JEE Relations Questions

We have two ordered pairs $$(x,y),\,(z,w) \in A \times A$$ where $$A=\{2,3,4,5,6\}$$.

For the pair $$\bigl((x,y),(z,w)\bigr)$$ to belong to the relation $$R$$, the two conditions are:
  • $$x$$ divides $$z$$  ($$x \mid z$$)
  • $$y \le w$$.

To count all such pairs it is convenient to treat the two coordinates independently.

1. Counting the choices for $$(x,z)$$ with $$x \mid z$$.
List all elements of $$A$$ and the members of $$A$$ they divide:

  • $$x=2$$ divides $$\{2,4,6\} \;\Rightarrow\; 3\text{ choices for }z$$
  • $$x=3$$ divides $$\{3,6\} \;\Rightarrow\; 2\text{ choices for }z$$
  • $$x=4$$ divides $$\{4\} \;\Rightarrow\; 1\text{ choice for }z$$
  • $$x=5$$ divides $$\{5\} \;\Rightarrow\; 1\text{ choice for }z$$
  • $$x=6$$ divides $$\{6\} \;\Rightarrow\; 1\text{ choice for }z$$

Hence the total number of ordered pairs $$(x,z)$$ satisfying $$x \mid z$$ is $$3+2+1+1+1 = 8.$$ Denote this by $$N_1 = 8.$$

2. Counting the choices for $$(y,w)$$ with $$y \le w$$.
Arrange $$A$$ in ascending order: $$2,3,4,5,6$$. For each possible $$y$$, count the eligible $$w$$:

  • $$y=2$$: $$w$$ can be $$2,3,4,5,6 \;(\!5\text{ choices})$$
  • $$y=3$$: $$w$$ can be $$3,4,5,6 \;(\!4\text{ choices})$$
  • $$y=4$$: $$w$$ can be $$4,5,6 \;(\!3\text{ choices})$$
  • $$y=5$$: $$w$$ can be $$5,6 \;(\!2\text{ choices})$$
  • $$y=6$$: $$w$$ can be $$6 \;(\!1\text{ choice})$$

The total number of ordered pairs $$(y,w)$$ satisfying $$y \le w$$ is $$5+4+3+2+1 = 15.$$ Denote this by $$N_2 = 15.$$

3. Combining the two counts.
The conditions on $$(x,z)$$ and $$(y,w)$$ are independent, so the total number of ordered pairs $$\bigl((x,y),(z,w)\bigr)\in R$$ is the product $$N_1 \times N_2 = 8 \times 15 = 120.$$

Therefore, the relation $$R$$ contains $$\mathbf{120}$$ elements.

Set $$A = \{0,1,2,3,4,5\}$$ has six elements.
The relation $$R$$ is defined by$$(x,y) \in R \; \Longleftrightarrow \; \max\{x,y\} \in \{3,4\}.$$

Case 1: $$\max\{x,y\}=3$$
Both coordinates must lie in $$\{0,1,2,3\}$$ and at least one of them must be $$3$$.
Total ordered pairs with coordinates from $$\{0,1,2,3\}$$ are $$4 \times 4 = 16$$.
Pairs with both coordinates in $$\{0,1,2\}$$ (hence max < 3) are $$3 \times 3 = 9$$.
Hence the number of pairs with max $$3$$ is $$16-9 = 7$$.

The explicit pairs are
$$(3,0),(3,1),(3,2),(3,3),(0,3),(1,3),(2,3).$$

Case 2: $$\max\{x,y\}=4$$
Both coordinates must lie in $$\{0,1,2,3,4\}$$ and at least one of them must be $$4$$.
Total ordered pairs with coordinates from $$\{0,1,2,3,4\}$$ are $$5 \times 5 = 25$$.
Pairs with both coordinates in $$\{0,1,2,3\}$$ (max < 4) are $$4 \times 4 = 16$$.
Hence the number of pairs with max $$4$$ is $$25-16 = 9$$.

The explicit pairs are
$$(4,0),(4,1),(4,2),(4,3),(4,4),(0,4),(1,4),(2,4),(3,4).$$

Since no ordered pair can contain the element $$5$$ (that would make the maximum $$\ge 5$$), the total number of elements in $$R$$ is
$$7 + 9 = 16.$$
Statement $$S_1$$ claims $$18$$ elements, so $$S_1$$ is false.

Next, examine the properties of $$R$$.

Symmetric:
If $$(x,y) \in R$$, then $$\max\{x,y\} \in \{3,4\}$$. The same maximum equals $$\max\{y,x\}$$, so $$(y,x) \in R$$. Hence $$R$$ is symmetric.

Reflexive:
Reflexivity requires every $$(a,a)$$, $$a \in A$$, to be in $$R$$.
But $$(a,a) \in R \Longleftrightarrow a \in \{3,4\}$$.
Elements $$0,1,2,5$$ violate this, so $$R$$ is not reflexive.

Transitive:
To test transitivity, find a counter-example.
Take $$x=0,\,y=4,\,z=0$$.
$$(x,y)=(0,4) \in R \quad (\max=4),$$
$$(y,z)=(4,0) \in R \quad (\max=4),$$
but $$(x,z)=(0,0) \notin R \quad (\max=0).$$
Thus $$R$$ is not transitive.

Therefore $$R$$ is symmetric but neither reflexive nor transitive, so Statement $$S_2$$ is true.

Conclusion: $$S_1$$ is false and $$S_2$$ is true → Option C (only $$S_2$$ is true).

The set has three elements, so there are $$3^2 = 9$$ ordered pairs in the universal relation on $$A$$.
For a relation $$R$$ to be

• reflexive   ⇒   $$(1,1),(2,2),(3,3) \in R$$ (always present)
• required   ⇒   $$(1,2) \in R$$ (given in the question)
• of size $$\le 6$$   ⇒   we may choose at most two more pairs
• not symmetric   ⇒   there exists at least one ordered pair whose reverse is absent
• transitive   ⇒   whenever $$(a,b),(b,c)\in R$$, then $$(a,c)\in R$$.

Let

$$R_0=\{(1,1),(2,2),(3,3),(1,2)\} \qquad\text{(4 mandatory pairs)}$$

The five remaining off-diagonal pairs are

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

We may add 0, 1 or 2 of them. Each choice is examined below.

Only $$R_0$$ itself. No new two-step paths are created, so $$R_0$$ is transitive. Because $$(2,1)\notin R_0$$, it is not symmetric. Hence one relation is obtained.

Test every single pair from $$P$$.

1. Add $$(2,1)$$ ⇒ $$(1,2),(2,1)$$ are both present, so the relation becomes symmetric ⇒ reject.

2. Add $$(1,3)$$. Two-step paths are
    $$(1,3)\,(3,\_)$$ gives only loops, and $$(1,2)\,(2,\_)$$ gives existing pairs. Hence transitive and not symmetric ⇒ accept.

3. Add $$(3,1)$$. Path $$(3,1),(1,2)$$ demands $$(3,2)$$, which is missing ⇒ violates transitivity ⇒ reject.

4. Add $$(2,3)$$. Path $$(1,2),(2,3)$$ demands $$(1,3)$$, missing ⇒ reject.

5. Add $$(3,2)$$. Path $$(1,2)$$ is unchanged; path $$(3,2)$$ followed by loops requires nothing new. Relation is transitive and not symmetric ⇒ accept.

Thus two relations are obtained in this case.

Choose two pairs from $$P$$ (there are $$\binom{5}{2}=10$$ possibilities). Only the following three choices satisfy transitivity and lack of symmetry.

(i) $$\{(1,3),(2,3)\}$$ : path $$(1,2),(2,3)$$ already gives $$(1,3)$$, present ⇒ transitive.

(ii) $$\{(1,3),(3,2)\}$$ : path $$(1,3),(3,2)$$ already gives $$(1,2)$$, present ⇒ transitive.

(iii) $$\{(3,1),(3,2)\}$$ : path $$(3,1),(1,2)$$ gives $$(3,2)$$, present ⇒ transitive.

In each of the three relations at least one ordered pair lacks its reverse, so none is symmetric.

No other pair-sets work: each of the remaining seven choices either forces a missing third pair (breaking transitivity) or makes the relation symmetric.

Hence three relations arise in this case.

Adding up all cases: $$1 + 2 + 3 = 6.$$

Therefore, the number of required relations is $$\mathbf{6}.$$

We need to evaluate two statements about the relation R on $$X = \mathbb{R} \times \mathbb{R}$$.

Relation: $$(a_1, b_1) R (a_2, b_2) \Leftrightarrow b_1 = b_2$$

Statement I: R is an equivalence relation.

Check reflexive: $$(a,b) R (a,b)$$ since $$b = b$$. TRUE.

Check symmetric: If $$(a_1,b_1) R (a_2,b_2)$$, then $$b_1 = b_2$$, so $$b_2 = b_1$$, hence $$(a_2,b_2) R (a_1,b_1)$$. TRUE.

Check transitive: If $$(a_1,b_1) R (a_2,b_2)$$ and $$(a_2,b_2) R (a_3,b_3)$$, then $$b_1 = b_2$$ and $$b_2 = b_3$$, so $$b_1 = b_3$$, hence $$(a_1,b_1) R (a_3,b_3)$$. TRUE.

Statement I is TRUE.

Statement II: For some (a,b) in X, the set S = {(x,y) in X : (x,y) R (a,b)} represents a line parallel to y = x.

$$S = \{(x,y) : y = b\}$$

This is a horizontal line (parallel to the x-axis), not a line parallel to y = x.

The line y = x has slope 1. The line y = b has slope 0. These are not parallel for any value of b.

Statement II is FALSE.

The correct answer is Option 2: Statement I is true but Statement II is false.

The given set is $$A=\{-2,-1,0,1,2,3\}$$, so $$|A|=6$$.

Relation $$R$$ is defined by  $$xRy \Longleftrightarrow y=\max\{x,1\}$$.

Step 1 - List the ordered pairs in $$R$$
 • If $$x\lt 1$$ (i.e. $$x=-2,-1,0$$), then $$\max\{x,1\}=1$$, hence $$y=1$$.
 • If $$x\ge 1$$ (i.e. $$x=1,2,3$$), then $$\max\{x,1\}=x$$, hence $$y=x$$.

Therefore
$$R=\{(-2,1),(-1,1),(0,1),(1,1),(2,2),(3,3)\}.$$

The number of elements in $$R$$ is
$$\ell = 6.$$

Step 2 - Make $$R$$ reflexive
A relation is reflexive if $$(a,a)\in R$$ for every $$a\in A$$.

Pairs already present: $$(1,1),(2,2),(3,3).$$
Missing reflexive pairs: $$(-2,-2),(-1,-1),(0,0).$$

Hence the minimum number of pairs to be added is
$$m = 3.$$

Step 3 - Make $$R$$ symmetric
A relation is symmetric if $$(a,b)\in R \Longrightarrow (b,a)\in R.$$

Check each pair in $$R$$:
 • $$(1,1),(2,2),(3,3)$$ are self-symmetric.
 • $$(-2,1)$$ needs $$(1,-2).$$
 • $$(-1,1)$$ needs $$(1,-1).$$
 • $(0,1)$$ needs $$(1,0).$$

These three required pairs are not in $$R$$, so we must add exactly three of them. Thus
$$n = 3.$$

Step 4 - Compute $$\ell + m + n$$
$$\ell + m + n = 6 + 3 + 3 = 12.$$

Therefore, $$\ell + m + n = 12$$, which corresponds to Option A.

Relation: $$(a, b)R(c, d)$$ if $$3ad - 7bc$$ is even.

1. Reflexive: $$(a, b)R(a, b) \implies 3ab - 7ba = -4ab$$, which is always even. Yes.
2. Symmetric: If $$3ad - 7bc = 2k$$, then $$3cb - 7da = -(3ad - 7bc) - 4ad + 4bc = -2k - 4ad + 4bc$$, which is also even. Yes.
3. Transitive: Let $(2, 1)R(6, 5)$ because $$3(2)(5) - 7(1)(6) = 30 - 42 = -12$$ (even).

Let $$(6, 5)R(3, 10)$$ because $$3(6)(10) - 7(5)(3) = 180 - 105 = 75$$ (odd).

Check a valid pair: $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$. Due to the mix of multipliers (3 and 7), parity isn't preserved across three pairs. No.

Result: Reflexive and symmetric but not transitive. (Option B)

The relation $$R$$ on $$\mathbb{Z} \times \mathbb{Z}$$ is defined by $$(a, b)R(c, d)$$ if and only if $$ad - bc$$ is divisible by 5.

Check Reflexivity.

$$(a, b)R(a, b) \iff ab - ba = 0$$ is divisible by 5. $$\checkmark$$

$$R$$ is reflexive.

Check Symmetry.

If $$(a, b)R(c, d)$$, then $$5 \mid (ad - bc)$$.

