Chapter 1: Relations and Functions

Recall that a relation $$R$$ in a set $$A$$ is the empty relation if $$R = \phi \subset A \times A$$, and the universal relation if $$R = A \times A$$.

Showing $$R$$ is the empty relation. Here $$A$$ is the set of all students of a boys school. Take any two students $$a, b \in A$$. Since every member of $$A$$ is a boy, $$a$$ can never be the sister of $$b$$. So no ordered pair $$(a, b)$$ satisfies the defining condition of $$R$$.

Hence $$R$$ contains no element, i.e. $$R = \phi$$. Therefore $$R$$ is the empty relation.

Showing $$R'$$ is the universal relation. The height of any student of the school is certainly less than $$3$$ metres. Hence for any two students $$a, b \in A$$, the difference between their heights is also less than $$3$$ metres.

So every ordered pair $$(a, b)$$ with $$a, b \in A$$ satisfies the defining condition, i.e. $$(a, b) \in R'$$ for all $$a, b \in A$$.

Therefore $$R' = A \times A$$, which means $$R'$$ is the universal relation.

We verify the three properties of an equivalence relation for $$R = \{(T_1, T_2) : T_1 \text{ is congruent to } T_2\}$$ on the set $$T$$ of all triangles.

Reflexive: Every triangle is congruent to itself, since it has exactly the same sides and angles as itself. So $$(T_1, T_1) \in R$$ for every $$T_1 \in T$$. Hence $$R$$ is reflexive.

Symmetric: Let $$(T_1, T_2) \in R$$. Then $$T_1$$ is congruent to $$T_2$$. If $$T_1$$ is congruent to $$T_2$$, then $$T_2$$ is congruent to $$T_1$$. So $$(T_2, T_1) \in R$$. Hence $$R$$ is symmetric.

Transitive: Let $$(T_1, T_2) \in R$$ and $$(T_2, T_3) \in R$$. Then $$T_1$$ is congruent to $$T_2$$ and $$T_2$$ is congruent to $$T_3$$. It follows that $$T_1$$ is congruent to $$T_3$$, i.e. $$(T_1, T_3) \in R$$. Hence $$R$$ is transitive.

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

Let $$R = \{(L_1, L_2) : L_1 \text{ is perpendicular to } L_2\}$$ on the set $$L$$ of all lines in a plane.

Symmetric: Let $$(L_1, L_2) \in R$$, so $$L_1$$ is perpendicular to $$L_2$$. Perpendicularity is a mutual property, so $$L_2$$ is also perpendicular to $$L_1$$. Hence $$(L_2, L_1) \in R$$, and $$R$$ is symmetric.

Not reflexive: A line cannot be perpendicular to itself, because perpendicular lines meet at $$90^\circ$$ while a line makes an angle $$0^\circ$$ with itself. So $$(L_1, L_1) \notin R$$, and $$R$$ is not reflexive.

Not transitive: Take three lines $$L_1, L_2, L_3$$ in the plane with $$L_1$$ perpendicular to $$L_2$$ and $$L_2$$ perpendicular to $$L_3$$. In a plane, if both $$L_1$$ and $$L_3$$ are perpendicular to the same line $$L_2$$, then $$L_1$$ is parallel to $$L_3$$, so $$L_1$$ is not perpendicular to $$L_3$$. Thus $$(L_1, L_2) \in R$$ and $$(L_2, L_3) \in R$$ but $$(L_1, L_3) \notin R$$. Hence $$R$$ is not transitive.

Therefore $$R$$ is symmetric but neither reflexive nor transitive.

The set is $$A = \{1, 2, 3\}$$ and $$R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$$.

Reflexive: For $$R$$ to be reflexive we need $$(a, a) \in R$$ for every $$a \in A$$. Here $$(1, 1), (2, 2)$$ and $$(3, 3)$$ all belong to $$R$$. Hence $$R$$ is reflexive.

Not symmetric: We have $$(1, 2) \in R$$, but the reverse pair $$(2, 1) \notin R$$. Since a required reverse pair is missing, $$R$$ is not symmetric.

Not transitive: We have $$(1, 2) \in R$$ and $$(2, 3) \in R$$. Transitivity would require $$(1, 3) \in R$$, but $$(1, 3) \notin R$$. Hence $$R$$ is not transitive.

Therefore $$R$$ is reflexive but neither symmetric nor transitive.

Let $$R = \{(a, b) : 2 \text{ divides } a - b\}$$ on the set $$\mathbf{Z}$$ of integers.

Reflexive: For any $$a \in \mathbf{Z}$$, $$a - a = 0$$ and $$2$$ divides $$0$$. So $$(a, a) \in R$$, and $$R$$ is reflexive.

Symmetric: Let $$(a, b) \in R$$. Then $$2$$ divides $$a - b$$, so $$a - b = 2k$$ for some integer $$k$$. Hence $$b - a = -(a - b) = -2k = 2(-k)$$, and since $$-k$$ is an integer, $$2$$ divides $$b - a$$. So $$(b, a) \in R$$, and $$R$$ is symmetric.

Transitive: Let $$(a, b) \in R$$ and $$(b, c) \in R$$. Then $$2$$ divides $$a - b$$ and $$2$$ divides $$b - c$$, so $$a - b = 2k$$ and $$b - c = 2l$$ for some integers $$k, l$$. Adding these, $$a - c = (a - b) + (b - c) = 2k + 2l = 2(k + l)$$. Since $$k + l$$ is an integer, $$2$$ divides $$a - c$$. So $$(a, c) \in R$$, and $$R$$ is transitive.

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

Here $$A = \{1, 2, 3, 4, 5, 6, 7\}$$ and $$R = \{(a, b) : a \text{ and } b \text{ are both odd or both even}\}$$.

Reflexive: For any $$a \in A$$, the elements $$a$$ and $$a$$ are obviously of the same parity (both odd or both even). So $$(a, a) \in R$$, and $$R$$ is reflexive.

Symmetric: If $$(a, b) \in R$$, then $$a$$ and $$b$$ are of the same parity. This is the same statement as "$$b$$ and $$a$$ are of the same parity". So $$(b, a) \in R$$, and $$R$$ is symmetric.

Transitive: Let $$(a, b) \in R$$ and $$(b, c) \in R$$. Then $$a, b$$ have the same parity and $$b, c$$ have the same parity. Hence $$a$$ and $$c$$ both share the parity of $$b$$, so they have the same parity. So $$(a, c) \in R$$, and $$R$$ is transitive.

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

Elements of $$\{1, 3, 5, 7\}$$: All elements of this subset are odd. Any two of them are both odd, hence related. So all elements of $$\{1, 3, 5, 7\}$$ are related to each other.

Elements of $$\{2, 4, 6\}$$: All elements of this subset are even. Any two of them are both even, hence related. So all elements of $$\{2, 4, 6\}$$ are related to each other.

Cross pairs: Take any element of $$\{1, 3, 5, 7\}$$ (odd) and any element of $$\{2, 4, 6\}$$ (even). One is odd and the other is even, so they are not of the same parity. Hence no element of $$\{1, 3, 5, 7\}$$ is related to any element of $$\{2, 4, 6\}$$.

Let $$R = \{(a, b) : a \leq b^2\}$$ on the set $$\mathbf{R}$$ of real numbers.

Not reflexive: Reflexivity would require $$a \leq a^2$$ for every real $$a$$. Take $$a = \dfrac{1}{2}$$. Then $$a^2 = \dfrac{1}{4}$$, and $$\dfrac{1}{2} \leq \dfrac{1}{4}$$ is false. So $$\left(\dfrac{1}{2}, \dfrac{1}{2}\right) \notin R$$, and $$R$$ is not reflexive.

Not symmetric: Take $$a = 1$$, $$b = 2$$. Then $$a \leq b^2$$ becomes $$1 \leq 4$$, which is true, so $$(1, 2) \in R$$. But $$(2, 1)$$ requires $$2 \leq 1^2 = 1$$, which is false, so $$(2, 1) \notin R$$. Hence $$R$$ is not symmetric.

