Relations and Functions

Given: $A = \{1, 2, 3, \dots, 14\}$, $R = \{(x, y): 3x - y = 0\}$, i.e., $y = 3x$.

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

Reflexive: For reflexivity, $(a,a) \in R$ for all $a \in A$, i.e., $3a = a \Rightarrow a = 0$, which is not in $A$. For example, $(1,1) \notin R$. Hence R is not reflexive.

Symmetric: $(1,3) \in R$ but $3(3) = 9 \neq 1$, so $(3,1) \notin R$. Hence R is not symmetric.

Transitive: We need: if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$. Check: $(1,3) \in R$ and $(3,9) \in R$. Is $(1,9) \in R$? $3(1) = 3 \neq 9$. So $(1,9) \notin R$. Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

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

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

Reflexive: $(1,1) \notin R$ since $1 \neq 1+5$. Hence R is not reflexive.

Symmetric: $(1,6) \in R$ but $(6,1) \notin R$ since 6 \not< 4. Hence R is not symmetric.

Transitive: We need $(x,y) \in R$ and $(y,z) \in R \Rightarrow (x,z) \in R$. The pairs in R have second elements $6,7,8$, none of which is less than 4, so no pair $(y,z)$ exists in R for any $(x,y) \in R$. The condition is vacuously satisfied. Hence R is transitive.

Conclusion: R is neither reflexive nor symmetric, but it is transitive.

Given: $R = \{(x,y): x \mid y\}$ on $A = \{1,2,3,4,5,6\}$.

Reflexive: Every element divides itself, so $(a,a) \in R$ for all $a \in A$. Hence R is reflexive.

Symmetric: $(1,2) \in R$ (since 2 is divisible by 1) but $(2,1) \notin R$ (since 1 is not divisible by 2). Hence R is not symmetric.

Transitive: Suppose $(x,y) \in R$ and $(y,z) \in R$, i.e., $x \mid y$ and $y \mid z$. Then $x \mid z$, so $(x,z) \in R$. Hence R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

Given: $R = \{(x,y): x - y \in \mathbf{Z}\}$ on $\mathbf{Z}$.

Reflexive: For any $a \in \mathbf{Z}$, $a - a = 0 \in \mathbf{Z}$, so $(a,a) \in R$. Hence R is reflexive.

Symmetric: If $(x,y) \in R$, then $x - y \in \mathbf{Z}$, so $y - x = -(x-y) \in \mathbf{Z}$, giving $(y,x) \in R$. Hence R is symmetric.

Transitive: If $(x,y) \in R$ and $(y,z) \in R$, then $x-y \in \mathbf{Z}$ and $y-z \in \mathbf{Z}$. So $x - z = (x-y)+(y-z) \in \mathbf{Z}$, giving $(x,z) \in R$. Hence R is transitive.

Conclusion: R is reflexive, symmetric and transitive, hence an equivalence relation.

Reflexive: Any person $x$ works at the same place as themselves, so $(x,x) \in R$. Hence R is reflexive.

Symmetric: If $x$ and $y$ work at the same place, then $y$ and $x$ work at the same place. So $(x,y) \in R \Rightarrow (y,x) \in R$. Hence R is symmetric.

Transitive: If $x$ and $y$ work at the same place, and $y$ and $z$ work at the same place, then $x$ and $z$ work at the same place. Hence R is transitive.

Conclusion: R is an equivalence relation.

Reflexive: $x$ lives in the same locality as $x$, so $(x,x) \in R$. Hence R is reflexive.

Symmetric: If $x$ and $y$ live in the same locality, then $y$ and $x$ live in the same locality. Hence R is symmetric.

Transitive: If $x,y$ live in the same locality and $y,z$ live in the same locality, then $x,z$ live in the same locality. Hence R is transitive.

Conclusion: R is an equivalence relation.

Reflexive: $x$ cannot be 7 cm taller than itself, so $(x,x) \notin R$. Hence R is not reflexive.

Symmetric: If $x$ is exactly 7 cm taller than $y$, then $y$ is 7 cm shorter than $x$, not taller. So $(x,y) \in R \not\Rightarrow (y,x) \in R$. Hence R 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. So $(x,y) \in R$ and $(y,z) \in R \not\Rightarrow (x,z) \in R$. Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Reflexive: $x$ cannot be the wife of herself/himself, so $(x,x) \notin R$. Hence R is not reflexive.