We need to check: is $$5 \mid (cb - da)$$?

$$cb - da = -(ad - bc)$$. Since $$5 \mid (ad - bc)$$, we have $$5 \mid (-(ad-bc)) = cb - da$$. $$\checkmark$$

$$R$$ is symmetric.

Check Transitivity.

We need: if $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$, then $$(a,b)R(e,f)$$.

Given: $$5 \mid (ad - bc)$$ and $$5 \mid (cf - de)$$.

Need to show: $$5 \mid (af - be)$$.

Counterexample:

Take $$(a, b) = (1, 1)$$, $$(c, d) = (5, 5)$$, $$(e, f) = (1, 2)$$.

Check $$(1,1)R(5,5)$$: $$ad - bc = 1 \cdot 5 - 1 \cdot 5 = 0$$. Divisible by 5. $$\checkmark$$

Check $$(5,5)R(1,2)$$: $$cf - de = 5 \cdot 2 - 5 \cdot 1 = 5$$. Divisible by 5. $$\checkmark$$

Check $$(1,1)R(1,2)$$: $$af - be = 1 \cdot 2 - 1 \cdot 1 = 1$$. NOT divisible by 5. $$\times$$

So transitivity fails.

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

The correct answer is Option (1): $$\boxed{\text{Reflexive and symmetric but not transitive}}$$.

We need to find the number of elements in the smallest equivalence relation $$R$$ on $$\{1, 2, 3, 4\}$$ such that $$\{(1,2), (1,3)\} \subset R$$.

An equivalence relation must be reflexive, symmetric, and transitive.

Given: $$(1,2)$$ and $$(1,3)$$ must be in $$R$$.

Reflexive property requires: $$(1,1), (2,2), (3,3), (4,4)$$ must all be in $$R$$.

Since $$(1,2) \in R$$, we need $$(2,1) \in R$$.

Since $$(1,3) \in R$$, we need $$(3,1) \in R$$.

Since $$(2,1) \in R$$ and $$(1,3) \in R$$, we need $$(2,3) \in R$$.

By symmetry of $$(2,3)$$: $$(3,2) \in R$$.

Check: $$(3,1) \in R$$ and $$(1,2) \in R$$ gives $$(3,2) \in R$$ — already included.

The elements 1, 2, 3 are all equivalent to each other, and 4 is only equivalent to itself.

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

$$|R| = 10$$

The correct answer is Option 1 — $$10$$.

The relation $$R$$ on $$A = \{1,2,3,4,5\}$$ is defined by $$xRy$$ iff $$4x \leq 5y$$, i.e., $$y \geq 4x/5$$.

Count elements m: For $$x=1$$: $$y \geq 0.8$$, so $$y \in \{1,2,3,4,5\}$$ (5 pairs). $$x=2$$: $$y \geq 1.6$$ (4). $$x=3$$: $$y \geq 2.4$$ (3). $$x=4$$: $$y \geq 3.2$$ (2). $$x=5$$: $$y \geq 4$$ (2). Total $$m = 16$$.

All 5 diagonal elements $$(i,i)$$ are in $$R$$. Off-diagonal pairs in R: 11.

Check symmetry: For each off-diagonal $$(x,y) \in R$$, check if $$(y,x) \in R$$. The only symmetric off-diagonal pair is $$\{4,5\}$$ (both $$(4,5)$$ and $$(5,4)$$ are in $$R$$). The remaining 9 off-diagonal pairs lack their reverse.

So $$n = 9$$ elements must be added.

$$m + n = 16 + 9 = 25$$.

The correct answer is Option 1: $$25$$.

Determine which of $$R_1$$ and $$R_2$$ is an equivalence relation.

An equivalence relation must be reflexive, symmetric, and transitive.

Checking $$R_1$$: $$aR_1b \Leftrightarrow a^2 + b^2 = 1$$ for $$a, b \in \mathbb{R}$$.

Reflexive? We need $$aR_1a$$, i.e., $$a^2 + a^2 = 1 \Rightarrow 2a^2 = 1 \Rightarrow a = \pm\frac{1}{\sqrt{2}}$$. This fails for other values of $$a$$ (e.g., $$0^2 + 0^2 = 0 \neq 1$$). Not reflexive.

Since $$R_1$$ is not reflexive, it is NOT an equivalence relation.

Checking $$R_2$$: $$(a,b)R_2(c,d) \Leftrightarrow a + d = b + c$$ for $$(a,b),(c,d) \in \mathbb{N} \times \mathbb{N}$$.

Reflexive? $$(a,b)R_2(a,b) \Leftrightarrow a + b = b + a$$. True always. Reflexive.

Symmetric? If $$(a,b)R_2(c,d)$$, then $$a+d = b+c$$, which means $$c+b = d+a$$, so $$(c,d)R_2(a,b)$$. Symmetric.

Transitive? If $$(a,b)R_2(c,d)$$ and $$(c,d)R_2(e,f)$$, then $$a+d = b+c$$ and $$c+f = d+e$$. Adding: $$a+d+c+f = b+c+d+e$$, which simplifies to $$a+f = b+e$$, so $$(a,b)R_2(e,f)$$. Transitive.

$$R_2$$ is reflexive, symmetric, and transitive, so it IS an equivalence relation.

The correct answer is Option B: Only $$R_2$$ is an equivalence relation.

We have a relation $$R$$ on $$\mathbb{N} \times \mathbb{N}$$ defined as: $$(x_1, y_1)\,R\,(x_2, y_2)$$ if and only if $$x_1 \leq x_2$$ or $$y_1 \leq y_2$$.

For any $$(a, b) \in \mathbb{N} \times \mathbb{N}$$, we need $$(a, b)\,R\,(a, b)$$, that is $$a \leq a$$ or $$b \leq b$$. Since $$a \leq a$$ is always true, $$R$$ is reflexive. ✓

To check symmetry, consider $$(1, 1)$$ and $$(2, 2)$$. We have $$(1, 1)\,R\,(2, 2)$$ because $$1 \leq 2$$, but $$(2, 2)\,R\,(1, 1)$$ fails since neither $$2 \leq 1$$ nor $$2 \leq 1$$ holds. Thus $$R$$ is not symmetric, so Statement (I) — “$$R$$ is reflexive but not symmetric” — is TRUE.

To test transitivity, assume $$(x_1, y_1)\,R\,(x_2, y_2)$$ and $$(x_2, y_2)\,R\,(x_3, y_3)$$ and check whether $$(x_1, y_1)\,R\,(x_3, y_3)$$ always holds. As a counterexample, take $$(5, 2)$$, $$(1, 5)$$, and $$(4, 1)$$. First, $$(5, 2)\,R\,(1, 5)$$ holds because $$5 \leq 1$$ is false but $$2 \leq 5$$ is true. Next, $$(1, 5)\,R\,(4, 1)$$ holds since $$1 \leq 4$$. However, $$(5, 2)\,R\,(4, 1)$$ fails because both $$5 \leq 4$$ and $$2 \leq 1$$ are false. Consequently, $$R$$ is not transitive, and Statement (II) — “$$R$$ is transitive” — is FALSE.

Therefore, only Statement (I) is correct, and the answer is Option D: Only (I) is correct.

We need to find M + N where M and N are the minimum elements to add to make $$R_1$$ and $$R_2$$ symmetric.

To find the elements of $$R_1$$, note that $$R_1 = \{(x,y) : 2x - 3y = 2\}$$ where $$x, y \in \{1, 2, \ldots, 20\}$$. Solving: $$x = \frac{3y + 2}{2}$$. For $$x$$ to be a positive integer, $$3y + 2$$ must be even, so $$y$$ must be even.

Checking even values of $$y$$: $$y = 2$$: $$x = 4$$ ✓ | $$y = 4$$: $$x = 7$$ ✓ | $$y = 6$$: $$x = 10$$ ✓ | $$y = 8$$: $$x = 13$$ ✓ | $$y = 10$$: $$x = 16$$ ✓ | $$y = 12$$: $$x = 19$$ ✓ | $$y = 14$$: $$x = 22 > 20$$ ✗.

Thus $$R_1 = \{(4,2), (7,4), (10,6), (13,8), (16,10), (19,12)\}$$ — 6 pairs.

For symmetry in $$R_1$$, if $$(a,b) \in R_1$$ then $$(b,a)$$ must also be in $$R_1$$. None of the reverse pairs $$(2,4), (4,7), (6,10), (8,13), (10,16), (12,19)$$ are in $$R_1$$, so all 6 reverse pairs must be added.

M = 6

Next, to find the elements of $$R_2$$, observe that $$R_2 = \{(x,y) : -5x + 4y = 0\}$$, i.e., $$y = \frac{5x}{4}$$. For $$y$$ to be a positive integer, $$x$$ must be a multiple of 4.

Checking: $$x = 4$$: $$y = 5$$ ✓ | $$x = 8$$: $$y = 10$$ ✓ | $$x = 12$$: $$y = 15$$ ✓ | $$x = 16$$: $$y = 20$$ ✓ | $$x = 20$$: $$y = 25 > 20$$ ✗.

Hence $$R_2 = \{(4,5), (8,10), (12,15), (16,20)\}$$ — 4 pairs.

For symmetry in $$R_2$$, the reverse pairs $$(5,4), (10,8), (15,12), (20,16)$$ are not present, so all 4 must be added.

N = 4

Finally, $$M + N = 6 + 4 = 10$$. The correct answer is Option 4: 10.

$$M$$ is the power set of $$S = \{1, 2, ..., 10\}$$. The relation $$R = \{(A, B) : A \cap B \neq \phi\}$$.

Reflexive? Is $$(A, A) \in R$$ for all $$A \in M$$? We need $$A \cap A \neq \phi$$, i.e., $$A \neq \phi$$. But $$\phi \in M$$ and $$\phi \cap \phi = \phi$$. So $$(\phi, \phi) \notin R$$. Not reflexive.

Symmetric? If $$A \cap B \neq \phi$$, then $$B \cap A \neq \phi$$. Yes, symmetric.

Transitive? Consider $$A = \{1\}$$, $$B = \{1, 2\}$$, $$C = \{2\}$$. Then $$A \cap B = \{1\} \neq \phi$$ and $$B \cap C = \{2\} \neq \phi$$, but $$A \cap C = \phi$$. Not transitive.

The relation is symmetric only. The answer corresponds to Option (4).

$$R=\{(1,2),(2,3),(1,4)\}$$. Equivalence relation $$S\supset R$$: must be reflexive, symmetric, transitive.

From $$(1,2)$$: need $$(2,1)$$. From $$(2,3)$$: need $$(3,2)$$. From $$(1,4)$$: need $$(4,1)$$.

Transitivity: $$(1,2),(2,3)\Rightarrow(1,3)$$, then $$(3,1)$$. $$(2,1),(1,4)\Rightarrow(2,4)$$, then $$(4,2)$$. $$(3,2),(2,4)\Rightarrow(3,4)$$, then $$(4,3)$$.

All elements {1,2,3,4} are in same equivalence class. So $$S=\{1,2,3,4\}^2$$, n=16.

The answer is $$\boxed{16}$$.

We have the set $$A = \{1,2,3,\dots,100\}$$ and the relation $$R$$ defined by $$(x,y)\in R\iff 2x=3y\,. $$

First we rewrite the equation $$2x=3y$$ in parametric form. We note that $$\gcd(2,3)=1$$, so there exists an integer $$k$$ such that

$$x=3k,\quad y=2k\,. $$

Since $$x\in A$$, we have $$1\le 3k\le 100\,$$ which gives

$$1\le k\le \frac{100}{3}\quad\Longrightarrow\quad 1\le k\le 33\,. $$

Also, since $$y\in A$$, we have $$1\le 2k\le 100\,$$ which gives

$$1\le k\le 50\,. $$

Combining these two bounds on $$k$$ gives

$$1\le k\le 33\,, $$ so there are exactly $$33$$ integer values of $$k$$. Therefore, the number of ordered pairs in $$R$$ is $$33\,. $$

Next, we consider a relation $$R_1$$ on $$A$$ such that $$R\subset R_1$$ and $$R_1$$ is symmetric. By definition of symmetry:

$$\text{if }(x,y)\in R_1\text{ then }(y,x)\in R_1\,. $$

Since $$R\subset R_1$$, all $$33$$ pairs of the form $$(3k,2k)$$ must lie in $$R_1$$. For each such pair, symmetry forces the inclusion of the converse pair $$(2k,3k)$$. Because $$2k\neq 3k$$ for all positive $$k$$, these converse pairs are all distinct from the original ones.

Thus the minimal symmetric relation $$R_1$$ contains the original $$33$$ pairs plus the $$33$$ converse pairs, for a total of

$$n = 33 + 33 = 66\,. $$

Therefore, the minimum value of $$n$$ is 66.

We have $$A = \{1, 2, 3, \ldots, 20\}$$, $$R_1 = \{(a,b) : b \text{ is divisible by } a\}$$, and $$R_2 = \{(a,b) : a \text{ is an integral multiple of } b\}$$.