Not transitive: Take $$a = 5$$, $$b = -3$$, $$c = 2$$. Then $$(5, -3) \in R$$ since $$5 \leq (-3)^2 = 9$$, and $$(-3, 2) \in R$$ since $$-3 \leq 2^2 = 4$$. But $$(5, 2)$$ requires $$5 \leq 2^2 = 4$$, which is false, so $$(5, 2) \notin R$$. Hence $$R$$ is not transitive.

Therefore $$R$$ is neither reflexive, nor symmetric, nor transitive.

Here the set is $$A = \{1, 2, 3, 4, 5, 6\}$$ and $$R = \{(a, b) : b = a + 1\}$$. Listing the pairs, $$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$$.

Reflexive: For $$(a, a) \in R$$ we would need $$a = a + 1$$, which is impossible. For example $$(1, 1) \notin R$$. So $$R$$ is not reflexive.

Symmetric: $$(1, 2) \in R$$ since $$2 = 1 + 1$$. But $$(2, 1) \notin R$$ since $$1 \neq 2 + 1$$. So $$R$$ is not symmetric.

Transitive: $$(1, 2) \in R$$ and $$(2, 3) \in R$$, but $$(1, 3) \notin R$$ since $$3 \neq 1 + 1$$. So $$R$$ is not transitive.

Therefore $$R$$ is neither reflexive, nor symmetric, nor transitive.

Let $$R = \{(a, b) : a \leq b\}$$ on the set $$\mathbf{R}$$ of real numbers.

Reflexive: For every real number $$a$$, $$a \leq a$$ is true. So $$(a, a) \in R$$, and $$R$$ is reflexive.

Transitive: Let $$(a, b) \in R$$ and $$(b, c) \in R$$. Then $$a \leq b$$ and $$b \leq c$$. Combining these inequalities gives $$a \leq c$$. So $$(a, c) \in R$$, and $$R$$ is transitive.

Not symmetric: Take $$a = 1$$, $$b = 2$$. Then $$1 \leq 2$$, so $$(1, 2) \in R$$. But $$(2, 1)$$ requires $$2 \leq 1$$, which is false, so $$(2, 1) \notin R$$. Hence $$R$$ is not symmetric.

Therefore $$R$$ is reflexive and transitive but not symmetric.

Let $$R = \{(a, b) : a \leq b^3\}$$ on the set $$\mathbf{R}$$.

Not reflexive: Reflexivity would require $$a \leq a^3$$ for every real $$a$$. Take $$a = \dfrac{1}{2}$$. Then $$a^3 = \dfrac{1}{8}$$, and $$\dfrac{1}{2} \leq \dfrac{1}{8}$$ is false. So $$\left(\dfrac{1}{2}, \dfrac{1}{2}\right) \notin R$$, and $$R$$ is not reflexive.

Not symmetric: Take $$a = 1$$, $$b = 2$$. Then $$1 \leq 2^3 = 8$$, so $$(1, 2) \in R$$. But $$(2, 1)$$ requires $$2 \leq 1^3 = 1$$, which is false, so $$(2, 1) \notin R$$. Hence $$R$$ is not symmetric.

Not transitive: Take $$a = 3$$, $$b = \dfrac{3}{2}$$, $$c = \dfrac{6}{5}$$. Then $$\left(3, \dfrac{3}{2}\right) \in R$$ since $$3 \leq \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8} = 3.375$$, and $$\left(\dfrac{3}{2}, \dfrac{6}{5}\right) \in R$$ since $$\dfrac{3}{2} = 1.5 \leq \left(\dfrac{6}{5}\right)^3 = \dfrac{216}{125} = 1.728$$. But $$\left(3, \dfrac{6}{5}\right)$$ requires $$3 \leq \left(\dfrac{6}{5}\right)^3 = 1.728$$, which is false, so $$\left(3, \dfrac{6}{5}\right) \notin R$$. Hence $$R$$ is not transitive.

Therefore $$R$$ is neither reflexive, nor symmetric, nor transitive.

The set is $$A = \{1, 2, 3\}$$ and $$R = \{(1, 2), (2, 1)\}$$.

Symmetric: $$(1, 2) \in R$$ and its reverse $$(2, 1) \in R$$; $$(2, 1) \in R$$ and its reverse $$(1, 2) \in R$$. Since every pair of $$R$$ has its reverse in $$R$$, $$R$$ is symmetric.

Not reflexive: For $$R$$ to be reflexive we would need $$(1, 1), (2, 2), (3, 3) \in R$$. But $$(1, 1) \notin R$$. So $$R$$ is not reflexive.

Not transitive: $$(1, 2) \in R$$ and $$(2, 1) \in R$$. Transitivity would require $$(1, 1) \in R$$, but $$(1, 1) \notin R$$. So $$R$$ is not transitive.

Therefore $$R$$ is symmetric but neither reflexive nor transitive.

Let $$R = \{(x, y) : x \text{ and } y \text{ have the same number of pages}\}$$ on the set $$A$$ of all books in the library.

Reflexive: Any book $$x$$ has the same number of pages as itself. So $$(x, x) \in R$$ for every $$x \in A$$, and $$R$$ is reflexive.

Symmetric: Let $$(x, y) \in R$$. Then $$x$$ and $$y$$ have the same number of pages, which is the same as saying $$y$$ and $$x$$ have the same number of pages. So $$(y, x) \in R$$, and $$R$$ is symmetric.

Transitive: Let $$(x, y) \in R$$ and $$(y, z) \in R$$. Then $$x$$ and $$y$$ have the same number of pages, and $$y$$ and $$z$$ have the same number of pages. Hence $$x$$ and $$z$$ have the same number of pages, so $$(x, z) \in R$$, and $$R$$ is transitive.

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

Here $$A = \{1, 2, 3, 4, 5\}$$ and $$R = \{(a, b) : |a - b| \text{ is even}\}$$. Note that $$|a - b|$$ is even exactly when $$a$$ and $$b$$ have the same parity (both odd or both even).

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

Symmetric: Let $$(a, b) \in R$$, so $$|a - b|$$ is even. Since $$|b - a| = |a - b|$$, $$|b - a|$$ is also even. So $$(b, a) \in R$$, and $$R$$ is symmetric.

Transitive: Let $$(a, b) \in R$$ and $$(b, c) \in R$$. Then $$|a - b|$$ is even and $$|b - c|$$ is even, so $$a, b$$ have the same parity and $$b, c$$ have the same parity. Hence $$a$$ and $$c$$ have the same parity, so $$|a - c|$$ is even, giving $$(a, c) \in R$$. So $$R$$ is transitive.

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

Elements of $$\{1, 3, 5\}$$: These are all odd. For any two of them the difference is even, so each pair is related. Hence all elements of $$\{1, 3, 5\}$$ are related to each other.

Elements of $$\{2, 4\}$$: These are both even, and $$|2 - 4| = 2$$ is even, so $$2$$ and $$4$$ are related to each other.

Cross pairs: Take an element of $$\{1, 3, 5\}$$ (odd) and an element of $$\{2, 4\}$$ (even). Their difference is odd, hence not even. So no element of $$\{1, 3, 5\}$$ is related to any element of $$\{2, 4\}$$.

Let $$O$$ denote the origin, and for a point $$P$$ let $$d(P)$$ be the distance of $$P$$ from $$O$$. The relation is $$R = \{(P, Q) : d(P) = d(Q)\}$$ on the set $$A$$ of all points of the plane.

Reflexive: For any point $$P$$, $$d(P) = d(P)$$. So $$(P, P) \in R$$, and $$R$$ is reflexive.

Symmetric: Let $$(P, Q) \in R$$, so $$d(P) = d(Q)$$. Then $$d(Q) = d(P)$$, so $$(Q, P) \in R$$, and $$R$$ is symmetric.

Transitive: Let $$(P, Q) \in R$$ and $$(Q, S) \in R$$. Then $$d(P) = d(Q)$$ and $$d(Q) = d(S)$$, hence $$d(P) = d(S)$$. So $$(P, S) \in R$$, and $$R$$ is transitive.

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

Set of points related to a fixed point $$P \neq (0, 0)$$: A point $$Q$$ is related to $$P$$ if and only if $$d(Q) = d(P)$$, i.e. $$Q$$ is at a fixed distance $$d(P)$$ from the origin.

