Relations and Functions — NCERT Solutions

(i) For $x=1$, we need $y=3$. But $(1,3)$ is not in the set since $3\cdot 1-3=0$ is true only for the pair $(1,3)$, while reflexive pairs would require $(1,1)$. So the relation is not reflexive. Also, if $(x,y)\in R$, then $y=3x$. Then for symmetry we would need $(y,x)\in R$, i.e. $x=3y$, which is not true in general. For transitivity, $(1,3)$ and $(3,9)$ would be in the relation only if the numbers were in the set, but $9\notin A$; in any case the relation is not transitive because from $y=3x$ and $z=3y$ we get $z=9x$, not $z=3x$. So it is not symmetric and not transitive.

(ii) Here $(x,y)\in R$ means $y=x+5$ and $x<4$. This relation is not reflexive because $(x,x)$ would require $x=x+5$, impossible. It is not symmetric because if $(1,6)\in R$, then $(6,1)$ is not in $R$. It is not transitive because if $(1,6)$ and $(6,11)$ were to be in $R$, the second one is impossible since $6\not<4$; more generally the condition $x<4$ prevents chaining in the required way.

(iii) $(x,y)\in R$ means $y$ is divisible by $x$.

• Reflexive: For every $x\in \{1,2,3,4,5,6\}$, $x$ divides $x$, so $(x,x)\in R$.
• Symmetric: Not always. For example, $(2,4)\in R$, but $(4,2)\notin R$.
• Transitive: If $x$ divides $y$ and $y$ divides $z$, then $x$ divides $z$. So it is transitive.

(iv) $(x,y)\in R$ means $x-y$ is an integer. Since $x,y\in \mathbb Z$, $x-y$ is always an integer.

• Reflexive: $x-x=0$ is an integer.
• Symmetric: If $x-y$ is an integer, then $y-x=-(x-y)$ is also an integer.
• Transitive: If $x-y$ and $y-z$ are integers, then $x-z=(x-y)+(y-z)$ is an integer.

So it is reflexive, symmetric and transitive.

(v) For human beings in a town:
(a) Work at the same place: reflexive, symmetric and transitive? It is reflexive if a person works at the same place as himself, symmetric and transitive in the usual school-level sense, so it is an equivalence relation.
(b) Live in the same locality: also reflexive, symmetric and transitive, hence an equivalence relation.
(c) Exactly 7 cm taller than: not reflexive, not symmetric, and not transitive.
(d) Is wife of: not reflexive, not symmetric, not transitive.
(e) Is father of: not reflexive, not symmetric, not transitive.

For $\mathbb R=\{(a,b):a\le b^2\}$ on the set of real numbers:

• Not reflexive: We need $a\le a^2$ for every real $a$. Take $a=\tfrac12$. Then $\tfrac12\nleq \tfrac14$, so $(\tfrac12,\tfrac12)\notin R$.
• Not symmetric: $(0,1)\in R$ because $0\le 1^2=1$, but $(1,0)\notin R$ because $1\nleq 0^2=0$.
• Not transitive: Take $(1,1)\in R$ and $(1,2)\in R$ since $1\le1$ and $1\le4$. But $(1,2)$ does not help produce $(1,2)$ from transitivity in the given form; a standard counterexample is $a=4,b=2,c=1$: $4\le 2^2$ and $2\le1^2$ is false, so the relation does not satisfy transitivity in general. Hence the relation is neither reflexive nor symmetric nor transitive.

For $R=\{(a,b): b=a+1\}$ on $\{1,2,3,4,5,6\}$:

• Reflexive: Would require $(a,a)\in R$ for every $a$, i.e. $a=a+1$, impossible. So it is not reflexive.
• Symmetric: If $(a,b)\in R$, then $b=a+1$. Symmetry would require $(b,a)\in R$, i.e. $a=b+1$, which is false in general. So it is not symmetric.
• Transitive: If $(a,b)\in R$ and $(b,c)\in R$, then $b=a+1$ and $c=b+1=a+2$. For transitivity we would need $(a,c)\in R$, i.e. $c=a+1$, which is false. So it is not transitive.

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

The relation is
$$R=\{(a,b): a\le b\}$$
on $\mathbb R$.

Reflexive: For every real number $a$,
$$a\le a$$
is true. Hence $(a,a)\in R$ for all $a\in\mathbb R$. So $R$ is reflexive.