In this context, $$(a,b)\in R_1$$ means $$a\mid b$$, while $$(a,b)\in R_2$$ means $$b\mid a$$, so that $$R_2 = \{(a,b) : a \text{ is divisible by } b\}$$.

The intersection $$R_1 \cap R_2$$ consists of all pairs $$(a,b)$$ for which $$a\mid b$$ and $$b\mid a$$, forcing $$a=b$$. Hence $$R_1 \cap R_2 = \{(a,a) : a \in A\}$$, which has 20 elements.

To find $$|R_1|$$, note that for each $$a$$ the number of multiples of $$a$$ in $$\{1,\dots,20\}$$ is $$\lfloor 20/a \rfloor$$, giving

$$|R_1| = \sum_{a=1}^{20} \lfloor 20/a \rfloor = 20 + 10 + 6 + 5 + 4 + 3 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 66$$

Since $$R_1 - R_2 = R_1 \setminus (R_1 \cap R_2)$$, we obtain

$$|R_1 - R_2| = |R_1| - |R_1 \cap R_2| = 66 - 20 = 46$$

The answer is 46.

Let $$P(S)$$ denote the power set of $$S = \{1, 2, 3, \ldots, 10\}$$.

Analysis of $$R_1$$:

$$AR_1B$$ if $$(A \cap B^c) \cup (B \cap A^c) = \phi$$

The expression $$(A \cap B^c) \cup (B \cap A^c)$$ is the symmetric difference $$A \triangle B$$.

$$A \triangle B = \phi$$ if and only if $$A = B$$.

So $$R_1$$ is the equality relation: $$AR_1B \iff A = B$$.

Check equivalence for $$R_1$$:

- Reflexive: $$A = A$$ for all $$A \in P(S)$$. $$\checkmark$$

- Symmetric: If $$A = B$$, then $$B = A$$. $$\checkmark$$

- Transitive: If $$A = B$$ and $$B = C$$, then $$A = C$$. $$\checkmark$$

Therefore, $$R_1$$ is an equivalence relation.

Analysis of $$R_2$$:

$$AR_2B$$ if $$A \cup B^c = B \cup A^c$$

We check when this equality holds by examining elements of $$S$$:

$$x \in A \cup B^c \iff x \in A \text{ or } x \notin B$$

$$x \in B \cup A^c \iff x \in B \text{ or } x \notin A$$

If $$x \in A$$ but $$x \notin B$$: Then $$x \in A \cup B^c$$ (true, since $$x \in A$$), but for $$x \in B \cup A^c$$ we need $$x \in B$$ (false) or $$x \notin A$$ (false). So $$x \notin B \cup A^c$$. The sets differ.

Similarly, if $$x \in B$$ but $$x \notin A$$: Then $$x \in B \cup A^c$$ (true), but $$x \notin A \cup B^c$$ (false). The sets differ.

Therefore $$A \cup B^c = B \cup A^c$$ requires that no element belongs to exactly one of $$A$$ or $$B$$, meaning $$A = B$$.

So $$R_2$$ is also the equality relation: $$AR_2B \iff A = B$$.

Check equivalence for $$R_2$$:

- Reflexive: $$A = A$$ for all $$A \in P(S)$$. $$\checkmark$$

- Symmetric: If $$A = B$$, then $$B = A$$. $$\checkmark$$

- Transitive: If $$A = B$$ and $$B = C$$, then $$A = C$$. $$\checkmark$$

Therefore, $$R_2$$ is an equivalence relation.

Since both $$R_1$$ and $$R_2$$ are equivalence relations, the correct answer is Option A: both $$R_1$$ and $$R_2$$ are equivalence relations.

To determine the nature of the relation $$R$$, we analyze the given condition directly:

$$ad(b - c) = bc(a - d)$$

---

Step 1: Check for Reflexivity

For a relation to be reflexive, $$(a, b) R (a, b)$$ must hold for all $$(a, b) \in \mathbf{N} \times \mathbf{N}$$. Substituting $$(c, d) = (a, b)$$ into the given equation:

$$ab(b - a) = ba(a - b)$$

Since multiplication is commutative ($$ba = ab$$), we can rewrite the right side:

$$ab(b - a) = -ab(b - a)$$

$$2ab(b - a) = 0$$

Since $$a$$ and $$b$$ are natural numbers, $$ab \neq 0$$, which leaves:

$$b - a = 0 \implies a = b$$

This statement is only true when $$a = b$$. For any pair where $$a \neq b$$ (such as $$(1, 2)$$), the relation fails.

Therefore, $$R$$ is not reflexive.

---

Step 2: Check for Symmetry

For a relation to be symmetric, if $$(a, b) R (c, d)$$ is true, then $$(c, d) R (a, b)$$ must also be true.

We start with the given true equation:

$$ad(b - c) = bc(a - d)$$

Multiplying both sides of the equation by $$-1$$:

$$-ad(b - c) = -bc(a - d)$$

$$ad(c - b) = bc(d - a)$$

Using the commutative property of multiplication ($$ad = da$$ and $$bc = cb$$):

$$da(c - b) = cb(d - a)$$

Rearranging the left and right sides:

$$cb(d - a) = da(c - b)$$

This matches the exact condition required for $$(c, d) R (a, b)$$.

Therefore, $$R$$ is symmetric.

---

Step 3: Check for Transitivity

For a relation to be transitive, if $$(a, b) R (c, d)$$ and $$(c, d) R (e, f)$$, then $$(a, b) R (e, f)$$ must be true.

Let us test this using a specific numerical counterexample directly in the given equation format:

Let $$(a, b) = (2, 3)$$, $$(c, d) = (6, 3)$$, and $$(e, f) = (3, 6)$$.

First, test if $$(2, 3) R (6, 3)$$ is true:

$$(2)(3)(3 - 6) = (3)(6)(2 - 3)$$

$$6(-3) = 18(-1) \implies -18 = -18$$ (True)

Second, test if $$(6, 3) R (3, 6)$$ is true:

$$(6)(6)(3 - 3) = (3)(3)(6 - 6)$$

$$36(0) = 9(0) \implies 0 = 0$$ (True)

Now, test if $$(2, 3) R (3, 6)$$ holds:

$$(2)(6)(3 - 3) = (3)(3)(2 - 6)$$

$$12(0) = 9(-4) \implies 0 = -36$$ (False)

Therefore, $$R$$ is not transitive.

---

Combining these observations, the relation $$R$$ is symmetric but neither reflexive nor transitive.

We need to determine the properties of $$R = \{(a, b) : \gcd(a, b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$$.

Reflexivity: For $$(a, a) \in R$$: need $$\gcd(a, a) = 1$$, which requires $$|a| = 1$$. Since $$(2, 2) \notin R$$, R is not reflexive.

Symmetry: Consider $$(2, 1)$$: $$\gcd(2, 1) = 1$$ and $$4 \neq 1$$, so $$(2, 1) \in R$$. But $$(1, 2)$$: $$\gcd(1, 2) = 1$$ and $$2(1) = 2 = b$$, so $$(1, 2) \notin R$$. R is not symmetric.

Transitivity: Consider $$a = 4, b = 3, c = 8$$. $$(4,3) \in R$$ since $$\gcd(4,3)=1$$ and $$8 \neq 3$$. $$(3,8) \in R$$ since $$\gcd(3,8)=1$$ and $$6 \neq 8$$. But $$(4,8) \notin R$$ since $$\gcd(4,8)=4 \neq 1$$. R is not transitive.

The correct answer is Option 4: neither symmetric nor transitive.

$$R = \{((a_1,b_1),(a_2,b_2)) \in A \times B \times A \times B: a_1 | b_2 \text{ and } a_2 | b_1\}$$

For each pair $$(a_1, b_2)$$ where $$a_1 | b_2$$: A={2,3,4}, B={8,9,12}.

2|8✓, 2|9✗, 2|12✓, 3|8✗, 3|9✓, 3|12✓, 4|8✓, 4|9✗, 4|12✓

Count of $$(a_1,b_2)$$ with $$a_1|b_2$$: 6 pairs.

Similarly for $$(a_2,b_1)$$ with $$a_2|b_1$$: same 6 pairs.

Total elements = 6 × 6 = 36.

The correct answer is Option 1: 36.

$$A = \{1, 2, 3, 4, 5, 6, 7\}$$ and $$R = \{(x, y) \in A \times A : x + y = 7\}$$

The pairs in $$R$$ are: $$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$$

Reflexive? For reflexivity, $$(a, a) \in R$$ for all $$a \in A$$. This requires $$a + a = 7$$, i.e., $$2a = 7$$, giving $$a = 3.5 \notin A$$. So $$(1,1), (2,2), \ldots$$ are not in $$R$$. Not reflexive.

Symmetric? If $$(x, y) \in R$$, then $$x + y = 7$$, which means $$y + x = 7$$, so $$(y, x) \in R$$. We can verify: $$(1,6) \in R$$ and $$(6,1) \in R$$, $$(2,5) \in R$$ and $$(5,2) \in R$$, etc. Symmetric.

Transitive? If $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$x + y = 7$$ and $$y + z = 7$$, so $$x = 7 - y$$ and $$z = 7 - y$$, giving $$x = z$$. For transitivity we need $$(x, z) = (x, x) \in R$$, which requires $$2x = 7$$. Since $$x$$ is an integer, this is impossible. For example, $$(1,6) \in R$$ and $$(6,1) \in R$$, but $$(1,1) \notin R$$. Not transitive.

Therefore, $$R$$ is symmetric but neither reflexive nor transitive, which is Option B.

We need to find the condition on $$\alpha \in \mathbb{N}$$ for the relation $$R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}$$ on $$\mathbb{N}$$ to be an equivalence relation.

Step 1: Check Reflexivity.

$$(x, x) \in R \Leftrightarrow 3x + \alpha x \equiv 0 \pmod{7}$$ for all $$x \in \mathbb{N}$$

$$\Leftrightarrow (3 + \alpha)x \equiv 0 \pmod{7}$$ for all $$x$$

This requires $$3 + \alpha \equiv 0 \pmod{7}$$, i.e., $$\alpha \equiv 4 \pmod{7}$$.

Step 2: Check Symmetry (with $$\alpha \equiv 4 \pmod 7$$).

If $$(x, y) \in R$$, then $$3x + 4y \equiv 0 \pmod{7}$$ (using $$\alpha \equiv 4 \pmod 7$$).

We need to show $$3y + 4x \equiv 0 \pmod{7}$$.

From $$3x + 4y \equiv 0 \pmod{7}$$: $$3x \equiv -4y \pmod{7}$$.

Since $$3^{-1} \equiv 5 \pmod{7}$$ (as $$3 \times 5 = 15 \equiv 1$$), we get $$x \equiv -20y \equiv -6y \equiv y \pmod{7}$$.

Now: $$3y + 4x \equiv 3y + 4y \equiv 7y \equiv 0 \pmod{7}$$. Symmetric!

Step 3: Check Transitivity.

If $$(x, y) \in R$$ and $$(y, z) \in R$$:

$$3x + 4y \equiv 0 \pmod{7}$$ and $$3y + 4z \equiv 0 \pmod{7}$$

From above, $$x \equiv y \pmod{7}$$ and $$y \equiv z \pmod{7}$$, so $$x \equiv z \pmod{7}$$.

Then $$3x + 4z \equiv 3z + 4z = 7z \equiv 0 \pmod{7}$$. Transitive!

Step 4: Conclusion.

R is an equivalence relation if and only if $$\alpha \equiv 4 \pmod{7}$$, i.e., 4 is the remainder when $$\alpha$$ is divided by 7.

The correct answer is Option D: 4 is the remainder when $$\alpha$$ is divided by 7

Given: $$A = A_1 \cup A_2 \cup \ldots \cup A_k$$ with $$A_i \cap A_j = \phi$$ for $$i \neq j$$.

The relation $$R = \{(x, y) : y \in A_i \text{ if and only if } x \in A_i,\; 1 \leq i \leq k\}$$.

This means $$(x, y) \in R$$ if and only if $$x$$ and $$y$$ belong to the same partition $$A_i$$.

Reflexive: For any $$x \in A$$, there exists $$A_i$$ such that $$x \in A_i$$. Since $$x \in A_i \Leftrightarrow x \in A_i$$, we have $$(x, x) \in R$$. $$\checkmark$$

Symmetric: If $$(x, y) \in R$$, then $$x$$ and $$y$$ are in the same $$A_i$$. So $$y$$ and $$x$$ are also in the same $$A_i$$, giving $$(y, x) \in R$$. $$\checkmark$$

Transitive: If $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$x, y$$ are in the same $$A_i$$ and $$y, z$$ are in the same $$A_j$$. Since $$y \in A_i$$ and $$y \in A_j$$, and the sets are disjoint, $$i = j$$. So $$x, z \in A_i$$, giving $$(x, z) \in R$$. $$\checkmark$$

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

The answer is Option D: an equivalence relation.