Let $$k = d(P) > 0$$ (since $$P \neq (0,0)$$). The set of all points $$Q$$ with $$d(Q) = k$$ is precisely the circle of radius $$k$$ with centre at the origin. Since $$d(P) = k$$, this circle passes through $$P$$.

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

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

Reflexive: Every triangle is similar to itself, so $$(T, T) \in R$$ for every triangle $$T$$. Hence $$R$$ is reflexive.

Symmetric: If $$T_1$$ is similar to $$T_2$$, then $$T_2$$ is similar to $$T_1$$. So $$(T_1, T_2) \in R \Rightarrow (T_2, T_1) \in R$$, and $$R$$ is symmetric.

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

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

Which of $$T_1, T_2, T_3$$ are related: Two triangles are similar if their corresponding sides are in the same ratio. Arrange the sides of each triangle in increasing order.

$$T_1 : 3, 4, 5$$ and $$T_3 : 6, 8, 10$$. The ratios of corresponding sides are $$\dfrac{3}{6} = \dfrac{4}{8} = \dfrac{5}{10} = \dfrac{1}{2}$$. All three ratios are equal, so $$T_1$$ is similar to $$T_3$$.

$$T_1 : 3, 4, 5$$ and $$T_2 : 5, 12, 13$$. The ratios are $$\dfrac{3}{5}, \dfrac{4}{12} = \dfrac{1}{3}, \dfrac{5}{13}$$. These are not all equal, so $$T_1$$ is not similar to $$T_2$$ (and likewise $$T_2$$ is not similar to $$T_3$$).

Hence $$T_1$$ and $$T_3$$ are related to each other, while $$T_2$$ is related to neither.

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

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

Symmetric: If $$P_1$$ and $$P_2$$ have the same number of sides, then $$P_2$$ and $$P_1$$ have the same number of sides. So $$(P_1, P_2) \in R \Rightarrow (P_2, P_1) \in R$$, and $$R$$ is symmetric.

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

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

Elements related to the triangle $$T$$: The triangle $$T$$ with sides $$3, 4, 5$$ has $$3$$ sides. A polygon $$P$$ is related to $$T$$ if and only if $$P$$ has the same number of sides as $$T$$, i.e. $$3$$ sides.

Hence the set of all elements related to $$T$$ is the set of all triangles in $$A$$ (all three-sided polygons).

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

Reflexive: Every line is parallel to itself, so $$(L_1, L_1) \in R$$. Hence $$R$$ is reflexive.

Symmetric: If $$L_1$$ is parallel to $$L_2$$, then $$L_2$$ is parallel to $$L_1$$. So $$(L_1, L_2) \in R \Rightarrow (L_2, L_1) \in R$$, and $$R$$ is symmetric.

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

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

Lines related to $$y = 2x + 4$$: The line $$y = 2x + 4$$ has slope $$2$$. A line is parallel to it precisely when it has the same slope $$2$$. Every line of slope $$2$$ can be written as $$y = 2x + c$$ for some real constant $$c$$.

Hence the set of all lines related to $$y = 2x + 4$$ is $$\{y = 2x + c : c \in \mathbf{R}\}$$, i.e. the family of all lines with slope $$2$$.

The set is $$A = \{1, 2, 3, 4\}$$ and $$R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$$.

Reflexive: $$(1, 1), (2, 2), (3, 3), (4, 4)$$ are all present in $$R$$. So $$R$$ is reflexive.

Symmetric: $$(1, 2) \in R$$, but $$(2, 1) \notin R$$. So $$R$$ is not symmetric.

Transitive: We check every chain $$(a, b), (b, c)$$ in $$R$$:

• $$(1, 1), (1, 2) \Rightarrow (1, 2) \in R$$ ✓
• $$(1, 1), (1, 3) \Rightarrow (1, 3) \in R$$ ✓
• $$(1, 2), (2, 2) \Rightarrow (1, 2) \in R$$ ✓
• $$(1, 3), (3, 3) \Rightarrow (1, 3) \in R$$ ✓
• $$(1, 3), (3, 2) \Rightarrow (1, 2) \in R$$ ✓
• $$(3, 3), (3, 2) \Rightarrow (3, 2) \in R$$ ✓
• $$(3, 2), (2, 2) \Rightarrow (3, 2) \in R$$ ✓

Every required pair is present, so $$R$$ is transitive.

Thus $$R$$ is reflexive and transitive but not symmetric. The correct answer is (B).

A pair $$(a, b)$$ belongs to $$R$$ if and only if both conditions hold: $$a = b - 2$$ and $$b > 6$$. We test each option.

(A) $$(2, 4)$$: Here $$b = 4$$, and $$b > 6$$ is false. So $$(2, 4) \notin R$$.

(B) $$(3, 8)$$: Here $$b = 8 > 6$$ ✓, but $$a = b - 2 = 8 - 2 = 6$$, while the given $$a = 3 \neq 6$$. So $$(3, 8) \notin R$$.

(C) $$(6, 8)$$: Here $$b = 8 > 6$$ ✓, and $$a = b - 2 = 8 - 2 = 6$$, which matches the given $$a = 6$$ ✓. So $$(6, 8) \in R$$.

(D) $$(8, 7)$$: Here $$b = 7 > 6$$ ✓, but $$a = b - 2 = 7 - 2 = 5$$, while the given $$a = 8 \neq 5$$. So $$(8, 7) \notin R$$.

The correct answer is (C).

Here $$A$$ has $$50$$ students and $$f : A \to \mathbf{N}$$ sends each student to his/her roll number.

One-one: No two different students of a class can be assigned the same roll number. So if $$x_1$$ and $$x_2$$ are two students with $$f(x_1) = f(x_2)$$, then they have the same roll number, which forces $$x_1 = x_2$$. Hence $$f$$ is one-one.

Not onto: The codomain is $$\mathbf{N}$$, which is infinite, but only $$50$$ roll numbers (those of the $$50$$ students) actually occur as images. So at least one natural number, for instance $$51$$, is not the roll number of any student in $$A$$. Thus there is an element of the codomain $$\mathbf{N}$$ that has no preimage, and $$f$$ is not onto.

Therefore $$f$$ is one-one but not onto.

The function is $$f : \mathbf{N} \to \mathbf{N}$$ with $$f(x) = 2x$$.

One-one: Let $$x_1, x_2 \in \mathbf{N}$$ with $$f(x_1) = f(x_2)$$. Then $$2x_1 = 2x_2$$, and dividing by $$2$$ gives $$x_1 = x_2$$. Hence $$f$$ is one-one.

Not onto: The image $$f(x) = 2x$$ is always an even natural number. Consider $$1 \in \mathbf{N}$$ (the codomain). If $$1$$ had a preimage, we would need $$2x = 1$$, i.e. $$x = \dfrac{1}{2}$$, which is not a natural number. So $$1$$ has no preimage in $$\mathbf{N}$$, and $$f$$ is not onto.

Therefore $$f$$ is one-one but not onto.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = 2x$$.

One-one: Let $$x_1, x_2 \in \mathbf{R}$$ with $$f(x_1) = f(x_2)$$. Then $$2x_1 = 2x_2$$, so $$x_1 = x_2$$. Hence $$f$$ is one-one.

Onto: Let $$y$$ be any element of the codomain $$\mathbf{R}$$. We must find $$x \in \mathbf{R}$$ with $$f(x) = y$$, i.e. $$2x = y$$. Choosing $$x = \dfrac{y}{2}$$, which is a real number, we get $$f\left(\dfrac{y}{2}\right) = 2 \cdot \dfrac{y}{2} = y$$. So every $$y \in \mathbf{R}$$ has a preimage, and $$f$$ is onto.

Therefore $$f$$ is one-one and onto, i.e. $$f$$ is a bijection.

The function $$f : \mathbf{N} \to \mathbf{N}$$ is defined by $$f(1) = f(2) = 1$$ and $$f(x) = x - 1$$ for every $$x > 2$$.

Not one-one: We have $$f(1) = 1$$ and $$f(2) = 1$$, so $$f(1) = f(2)$$, yet $$1 \neq 2$$. Hence $$f$$ is not one-one.