Symmetric: If $x$ is the wife of $y$, then $y$ is the husband of $x$, not the wife. So $(x,y) \in R \not\Rightarrow (y,x) \in R$. Hence R is not symmetric.

Transitive: If $x$ is the wife of $y$, then $y$ is male, so $y$ cannot be the wife of anyone. The condition $(x,y) \in R$ and $(y,z) \in R$ never holds. Hence the condition is vacuously true, so R is transitive.

Conclusion: R is transitive but neither reflexive nor symmetric.

Reflexive: $x$ cannot be the father of himself, so $(x,x) \notin R$. Hence R is not reflexive.

Symmetric: If $x$ is the father of $y$, then $y$ is the child of $x$, not the father. So $(x,y) \in R \not\Rightarrow (y,x) \in R$. Hence R is not symmetric.

Transitive: If $x$ is the father of $y$, and $y$ is the father of $z$, then $x$ is the grandfather of $z$, not the father. So $(x,y) \in R$ and $(y,z) \in R \not\Rightarrow (x,z) \in R$. Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Not Reflexive: Take $a = \dfrac{1}{2}$. Then $a \leq a^2$ requires $\dfrac{1}{2} \leq \dfrac{1}{4}$, which is false. So $\left(\dfrac{1}{2}, \dfrac{1}{2}\right) \notin R$. Hence R is not reflexive.

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

Not Transitive: Take $a = 2, b = -2, c = -\dfrac{1}{2}$.
- $(2,-2) \in R$? $2 \leq (-2)^2 = 4$. Yes.
- $(-2, -\frac{1}{2}) \in R$? $-2 \leq \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$. Yes.
- $(2, -\frac{1}{2}) \in R$? $2 \leq \frac{1}{4}$. No.

So $(2,-2) \in R$ and $(-2,-\frac{1}{2}) \in R$ but $(2,-\frac{1}{2}) \notin R$. Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Given: $A = \{1,2,3,4,5,6\}$, $R = \{(a,b): b = a+1\}$.

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

Reflexive: $(1,1) \notin R$ since $1 \neq 1+1$. Hence R is not reflexive.

Symmetric: $(1,2) \in R$ but $(2,1) \notin R$ since $1 \neq 2+1$. Hence R is not symmetric.

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

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Reflexive: For any $a \in \mathbf{R}$, $a \leq a$ is always true. So $(a,a) \in R$ for all $a$. Hence R is reflexive.

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

Transitive: Let $(a,b) \in R$ and $(b,c) \in R$, i.e., $a \leq b$ and $b \leq c$. Then $a \leq c$, so $(a,c) \in R$. Hence R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

Not Reflexive: Take $a = \dfrac{1}{2}$. Then $a \leq a^3$ requires $\dfrac{1}{2} \leq \dfrac{1}{8}$, which is false. So $\left(\dfrac{1}{2},\dfrac{1}{2}\right) \notin R$. Hence R is not reflexive.

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

Not Transitive: Take $a = 3, b = \dfrac{3}{2}, c = \dfrac{6}{5}$.
- $(3, \frac{3}{2}) \in R$? $3 \leq \left(\frac{3}{2}\right)^3 = \frac{27}{8} = 3.375$. Yes.
- $(\frac{3}{2}, \frac{6}{5}) \in R$? $\frac{3}{2} \leq \left(\frac{6}{5}\right)^3 = \frac{216}{125} = 1.728$. Yes.
- $(3, \frac{6}{5}) \in R$? $3 \leq \left(\frac{6}{5}\right)^3 = 1.728$. No.

Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Given: $A = \{1,2,3\}$, $R = \{(1,2),(2,1)\}$.

Not Reflexive: $(1,1) \notin R$. Hence R is not reflexive.

Symmetric: $(1,2) \in R \Rightarrow (2,1) \in R$, and $(2,1) \in R \Rightarrow (1,2) \in R$. Hence R is symmetric.

Not Transitive: $(1,2) \in R$ and $(2,1) \in R$, but $(1,1) \notin R$. Hence R is not transitive.

Conclusion: R is symmetric but neither reflexive nor transitive.

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

Symmetric: If $x$ and $y$ have the same number of pages, then $y$ and $x$ have the same number of pages. So $(x,y) \in R \Rightarrow (y,x) \in R$. Hence R is symmetric.

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

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

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

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

Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $|a-b|$ and $|b-c|$ are both even, meaning $a-b$ and $b-c$ are both even. Then $a - c = (a-b)+(b-c)$ is even, so $|a-c|$ is even, giving $(a,c) \in R$. Hence R is transitive.

