexive.

2. Symmetric:

A relation is symmetric if

(a,b)\in R \Rightarrow (b,a)\in R

Now,

(1,2)\in R

but

(2,1)\notin R

Therefore, R is not symmetric.

3. Transitive:

A relation is transitive if

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

Checking important chains:

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

So, we must have

(1,2)\in R

which is present in R .

Also,

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

implies

(1,2)\in R

which is true.

Similarly, all other required transitive conditions are satisfied.

Therefore, R is transitive.

Conclusion:

The relation R is reflexive and transitive but not symmetric.

✅️ Correct Option:

(B)

16. Let R be the relation in the set N given by

R=\{(a,b): a=b-2,\; b>6\}

Choose the correct option.

• (A) (2,4)\in R
• (B) (3,8)\in R
• (C) (6,8)\in R
• (D) (8,7)\in R

Solution

We check each ordered pair using the condition:

a=b-2 \quad \text{and} \quad b>6

(A) (2,4)

Here, b=4 , but

4\not>6

Therefore, (2,4)\notin R .

(B) (3,8)

Here, b=8>6 .

Now,

b-2=8-2=6

But a=3\neq6 .

Therefore, (3,8)\notin R .

(C) (6,8)

Here, b=8>6 .

Also,

b-2=8-2=6=a

Therefore, (6,8)\in R .

(D) (8,7)

Here, b=7>6 .

But

b-2=7-2=5\neq8

Therefore, (8,7)\notin R .

✅️ Final Answer:

(C) (6,8)\in R