Onto: Let $$y$$ be any element of the codomain $$\mathbf{N}$$.

• If $$y = 1$$, then $$f(1) = 1 = y$$, so $$y$$ has the preimage $$1$$.
• If $$y \geq 2$$, take $$x = y + 1$$. Then $$x = y + 1 > 2$$, so $$f(x) = x - 1 = (y + 1) - 1 = y$$. Thus $$y$$ has the preimage $$y + 1$$.

So every $$y \in \mathbf{N}$$ has a preimage in $$\mathbf{N}$$, and $$f$$ is onto.

Therefore $$f$$ is onto but not one-one.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = x^2$$.

Not one-one: Take $$x_1 = 1$$ and $$x_2 = -1$$. Then $$f(1) = 1^2 = 1$$ and $$f(-1) = (-1)^2 = 1$$, so $$f(1) = f(-1)$$, but $$1 \neq -1$$. Hence $$f$$ is not one-one.

Not onto: For every real $$x$$, $$f(x) = x^2 \geq 0$$. Consider the element $$-2$$ of the codomain $$\mathbf{R}$$. If $$-2$$ had a preimage, we would need $$x^2 = -2$$, which is impossible for a real $$x$$. So $$-2$$ has no preimage, and $$f$$ is not onto.

Therefore $$f$$ is neither one-one nor onto.

The function $$f : \mathbf{N} \to \mathbf{N}$$ is $$f(x) = x + 1$$ if $$x$$ is odd, and $$f(x) = x - 1$$ if $$x$$ is even. Note that if $$x$$ is odd then $$x + 1$$ is even, and if $$x$$ is even then $$x - 1$$ is odd. So $$f$$ sends odd numbers to even numbers and even numbers to odd numbers.

One-one: Suppose $$f(x_1) = f(x_2)$$.

• If one of $$x_1, x_2$$ is odd and the other is even, then one image is even and the other is odd, so they cannot be equal. Hence $$x_1$$ and $$x_2$$ must have the same parity.
• If both are odd: $$f(x_1) = f(x_2)$$ gives $$x_1 + 1 = x_2 + 1$$, so $$x_1 = x_2$$.
• If both are even: $$f(x_1) = f(x_2)$$ gives $$x_1 - 1 = x_2 - 1$$, so $$x_1 = x_2$$.

In every case $$x_1 = x_2$$, so $$f$$ is one-one.

Onto: Let $$y$$ be any element of the codomain $$\mathbf{N}$$.

• If $$y$$ is odd, then $$y + 1$$ is even, and $$f(y + 1) = (y + 1) - 1 = y$$.
• If $$y$$ is even, then $$y \geq 2$$, so $$y - 1 \in \mathbf{N}$$ and $$y - 1$$ is odd; then $$f(y - 1) = (y - 1) + 1 = y$$.

So every $$y \in \mathbf{N}$$ has a preimage, and $$f$$ is onto.

Therefore $$f$$ is both one-one and onto.

Let $$f : \{1, 2, 3\} \to \{1, 2, 3\}$$ be an onto function.

Suppose, on the contrary, that $$f$$ is not one-one. Then at least two distinct elements of the domain have the same image. So among the three values $$f(1), f(2), f(3)$$, at least two are equal, which means the set of images $$\{f(1), f(2), f(3)\}$$ contains at most $$2$$ distinct elements.

Hence the range of $$f$$ has at most $$2$$ elements, so it cannot equal the codomain $$\{1, 2, 3\}$$ which has $$3$$ elements. This contradicts the assumption that $$f$$ is onto.

Therefore our supposition is wrong, and $$f$$ must be one-one. Hence every onto function $$f : \{1, 2, 3\} \to \{1, 2, 3\}$$ is one-one.

Let $$f : \{1, 2, 3\} \to \{1, 2, 3\}$$ be a one-one function.

Since $$f$$ is one-one, the three elements $$1, 2, 3$$ of the domain have three distinct images $$f(1), f(2), f(3)$$.

So the range $$\{f(1), f(2), f(3)\}$$ contains exactly $$3$$ distinct elements, all lying in the codomain $$\{1, 2, 3\}$$. But the codomain itself has only $$3$$ elements. A $$3$$-element subset of a $$3$$-element set must be the whole set.

Hence the range equals the codomain $$\{1, 2, 3\}$$, which means $$f$$ is onto.

Therefore every one-one function $$f : \{1, 2, 3\} \to \{1, 2, 3\}$$ must be onto.

The function is $$f : \mathbf{R}_* \to \mathbf{R}_*$$ with $$f(x) = \dfrac{1}{x}$$, where $$\mathbf{R}_*$$ is the set of all non-zero real numbers.

One-one: Let $$x_1, x_2 \in \mathbf{R}_*$$ with $$f(x_1) = f(x_2)$$. Then $$\dfrac{1}{x_1} = \dfrac{1}{x_2}$$. Taking reciprocals, $$x_1 = x_2$$. Hence $$f$$ is one-one.

Onto: Let $$y \in \mathbf{R}_*$$ be any element of the codomain. Take $$x = \dfrac{1}{y}$$, which is a non-zero real number, so $$x \in \mathbf{R}_*$$. Then $$f(x) = \dfrac{1}{x} = \dfrac{1}{1/y} = y$$. So every $$y$$ has a preimage, and $$f$$ is onto.

Hence $$f$$ is one-one and onto.

When the domain is replaced by $$\mathbf{N}$$: Consider $$g : \mathbf{N} \to \mathbf{R}_*$$ defined by $$g(x) = \dfrac{1}{x}$$.

One-one: The same argument as above shows $$g(x_1) = g(x_2) \Rightarrow x_1 = x_2$$, so $$g$$ is still one-one.

Onto: Take $$y = 2 \in \mathbf{R}_*$$. If $$2$$ had a preimage, we would need $$\dfrac{1}{x} = 2$$, i.e. $$x = \dfrac{1}{2}$$, which is not a natural number. So $$2$$ has no preimage in $$\mathbf{N}$$, and $$g$$ is not onto.

Therefore the result is not true when the domain is replaced by $$\mathbf{N}$$: $$g$$ is one-one but not onto.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = [x]$$, where $$[x]$$ is the greatest integer not exceeding $$x$$.

Not one-one: Take $$x_1 = 1.2$$ and $$x_2 = 1.7$$. Then $$f(1.2) = [1.2] = 1$$ and $$f(1.7) = [1.7] = 1$$, so $$f(1.2) = f(1.7)$$, but $$1.2 \neq 1.7$$. Hence $$f$$ is not one-one. (In fact every $$x$$ in $$1 \leq x < 2$$ has image $$1$$.)

Not onto: The value $$[x]$$ is always an integer. So the range of $$f$$ is the set of integers $$\mathbf{Z}$$. Consider a non-integer element of the codomain, say $$0.5 \in \mathbf{R}$$. Since $$[x]$$ can never equal $$0.5$$, the number $$0.5$$ has no preimage. Hence $$f$$ is not onto.

Therefore the greatest integer function is neither one-one nor onto.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = |x|$$, where $$|x| = x$$ for $$x \geq 0$$ and $$|x| = -x$$ for $$x < 0$$.

Not one-one: Take $$x_1 = -1$$ and $$x_2 = 1$$. Then $$f(-1) = |-1| = 1$$ and $$f(1) = |1| = 1$$, so $$f(-1) = f(1)$$, but $$-1 \neq 1$$. Hence $$f$$ is not one-one.

Not onto: For every real $$x$$, $$f(x) = |x| \geq 0$$. So the range of $$f$$ consists only of non-negative reals. Consider $$-1 \in \mathbf{R}$$ (the codomain). Since $$|x|$$ can never be negative, $$-1$$ has no preimage. Hence $$f$$ is not onto.

Therefore the modulus function is neither one-one nor onto.

The signum function $$f : \mathbf{R} \to \mathbf{R}$$ takes the value $$1$$ for $$x > 0$$, $$0$$ for $$x = 0$$, and $$-1$$ for $$x < 0$$.