Therefore, R is an equivalence relation.

Elements of $\{1,3,5\}$: $|1-3|=2$ (even), $|1-5|=4$ (even), $|3-5|=2$ (even). So all elements of $\{1,3,5\}$ are related to each other.

Elements of $\{2,4\}$: $|2-4|=2$ (even). So 2 and 4 are related to each other.

Cross-check: For $a \in \{1,3,5\}$ and $b \in \{2,4\}$: $|1-2|=1, |1-4|=3, |3-2|=1, |3-4|=1, |5-2|=3, |5-4|=1$ — all odd. So no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

Given: $A = \{0,1,2,3,4,5,6,7,8,9,10,11,12\}$, $R = \{(a,b): 4 \mid |a-b|\}$.

Reflexive: For any $a \in A$, $|a-a| = 0 = 4 \times 0$, a multiple of 4. So $(a,a) \in R$. Hence R is reflexive.

Symmetric: If $(a,b) \in R$, then $4 \mid |a-b|$. Since $|a-b| = |b-a|$, we have $4 \mid |b-a|$, so $(b,a) \in R$. Hence R is symmetric.

Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $4 \mid (a-b)$ and $4 \mid (b-c)$. So $4 \mid [(a-b)+(b-c)] = (a-c)$, giving $4 \mid |a-c|$, so $(a,c) \in R$. Hence R is transitive.

Therefore, R is an equivalence relation.

Set of elements related to 1: We need $|a - 1|$ to be a multiple of 4, i.e., $a - 1 \in \{0, \pm4, \pm8, \pm12,\ldots\}$, so $a \in \{1, 5, 9\}$ (within $A$).

$$\text{Required set} = \{1, 5, 9\}$$

Given: $R = \{(a,b): a = b\}$ on $A = \{0,1,2,\ldots,12\}$.

Reflexive: For any $a \in A$, $a = a$, so $(a,a) \in R$. Hence R is reflexive.

Symmetric: If $(a,b) \in R$, then $a = b$, so $b = a$, giving $(b,a) \in R$. Hence R is symmetric.

Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $a = b$ and $b = c$, so $a = c$, giving $(a,c) \in R$. Hence R is transitive.

Therefore, R is an equivalence relation.

Set of elements related to 1: We need $a = 1$. So the only element related to 1 is 1 itself.

$$\text{Required set} = \{1\}$$

(i) Symmetric but neither reflexive nor transitive:

Let $A = \{1,2,3\}$ and $R = \{(1,2),(2,1)\}$.
- Not reflexive: $(1,1) \notin R$.
- Symmetric: $(1,2) \in R \Rightarrow (2,1) \in R$. ✓
- Not transitive: $(1,2) \in R$ and $(2,1) \in R$ but $(1,1) \notin R$.

(ii) Transitive but neither reflexive nor symmetric:

Let $A = \{1,2,3\}$ and $R = \{(1,2),(2,3),(1,3)\}$.
- Not reflexive: $(1,1) \notin R$.
- Not symmetric: $(1,2) \in R$ but $(2,1) \notin R$.
- Transitive: $(1,2) \in R, (2,3) \in R \Rightarrow (1,3) \in R$. ✓

(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: $(1,1),(2,2),(3,3) \in R$. ✓
- Symmetric: For every $(a,b) \in R$, $(b,a) \in R$. ✓
- Not transitive: $(1,2) \in R$ and $(2,3) \in R$ but $(1,3) \notin R$.

(iv) Reflexive and transitive but not symmetric:

Let $R = \{(a,b): a \leq b\}$ on $\mathbf{R}$.
- Reflexive: $a \leq a$. ✓
- Transitive: $a \leq b$ and $b \leq c \Rightarrow a \leq c$. ✓
- Not symmetric: $1 \leq 2$ but $2 \not\leq 1$.

(v) Symmetric and transitive but not reflexive:

Let $A = \{1,2,3\}$ and $R = \{(1,2),(2,1),(1,1),(2,2)\}$.
- Not reflexive: $(3,3) \notin R$.
- Symmetric: $(1,2) \in R \Rightarrow (2,1) \in R$. ✓
- Transitive: $(1,2) \in R,(2,1) \in R \Rightarrow (1,1) \in R$ ✓; $(2,1) \in R,(1,2) \in R \Rightarrow (2,2) \in R$ ✓.