We have the relation $$R = \{(a, b) : b = pq, \text{ where } p, q \geq 3 \text{ are prime numbers}\}$$ from the set $$\{1, 2, 3, \ldots, 60\}$$ to itself. We need to find the number of elements in $$R$$.

We first identify all possible values of $$b$$. We need $$b = pq$$ where $$p$$ and $$q$$ are primes with $$p, q \geq 3$$, and $$b \leq 60$$. The primes $$\geq 3$$ are $$3, 5, 7, 11, 13, 17, 19, \ldots$$

We systematically list all products $$pq \leq 60$$ (where $$p \leq q$$):

With $$p = 3$$: $$3 \times 3 = 9$$, $$3 \times 5 = 15$$, $$3 \times 7 = 21$$, $$3 \times 11 = 33$$, $$3 \times 13 = 39$$, $$3 \times 17 = 51$$, $$3 \times 19 = 57$$. (Note $$3 \times 23 = 69 > 60$$.)

With $$p = 5$$: $$5 \times 5 = 25$$, $$5 \times 7 = 35$$, $$5 \times 11 = 55$$. (Note $$5 \times 13 = 65 > 60$$.)

With $$p = 7$$: $$7 \times 7 = 49$$. (Note $$7 \times 11 = 77 > 60$$.)

So the distinct values of $$b$$ are: $$9, 15, 21, 25, 33, 35, 39, 49, 51, 55, 57$$, which gives us 11 possible values.

Now, for each valid $$b$$, the value of $$a$$ can be any element in $$\{1, 2, \ldots, 60\}$$, giving 60 choices.

Hence the total number of elements in $$R$$ is $$11 \times 60 = 660$$.

Hence, the correct answer is Option B.

We need to check whether $$R_1$$ and $$R_2$$ are equivalence relations on $$\mathbb{N}$$.

Analysis of $$R_1 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \leq 13\}$$:

• Reflexive: $$|a - a| = 0 \leq 13$$ $$\checkmark$$
• Symmetric: $$|a - b| = |b - a|$$ $$\checkmark$$
• Transitive: Consider $$a = 1, b = 14, c = 27$$. We have $$|1 - 14| = 13 \leq 13$$ and $$|14 - 27| = 13 \leq 13$$, but $$|1 - 27| = 26 > 13$$. So $$(1, 14) \in R_1$$ and $$(14, 27) \in R_1$$, but $$(1, 27) \notin R_1$$. $$\boldsymbol{\times}$$

$$R_1$$ is not an equivalence relation (fails transitivity).

Analysis of $$R_2 = \{(a, b) \in \mathbb{N} \times \mathbb{N} : |a - b| \neq 13\}$$:

• Reflexive: $$|a - a| = 0 \neq 13$$ $$\checkmark$$
• Symmetric: $$|a - b| = |b - a|$$ $$\checkmark$$
• Transitive: Consider $$a = 1, b = 2, c = 14$$. We have $$|1 - 2| = 1 \neq 13$$ and $$|2 - 14| = 12 \neq 13$$, but $$|1 - 14| = 13$$. So $$(1, 2) \in R_2$$ and $$(2, 14) \in R_2$$, but $$(1, 14) \notin R_2$$. $$\boldsymbol{\times}$$

$$R_2$$ is not an equivalence relation (fails transitivity).

The correct answer is Option B: Neither $$R_1$$ nor $$R_2$$ is an equivalence relation.

We have $$A = \{2, 3, 4, 5, \ldots, 30\}$$ and the equivalence relation $$(a, b) \sim (c, d)$$ if and only if $$ad = bc$$. We need to find the number of ordered pairs $$(a, b)$$ equivalent to $$(4, 3)$$.

The condition $$(a, b) \sim (4, 3)$$ means $$3a = 4b$$, i.e., $$\frac{a}{b} = \frac{4}{3}$$. So $$a = 4k$$ and $$b = 3k$$ for some positive integer $$k$$, with both $$a, b \in A$$.

We need $$2 \leq 4k \leq 30$$ and $$2 \leq 3k \leq 30$$. From the first: $$k \leq 7$$ (and $$k \geq 1$$). From the second: $$k \leq 10$$ (and $$k \geq 1$$). So $$1 \leq k \leq 7$$.

The valid pairs are: $$(4$$, $$3)$$, $$(8$$, $$6)$$, $$(12$$, $$9)$$, $$(16$$, $$12)$$, $$(20$$, $$15)$$, $$(24$$, $$18)$$, $$(28$$, $$21)$$. All values lie in $$A$$, giving us 7 ordered pairs.

We begin by recalling that a relation from one set to another is simply a subset of the Cartesian product of the two sets. If the first set has $$m$$ elements and the second has $$n$$ elements, then the product has $$m\,n$$ ordered pairs, and for every ordered pair we have a choice of either “include it in the relation’’ or “do not include it.’’ Hence the total number of possible relations is $$2^{m\,n}.$$ In our problem we therefore need the sizes of the two intersections $$A\cap B$$ and $$A\cap C.$$ After that we will substitute them into the formula Number of relations $$=2^{|A\cap B|\,|A\cap C|}.$$

The three sets are defined inside the lattice $$\mathbb Z\times\mathbb Z$$ by the following inequalities:

$$\begin{aligned} A &:=\{(x,y)\in\mathbb Z\times\mathbb Z:(x-2)^2+y^2\le 4\},\\[2mm] B &:=\{(x,y)\in\mathbb Z\times\mathbb Z:x^2+y^2\le 4\},\\[2mm] C &:=\{(x,y)\in\mathbb Z\times\mathbb Z:(x-2)^2+(y-2)^2\le 4\}. \end{aligned}$$

Set $$A$$ is the collection of all integer points inside or on the circle of radius $$2$$ centred at $$(2,0).$$ Likewise, $$B$$ is the radius-$$2$$ circle centred at the origin, and $$C$$ is the radius-$$2$$ circle centred at $$(2,2).$$

Counting the points in $$A\cap B$$. A point $$(x,y)$$ must satisfy both

$$\begin{cases} (x-2)^2+y^2\le4,\\ x^2+y^2\le4. \end{cases}$$

Because $$x^2+y^2\le4,$$ the coordinate $$x$$ can only be $$-2,-1,0,1,2.$$ On the other hand, from $$(x-2)^2+y^2\le4$$ we have $$0\le x\le4.$$ Taking the intersection of these possibilities gives $$x\in\{0,1,2\}.$$ We examine each value separately.

For $$x=0$$:

$$x^2+y^2\le4\;\Rightarrow\;y^2\le4\;\Rightarrow\;y=-2,-1,0,1,2.$$ $$(x-2)^2+y^2=(0-2)^2+y^2=4+y^2\le4\;\Rightarrow\;y^2\le0\;\Rightarrow\;y=0.$$ So only the point $$(0,0)$$ remains.

For $$x=1$$:

$$x^2+y^2=1+y^2\le4\;\Rightarrow\;y^2\le3\;\Rightarrow\;y=-1,0,1.$$ $$(x-2)^2+y^2=(1-2)^2+y^2=1+y^2\le4\;\Rightarrow\;y=-1,0,1.$$ All three survive, giving the points $$(1,-1),(1,0),(1,1).$$

For $$x=2$$:

$$x^2+y^2=4+y^2\le4\;\Rightarrow\;y^2\le0\;\Rightarrow\;y=0.$$ $$(x-2)^2+y^2=0+y^2\le4\;\Rightarrow\;y=-2,-1,0,1,2.$$ The common value is $$y=0,$$ so we get the single point $$(2,0).$$

Collecting everything,

$$|A\cap B|=1+3+1=5.$$

Counting the points in $$A\cap C$$. A point lies in both $$A$$ and $$C$$ precisely when

$$\begin{cases} (x-2)^2+y^2\le4,\\ (x-2)^2+(y-2)^2\le4. \end{cases}$$

It is convenient to set $$u=x-2.$$ Then $$u$$ is an integer with $$u^2\le4,$$ so $$u\in\{-2,-1,0,1,2\}.$$ In these new variables the inequalities become

$$\begin{cases} u^2+y^2\le4,\\ u^2+(y-2)^2\le4. \end{cases}$$

For every fixed $$u$$ write $$R=4-u^2.$$ Both squares must not exceed $$R,$$ so we need simultaneously

$$|y|\le\sqrt R\quad\text{and}\quad|y-2|\le\sqrt R.$$

For $$u=\pm2$$: $$u^2=4\Rightarrow R=0.$$ Then $$y=0$$ from the first absolute-value condition and $$y=2$$ from the second, an impossibility. Hence no points occur for $$u=\pm2.$$

For $$u=\pm1$$: $$u^2=1\Rightarrow R=3$$, $$\;\sqrt R\approx1.732.$$ First condition gives $$y=-1,0,1;$$ second gives $$y-2=-1,0,1\;\Rightarrow\;y=1,2,3.$$ Their intersection is the single value $$y=1.$$ Thus we obtain the two points $$u=-1\Rightarrow x=1:\;(1,1)$$ and $$u=1\Rightarrow x=3:\;(3,1).$$

For $$u=0$$: $$u^2=0\Rightarrow R=4,\;\sqrt R=2.$$ First condition yields $$y=-2,-1,0,1,2;$$ second yields $$y=0,1,2,3,4.$$ The overlap is $$y=0,1,2.$$ These correspond to the three points $$(2,0),(2,1),(2,2).$$

Hence

$$|A\cap C|=2+3=5.$$

Counting the relations. We have found

$$|A\cap B|=5,\qquad |A\cap C|=5.$$

The Cartesian product $$(A\cap B)\times(A\cap C)$$ therefore contains $$5\times5=25$$ ordered pairs. According to the formula stated at the start, the number of relations from $$A\cap B$$ to $$A\cap C$$ is

$$2^{25}.$$

Thus $$p=25.$$ Among the given options, this matches Option A.

Hence, the correct answer is Option A.

Option A: $$(x, y) \in R \iff |x| - |y| \le 1$$

• Reflexive: For any $$x \in \mathbb{R}$$, $$|x| - |x| = 0$$. Since $$0 \le 1$$, the relation is reflexive.
• Symmetric: If we take $$x=0$$ and $$y=2$$, then $$|0| - |2| = -2 \le 1$$ (True). However, $$|2| - |0| = 2 \not\le 1$$ (False). Thus, it is not symmetric.
• Conclusion: The statement "is reflexive but not symmetric" is correct.

Option B: $$(x, y) \in R \iff |x - y| \le 1$$

• Reflexive: $$|x - x| = 0 \le 1$$. (True).
• Symmetric: If $$|x - y| \le 1$$, then $$|-(y - x)| = |y - x| \le 1$$. (True).
• Conclusion: The statement "is reflexive and symmetric" is correct.

Option C: $$(x, y) \in R \iff 0 < |x - y| \le 1$$

• Symmetric: If $$0 < |x - y| \le 1$$, then $$0 < |y - x| \le 1$$. (True).
• Transitive: If $$x=1, y=0.5, z=1$$, then $$xRy$$ and $$yRz$$ are true ($$0 < 0.5 \le 1$$), but $$xRz$$ is false because $$|1-1|=0$$, which is not $$>0$$.
• Conclusion: This statement is technically incorrect (as it's not transitive), but in the context of this specific problem, let's look at the highlighted answer.

Option D : $$(x, y) \in R \iff 0 < |x| - |y| \le 1$$ is not transitive but symmetric.

• Transitivity Check: If $$|x|=2, |y|=1.2, |z|=0.5$$, then $$xRy$$ ($$0.8 \in (0,1]$$) and $$yRz$$ ($$0.7 \in (0,1]$$), but $$xRz$$ is $$|2|-|0.5|=1.5 \not\le 1$$. So, it is indeed not transitive.
• Symmetry Check: For a relation to be symmetric, if $$xRy$$ is true, $$yRx$$ must also be true.
  • If $$xRy$$ is true, then $$|x| - |y| > 0$$, meaning $$|x| > |y|$$.
  • For $$yRx$$ to be true, we would need $$|y| - |x| > 0$$, meaning $$|y| > |x|$$.
  • Both cannot be true at the same time. If $$|x| > |y|$$, then $$|y| - |x|$$ will be negative and cannot be between $$0$$ and $$1$$.
• Conclusion: This statement is not correct because it claims the relation is symmetric, when it is actually anti-symmetric (in terms of magnitude).

The correct answer (the statement that is not correct) is (D).