Not one-one: Take $$x_1 = 1$$ and $$x_2 = 2$$. Both are positive, so $$f(1) = 1$$ and $$f(2) = 1$$, giving $$f(1) = f(2)$$, but $$1 \neq 2$$. Hence $$f$$ is not one-one. (Indeed, all positive reals share the image $$1$$.)

Not onto: The function takes only three values, so its range is $$\{-1, 0, 1\}$$. Consider $$2 \in \mathbf{R}$$ (the codomain). Since $$f(x)$$ is never equal to $$2$$, the number $$2$$ has no preimage. Hence $$f$$ is not onto.

Therefore the signum function is neither one-one nor onto.

The function $$f : A \to B$$ is given by the pairs $$f = \{(1, 4), (2, 5), (3, 6)\}$$, that is, $$f(1) = 4$$, $$f(2) = 5$$ and $$f(3) = 6$$.

A function is one-one if distinct elements of the domain have distinct images. The three images are

$$f(1) = 4, \qquad f(2) = 5, \qquad f(3) = 6.$$

These three values $$4, 5, 6$$ are all different from one another. So no two distinct elements of $$A$$ have the same image; equivalently, if $$f(x_1) = f(x_2)$$ then it must be the same pair, forcing $$x_1 = x_2$$.

Hence $$f$$ is one-one.

The function $$f : A \times B \to B \times A$$ is defined by $$f(a, b) = (b, a)$$ for every $$(a, b) \in A \times B$$.

One-one: Let $$(a_1, b_1)$$ and $$(a_2, b_2)$$ be elements of $$A \times B$$ with $$f(a_1, b_1) = f(a_2, b_2)$$. Then $$(b_1, a_1) = (b_2, a_2)$$. Two ordered pairs are equal exactly when their corresponding components are equal, so $$b_1 = b_2$$ and $$a_1 = a_2$$. Hence $$(a_1, b_1) = (a_2, b_2)$$, and $$f$$ is one-one.

Onto: Let $$(b, a)$$ be any element of the codomain $$B \times A$$, so $$b \in B$$ and $$a \in A$$. Then $$(a, b) \in A \times B$$, and $$f(a, b) = (b, a)$$. So every element of $$B \times A$$ has a preimage, and $$f$$ is onto.

Since $$f$$ is both one-one and onto, $$f$$ is a bijective function.

The function $$f : \mathbf{N} \to \mathbf{N}$$ is $$f(n) = \dfrac{n+1}{2}$$ if $$n$$ is odd, and $$f(n) = \dfrac{n}{2}$$ if $$n$$ is even. Some values: $$f(1) = \dfrac{2}{2} = 1$$, $$f(2) = \dfrac{2}{2} = 1$$, $$f(3) = \dfrac{4}{2} = 2$$, $$f(4) = \dfrac{4}{2} = 2$$, and so on.

One-one: We have $$f(1) = 1$$ and $$f(2) = 1$$, so $$f(1) = f(2)$$, but $$1 \neq 2$$. Hence $$f$$ is not one-one.

Onto: Let $$y$$ be any element of the codomain $$\mathbf{N}$$. Take $$n = 2y$$, which is an even natural number. Then $$f(2y) = \dfrac{2y}{2} = y$$. So every $$y \in \mathbf{N}$$ has a preimage, and $$f$$ is onto.

Since $$f$$ is onto but not one-one, $$f$$ is not bijective.

Here $$A = \mathbf{R} - \{3\}$$, $$B = \mathbf{R} - \{1\}$$ and $$f(x) = \dfrac{x - 2}{x - 3}$$.

One-one: Let $$x_1, x_2 \in A$$ with $$f(x_1) = f(x_2)$$. Then $$\dfrac{x_1 - 2}{x_1 - 3} = \dfrac{x_2 - 2}{x_2 - 3}$$. Cross-multiplying,

$$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3).$$

Expanding both sides: $$x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6$$. Cancelling $$x_1 x_2$$ and $$6$$,

$$-3x_1 - 2x_2 = -3x_2 - 2x_1 \;\Rightarrow\; -3x_1 + 2x_1 = -3x_2 + 2x_2 \;\Rightarrow\; -x_1 = -x_2,$$

so $$x_1 = x_2$$. Hence $$f$$ is one-one.

Onto: Let $$y \in B$$, so $$y \neq 1$$. We solve $$y = \dfrac{x - 2}{x - 3}$$ for $$x$$:

$$y(x - 3) = x - 2 \;\Rightarrow\; yx - 3y = x - 2 \;\Rightarrow\; yx - x = 3y - 2 \;\Rightarrow\; x(y - 1) = 3y - 2.$$

Since $$y \neq 1$$, we may divide: $$x = \dfrac{3y - 2}{y - 1}$$, a real number. We must check $$x \in A$$, i.e. $$x \neq 3$$. If $$x = 3$$ then $$\dfrac{3y - 2}{y - 1} = 3$$, giving $$3y - 2 = 3y - 3$$, i.e. $$-2 = -3$$, which is false. So $$x \neq 3$$ and $$x \in A$$.

Substituting back confirms $$f(x) = y$$. So every $$y \in B$$ has a preimage in $$A$$, and $$f$$ is onto.

Therefore $$f$$ is both one-one and onto.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = x^4$$.

Check one-one: Take $$x_1 = 1$$ and $$x_2 = -1$$. Then $$f(1) = 1^4 = 1$$ and $$f(-1) = (-1)^4 = 1$$, so $$f(1) = f(-1)$$ but $$1 \neq -1$$. Hence $$f$$ is not one-one (it is many-one).

Check onto: For every real $$x$$, $$f(x) = x^4 \geq 0$$. Consider $$-1 \in \mathbf{R}$$ (the codomain). Since $$x^4 = -1$$ has no real solution, $$-1$$ has no preimage. Hence $$f$$ is not onto.

So $$f$$ is neither one-one nor onto. The correct answer is (D).

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = 3x$$.

Check one-one: Let $$x_1, x_2 \in \mathbf{R}$$ with $$f(x_1) = f(x_2)$$. Then $$3x_1 = 3x_2$$, so $$x_1 = x_2$$. Hence $$f$$ is one-one.

Check onto: Let $$y \in \mathbf{R}$$ be any element of the codomain. Take $$x = \dfrac{y}{3} \in \mathbf{R}$$. Then $$f\left(\dfrac{y}{3}\right) = 3 \cdot \dfrac{y}{3} = y$$. So every $$y$$ has a preimage, and $$f$$ is onto.

So $$f$$ is one-one and onto. The correct answer is (A).

The composite $$gof$$ is defined by $$(gof)(x) = g(f(x))$$, and its domain is the domain of $$f$$, namely $$\{2, 3, 4, 5\}$$. We compute it at each element.

$$(gof)(2) = g(f(2)) = g(3) = 7$$

$$(gof)(3) = g(f(3)) = g(4) = 7$$

$$(gof)(4) = g(f(4)) = g(5) = 11$$

$$(gof)(5) = g(f(5)) = g(5) = 11$$

Collecting these results,

$$gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\}.$$

We are given $$f(x) = \cos x$$ and $$g(x) = 3x^2$$, both from $$\mathbf{R}$$ to $$\mathbf{R}$$.

Computing $$gof$$: $$(gof)(x) = g(f(x)) = g(\cos x) = 3(\cos x)^2 = 3\cos^2 x$$.

Computing $$fog$$: $$(fog)(x) = f(g(x)) = f(3x^2) = \cos(3x^2)$$.

Showing $$gof \neq fog$$: It is enough to find one value of $$x$$ at which the two composites differ. Take $$x = 0$$:

$$(gof)(0) = 3\cos^2 0 = 3(1)^2 = 3,$$

$$(fog)(0) = \cos(3 \cdot 0^2) = \cos 0 = 1.$$

Since $$(gof)(0) = 3 \neq 1 = (fog)(0)$$, the two functions are not equal. Hence $$gof \neq fog$$.

The function is $$f : \mathbf{N} \to Y$$ with $$f(x) = 4x + 3$$, and $$Y$$ is the range of $$f$$, i.e. $$Y = \{y \in \mathbf{N} : y = 4x + 3 \text{ for some } x \in \mathbf{N}\}$$. A function is invertible if and only if it is both one-one and onto.