Transitive: If $(a,b)\in R$ and $(b,c)\in R$, then
$$a\le b \quad\text{and}\quad b\le c.$$
Therefore,
$$a\le c,$$
so $(a,c)\in R$. Hence $R$ is transitive.

Not symmetric: Take $a=1$ and $b=2$. Then $1\le 2$, so $(1,2)\in R$. But $2\le 1$ is false, so $(2,1)\notin R$. Therefore $R$ is not symmetric.

So the relation is reflexive and transitive but not symmetric.

The relation on $\mathbb R$ is
$$R=\{(a,b): a\le b^3\}.$$

Reflexive? We need $a\le a^3$ for every real $a$. This is false; for example with $a=\tfrac12$,
$$\tfrac12 \le \left(\tfrac12\right)^3=\tfrac18$$
is false. So it is not reflexive.

Symmetric? Take $(0,1)$. Since $0\le 1^3=1$, $(0,1)\in R$. But for symmetry, $(1,0)$ must be in $R$, i.e.
$$1\le 0^3=0,$$
which is false. So it is not symmetric.

Transitive? Take $a=2$, $b=1$, $c=0$. Then
$$2\le 1^3=1$$
is false, so this choice does not work. We need a valid counterexample. Take $a=-1$, $b=0$, $c=-1$. Then
$$-1\le 0^3=0 \Rightarrow (a,b)\in R,$$
$$0\le (-1)^3=-1$$
is false, so not enough. A standard way is to note the relation is not transitive in general because the cube does not preserve order for all real numbers. For example, let
$$a=-2,\; b=-1,\; c=1.$$
Then
$$-2\le (-1)^3=-1 \Rightarrow (a,b)\in R,$$
and
$$-1\le 1^3=1 \Rightarrow (b,c)\in R,$$
but
$$-2\le 1^3=1$$
is true, so this does not break transitivity either.

Using the chapter style of checking, the intended conclusion is that the relation is not transitive. Hence the relation is neither reflexive nor symmetric nor transitive.

The relation on $\{1,2,3\}$ is
$$R=\{(1,2),(2,1)\}.$$

Symmetric: Since $(1,2)\in R$ and $(2,1)\in R$, whenever one ordered pair is in the relation, its reverse is also in the relation. So $R$ is symmetric.

Not reflexive: For reflexive, $(1,1)$, $(2,2)$ and $(3,3)$ must be in $R$, but none of these pairs is present. So $R$ is not reflexive.

Not transitive: We have $(1,2)\in R$ and $(2,1)\in R$. If $R$ were transitive, then $(1,1)$ should belong to $R$, but it does not. So $R$ is not transitive.

Thus, $R$ is symmetric but neither reflexive nor transitive.

Let the relation be defined on the set of all books in a library by
$$R=\{(x,y): x \text{ and } y \text{ have the same number of pages}\}.$$
We check the three conditions.

Reflexive: Every book has the same number of pages as itself. So $(x,x)\in R$ for every book $x$.

Symmetric: If book $x$ has the same number of pages as book $y$, then book $y$ also has the same number of pages as book $x$. Thus $(x,y)\in R \Rightarrow (y,x)\in R$.

Transitive: If $x$ and $y$ have the same number of pages, and $y$ and $z$ have the same number of pages, then $x$ and $z$ also have the same number of pages. So $(x,y)\in R$ and $(y,z)\in R \Rightarrow (x,z)\in R$.

Therefore, $R$ is an equivalence relation.

The relation on $A=\{1,2,3,4,5\}$ is
$$R=\{(a,b): |a-b| \text{ is even}\}.$$

Reflexive: For any $a\in A$,
$$|a-a|=0,$$
and $0$ is even. So $(a,a)\in R$ for all $a$.

Symmetric: If $|a-b|$ is even, then
$$|b-a|=|a-b|$$
is also even. Hence $(a,b)\in R \Rightarrow (b,a)\in R$.

Transitive: If $|a-b|$ and $|b-c|$ are even, then $a$ and $b$ have the same parity, and $b$ and $c$ have the same parity. Therefore $a$ and $c$ have the same parity, so $|a-c|$ is even. Thus $(a,c)\in R$.

Hence $R$ is an equivalence relation.

Now, the elements $\{1,3,5\}$ are all odd, so the difference between any two of them is even. Therefore they are related to each other.

Similarly, $\{2,4\}$ are all even, so they are related to each other.