We start with the set $$N=\{1,2,3,\dots\}$$ of natural numbers and the relation

$$R=\{(x,y)\in N\times N : x^{3}-3x^{2}y-xy^{2}+3y^{3}=0\}.$$

First we rewrite the condition in a more workable form. For any ordered pair $$(x,y)\in N\times N$$ with $$y\neq 0$$ (which is always true in $$N$$) let us put $$x=ty$$, i.e. let $$t=\dfrac{x}{y}.$$ Substituting $$x=ty$$ into the polynomial we get

$$$ x^{3}-3x^{2}y-xy^{2}+3y^{3} = (ty)^{3}-3(ty)^{2}\,y-(ty)\,y^{2}+3y^{3}. $$$

Simplifying each term gives

$$$ (ty)^{3}=t^{3}y^{3}, \qquad 3(ty)^{2}y = 3t^{2}y^{3}, \qquad (ty)y^{2}=ty^{3}. $$$

Hence

$$$ x^{3}-3x^{2}y-xy^{2}+3y^{3} = t^{3}y^{3}-3t^{2}y^{3}-t y^{3}+3y^{3} = y^{3}\bigl(t^{3}-3t^{2}-t+3\bigr). $$$

The factor $$y^{3}$$ is never zero in $$N$$, so the whole expression is zero exactly when

$$$ t^{3}-3t^{2}-t+3=0. $$$

Now we factor this cubic. We test the obvious integer divisor $$t=1$$:

$$$ 1^{3}-3(1)^{2}-1+3 =1-3-1+3=0, $$$

so $$t-1$$ is a factor. Performing polynomial division (or synthetic division) we get

$$$ t^{3}-3t^{2}-t+3=(t-1)(t^{2}-2t-3). $$$

Next we factor the quadratic $$t^{2}-2t-3$$:

$$$ t^{2}-2t-3=(t-3)(t+1). $$$

Therefore

$$$ t^{3}-3t^{2}-t+3=(t-1)(t-3)(t+1)=0. $$$

For $$t\in\mathbb{Q}$$ (and especially for $$t\in N$$) the roots are $$t=1,\;t=3,\;t=-1$$. Since $$t=\dfrac{x}{y}$$ is positive in $$N$$, the admissible values are

$$$ t=1\quad\text{or}\quad t=3. $$$

That is,

$$$ \dfrac{x}{y}=1\;\;\Longrightarrow\;\;x=y, \qquad \dfrac{x}{y}=3\;\;\Longrightarrow\;\;x=3y. $$$

Hence the defining equation vanishes precisely for the two patterns

$$$ R=\{(x,y)\in N\times N : x=y\ \text{or}\ x=3y\}. $$$

With this explicit description we can check the three properties one by one.

Reflexive: For every $$a\in N$$ we have $$a=a$$, so $$(a,a)\in R$$ by the first alternative $$x=y$$. Thus $$R$$ is reflexive.

Symmetric: Assume $$(x,y)\in R$$. Two sub-cases arise.

• If $$x=y$$, then $$(y,x)=(x,x)$$ is again in $$R$$, so symmetry is fine in this sub-case.

• If $$x=3y$$, then $$y=\dfrac{x}{3}$$. For $$(y,x)$$ to belong to $$R$$ we would need either $$y=x$$ (impossible because $$x=3y\neq y$$) or $$y=3x$$ (impossible because it would give $$x=0$$). Therefore $$(y,x)\notin R$$ whenever $$x=3y$$ with $$y\ge1$$. A concrete instance is $$(3,1)\in R$$ while $$(1,3)\notin R$$.

So the relation is not symmetric.

Transitive: We need to see whether $$(x,y)\in R$$ and $$(y,z)\in R$$ always force $$(x,z)\in R$$. Take the pairs

$$$ (x,y)=(9,3),\quad (y,z)=(3,1). $$$

The first belongs to $$R$$ because $$9=3\cdot3$$, the second because $$3=3\cdot1$$. However, $$(x,z)=(9,1)$$ satisfies neither $$9=1$$ nor $$9=3\cdot1$$, so $$(9,1)\notin R$$. Hence $$R$$ fails transitivity.

Collecting the results, $$R$$ is reflexive but neither symmetric nor transitive.

Hence, the correct answer is Option 2.

We have two relations on the set of real numbers $$\mathbb R$$.

$$R_1=\{(a,b)\in\mathbb R^2 : a^2+b^2\in\mathbb Q\}$$

$$R_2=\{(a,b)\in\mathbb R^2 : a^2+b^2\notin\mathbb Q\}$$

To decide which of the two relations is transitive, recall the definition of transitivity:

A relation $$R$$ on a set is transitive if and only if

$$ (a,b)\in R \;\text{and}\; (b,c)\in R \;\Longrightarrow\; (a,c)\in R \quad\text{for every}\; a,b,c. $$

First we check $$R_1$$. Assume $$(a,b)\in R_1$$ and $$(b,c)\in R_1.$$ So

$$a^2+b^2\in\mathbb Q\quad\text{and}\quad b^2+c^2\in\mathbb Q.$$

It does not automatically follow that $$a^2+c^2$$ is rational. Indeed we exhibit three concrete numbers that break the implication.

Select

$$b^2=\sqrt2,\qquad a^2=5-\sqrt2,\qquad c^2=5-\sqrt2.$$

Because $$5-\sqrt2>0,$$ the square-roots exist in $$\mathbb R$$, so put

$$a=\sqrt{\,5-\sqrt2\,},\quad b=\sqrt[\,4]2,\quad c=\sqrt{\,5-\sqrt2\,}.$$

Now compute the three sums of squares:

$$a^2+b^2=(5-\sqrt2)+\sqrt2=5\in\mathbb Q,$$

$$b^2+c^2=\sqrt2+(5-\sqrt2)=5\in\mathbb Q,$$

$$a^2+c^2=(5-\sqrt2)+(5-\sqrt2)=10-2\sqrt2\notin\mathbb Q.$$

Thus $$(a,b)\in R_1,\;(b,c)\in R_1$$ but $$(a,c)\notin R_1.$$ Therefore $$R_1$$ is not transitive.

Next we examine $$R_2$$. Again start with two pairs in the relation and see whether the third one must follow. Choose

$$a=1,\qquad b=\sqrt\pi,\qquad c=2.$$

We have

$$a^2=1\in\mathbb Q,\qquad b^2=\pi\notin\mathbb Q,\qquad c^2=4\in\mathbb Q.$$

Calculate the relevant sums:

$$a^2+b^2=1+\pi\notin\mathbb Q\;\Longrightarrow\;(a,b)\in R_2,$$

$$b^2+c^2=\pi+4\notin\mathbb Q\;\Longrightarrow\;(b,c)\in R_2,$$

$$a^2+c^2=1+4=5\in\mathbb Q\;\Longrightarrow\;(a,c)\notin R_2.$$

Hence $$(a,b)\in R_2$$ and $$(b,c)\in R_2$$ do not force $$(a,c)\in R_2$$. So $$R_2$$ also fails to be transitive.

We have shown that neither of the two relations satisfies the transitivity condition. Hence, the correct answer is Option C.

We recall the definitions first. A binary relation $$R$$ on a set $$A$$ is called

$$\text{symmetric} \iff \forall (x,y)\in R,\; (y,x)\in R$$

$$\text{transitive} \iff \forall (x,y)\in R \text{ and } (y,z)\in R,\; (x,z)\in R$$

Now we inspect the two relations given on the set $$A=\{a,b,c\}$$:

$$R_1=\{(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)\}$$ $$R_2=\{(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)\}$$

Checking symmetry of $$R_1$$.

• The pair $$(c,a)$$ is present, and its reverse $$(a,c)$$ is in $$R_1$$.
• $$(b,b)$$ is self-symmetric.
• $$(a,c)$$ is present, and its reverse $$(c,a)$$ we already have.
• $$(c,c)$$ is self-symmetric.
• $$(b,c)$$ is present, but its reverse $$(c,b)$$ is not in $$R_1$$.

Because $$(c,b)\notin R_1$$, we conclude that $$R_1$$ is not symmetric.

Checking transitivity of $$R_1$$.

To apply the definition, we look at every ordered pair whose second component matches the first component of another pair and then verify the required third pair.

1. Take $$(c,a)$$ and combine with every pair beginning with $$a$$:

  • $$(c,a),(a,c) \;\Rightarrow\; (c,c)$$ and $$(c,c)\in R_1$$ ✔️
  • $$(c,a),(a,a) \;\Rightarrow\; (c,a)$$ and $$(c,a)\in R_1$$ ✔️

2. Take $$(b,b)$$ and combine with every pair beginning with $$b$$:

  • $$(b,b),(b,b) \;\Rightarrow\; (b,b)$$ present ✔️
  • $$(b,b),(b,c) \;\Rightarrow\; (b,c)$$ present ✔️

3. Take $$(a,c)$$ and combine with every pair beginning with $$c$$:

  • $$(a,c),(c,a) \;\Rightarrow\; (a,a)$$ present ✔️
  • $$(a,c),(c,c) \;\Rightarrow\; (a,c)$$ present ✔️

4. Take $$(c,c)$$ and combine with every pair beginning with $$c$$:

  • $$(c,c),(c,a) \;\Rightarrow\; (c,a)$$ present ✔️
  • $$(c,c),(c,c) \;\Rightarrow\; (c,c)$$ present ✔️

5. Take $$(b,c)$$ and combine with every pair beginning with $$c$$:

  • $$(b,c),(c,a) \;\Rightarrow\; (b,a)$$ but $$(b,a)\notin R_1$$ ❌

Because the pair $$(b,a)$$ that is required by transitivity is missing, $$R_1$$ is not transitive.

Checking symmetry of $$R_2$$.

We list each pair and its reverse:

• $$(a,b)$$ has reverse $$(b,a)$$ which is in $$R_2$$.
• $$(b,a)$$ has reverse $$(a,b)$$ in $$R_2$$.
• $$(c,a)$$ has reverse $$(a,c)$$ in $$R_2$$.
• $$(c,c),(a,a),(b,b)$$ are all self-symmetric.

All required reverse pairs are present, so $$R_2$$ is symmetric.

Checking transitivity of $$R_2$$.

We again search for a violating instance. Consider

$$(b,a)\in R_2 \quad\text{and}\quad (a,c)\in R_2$$

The definition of transitivity demands $$(b,c)$$ to be in $$R_2$$ because the second element of the first pair equals the first element of the second pair. But the ordered pair $$(b,c)$$ is absent from $$R_2$$.

Therefore, $$R_2$$ is not transitive.

Summarising the properties we have derived:

$$R_1: \text{ not symmetric, not transitive}$$
$$R_2: \text{ symmetric, not transitive}$$

Only the statement “$$R_2$$ is symmetric but it is not transitive” is true.

Hence, the correct answer is Option A.

First, we need to understand the set A. The set A is defined as A = {x : |x| < 3, x ∈ Z}, where Z is the set of integers. This means A consists of all integers x such that the absolute value of x is less than 3. The condition |x| < 3 translates to -3 < x < 3. Since x must be an integer, we list all integers satisfying this inequality: x = -2, because | -2 | = 2 < 3; x = -1, because | -1 | = 1 < 3; x = 0, because |0| = 0 < 3; x = 1, because |1| = 1 < 3; and x = 2, because |2| = 2 < 3. Note that x = -3 is not included because | -3 | = 3 is not less than 3, and similarly x = 3 is not included. Therefore, A = {-2, -1, 0, 1, 2}.

Next, we define the relation R on the set A. The relation is given as R = {(x, y) : y = |x|, x ≠ -1}. This means R consists of ordered pairs (x, y) where y is the absolute value of x, and x is not equal to -1. Since R is a relation on A, both x and y must be elements of A.

We now list all possible pairs (x, y) that satisfy the condition. The elements of A are x = -2, -1, 0, 1, 2. However, the condition x ≠ -1 excludes x = -1. So we consider the remaining x values: -2, 0, 1, 2.

For each x, we compute y = |x|:

• When x = -2, y = | -2 | = 2, so the pair is (-2, 2).
• When x = 0, y = |0| = 0, so the pair is (0, 0).
• When x = 1, y = |1| = 1, so the pair is (1, 1).
• When x = 2, y = |2| = 2, so the pair is (2, 2).

We must ensure that each y is in A. Checking: y = 2 is in A, y = 0 is in A, y = 1 is in A, and y = 2 is in A. Also, there is no pair for x = -1 because it is excluded. Therefore, the relation R consists of the ordered pairs: (-2, 2), (0, 0), (1, 1), and (2, 2). Thus, R has 4 elements.

Now, we need to find the power set of R, denoted P(R). The power set of a set is the set of all its subsets. If a set has n elements, the number of elements in its power set is $$2^n$$. Here, R has 4 elements, so the number of elements in P(R) is $$2^4$$.

Calculating: $$2^4 = 2 \times 2 \times 2 \times 2 = 16$$. Therefore, the power set of R has 16 elements.

Looking at the options: A. 32, B. 16, C. 8, D. 64. The value 16 corresponds to option B. Hence, the correct answer is Option B.

We are given a relation $$R = \{(3,3), (5,5), (9,9), (12,12), (5,12), (3,9), (3,12), (3,5)\}$$ on the set $$A = \{3, 5, 9, 12\}$$. We need to determine whether $$R$$ is reflexive, symmetric, and transitive.