One-one: Let $$x_1, x_2 \in \mathbf{N}$$ with $$f(x_1) = f(x_2)$$. Then $$4x_1 + 3 = 4x_2 + 3$$, so $$4x_1 = 4x_2$$, giving $$x_1 = x_2$$. Hence $$f$$ is one-one.

Onto: Let $$y \in Y$$. By the definition of $$Y$$, there exists some $$x \in \mathbf{N}$$ with $$y = 4x + 3 = f(x)$$. So every $$y \in Y$$ has a preimage, and $$f$$ is onto.

Since $$f$$ is one-one and onto, it is invertible.

Finding the inverse: To invert, solve $$y = 4x + 3$$ for $$x$$: $$4x = y - 3$$, so $$x = \dfrac{y - 3}{4}$$. Define $$g : Y \to \mathbf{N}$$ by $$g(y) = \dfrac{y - 3}{4}$$.

We verify that $$g$$ is the inverse of $$f$$:

$$(gof)(x) = g(f(x)) = g(4x + 3) = \dfrac{(4x + 3) - 3}{4} = \dfrac{4x}{4} = x,$$

$$(fog)(y) = f(g(y)) = f\!\left(\dfrac{y - 3}{4}\right) = 4 \cdot \dfrac{y - 3}{4} + 3 = (y - 3) + 3 = y.$$

So $$gof = I_{\mathbf{N}}$$ and $$fog = I_Y$$. Hence $$f$$ is invertible with inverse $$f^{-1}(y) = \dfrac{y - 3}{4}$$.

Let $$R_1$$ and $$R_2$$ be equivalence relations on the set $$A$$. We show $$R_1 \cap R_2$$ is reflexive, symmetric and transitive. Recall $$(a, b) \in R_1 \cap R_2$$ means $$(a, b) \in R_1$$ and $$(a, b) \in R_2$$.

Reflexive: Let $$a \in A$$. Since $$R_1$$ is reflexive, $$(a, a) \in R_1$$; since $$R_2$$ is reflexive, $$(a, a) \in R_2$$. Hence $$(a, a) \in R_1 \cap R_2$$, so $$R_1 \cap R_2$$ is reflexive.

Symmetric: Let $$(a, b) \in R_1 \cap R_2$$. Then $$(a, b) \in R_1$$ and $$(a, b) \in R_2$$. As $$R_1$$ is symmetric, $$(b, a) \in R_1$$; as $$R_2$$ is symmetric, $$(b, a) \in R_2$$. Hence $$(b, a) \in R_1 \cap R_2$$, so $$R_1 \cap R_2$$ is symmetric.

Transitive: Let $$(a, b) \in R_1 \cap R_2$$ and $$(b, c) \in R_1 \cap R_2$$. Then $$(a, b), (b, c) \in R_1$$, and since $$R_1$$ is transitive, $$(a, c) \in R_1$$. Likewise $$(a, b), (b, c) \in R_2$$, and since $$R_2$$ is transitive, $$(a, c) \in R_2$$. Hence $$(a, c) \in R_1 \cap R_2$$, so $$R_1 \cap R_2$$ is transitive.

Since $$R_1 \cap R_2$$ is reflexive, symmetric and transitive, it is an equivalence relation.

Here $$A$$ is the set of ordered pairs of positive integers, and $$(x, y) \, R \, (u, v)$$ means $$xv = yu$$.

Reflexive: For any $$(x, y) \in A$$, we have $$xy = yx$$ (multiplication is commutative). This is exactly the condition $$(x, y) \, R \, (x, y)$$. Hence $$R$$ is reflexive.

Symmetric: Let $$(x, y) \, R \, (u, v)$$, so $$xv = yu$$. This can be rewritten as $$uy = vx$$, i.e. $$u y = v x$$, which is precisely the condition $$(u, v) \, R \, (x, y)$$. Hence $$R$$ is symmetric.

Transitive: Let $$(x, y) \, R \, (u, v)$$ and $$(u, v) \, R \, (a, b)$$. Then

$$xv = yu \qquad \text{and} \qquad ub = va.$$

Multiply the first equation by $$b$$: $$xvb = yub$$. Now substitute $$ub = va$$ on the right side: $$xvb = y(va) = yva$$. So $$xvb = yva$$, which we write as $$v(xb) = v(ya)$$.

Since $$v$$ is a positive integer, $$v \neq 0$$, so we may cancel it to get $$xb = ya$$. This is exactly the condition $$(x, y) \, R \, (a, b)$$. Hence $$R$$ is transitive.

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

Group the elements of $$X$$ by their remainder on division by $$3$$:

• remainder $$1$$: $$\{1, 4, 7\}$$,
• remainder $$2$$: $$\{2, 5, 8\}$$,
• remainder $$0$$: $$\{3, 6, 9\}$$.

The key fact is: $$x - y$$ is divisible by $$3$$ if and only if $$x$$ and $$y$$ leave the same remainder on division by $$3$$, i.e. if and only if $$x$$ and $$y$$ belong to the same one of the three subsets above.

To show $$R_1 = R_2$$, we prove $$R_1 \subseteq R_2$$ and $$R_2 \subseteq R_1$$.

$$R_1 \subseteq R_2$$: Let $$(x, y) \in R_1$$, so $$x - y$$ is divisible by $$3$$. Then $$x$$ and $$y$$ have the same remainder mod $$3$$, so they lie in the same subset $$\{1,4,7\}$$, $$\{2,5,8\}$$ or $$\{3,6,9\}$$. Hence $$\{x, y\}$$ is a subset of one of these, so $$(x, y) \in R_2$$.

$$R_2 \subseteq R_1$$: Let $$(x, y) \in R_2$$. Then $$\{x, y\}$$ lies entirely within one of the subsets, so $$x$$ and $$y$$ have the same remainder mod $$3$$. Therefore $$x - y$$ is divisible by $$3$$, giving $$(x, y) \in R_1$$.

Since each relation is contained in the other, $$R_1 = R_2$$.

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

Reflexive: For any $$a \in X$$, clearly $$f(a) = f(a)$$. So $$(a, a) \in R$$, and $$R$$ is reflexive.

Symmetric: Let $$(a, b) \in R$$, so $$f(a) = f(b)$$. Then $$f(b) = f(a)$$, which means $$(b, a) \in R$$. Hence $$R$$ is symmetric.

Transitive: Let $$(a, b) \in R$$ and $$(b, c) \in R$$. Then $$f(a) = f(b)$$ and $$f(b) = f(c)$$. Combining, $$f(a) = f(c)$$, so $$(a, c) \in R$$. Hence $$R$$ is transitive.

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

A one-one function $$f : A \to A$$ with $$A = \{1, 2, 3\}$$ must assign distinct images to the three elements $$1, 2, 3$$. We count the choices step by step.

Choice of $$f(1)$$: it can be any of the $$3$$ elements of $$A$$.

Choice of $$f(2)$$: to keep $$f$$ one-one it must differ from $$f(1)$$, so there are $$2$$ choices left.

Choice of $$f(3)$$: it must differ from both $$f(1)$$ and $$f(2)$$, leaving $$1$$ choice.

By the multiplication principle, the total number of one-one functions is

$$3 \times 2 \times 1 = 6.$$

(These are exactly the $$3! = 6$$ arrangements of $$\{1, 2, 3\}$$.)

We need relations $$R$$ on $$A = \{1, 2, 3\}$$ that contain $$(1, 2)$$ and $$(2, 3)$$, and are reflexive and transitive but not symmetric.

Forced pairs. Being reflexive, $$R$$ must contain $$(1, 1), (2, 2), (3, 3)$$. It must contain the given $$(1, 2)$$ and $$(2, 3)$$. By transitivity, $$(1, 2) \in R$$ and $$(2, 3) \in R$$ force $$(1, 3) \in R$$.

So every such relation contains at least

$$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}.$$

One checks $$R_1$$ is reflexive and transitive, and it is not symmetric (e.g. $$(1, 2) \in R_1$$ but $$(2, 1) \notin R_1$$). So $$R_1$$ is one such relation.