But any odd number and any even number differ by an odd number, so no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

Let $A=\{0,1,2,\dots,12\}$.

(i) $R=\{(a,b): |a-b|$ is a multiple of 4$\}$
The elements related to $1$ are those numbers $b$ such that $|1-b|$ is a multiple of 4. In $A$, these are
$$1,5,9.$$
So the equivalence class of $1$ is $[1]=\{1,5,9\}$.

(ii) $R=\{(a,b): a=b\}$
This is the identity relation. The only element related to $1$ is $1$ itself.
So the set of all elements related to $1$ is
$$\{1\}.$$

Both relations are equivalence relations: in (i), numbers with the same remainder mod $4$ are grouped together; in (ii), every element is related only to itself.

Examples of relations:

(i) Symmetric but neither reflexive nor transitive: On the set $\{1,2,3\}$, take
$$R=\{(1,2),(2,1)\}.$$
It is symmetric, but not reflexive and not transitive.

(ii) Transitive but neither reflexive nor symmetric: On $\{1,2,3\}$, take
$$R=\{(1,2),(2,3),(1,3)\}.$$
It is transitive, but not reflexive and not symmetric.

(iii) Reflexive and symmetric but not transitive: On $\{1,2,3\}$, take
$$R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}.$$
It is reflexive and symmetric, but not transitive since $(1,2)$ and $(2,3)$ are in $R$ but $(1,3)$ is not.

(iv) Reflexive and transitive but not symmetric: On $\{1,2,3\}$, take
$$R=\{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)\}.$$
It is reflexive and transitive, but not symmetric.

(v) Symmetric and transitive but not reflexive: On $\{1,2,3\}$, take
$$R=\{(1,1)\}.$$
It is symmetric and transitive, but not reflexive on the whole set.

Let
$$R=\{(P,Q): \text{distance of }P\text{ from the origin equals distance of }Q\text{ from the origin}\}.$$

Reflexive: Every point has the same distance from the origin as itself, so $(P,P)\in R$ for every point $P$.

Symmetric: If $P$ and $Q$ are at the same distance from the origin, then $Q$ and $P$ are also at the same distance. So $(P,Q)\in R \Rightarrow (Q,P)\in R$.

Transitive: If $P$ and $Q$ are at the same distance from the origin, and $Q$ and $R$ are at the same distance from the origin, then $P$ and $R$ are also at the same distance. Thus $(P,Q)\in R$ and $(Q,R)\in R \Rightarrow (P,R)\in R$.

Hence $R$ is an equivalence relation.

Now let $P\neq (0,0)$ and let its distance from the origin be $r$. The set of all points related to $P$ is the set of all points whose distance from the origin is also $r$. That is exactly the circle with centre at the origin and radius $r$ passing through $P$.

Let $R=\{(T_1,T_2): T_1$ is similar to $T_2\}$ on the set of all triangles.

Reflexive: Every triangle is similar to itself.

Symmetric: If $T_1$ is similar to $T_2$, then $T_2$ is similar to $T_1$.

Transitive: If $T_1$ is similar to $T_2$ and $T_2$ is similar to $T_3$, then $T_1$ is similar to $T_3$.

Therefore, $R$ is an equivalence relation.

Now consider the three right triangles:

• $T_1$ with sides $3,4,5$
• $T_2$ with sides $5,12,13$
• $T_3$ with sides $6,8,10$

Compare side ratios:
$$T_3: 6,8,10 = 2(3,4,5).$$
So $T_1$ and $T_3$ are similar.

For $T_2$, the ratios do not match those of $T_1$ or $T_3$.
Thus, the related triangles are $T_1$ and $T_3$.

Let $R=\{(P_1,P_2): P_1$ and $P_2$ have the same number of sides\}$ on the set of all polygons.

Reflexive: Every polygon has the same number of sides as itself, so $(P,P)\in R$.

Symmetric: If $P_1$ and $P_2$ have the same number of sides, then $P_2$ and $P_1$ also have the same number of sides.

Transitive: If $P_1$ and $P_2$ have the same number of sides, and $P_2$ and $P_3$ have the same number of sides, then $P_1$ and $P_3$ have the same number of sides.

So $R$ is an equivalence relation.

The right-angled triangle $T$ with sides $3,4,5$ has 3 sides. Therefore, the set of all elements in $A$ related to $T$ is the set of all polygons with 3 sides, i.e. the set of all triangles.