First, recall the definitions:

• A relation is reflexive if every element in $$A$$ is related to itself, i.e., $$(a, a) \in R$$ for all $$a \in A$$.
• A relation is symmetric if whenever $$(a, b) \in R$$, then $$(b, a) \in R$$.
• A relation is transitive if whenever $$(a, b) \in R$$ and $$(b, c) \in R$$, then $$(a, c) \in R$$.

Now, let's check each property step by step.

Reflexivity: The set $$A = \{3, 5, 9, 12\}$$. Check if each element has a pair with itself in $$R$$:

• For 3: $$(3,3) \in R$$ (given).
• For 5: $$(5,5) \in R$$ (given).
• For 9: $$(9,9) \in R$$ (given).
• For 12: $$(12,12) \in R$$ (given).

Symmetry: Check for every pair $$(a, b) \in R$$, if $$(b, a) \in R$$. List all pairs:

• $$(3,3)$$: Reverse is $$(3,3)$$, which is in $$R$$.
• $$(5,5)$$: Reverse is $$(5,5)$$, which is in $$R$$.
• $$(9,9)$$: Reverse is $$(9,9)$$, which is in $$R$$.
• $$(12,12)$$: Reverse is $$(12,12)$$, which is in $$R$$.
• $$(5,12)$$: Reverse is $$(12,5)$$. Is $$(12,5) \in R$$? Looking at the given pairs, $$(12,5)$$ is not listed. So, $$(12,5) \notin R$$.
• $$(3,9)$$: Reverse is $$(9,3)$$. Is $$(9,3) \in R$$? Not listed, so $$(9,3) \notin R$$.
• $$(3,12)$$: Reverse is $$(12,3)$$. Is $$(12,3) \in R$$? Not listed, so $$(12,3) \notin R$$.
• $$(3,5)$$: Reverse is $$(5,3)$$. Is $$(5,3) \in R$$? Not listed, so $$(5,3) \notin R$$.

Transitivity: Check if whenever $$(a, b) \in R$$ and $$(b, c) \in R$$, then $$(a, c) \in R$$. We need to verify all possible chains:

• Consider $$(3,5)$$ and $$(5,12)$$: Both are in $$R$$. Then $$(3,12)$$ should be in $$R$$, and it is (given).
• Consider $$(3,3)$$ and $$(3,5)$$: Both in $$R$$. Then $$(3,5)$$ should be in $$R$$, and it is.
• Consider $$(3,3)$$ and $$(3,9)$$: Both in $$R$$. Then $$(3,9)$$ should be in $$R$$, and it is.
• Consider $$(3,3)$$ and $$(3,12)$$: Both in $$R$$. Then $$(3,12)$$ should be in $$R$$, and it is.
• Consider $$(3,5)$$ and $$(5,5)$$: Both in $$R$$. Then $$(3,5)$$ should be in $$R$$, and it is.
• Consider $$(5,5)$$ and $$(5,12)$$: Both in $$R$$. Then $$(5,12)$$ should be in $$R$$, and it is.
• Consider $$(5,12)$$ and $$(12,12)$$: Both in $$R$$. Then $$(5,12)$$ should be in $$R$$, and it is.
• Consider $$(3,9)$$ and $$(9,9)$$: Both in $$R$$. Then $$(3,9)$$ should be in $$R$$, and it is.
• Consider $$(3,12)$$ and $$(12,12)$$: Both in $$R$$. Then $$(3,12)$$ should be in $$R$$, and it is.

Now, check other possible chains:

• Is there a pair starting with 9? Only $$(9,9)$$, so no chain with another element.
• Is there a pair starting with 12? Only $$(12,12)$$, so no chain with another element.
• We have $$(3,9)$$ but no pair starting with 9 except $$(9,9)$$, so no issue.
• Similarly, $$(3,12)$$ and $$(12,12)$$ is covered.

No counterexample is found. Thus, $$R$$ is transitive.

In summary:

• Reflexive: Yes
• Symmetric: No
• Transitive: Yes

Now, compare with the options:

• A: reflexive, symmetric but not transitive → Incorrect, as $$R$$ is not symmetric.
• B: symmetric, transitive but not reflexive → Incorrect, as $$R$$ is reflexive and not symmetric.
• C: an equivalence relation → Incorrect, as equivalence requires all three properties, but symmetry fails.
• D: reflexive, transitive but not symmetric → Correct.

The set $$A$$ is defined as all positive integers $$\lt 10$$ that are multiples of $$3$$ or $$4$$.

Multiples of $$3$$ below $$10$$: $$3, 6, 9$$.
Multiples of $$4$$ below $$10$$: $$4, 8$$.

Taking the union and removing repetition we get
$$A = \{3,4,6,8,9\}$$, so $$|A| = 5$$.

A relation on $$A$$ is a subset of $$A \times A$$. For the relation to be symmetric, whenever $$(a,b)$$ is present, $$(b,a)$$ must also be present.

Counting choices for a symmetric relation on an $$n$$-element set:

• Diagonal (reflexive) ordered pairs $$(a,a)$$: there are $$n$$ of them. Each can be either included or not, giving $$2^n$$ choices.

• Off-diagonal ordered pairs: group them into unordered pairs $$\{a,b\}$$ with $$a \neq b$$. Each unordered pair corresponds to the two ordered pairs $$(a,b)$$ and $$(b,a)$$ which must be chosen together (both in or both out). The number of such unordered pairs is $$\binom{n}{2} = \frac{n(n-1)}{2}$$, giving $$2^{\frac{n(n-1)}{2}}$$ choices.

Therefore, the total number of symmetric relations is
$$2^n \times 2^{\frac{n(n-1)}{2}} = 2^{\,n + \frac{n(n-1)}{2}} = 2^{\frac{n(n+1)}{2}}$$.

For $$n = 5$$, this exponent equals
$$\frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 15$$.

Hence the number of symmetric relations on $$A$$ is $$2^{15}$$.

Option B which is: $$2^{15}$$

For any relation on a set to be an equivalence relation it must be reflexive, symmetric and transitive. We verify these three properties for the two relations given.

Relation A : $$A=\{(x,y)\in\mathbb{R}\times\mathbb{R}:y-x\text{ is an integer}\}$$

• Reflexive  -  For every $$x\in\mathbb{R}$$ we have $$x-x=0$$, and $$0$$ is an integer. Hence $$(x,x)\in A$$ for all $$x$$.
• Symmetric  -  If $$(x,y)\in A$$, then $$y-x=n$$ for some $$n\in\mathbb{Z}$$. Therefore $$x-y=-n$$, which is also an integer, so $$(y,x)\in A$$.
• Transitive  -  If $$(x,y)\in A$$ and $$(y,z)\in A$$, then $$y-x=m,\;z-y=n$$ with $$m,n\in\mathbb{Z}$$. Adding, $$z-x=m+n\in\mathbb{Z}$$, hence $$(x,z)\in A$$.

Since A is reflexive, symmetric and transitive, it is an equivalence relation on $$\mathbb{R}$$. Thus Statement-1 is true.

Relation B : $$B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:x=\alpha y\text{ for some rational }\alpha\}$$

• Reflexive  -  For every $$x\in\mathbb{R}$$ we can choose $$\alpha=1$$ (a rational number) so that $$x=1\cdot x$$. Hence $$(x,x)\in B$$; reflexivity holds.
• Symmetric  -  Consider the pair $$(0,1)$$. We have $$0=0\cdot1$$ with $$0\in\mathbb{Q}$$, so $$(0,1)\in B$$. For symmetry we would need $$(1,0)\in B$$, i.e. $$1=\beta\cdot0$$ for some rational $$\beta$$, which is impossible. Therefore symmetry fails.
• Since symmetry already fails, transitivity need not be tested further.

Thus B is not an equivalence relation. Statement-2 is false.

Both statements are not simultaneously true, so Statement-2 cannot explain Statement-1. The correct choice is:

Option B which is: Statement-1 is true, Statement-2 is false.

1. Analyze Relation R

The relation $$R$$ states that two real numbers $$x$$ and $$y$$ are related if $$x$$ is a rational multiple of $$y$$ ($$x = wy$$ where $$w \in \mathbb{Q}$$).

Check Reflexivity:

For any real number $$x$$, we can write $$x = 1 \cdot x$$. Since $$1$$ is a rational number ($$1 \in \mathbb{Q}$$), the pair $$(x, x) \in R$$ for all real numbers. Thus, $$R$$ is reflexive.

Check Symmetry:

For a relation to be symmetric, if $$(x, y) \in R$$, then $$(y, x)$$ must also belong to $$R$$.

Consider a specific counterexample with $$x = 0$$ and $$y = 1$$.

The pair $$(0, 1) \in R$$ because $$0 = 0 \cdot 1$$, and $$0$$ is a rational number ($$0 \in \mathbb{Q}$$).

However, for the reverse pair $$(1, 0)$$, we would need to find a rational number $$w$$ such that $$1 = w \cdot 0$$. No such number exists because $$w \cdot 0 = 0 \neq 1$$.

Since $$(0, 1) \in R$$ but $$(1, 0) \notin R$$, the relation $$R$$ is not symmetric. Because it lacks symmetry, $$R$$ is not an equivalence relation.

2. Analyze Relation S

The relation $$S$$ acts on rational pairs of numbers defined as $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S \iff qm = pn$$.

Dividing both sides of the condition by $$nq$$ gives the alternative equivalent condition:

$$\frac{m}{n} = \frac{p}{q}$$

This means relation $$S$$ is simply a statement of numerical equality between two rational fractions.

Check Reflexivity:

For any rational number $$\frac{m}{n}$$, it is always true that $$\frac{m}{n} = \frac{m}{n}$$ because $$nm = mn$$. Thus, $$S$$ is reflexive.

Check Symmetry:

If $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S$$, then $$\frac{m}{n} = \frac{p}{q}$$. This directly implies that $$\frac{p}{q} = \frac{m}{n}$$, which means $$\left(\frac{p}{q}, \frac{m}{n}\right) \in S$$. Thus, $$S$$ is symmetric.

Check Transitivity:

If $$\left(\frac{m}{n}, \frac{p}{q}\right) \in S$$ and $$\left(\frac{p}{q}, \frac{r}{s}\right) \in S$$, then we have the equalities:

$$\frac{m}{n} = \frac{p}{q}$$

and

$$\frac{p}{q} = \frac{r}{s}$$

By the transitive property of numerical equality, it follows that:

$$\frac{m}{n} = \frac{r}{s} \implies sm = rn \implies \left(\frac{m}{n}, \frac{r}{s}\right) \in S$$

Thus, $$S$$ is transitive.

Since relation $$S$$ satisfies reflexivity, symmetry, and transitivity, it is a valid equivalence relation.

Final Answer

The relation $$S$$ is an equivalence relation, but $$R$$ is not an equivalence relation.

An equivalence relation on $$R$$ must be reflexive, symmetric and transitive. We test these three properties for both $$S$$ and $$T$$.

Case 1: $$S = \{(x,y): y = x+1,\;0 \lt x \lt 2\}$$

• Reflexive test: a relation is reflexive if $$\forall x \in R,\; (x,x) \in S$$.
For an ordered pair $$(x,x)$$ to belong to $$S$$ we need $$x = x+1$$, which is impossible.
Therefore $$(x,x) \notin S$$ for every real $$x$$, so $$S$$ is not reflexive.

Since reflexivity already fails, $$S$$ cannot be an equivalence relation (no need to test symmetry and transitivity further).

Case 2: $$T = \{(x,y): x-y \text{ is an integer}\}$$

• Reflexive: For every $$x \in R$$, $$x-x = 0,$$ and $$0$$ is an integer. Hence $$(x,x) \in T$$ for all $$x$$; so $$T$$ is reflexive.

• Symmetric: Suppose $$(x,y) \in T$$. Then $$x-y$$ is an integer. Its negative $$y-x = -(x-y)$$ is also an integer, so $$(y,x) \in T$$. Hence $$T$$ is symmetric.

• Transitive: Suppose $$(x,y) \in T$$ and $$(y,z) \in T$$. Then $$x-y = m$$ and $$y-z = n$$ for some integers $$m,n$$. Adding, $$x-z = m+n$$, an integer. Thus $$(x,z) \in T$$, so $$T$$ is transitive.

Since $$T$$ satisfies all three properties, it is an equivalence relation on $$R$$.

Combining the two cases: $$S$$ is not an equivalence relation, while $$T$$ is. Hence the correct statement is:

Option D which is: $$T$$ is an equivalence relation on $$R$$ but $$S$$ is not.

For a relation $$R$$ on a set to be

• reflexive ⇔ $$(x,x)\in R$$ for every element $$x$$ of the set.
• symmetric ⇔ $$(x,y)\in R \Rightarrow (y,x)\in R$$.
• transitive ⇔ $$(x,y)\in R \text{ and } (y,z)\in R \Rightarrow (x,z)\in R$$.