Adding more pairs. The only pairs not yet in $$R_1$$ are $$(2, 1), (3, 2), (3, 1)$$.

• $$R_2 = R_1 \cup \{(2, 1)\}$$: checking all chains, $$R_2$$ stays transitive (for instance $$(2,1),(1,3) \Rightarrow (2,3) \in R_2$$), and it is still not symmetric since $$(2, 3) \in R_2$$ but $$(3, 2) \notin R_2$$. Valid.
• $$R_3 = R_1 \cup \{(3, 2)\}$$: checking all chains, $$R_3$$ stays transitive (for instance $$(3,2),(2,3) \Rightarrow (3,3) \in R_3$$), and it is still not symmetric since $$(1, 2) \in R_3$$ but $$(2, 1) \notin R_3$$. Valid.
• If we add both $$(2, 1)$$ and $$(3, 2)$$, then transitivity ($$(3,2),(2,1) \Rightarrow (3,1)$$) also forces $$(3, 1)$$; the relation then contains every pair, making it symmetric. Adding $$(3, 1)$$ alone forces (via $$(3,1),(1,2)$$) the pair $$(3, 2)$$, and then $$(2,3),(3,1) \Rightarrow (2,1)$$, again leading to the symmetric universal relation.

So no further pairs can be added while remaining "not symmetric". Hence exactly three relations work: $$R_1, R_2, R_3$$.

We need equivalence relations $$R$$ on $$A = \{1, 2, 3\}$$ containing $$(1, 2)$$ and $$(2, 1)$$.

Forced pairs. Being reflexive, $$R$$ must contain $$(1, 1), (2, 2), (3, 3)$$. It must contain the given $$(1, 2)$$ and $$(2, 1)$$. So $$R$$ contains at least

$$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}.$$

$$R_1$$ is an equivalence relation: it is reflexive; it is symmetric (the only non-diagonal pairs $$(1,2)$$ and $$(2,1)$$ are reverses of each other); and it is transitive — the possible chains all close up, e.g. $$(1,2),(2,1) \Rightarrow (1,1) \in R_1$$ and $$(2,1),(1,2) \Rightarrow (2,2) \in R_1$$. So $$R_1$$ is one such relation.

Any larger relation. Suppose an equivalence relation $$R$$ containing $$(1, 2), (2, 1)$$ is strictly larger than $$R_1$$. Then it must contain a pair linking $$3$$ to $$1$$ or $$2$$, say $$(1, 3)$$ (or $$(2, 3)$$, etc.). Then:

• symmetry forces $$(3, 1) \in R$$;
• transitivity with $$(2, 1), (1, 3)$$ forces $$(2, 3) \in R$$, and symmetry forces $$(3, 2) \in R$$.

Now $$R$$ contains all $$9$$ ordered pairs of $$A$$, i.e. $$R = A \times A$$, the universal relation $$R_2$$, which is indeed an equivalence relation.

So an equivalence relation containing $$(1, 2)$$ and $$(2, 1)$$ is either $$R_1$$ (keeping $$3$$ separate) or $$R_2 = A \times A$$ (merging $$3$$ in). Hence there are exactly two such equivalence relations.

$$I_{\mathbf{N}}$$ is onto. The identity function satisfies $$I_{\mathbf{N}}(x) = x$$ for every $$x \in \mathbf{N}$$. Given any $$y \in \mathbf{N}$$ in the codomain, the element $$x = y$$ of the domain gives $$I_{\mathbf{N}}(y) = y$$. So every $$y$$ has a preimage, and $$I_{\mathbf{N}}$$ is onto.

$$I_{\mathbf{N}} + I_{\mathbf{N}}$$ is not onto. By definition $$(I_{\mathbf{N}} + I_{\mathbf{N}})(x) = I_{\mathbf{N}}(x) + I_{\mathbf{N}}(x) = x + x = 2x$$.

So this function always outputs an even natural number; its range is the set of even natural numbers.

Consider $$1 \in \mathbf{N}$$ (the codomain). If $$1$$ had a preimage, we would need $$2x = 1$$, i.e. $$x = \dfrac{1}{2}$$, which is not a natural number. So $$1$$ has no preimage.

Hence $$I_{\mathbf{N}} + I_{\mathbf{N}}$$ is not onto, even though $$I_{\mathbf{N}}$$ itself is onto.

Both functions have domain $$\left[0, \dfrac{\pi}{2}\right]$$.

$$f(x) = \sin x$$ is one-one. On the interval $$\left[0, \dfrac{\pi}{2}\right]$$ the sine function is strictly increasing: if $$0 \leq x_1 < x_2 \leq \dfrac{\pi}{2}$$ then $$\sin x_1 < \sin x_2$$. A strictly increasing function takes distinct values at distinct points, so $$f(x_1) = f(x_2)$$ forces $$x_1 = x_2$$. Hence $$f$$ is one-one.

$$g(x) = \cos x$$ is one-one. On the interval $$\left[0, \dfrac{\pi}{2}\right]$$ the cosine function is strictly decreasing: if $$0 \leq x_1 < x_2 \leq \dfrac{\pi}{2}$$ then $$\cos x_1 > \cos x_2$$. A strictly decreasing function also takes distinct values at distinct points, so $$g(x_1) = g(x_2)$$ forces $$x_1 = x_2$$. Hence $$g$$ is one-one.

$$f + g$$ is not one-one. Here $$(f + g)(x) = \sin x + \cos x$$. Evaluate at the two endpoints of the domain:

$$(f + g)(0) = \sin 0 + \cos 0 = 0 + 1 = 1,$$

$$(f + g)\!\left(\dfrac{\pi}{2}\right) = \sin \dfrac{\pi}{2} + \cos \dfrac{\pi}{2} = 1 + 0 = 1.$$

So $$(f + g)(0) = (f + g)\!\left(\dfrac{\pi}{2}\right) = 1$$, yet $$0 \neq \dfrac{\pi}{2}$$. Hence $$f + g$$ is not one-one.

Let the codomain be $$Y = \{x \in \mathbf{R} : -1 < x < 1\}$$. Since $$|x|$$ depends on the sign of $$x$$, write $$f$$ piecewise:

$$f(x) = \dfrac{x}{1 + x} \ \text{ if } x \geq 0, \qquad f(x) = \dfrac{x}{1 - x} \ \text{ if } x < 0.$$

Note that for $$x \geq 0$$, $$f(x) = \dfrac{x}{1+x} \geq 0$$, and for $$x < 0$$, $$f(x) = \dfrac{x}{1-x} < 0$$.

One-one. Suppose $$f(x_1) = f(x_2)$$. If one of $$x_1, x_2$$ were $$\geq 0$$ and the other $$< 0$$, then one image is $$\geq 0$$ and the other $$< 0$$, so they could not be equal. Hence $$x_1$$ and $$x_2$$ have the same sign.

• Both $$\geq 0$$: $$\dfrac{x_1}{1 + x_1} = \dfrac{x_2}{1 + x_2} \Rightarrow x_1(1 + x_2) = x_2(1 + x_1) \Rightarrow x_1 + x_1 x_2 = x_2 + x_1 x_2 \Rightarrow x_1 = x_2$$.
• Both $$< 0$$: $$\dfrac{x_1}{1 - x_1} = \dfrac{x_2}{1 - x_2} \Rightarrow x_1(1 - x_2) = x_2(1 - x_1) \Rightarrow x_1 - x_1 x_2 = x_2 - x_1 x_2 \Rightarrow x_1 = x_2$$.

In every case $$x_1 = x_2$$, so $$f$$ is one-one.

Onto. Let $$y \in Y$$, so $$-1 < y < 1$$.

• If $$0 \leq y < 1$$: solve $$y = \dfrac{x}{1 + x}$$ with $$x \geq 0$$. Then $$y(1 + x) = x \Rightarrow y = x(1 - y) \Rightarrow x = \dfrac{y}{1 - y}$$. Since $$0 \leq y < 1$$, both $$y \geq 0$$ and $$1 - y > 0$$, so $$x \geq 0$$ and $$f(x) = y$$.
• If $$-1 < y < 0$$: solve $$y = \dfrac{x}{1 - x}$$ with $$x < 0$$. Then $$y(1 - x) = x \Rightarrow y = x(1 + y) \Rightarrow x = \dfrac{y}{1 + y}$$. Since $$-1 < y < 0$$, we have $$1 + y > 0$$ and $$y < 0$$, so $$x < 0$$ and $$f(x) = y$$.