Let $O$ denote the origin. For any point $P$ in the plane, let $d(P)$ denote the distance of $P$ from $O$.

Reflexive: $d(P) = d(P)$ for any point $P$, so $(P,P) \in R$. Hence R is reflexive.

Symmetric: If $(P,Q) \in R$, then $d(P) = d(Q)$, so $d(Q) = d(P)$, giving $(Q,P) \in R$. Hence R is symmetric.

Transitive: If $(P,Q) \in R$ and $(Q,S) \in R$, then $d(P) = d(Q)$ and $d(Q) = d(S)$, so $d(P) = d(S)$, giving $(P,S) \in R$. Hence R is transitive.

Therefore, R is an equivalence relation.

Set of all points related to $P \neq (0,0)$: The set of all points $Q$ such that $(P,Q) \in R$ is $\{Q : d(Q) = d(P)\}$, i.e., all points at distance $d(P)$ from the origin. This is precisely the circle with centre at the origin and radius $d(P)$, which passes through $P$.

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

Symmetric: If $T_1$ is similar to $T_2$, then $T_2$ is similar to $T_1$. So $(T_1,T_2) \in R \Rightarrow (T_2,T_1) \in R$. Hence 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 $T_3$. Hence R is transitive.

Therefore, R is an equivalence relation.

Which triangles are related?

Check if $T_1$ (sides 3,4,5) and $T_3$ (sides 6,8,10) are similar:
$$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$$
The ratios of corresponding sides are equal, so $T_1 \sim T_3$. Hence $(T_1, T_3) \in R$.

Check $T_1$ and $T_2$ (sides 5,12,13): $\dfrac{3}{5} \neq \dfrac{4}{12}$, so $T_1$ and $T_2$ are not similar.

Check $T_2$ and $T_3$: $\dfrac{5}{6} \neq \dfrac{12}{8}$, so $T_2$ and $T_3$ are not similar.

Conclusion: $T_1$ and $T_3$ are related to each other.

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

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. Hence R is transitive.

Therefore, R is an equivalence relation.

Set of elements related to triangle T: The triangle T has 3 sides. The set of all elements related to T is the set of all polygons having 3 sides, i.e., the set of all triangles in A.

Reflexive: Every line is parallel to itself (a line is considered parallel to itself). So $(L,L) \in R$ for all $L \in$ L. Hence R is reflexive.

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

Transitive: If $L_1 \parallel L_2$ and $L_2 \parallel L_3$, then $L_1 \parallel L_3$. Hence R is transitive.

Therefore, R is an equivalence relation.

Set of lines related to $y = 2x + 4$: The given line has slope 2. All lines parallel to it also have slope 2 and are of the form $y = 2x + c$, where $c \in \mathbf{R}$.

$$\text{Required set} = \{y = 2x + c : c \in \mathbf{R}\}$$

Correct Answer: (B) R is reflexive and transitive but not symmetric.

Reflexive: $(1,1),(2,2),(3,3),(4,4) \in R$. ✓ R is reflexive.

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

Transitive: Check all pairs:
- $(1,2) \in R, (2,2) \in R \Rightarrow (1,2) \in R$ ✓
- $(1,3) \in R, (3,2) \in R \Rightarrow (1,2) \in R$ ✓
- $(1,3) \in R, (3,3) \in R \Rightarrow (1,3) \in R$ ✓
- $(3,2) \in R, (2,2) \in R \Rightarrow (3,2) \in R$ ✓

All required pairs are present. R is transitive.

Hence option (B) is correct.

Correct Answer: (C) $(6,8) \in R$.

Verification of each option:

For $(a,b) \in R$: we need $a = b-2$ and b > 6.

(A) $(2,4)$: $b = 4$, but 4 \not> 6. ✗

(B) $(3,8)$: $a = 3$, $b-2 = 6 \neq 3$. ✗

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

(D) $(8,7)$: b = 7 > 6 ✓, but $b-2 = 5 \neq 8$. ✗

Hence option (C) is correct.

**Part 1: $f: \mathbf{R}_* \to \mathbf{R}_*$, $f(x) = \dfrac{1}{x}$

One-one (Injective): Let $f(x_1) = f(x_2)$.
$$\frac{1}{x_1} = \frac{1}{x_2} \Rightarrow x_1 = x_2$$
So $f$ is one-one.