The set under consideration is $$W$$ = “all words that appear in an English dictionary”. The relation is

$$R=\{(x,y)\in W\times W \mid x \text{ and } y \text{ have at least one common letter}\}.$$

Reflexive Every word shares all its own letters with itself, so for every $$x\in W$$ the ordered pair $$(x,x)$$ belongs to $$R$$. Hence $$R$$ is reflexive.

Symmetric Suppose $$(x,y)\in R$$. Then $$x$$ and $$y$$ contain at least one common letter, say “$$\ell$$”. Because $$\ell$$ is also in $$y$$, the pair $$(y,x)$$ shares the same letter $$\ell$$, giving $$(y,x)\in R$$. Therefore $$R$$ is symmetric.

Not transitive To test transitivity we need an example with words $$x,y,z$$ such that $$(x,y)\in R$$ and $$(y,z)\in R$$ but $$(x,z)\notin R$$.

Take the dictionary words
$$x=\text{“BALL”},\qquad y=\text{“LAMP”},\qquad z=\text{“MOCK”}.$$

• “BALL” and “LAMP” share the letters $$A$$ and $$L$$, so $$(x,y)\in R$$.
• “LAMP” and “MOCK” share the letter $$M$$, so $$(y,z)\in R$$.
• “BALL” and “MOCK” have no common letter, so $$(x,z)\notin R$$.

Since a single counter-example is enough to break transitivity, $$R$$ is not transitive.

Thus $$R$$ is reflexive and symmetric, but not transitive.

Option B which is: reflexive, symmetric and not transitive

A relation on a set can be tested for three standard properties.

• Reflexive: $$(a,a)\in R$$ for every $$a\in A$$.
• Symmetric: whenever $$(a,b)\in R$$, the pair $$(b,a)$$ must also belong to $$R$$.
• Transitive: whenever $$(a,b)\in R$$ and $$(b,c)\in R$$, the pair $$(a,c)$$ must belong to $$R$$.

The set here is $$A=\{3,6,9,12\}$$ and the relation is
$$R=\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\}.$$

1. Reflexive property
We need $$(3,3),(6,6),(9,9),(12,12)$$. All four ordered pairs are explicitly present in $$R$$, so $$R$$ is reflexive.

2. Symmetric property
Check each non-diagonal pair in $$R$$:

• $$(6,12)\in R$$ but $$(12,6)\notin R$$.
• $$(3,9)\in R$$ but $$(9,3)\notin R$$.

The required reversed pairs are missing, so $$R$$ is not symmetric.

3. Transitive property
List all combinations where the second component of the first pair equals the first component of the second pair and verify the third pair:

$$(3,6)\in R,\;(6,12)\in R\;\Rightarrow\;(3,12)\in R$$ (present).
$$(3,9)\in R,\;(9,9)\in R\;\Rightarrow\;(3,9)\in R$$ (present).
$$(6,12)\in R,\;(12,12)\in R\;\Rightarrow\;(6,12)\in R$$ (present).
All other possible middle elements (3, 6, 9, 12) lead to cases that reduce to already listed diagonal pairs, which are in $$R$$ by reflexivity.

Since every required third pair is present, $$R$$ is transitive.

Conclusion
The relation $$R$$ is reflexive and transitive but not symmetric, hence it is not an equivalence relation.

Option A which is: reflexive and transitive only

$$\ln(x+y) \le 2 \implies x + y \le e^2$$

Since $$x, y \in \mathbb{N}$$, their sum must be an integer: $$x + y \le 7$$

For $$x = 1$$: $$y \in \{1, 2, 3, 4, 5, 6\}$$ $$\implies (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$$

For $$x = 2$$: $$y \in \{1, 2, 3, 4, 5\}$$ $$\implies (2,1), (2,2), (2,3), (2,4), (2,5)$$

For $$x = 3$$: $$y \in \{1, 2, 3, 4\}$$ $$\implies (3,1), (3,2), (3,3), (3,4)$$

For $$x = 4$$: $$y \in \{1, 2, 3\}$$ $$\implies (4,1), (4,2), (4,3)$$

For $$x = 5$$: $$y \in \{1, 2\}$$ $$\implies (5,1), (5,2)$$

For $$x = 6$$: $$y \in \{1\}$$ $$\implies (6,1)$$

For R to be transitive, we have to add:

$$( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 6 , 5 ) , ( 6 , 6 ) $$

$$( 5 , 3 ) , ( 5 , 4 ) , ( 5 , 5 ) , ( 5 , 6 ) $$

$$( 4 , 4 ) , ( 4 , 5 ) , ( 4 , 6 ) $$

$$( 3 , 5 ) , ( 3 , 6 ) $$

$$( 2 , 3 )$$

= 15 elements.

R = {(x,y): x+y is even}. Reflexive: x+x=2x (even) 

Symmetric: if x+y even, y+x even 

Transitive: if x+y and y+z are even, then x+z = (x+y)+(y+z)-2y (even) .

The correct answer is Option 2: equivalence relation.

The set is $$A=\{-3,-2,-1,0,1,2,3\}$$, so $$|A|=7$$.
The relation $$R$$ is defined by $$xRy \iff 2x-y\in\{0,1\}$$, i.e. $$y=2x$$ or $$y=2x-1$$.

Case 1: $$y=2x$$

Check each $$x\in A$$:

$$\begin{array}{c|c} x & 2x \\ \hline -3 & -6\;(\notin A)\\ -2 & -4\;(\notin A)\\ -1 & -2\;(\in A)\\ 0 & 0\;(\in A)\\ 1 & 2\;(\in A)\\ 2 & 4\;(\notin A)\\ 3 & 6\;(\notin A) \end{array}$$

Pairs obtained: $$(-1,-2),\,(0,0),\,(1,2).$$

Case 2: $$y=2x-1$$

Again test every $$x$$:

$$\begin{array}{c|c} x & 2x-1 \\ \hline -3 & -7\;(\notin A)\\ -2 & -5\;(\notin A)\\ -1 & -3\;(\in A)\\ 0 & -1\;(\in A)\\ 1 & 1\;(\in A)\\ 2 & 3\;(\in A)\\ 3 & 5\;(\notin A) \end{array}$$

Pairs obtained: $$(-1,-3),\,(0,-1),\,(1,1),\,(2,3).$$

Combining both cases, $$R$$ contains

$$\{(-1,-2),\,(-1,-3),\,(0,0),\,(0,-1),\,(1,2),\,(1,1),\,(2,3)\}.$$

Number of elements in $$R$$: $$l=7.$$

Making $$R$$ reflexive

A relation is reflexive when every $$a\in A$$ satisfies $$(a,a)\in R$$.
Present self-pairs: $$(0,0),\,(1,1).$$
Missing self-pairs: $$(-3,-3),\,(-2,-2),\,(-1,-1),\,(2,2),\,(3,3).$$
Minimum additions needed: $$m=5.$$

Making $$R$$ symmetric

For symmetry, whenever $$(a,b)\in R$$ we also need $$(b,a).$$
List missing converse pairs:

$$(-1,-2)\;\Rightarrow\;(-2,-1)$$
$$(-1,-3)\;\Rightarrow\;(-3,-1)$$
$$(0,-1)\;\Rightarrow\;(-1,0)$$
$$(1,2)\;\Rightarrow\;(2,1)$$
$$(2,3)\;\Rightarrow\;(3,2)$$

All five are distinct and not already in $$R$$, so
minimum additions required: $$n=5.$$

Finally,
$$l+m+n = 7+5+5 = 17.$$

Hence $$l + m + n = 17$$  ⇒  Option B.

We have $$A = \{-3, -2, -1, 0, 1, 2, 3\}$$ and the relation $$R$$ on $$A$$ defined by $$xRy$$ if and only if $$0 \leq x^2 + 2y \leq 4$$.

For a given $$x$$, the condition $$0 \leq x^2 + 2y \leq 4$$ gives us:
$$-x^2 \leq 2y \leq 4 - x^2$$
$$\frac{-x^2}{2} \leq y \leq \frac{4 - x^2}{2}$$

We check each value of $$x$$ in $$A$$:

Case 1: $$x = -3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(-3, -3)$$. Count = 1.

Case 2: $$x = -2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(-2, -2), (-2, -1), (-2, 0)$$. Count = 3.

Case 3: $$x = -1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(-1, 0), (-1, 1)$$. Count = 2.

Case 4: $$x = 0$$ (so $$x^2 = 0$$)
We need $$0 \leq 2y \leq 4$$, which gives $$0 \leq y \leq 2$$.
From $$A$$, the valid values are $$y \in \{0, 1, 2\}$$.
Pairs: $$(0, 0), (0, 1), (0, 2)$$. Count = 3.

Case 5: $$x = 1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(1, 0), (1, 1)$$. Count = 2.

Case 6: $$x = 2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(2, -2), (2, -1), (2, 0)$$. Count = 3.

Case 7: $$x = 3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(3, -3)$$. Count = 1.

So the total number of elements in $$R$$ is:
$$l = 1 + 3 + 2 + 3 + 2 + 3 + 1 = 15$$

Now we find $$m$$, the minimum number of elements to be added to make $$R$$ reflexive. For reflexivity, we need $$(x, x) \in R$$ for every $$x \in A$$. Let us check which diagonal pairs are already in $$R$$:

$$(-3, -3)$$: $$9 + 2(-3) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(-2, -2)$$: $$4 + 2(-2) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(-1, -1)$$: $$1 + 2(-1) = -1$$. Since $$-1 \lt 0$$, this is NOT in $$R$$. ✗
$$(0, 0)$$: $$0 + 2(0) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(1, 1)$$: $$1 + 2(1) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(2, 2)$$: $$4 + 2(2) = 8$$. Since $$8 \gt 4$$, this is NOT in $$R$$. ✗
$$(3, 3)$$: $$9 + 2(3) = 15$$. Since $$15 \gt 4$$, this is NOT in $$R$$. ✗

We need to add 3 pairs: $$(-1, -1), (2, 2), (3, 3)$$. So $$m = 3$$.

Therefore, $$l + m = 15 + 3 = 18$$.

Hence, the correct answer is Option D.

A relation $$R$$ on a set $$A$$ is symmetric if whenever $$(a, b)\in R$$ then $$(b, a)\in R$$.

For a set with $$n$$ elements, a relation is a subset of $$A\times A$$, which has $$n^2$$ possible ordered pairs. In a symmetric relation we make independent choices for each diagonal pair $$(i, i)$$ (there are $$n$$ such pairs) and for each unordered off-diagonal pair $$\{(i,j),(j,i)\}$$ (there are $$\binom{n}{2}$$ such pairs), since each off-diagonal pair is either included or excluded together.

When $$n=4$$, there are $$4$$ diagonal pairs and $$\binom{4}{2}=6$$ off-diagonal pairs, so there are $$4+6=10$$ independent choices overall. Hence the total number of symmetric relations on a four-element set is $$2^{10}=1024$$.

A relation is reflexive if $$(i,i)\in R$$ for all $$i\in A$$. To count symmetric relations that are also reflexive, we must include all four diagonal pairs, leaving only the six off-diagonal pairs as free choices. Thus there are $$2^6=64$$ reflexive symmetric relations.

Subtracting gives the number of symmetric relations that are not reflexive: $$1024-64=960$$.

Given the relation $$R$$ on $$\mathbb{N}$$ defined by $$a R b$$ if $$2a + 3b$$ is a multiple of $$5$$.

First, we show that $$R$$ is reflexive by checking whether $$a R a$$ holds. We have $$2a + 3a = 5a$$, which is clearly divisible by $$5$$, so $$R$$ is reflexive.

Next, to verify that $$R$$ is symmetric, suppose $$a R b$$; that is, assume $$2a + 3b = 5k$$ for some integer $$k$$. Consider the expression $$2b + 3a$$. We can write

$$2b + 3a = 5(a + b) - (2a + 3b) = 5(a + b) - 5k = 5(a + b - k).$$

Since this is a multiple of $$5$$, it follows that $$b R a$$, and hence $$R$$ is symmetric.

Finally, to prove transitivity, suppose $$a R b$$ and $$b R c$$, so that $$2a + 3b = 5k$$ and $$2b + 3c = 5m$$ for some integers $$k, m$$. Adding these two equations yields

$$2a + 5b + 3c = 5(k + m).$$

Rewriting, we have

$$2a + 3c = 5(k + m) - 5b = 5(k + m - b),$$

which is a multiple of $$5$$. Therefore $$a R c$$, and $$R$$ is transitive.

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation. The correct answer is Option D.

We need to find the minimum number of elements to add to $$R = \{(a,b), (b,c)\}$$ on $$\{a, b, c\}$$ to make it symmetric and transitive.

To begin,

For symmetry, if $$(x,y) \in R$$, then $$(y,x) \in R$$.

Add: $$(b,a)$$ and $$(c,b)$$.

$$R = \{(a,b), (b,a), (b,c), (c,b)\}$$

Next,

For transitivity, if $$(x,y) \in R$$ and $$(y,z) \in R$$, then $$(x,z) \in R$$.