So every $$y \in Y$$ has a preimage, and $$f$$ is onto.

Therefore $$f$$ is one-one and onto.

The function is $$f : \mathbf{R} \to \mathbf{R}$$ with $$f(x) = x^3$$.

Let $$x_1, x_2 \in \mathbf{R}$$ with $$f(x_1) = f(x_2)$$. Then $$x_1^3 = x_2^3$$, so

$$x_1^3 - x_2^3 = 0.$$

Factorising the difference of cubes,

$$(x_1 - x_2)\left(x_1^2 + x_1 x_2 + x_2^2\right) = 0.$$

Examine the second factor. Completing the square,

$$x_1^2 + x_1 x_2 + x_2^2 = \left(x_1 + \dfrac{x_2}{2}\right)^2 + \dfrac{3x_2^2}{4}.$$

Being a sum of two squares, this is $$\geq 0$$, and it equals $$0$$ only when $$x_1 + \dfrac{x_2}{2} = 0$$ and $$x_2 = 0$$, i.e. only when $$x_1 = x_2 = 0$$.

• If $$x_1$$ and $$x_2$$ are not both zero, the second factor is strictly positive, so $$x_1 - x_2 = 0$$, giving $$x_1 = x_2$$.
• If $$x_1 = x_2 = 0$$, then trivially $$x_1 = x_2$$.

In every case $$x_1 = x_2$$. Hence $$f$$ is injective (one-one).

The relation $$R$$ on $$P(X)$$ is: $$A R B$$ if and only if $$A \subset B$$.

Reflexive: Every set is a subset of itself, so $$A \subset A$$ for all $$A \in P(X)$$. Hence $$A R A$$, and $$R$$ is reflexive.

Symmetric: We test whether $$A \subset B$$ forces $$B \subset A$$. Since $$X$$ is non-empty, pick an element $$x \in X$$ and take $$A = \phi$$ and $$B = \{x\}$$. Then $$A \subset B$$, so $$A R B$$. But $$B = \{x\}$$ is not a subset of $$A = \phi$$, so $$B R A$$ is false. Hence $$R$$ is not symmetric.

Transitive: Let $$A R B$$ and $$B R C$$, i.e. $$A \subset B$$ and $$B \subset C$$. Then every element of $$A$$ is in $$B$$, and every element of $$B$$ is in $$C$$, so every element of $$A$$ is in $$C$$, i.e. $$A \subset C$$. Hence $$A R C$$, and $$R$$ is transitive.

Conclusion: $$R$$ is reflexive and transitive but not symmetric. Since an equivalence relation must be symmetric, $$R$$ is not an equivalence relation on $$P(X)$$.

Let $$A = \{1, 2, 3, \ldots, n\}$$ and let $$f : A \to A$$ be an onto function.

An onto function from $$A$$ to itself is automatically one-one. If $$f$$ were not one-one, two distinct elements of $$A$$ would share an image, so the range $$\{f(1), f(2), \ldots, f(n)\}$$ would contain fewer than $$n$$ distinct elements. Then the range could not be all of the $$n$$-element codomain $$A$$, contradicting that $$f$$ is onto. Hence $$f$$ must be one-one, so $$f$$ is a bijection.

Counting the bijections. A bijection $$f : A \to A$$ is just an arrangement (permutation) of the $$n$$ elements. Count the choices of images:

• $$f(1)$$ can be any of the $$n$$ elements;
• $$f(2)$$ must differ from $$f(1)$$: $$n - 1$$ choices;
• $$f(3)$$ must differ from $$f(1), f(2)$$: $$n - 2$$ choices;
• and so on, until $$f(n)$$ has just $$1$$ choice.

By the multiplication principle, the number of onto functions is

$$n \times (n - 1) \times (n - 2) \times \cdots \times 2 \times 1 = n!.$$

Two functions $$f, g : A \to B$$ are equal if $$f(a) = g(a)$$ for every $$a \in A$$. We evaluate both functions at each element of $$A = \{-1, 0, 1, 2\}$$.

Values of $$f(x) = x^2 - x$$:

$$f(-1) = (-1)^2 - (-1) = 1 + 1 = 2$$

$$f(0) = 0^2 - 0 = 0$$

$$f(1) = 1^2 - 1 = 0$$

$$f(2) = 2^2 - 2 = 4 - 2 = 2$$

Values of $$g(x) = 2\left|x - \dfrac{1}{2}\right| - 1$$:

$$g(-1) = 2\left|-1 - \tfrac{1}{2}\right| - 1 = 2 \cdot \tfrac{3}{2} - 1 = 3 - 1 = 2$$

$$g(0) = 2\left|0 - \tfrac{1}{2}\right| - 1 = 2 \cdot \tfrac{1}{2} - 1 = 1 - 1 = 0$$

$$g(1) = 2\left|1 - \tfrac{1}{2}\right| - 1 = 2 \cdot \tfrac{1}{2} - 1 = 1 - 1 = 0$$

$$g(2) = 2\left|2 - \tfrac{1}{2}\right| - 1 = 2 \cdot \tfrac{3}{2} - 1 = 3 - 1 = 2$$

Comparison:

Since $$f(x) = g(x)$$ for every $$x \in A$$, the functions $$f$$ and $$g$$ are equal.

We want relations $$R$$ on $$A = \{1, 2, 3\}$$ containing $$(1, 2)$$ and $$(1, 3)$$ that are reflexive and symmetric but not transitive.

Forced pairs. Reflexivity forces $$(1, 1), (2, 2), (3, 3) \in R$$. The pairs $$(1, 2)$$ and $$(1, 3)$$ are given. Symmetry then forces $$(2, 1)$$ and $$(3, 1)$$ into $$R$$. So $$R$$ must contain

$$R_0 = \{(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)\}.$$

Is $$R_0$$ valid? It is reflexive and symmetric. It is not transitive: $$(2, 1) \in R_0$$ and $$(1, 3) \in R_0$$, but $$(2, 3) \notin R_0$$. So $$R_0$$ itself is a relation that is reflexive and symmetric but not transitive.

Can we add more? The only pairs not in $$R_0$$ are $$(2, 3)$$ and $$(3, 2)$$. To keep the relation symmetric we would have to add both. But $$R_0 \cup \{(2,3),(3,2)\}$$ is the whole set $$A \times A$$, which is transitive, so it does not satisfy "not transitive". Adding only one of $$(2,3), (3,2)$$ destroys symmetry.

Hence $$R_0$$ is the only relation satisfying all the conditions. The correct answer is (A) $$1$$.

We count equivalence relations $$R$$ on $$A = \{1, 2, 3\}$$ that contain $$(1, 2)$$.

Forced pairs. Reflexivity forces $$(1, 1), (2, 2), (3, 3) \in R$$. The pair $$(1, 2)$$ is given, and symmetry forces $$(2, 1) \in R$$. So $$R$$ must contain

$$R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}.$$

$$R_1$$ is an equivalence relation: it is reflexive, symmetric, and transitive (every chain closes, e.g. $$(1,2),(2,1) \Rightarrow (1,1) \in R_1$$). So $$R_1$$ is one valid relation.

Any larger one. If the equivalence relation also relates $$3$$ to $$1$$ or $$2$$ — say it contains $$(2, 3)$$ — then symmetry gives $$(3, 2)$$, and transitivity with $$(1, 2)$$ gives $$(1, 3)$$ and hence $$(3, 1)$$. The relation then contains all $$9$$ pairs, i.e. $$R_2 = A \times A$$, the universal relation, which is also an equivalence relation.

So an equivalence relation containing $$(1, 2)$$ either keeps $$3$$ in a class by itself ($$R_1$$) or puts all three elements in one class ($$R_2 = A \times A$$). There is no other possibility.

Hence there are exactly $$2$$ such equivalence relations. The correct answer is (B) $$2$$.