Onto (Surjective):** For any $y \in \mathbf{R}_*$, we need $x \in \mathbf{R}_*$ such that $f(x) = y$, i.e., $\dfrac{1}{x} = y \Rightarrow x = \dfrac{1}{y}$. Since $y \neq 0$, $x = \dfrac{1}{y} \in \mathbf{R}_*$ and $f\left(\dfrac{1}{y}\right) = y$. So $f$ is onto.

Hence $f$ is bijective.

**Part 2: $f: \mathbf{N} \to \mathbf{R}_*$, $f(x) = \dfrac{1}{x}$

One-one: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$. So $f$ is still one-one.

Onto:** Consider $y = \dfrac{1}{2} \in \mathbf{R}_*$. We need $x \in \mathbf{N}$ such that $\dfrac{1}{x} = \dfrac{1}{2}$, giving $x = 2 \in \mathbf{N}$. But consider $y = \dfrac{1}{2.5} = 0.4 \in \mathbf{R}_*$; we need $x = 2.5 \notin \mathbf{N}$. So $f$ is not onto.

Conclusion: The result is not true when the domain is replaced by $\mathbf{N}$; $f$ is one-one but not onto.

Injective: Let $f(x_1) = f(x_2)$, i.e., $x_1^2 = x_2^2$. Since $x_1, x_2 \in \mathbf{N}$, both are positive, so $x_1 = x_2$. Hence $f$ is injective.

Surjective: Consider $y = 2 \in \mathbf{N}$. We need $x \in \mathbf{N}$ such that $x^2 = 2$, i.e., $x = \sqrt{2} \notin \mathbf{N}$. So $f$ is not surjective.

Conclusion: $f$ is injective but not surjective.

Injective: $f(-1) = 1 = f(1)$, but $-1 \neq 1$. So $f$ is not injective.

Surjective: Consider $y = -1 \in \mathbf{Z}$. We need $x^2 = -1$, which has no real solution. So $f$ is not surjective.

Conclusion: $f$ is neither injective nor surjective.

Injective: $f(-1) = 1 = f(1)$, but $-1 \neq 1$. So $f$ is not injective.

Surjective: Consider $y = -1 \in \mathbf{R}$. We need $x^2 = -1$, which has no real solution. So $f$ is not surjective.

Conclusion: $f$ is neither injective nor surjective.

Injective: Let $f(x_1) = f(x_2)$, i.e., $x_1^3 = x_2^3$. This gives $x_1 = x_2$. So $f$ is injective.

Surjective: Consider $y = 2 \in \mathbf{N}$. We need $x^3 = 2$, i.e., $x = 2^{1/3} \notin \mathbf{N}$. So $f$ is not surjective.

Conclusion: $f$ is injective but not surjective.

Injective: Let $f(x_1) = f(x_2)$, i.e., $x_1^3 = x_2^3$. This gives $x_1 = x_2$ (cube function is strictly monotone on $\mathbf{Z}$). So $f$ is injective.

Surjective: Consider $y = 2 \in \mathbf{Z}$. We need $x^3 = 2$, i.e., $x = 2^{1/3} \notin \mathbf{Z}$. So $f$ is not surjective.

Conclusion: $f$ is injective but not surjective.

Not one-one: $f(1.2) = [1.2] = 1$ and $f(1.5) = [1.5] = 1$, but $1.2 \neq 1.5$. So $f$ is not one-one.

Not onto: Consider $y = 1.5 \in \mathbf{R}$. We need $x \in \mathbf{R}$ such that $[x] = 1.5$. But $[x]$ is always an integer for any $x \in \mathbf{R}$, so $[x] = 1.5$ has no solution. Hence $f$ is not onto.

Conclusion: The greatest integer function is neither one-one nor onto.

Not one-one: $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$, but $-1 \neq 1$. So $f$ is not one-one.

Not onto: Consider $y = -1 \in \mathbf{R}$. We need $|x| = -1$, but $|x| \geq 0$ for all $x \in \mathbf{R}$, so no such $x$ exists. Hence $f$ is not onto.

Conclusion: The modulus function is neither one-one nor onto.

Not one-one: $f(1) = 1$ and $f(2) = 1$, but $1 \neq 2$. So $f$ is not one-one.

Not onto: The range of $f$ is $\{-1, 0, 1\}$, which is a proper subset of $\mathbf{R}$. For example, $y = 2 \in \mathbf{R}$ has no pre-image. Hence $f$ is not onto.

Conclusion: The signum function is neither one-one nor onto.