Relations and Functions

From the chapter, these are the relations in Exercise 1.1, Q1. We check each using the definitions of reflexive, symmetric and transitive.

(i) $\mathbb R=\{(x,y):3x-y=0\}$ on $\{1,2,\dots,14\}$.
- For reflexive, we need $(x,x)$ for every $x$, i.e. $3x-x=0\Rightarrow 2x=0$, impossible for positive integers. So it is not reflexive.
- If $(x,y)\in R$, then $y=3x$. For symmetry we would need $(y,x)\in R$, i.e. $x=3y=9x$, impossible in general. So not symmetric.
- If $(x,y)\in R$ and $(y,z)\in R$, then $y=3x$ and $z=3y=9x$. For transitivity we would need $z=3x$, which is not true in general. So not transitive.

(ii) $\mathbb R=\{(x,y):y=x+5\text{ and }x<4\}$ on $\mathbb N$.
- Not reflexive, since $(x,x)$ would require $x=x+5$, impossible.
- Not symmetric, since if $y=x+5$ then generally $x\neq y+5$.
- Not transitive. For example, $(1,6)\in R$ because $6=1+5$ and $1<4$, but there is no pair from $6$ to another number in the relation, so the transitive condition fails in general.

(iii) $\mathbb R=\{(x,y):y$ is divisible by $x\}$ on $\{1,2,3,4,5,6\}$.
- Reflexive: every number divides itself, so $(x,x)\in R$ for all $x$.
- Symmetric: not true. For example, $2$ divides $4$, so $(2,4)\in R$, but $4$ does not divide $2$.
- Transitive: if $x$ divides $y$ and $y$ divides $z$, then $x$ divides $z$. So it is transitive.

(iv) $\mathbb R=\{(x,y):x-y\text{ is an integer}\}$ on $\mathbb Z$.
For any integers $x,y$, the difference $x-y$ is always an integer. Hence this relation is actually the universal relation on $\mathbb Z$.
- So it is reflexive, symmetric, and transitive.

(v) On human beings in a town:
- (a) same place of work: generally not reflexive? Actually a person works at the same place as himself, so reflexive. Symmetric is true. Transitive is true if “same place” means exactly the same workplace. Hence it is an equivalence relation.
- (b) live in the same locality: reflexive, symmetric, and transitive, so an equivalence relation.
- (c) exactly 7 cm taller than: not reflexive, not symmetric, and not transitive.
- (d) x is wife of y: not reflexive, not symmetric, and not transitive.
- (e) x is father of y: not reflexive, not symmetric, and not transitive.

The relation on $\mathbb R$ is
$$R=\{(a,b):a\le b^2\}.$$
We check the three properties.

Not reflexive: For reflexive, we need $a\le a^2$ for every real $a$. Take $a=\tfrac12$. Then
$$\tfrac12 \le \left(\tfrac12\right)^2=\tfrac14$$
is false. So $R$ is not reflexive.

Not symmetric: Take $(0,1)$. Since $0\le 1^2=1$, we have $(0,1)\in R$. But for symmetry we would need $(1,0)\in R$, i.e. $1\le 0^2=0$, which is false. So $R$ is not symmetric.

Not transitive: Take $a=1$, $b=-1$, $c=0$. Then
$$1\le (-1)^2=1 \quad\Rightarrow\quad (1,-1)\in R,$$
and
$$-1\le 0^2=0 \quad\Rightarrow\quad (-1,0)\in R.$$
But
$$1\le 0^2=0$$
is false, so $(1,0)\notin R$. Hence $R$ is not transitive.

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

The relation on $\{1,2,3,4,5,6\}$ is
$$R=\{(a,b): b=a+1\}.$$
We test each property.

Reflexive: We would need $(a,a)\in R$ for every $a$, i.e. $a=a+1$, impossible. So it is not reflexive.

Symmetric: If $(1,2)\in R$ because $2=1+1$, then symmetry would require $(2,1)\in R$, but $1\neq 2+1$. So it is not symmetric.

Transitive: If $(1,2)\in R$ and $(2,3)\in R$, transitivity would require $(1,3)\in R$. But $3\neq 1+1$. 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 $\{2,4\}$.

For $A=\{x\in \mathbb Z:0\le x\le 12\}$, the two relations are from Exercise 1.1, Q9.

(i) $R=\{(a,b): |a-b|$ is a multiple of $4\}$.
- Reflexive: $|a-a|=0$, and $0$ is a multiple of $4$.
- Symmetric: $|a-b|=|b-a|$.
- Transitive: if $a-b$ and $b-c$ are multiples of $4$, then $(a-c)=(a-b)+(b-c)$ is also a multiple of $4$.
So it is an equivalence relation.

Elements related to $1$ are those $x$ such that $|x-1|$ is a multiple of $4$.
Within $0\le x\le 12$, these are
$$1,5,9.$$
So the equivalence class of $1$ is $\{1,5,9\}$.

(ii) $R=\{(a,b): a=b\}$.
- Reflexive: every $a$ equals itself.
- Symmetric: if $a=b$, then $b=a$.
- Transitive: if $a=b$ and $b=c$, then $a=c$.
So it is an equivalence relation.

Elements related to $1$ are only those equal to $1$.
So the set is
$$\{1\}.$$

We can give standard examples from the chapter.

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

(ii) Transitive but neither reflexive nor symmetric: On $\{1,2,3\}$, take
$$R=\{(1,2),(2,3),(1,3)\}.$$
This 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 because $(1,2)$ and $(2,3)$ are in $R$ while $(1,3)$ is not.

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

(v) Symmetric and transitive but not reflexive: The empty relation on a non-empty set is symmetric and transitive, but not reflexive.

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 $\mathbf N$.

The visible part of Exercise 1.2, Q2 asks to check injectivity and surjectivity of the listed functions:

(i) $f:\mathbf N\to\mathbf N$, $f(x)=x^2$.
- One-one: yes, because if $x_1^2=x_2^2$ for natural numbers, then $x_1=x_2$.
- Onto: no, because not every natural number is a perfect square.

(ii) $f:\mathbf Z\to\mathbf Z$, $f(x)=x^2$.
- One-one: no, because $f(1)=1=f(-1)$.
- Onto: no, because negative integers are not squares.

(iii) $f:\mathbf R\to\mathbf R$, $f(x)=x^2$.
- One-one: no, because $f(1)=f(-1)=1$.
- Onto: no, because negative real numbers are not images.

(iv) $f:\mathbf N\to\mathbf N$, $f(x)=x^3$.
- One-one: yes, because $x^3$ is strictly increasing on $\mathbf N$.
- Onto: no, because not every natural number is a perfect cube.

(v) $f:\mathbf Z\to\mathbf Z$, $f(x)=x^3$.
- One-one: yes, because $x^3$ is strictly increasing on integers.
- Onto: no, because not every integer is a perfect cube.

So the answers are: (i) one-one only; (ii) neither; (iii) neither; (iv) one-one only; (v) one-one only.