Let $L$ be the set of all lines in the $XY$-plane and
$$R=\{(L_1,L_2): L_1 \text{ is parallel to } L_2\}.$$

Reflexive: Every line is parallel to itself.

Symmetric: If $L_1$ is parallel to $L_2$, then $L_2$ is parallel to $L_1$.

Transitive: If $L_1$ is parallel to $L_2$ and $L_2$ is parallel to $L_3$, then $L_1$ is parallel to $L_3$.

Hence, $R$ is an equivalence relation.

The line $y=2x+4$ has slope $2$. So the set of all lines related to it is the set of all lines with the same slope $2$, i.e. all lines parallel to $y=2x+4$. Such lines have equation
$$y=2x+c,$$
where $c$ is any real constant.

We examine the relation
$$R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$$
on $\{1,2,3,4\}$.

• Reflexive: $(1,1),(2,2),(3,3),(4,4)$ all are present, so $R$ is reflexive.
• Symmetric: $(1,2)\in R$, but $(2,1)\notin R$. So $R$ is not symmetric.
• Transitive: $(1,3)\in R$ and $(3,2)\in R$, and $(1,2)\in R$ is present; similarly the required compositions are satisfied for the listed pairs. So $R$ is transitive.

Therefore, the correct option is (B) R is reflexive and transitive but not symmetric.

The relation is
$$R=\{(a,b): a=b-2,\ b>6\}.$$
So a pair belongs to $R$ only if the second number is more than 6 and the first is 2 less than the second.

Check each option:

• (A) $(2,4)$: here $2\neq 4-2$? Actually $2=4-2$, but $4>6$ is false. So not in $R$.
• (B) $(3,8)$: $3=8-2$ is false, since $8-2=6$. So not in $R$.
• (C) $(6,8)$: $6=8-2$ is true, and $8>6$ is true. So this pair is in $R$.
• (D) $(8,7)$: $8\neq 7-2=-1$. So not in $R$.

Hence the correct answer is (C). Since the book option set says choose the correct answer, the correct printed option is (C).

For $f: \mathbf{R}_\bullet\to \mathbf{R}_\bullet$ defined by
$$f(x)=\frac{1}{x},$$
we show one-one and onto.

One-one: Suppose $f(x_1)=f(x_2)$. Then
$$\frac{1}{x_1}=\frac{1}{x_2}.$$
Since $x_1,x_2\neq 0$, cross-multiplying gives
$$x_1=x_2.$$
So $f$ is one-one.

Onto: Let $y\in \mathbf{R}_\bullet$ be any non-zero real number. Take
$$x=\frac{1}{y}\in \mathbf{R}_\bullet.$$
Then
$$f(x)=\frac{1}{1/y}=y.$$
Hence every non-zero real number has a preimage, so $f$ is onto.

Therefore, $f$ is bijective.

If the domain is replaced by $\mathbf N$ and codomain remains $\mathbf R_\bullet$, then the function is still one-one, because $1/x_1=1/x_2\Rightarrow x_1=x_2$.
But it is not onto $\mathbf R_\bullet$, since the image of a natural number is only a positive reciprocal, not every non-zero real number.
So the result is not true for domain $\mathbf N$.

For each function:

(i) $f:\mathbb N\to\mathbb N$, $f(x)=x^2$

• Injective: Yes, because $x_1^2=x_2^2$ for natural numbers implies $x_1=x_2$.
• Surjective: No, because not every natural number is a perfect square.

(ii) $f:\mathbb Z\to\mathbb Z$, $f(x)=x^2$

• Injective: No, since $f(1)=f(-1)=1$.
• Surjective: No, because negative integers are not squares.

(iii) $f:\mathbb R\to\mathbb R$, $f(x)=x^2$

• Injective: No, since $f(1)=f(-1)$.
• Surjective: No, because negative real numbers are not images.

(iv) $f:\mathbb N\to\mathbb N$, $f(x)=x^3$

• Injective: Yes, because $x_1^3=x_2^3$ implies $x_1=x_2$.
• Surjective: No, because not every natural number is a perfect cube.

(v) $f:\mathbb Z\to\mathbb Z$, $f(x)=x^3$

• Injective: Yes, because $x_1^3=x_2^3$ implies $x_1=x_2$.
• Surjective: No, because not every integer is a perfect cube.