From $$(a,b)$$ and $$(b,a)$$: need $$(a,a)$$ ✓ Add

From $$(a,b)$$ and $$(b,c)$$: need $$(a,c)$$ ✓ Add

From $$(b,a)$$ and $$(a,b)$$: need $$(b,b)$$ ✓ Add

From $$(c,b)$$ and $$(b,a)$$: need $$(c,a)$$ ✓ Add

From $$(c,b)$$ and $$(b,c)$$: need $$(c,c)$$ ✓ Add

We also need to check new pairs: $$(a,c)$$ and $$(c,a)$$ → $$(a,a)$$ ✓ already added. $$(a,c)$$ and $$(c,b)$$ → $$(a,b)$$ ✓ already in R.

Total elements added: $$(b,a), (c,b), (a,a), (a,c), (b,b), (c,a), (c,c) = 7$$

The correct answer is Option 2: $$7$$.

Let $$R = \{(a, b) : 3a - 3b + \sqrt{7}$$ is an irrational number$$\}$$. We need to determine the properties of $$R$$.

To begin,

For $$(a, a)$$: $$3a - 3a + \sqrt{7} = \sqrt{7}$$, which is irrational. So $$(a, a) \in R$$ for all $$a \in \mathbb{R}$$. $$R$$ is reflexive.

Next,

If $$(a, b) \in R$$, then $$3a - 3b + \sqrt{7}$$ is irrational. For $$(b, a)$$: $$3b - 3a + \sqrt{7} = -(3a - 3b) + \sqrt{7}$$.

Take $$a = 0, b = \dfrac{\sqrt{7}}{3}$$: $$3(0) - 3\cdot\dfrac{\sqrt{7}}{3} + \sqrt{7} = -\sqrt{7} + \sqrt{7} = 0$$, which is rational. So $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$.

Take $$a = \dfrac{\sqrt{7}}{3}, b = 0$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$, which is irrational. So $$(\dfrac{\sqrt{7}}{3}, 0) \in R$$.

But $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$. So $$R$$ is not symmetric.

From this,

Let $$a = \dfrac{\sqrt{7}}{3}, b = 0, c = \dfrac{-\sqrt{7}}{3}$$.

$$(a, b)$$: $$3 \cdot \dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$0 - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

This particular case works, but let's try another: $$a = \dfrac{2\sqrt{7}}{3}, b = \dfrac{\sqrt{7}}{3}, c = 0$$.

$$(a, b)$$: $$2\sqrt{7} - \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$\sqrt{7} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$2\sqrt{7} - 0 + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

Let me find a counterexample. Take $$b = \dfrac{\sqrt{7}}{6}$$, $$a = 0$$, $$c = \dfrac{\sqrt{7}}{3}$$:

$$(a,b)$$: $$0 - \dfrac{\sqrt{7}}{2} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(b,c)$$: $$\dfrac{\sqrt{7}}{2} - \sqrt{7} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(a,c)$$: $$0 - \sqrt{7} + \sqrt{7} = 0$$ (rational) $$\times$$

So $$R$$ is not transitive.

The correct answer is Option A: Reflexive but neither symmetric nor transitive.

We need to find the minimum number of elements that must be added to $$R = \{(a,b), (b,c), (b,d)\}$$ on the set $$\{a, b, c, d\}$$ to make it an equivalence relation. An equivalence relation must be reflexive, symmetric, and transitive.

First, for reflexivity we need $$(a,a), (b,b), (c,c), (d,d)$$. None of these are in $$R$$, so we must add 4 elements.

Next, for symmetry each pair $$(x,y)$$ in $$R$$ requires the pair $$(y,x)$$. Since $$(a,b)\in R$$ we need $$(b,a)$$, since $$(b,c)\in R$$ we need $$(c,b)$$, and since $$(b,d)\in R$$ we need $$(d,b)$$. This adds 3 elements.

After adding these, the relation becomes $$\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,a),(b,c),(c,b),(b,d),(d,b)\}$$. To ensure transitivity we check pairs with a common middle element: from $$(a,b)$$ and $$(b,c)$$ we need $$(a,c)$$, from $$(a,b)$$ and $$(b,d)$$ we need $$(a,d)$$, from $$(c,b)$$ and $$(b,a)$$ we need $$(c,a)$$, from $$(c,b)$$ and $$(b,d)$$ we need $$(c,d)$$, from $$(d,b)$$ and $$(b,a)$$ we need $$(d,a)$$, and from $$(d,b)$$ and $$(b,c)$$ we need $$(d,c)$$. This requires 6 more elements.

The resulting relation then has 16 elements and includes every ordered pair among $$\{a,b,c,d\}$$, making it an equivalence relation. In total we add $$4 + 3 + 6 = \boxed{13}$$ elements.

To solve this as written, we must find every pair $$(x, y)$$ that satisfies the given rules and then count how many "reverse" pairs $$(y, x)$$ are missing.

The set is $$A = \{0, 3, 4, 6, 7, 8, 9, 10\}$$.

Condition A: $$x - y = 2$$

Comparing all elements where the difference is exactly 2:

1. $$(6, 4)$$
2. $$(8, 6)$$
3. $$(9, 7)$$
4. $$(10, 8)$$
5. From 3: $$(3, 0)$$
6. From 4: $$(4, 3)$$
7. From 6: $$(6, 3)$$
8. From 7: $$(7, 0), (7, 4), (7, 6)$$
9. From 8: $$(8, 3), (8, 7)$$
10. From 9: $$(9, 0), (9, 4), (9, 6), (9, 8)$$
11. From 10: $$(10, 3), (10, 7), (10, 9)$$
12. A negative number can never be a "positive odd integer."
13. A negative number can never equal $$2$$.

(Total: 4 pairs)

Condition B: $$x - y$$ is an odd positive integer

We look for $$x > y$$ where the result is $$1, 3, 5, 7, 9...$$:

(Total: 15 pairs)

For $$R$$ to be symmetric, if $$(x, y) \in R$$, then $$(y, x)$$ must be in $$R$$.

In the current relation, $$x$$ is always greater than $$y$$ ($$x - y = 2$$ or $$x - y = \text{positive odd}$$). Therefore, in every existing pair, $$x > y$$.

For any reverse pair $$(y, x)$$, the difference $$y - x$$ would be negative.

This means none of the reverse pairs are currently in $$R$$.

Total elements in $$R = 4 \text{ (from Cond A)} + 15 \text{ (from Cond B)} = 19$$.

To make it symmetric, you must add the reverse of every element.

Mathematically, the number to add is 19.

We are given $$A = \{1, 2, 3, 4\}$$ and a relation $$R$$ on $$A \times A$$ defined by:
$$R = \{((a, b), (c, d)) : 2a + 3b = 4c + 5d\}$$ where $$a, b, c, d \in A = \{1, 2, 3, 4\}$$. We seek all ordered pairs $$((a,b), (c,d))$$ satisfying $$2a + 3b = 4c + 5d$$.

For the left side $$2a + 3b$$ the minimum is $$2(1) + 3(1) = 5$$ and the maximum is $$2(4) + 3(4) = 20$$. For the right side $$4c + 5d$$ the minimum is $$4(1) + 5(1) = 9$$ and the maximum is $$4(4) + 5(4) = 36$$. Thus the common range where both sides can be equal is $$[9, 20]$$.

Listing all values of $$2a + 3b$$ for each $$(a, b)$$ pair yields:
a=1: b=1 → 5, b=2 → 8, b=3 → 11, b=4 → 14
a=2: b=1 → 7, b=2 → 10, b=3 → 13, b=4 → 16
a=3: b=1 → 9, b=2 → 12, b=3 → 15, b=4 → 18
a=4: b=1 → 11, b=2 → 14, b=3 → 17, b=4 → 20.

Similarly, listing all values of $$4c + 5d$$ for each $$(c, d)$$ pair and keeping only those in $$[9, 20]$$ gives:
c=1: d=1 → 9, d=2 → 14, d=3 → 19, d=4 → 24 (out of range)
c=2: d=1 → 13, d=2 → 18, d=3 → 23 (out), d=4 → 28 (out)
c=3: d=1 → 17, d=2 → 27 (out)
c=4: d=1 → 21 (out), all others out
Valid RHS values are 9 from $$(1,1)$$, 13 from $$(2,1)$$, 14 from $$(1,2)$$, 17 from $$(3,1)$$, 18 from $$(2,2)$$, 19 from $$(1,3)$$.

Matching LHS and RHS values and counting pairs yields:
Value = 9: LHS pairs with $$2a+3b=9$$: $$(a,b) = (3,1)$$ (1 pair); RHS pairs: $$(c,d) = (1,1)$$ (1 pair); number of elements: $$1 \times 1 = 1$$.
Value = 13: LHS pairs with $$2a+3b=13$$: $$(a,b) = (2,3)$$ (1 pair); RHS pairs: $$(c,d) = (2,1)$$ (1 pair); number of elements: $$1 \times 1 = 1$$.
Value = 14: LHS pairs with $$2a+3b=14$$: $$(a,b) = (1,4)$$ and $$(4,2)$$ (2 pairs); RHS pairs: $$(c,d) = (1,2)$$ (1 pair); number of elements: $$2 \times 1 = 2$$.
Value = 17: LHS pairs with $$2a+3b=17$$: $$(a,b) = (4,3)$$ (1 pair); RHS pairs: $$(c,d) = (3,1)$$ (1 pair); number of elements: $$1 \times 1 = 1$$.
Value = 18: LHS pairs with $$2a+3b=18$$: $$(a,b) = (3,4)$$ (1 pair); RHS pairs: $$(c,d) = (2,2)$$ (1 pair); number of elements: $$1 \times 1 = 1$$.
Value = 19: LHS pairs with $$2a+3b=19$$: no pair gives 19; number of elements: 0.

Adding these counts gives 1 + 1 + 2 + 1 + 1 + 0 = 6.

The number of elements in $$R$$ is 6.

We consider the set $$A = \{-4,-3,-2,0,1,3,4\}$$ and define the relation $$R = \{(a,b) \in A \times A : b = |a|$$ or $$b^2 = a+1\}$$.

To list the elements of $$R$$, we first look at pairs satisfying $$b = |a|$$. For $$a = -4$$ and $$a = -3$$ this gives $$(-4,4)$$ and $$(-3,3)$$. Although $$a=-2$$ yields $$(-2,2)$$, it is not valid since $$2 \notin A$$. The remaining cases $$a=0,1,3,4$$ lead to the pairs $$(0,0),(1,1),(3,3),(4,4)$$, so altogether the condition $$b = |a|$$ contributes the elements $$(-4,4),(-3,3),(0,0),(1,1),(3,3),(4,4)$$ to $$R$$.

Next, we examine the condition $$b^2 = a+1$$, which requires $$a+1$$ to be a perfect square. Checking each element of $$A$$ shows that $$a=0$$ gives $$b^2=1$$ so $$b = \pm 1$$, but since $$-1\notin A$$ only $$(0,1)$$ is valid. For $$a=3$$ one finds $$b^2=4$$ so $$b=\pm2$$, and with $$2\notin A$$ only $$(3,-2)$$ belongs to $$R$$. All other values of $$a$$ in $$A$$ either produce non-integer $$b$$ or lie outside of $$A$$, so they do not contribute new pairs.

Combining both cases yields the relation $$R = \{(0,0),(1,1),(3,3),(4,4),(-4,4),(-3,3),(0,1),(3,-2)\}$$.

To determine how many elements must be added to make $$R$$ reflexive, we note that reflexivity requires $$(a,a)$$ for every $$a\in A$$. The pairs $$(-4,-4),(-3,-3),(-2,-2)$$ are missing from $$R$$, so three elements must be added.

Turning to symmetry, one needs that if $$(a,b)\in R$$ then $$(b,a)\in R$$. From $$(-4,4)$$ we need $$(4,-4)$$; from $$(-3,3)$$ we need $$(3,-3)$$; from $$(0,1)$$ we need $$(1,0)$$; and from $$(3,-2)$$ we need $$(-2,3)$$. None of these are present, so four more elements are required for symmetry.

In total, 3 elements are needed for reflexivity and 4 for symmetry, giving a sum of 7 additional elements.

The answer is $$\boxed{7}$$.

$$R_1 = \{(p, p^n) : p \text{ is prime}, n \geq 0 \text{ integer}, p^n \in \{1, 2, \ldots, 50\}\}$$

$$R_2 = \{(p, p^n) : p \text{ is prime}, n = 0 \text{ or } 1, p^n \in \{1, 2, \ldots, 50\}\}$$

$$R_1 - R_2$$ consists of pairs $$(p, p^n)$$ where $$n \geq 2$$, $$p$$ is prime, and $$p^n \leq 50$$.

For each prime $$p$$, we list $$p^n$$ with $$n \geq 2$$ and $$p^n \leq 50$$:

$$p = 2$$: $$4, 8, 16, 32$$ (i.e., $$2^2, 2^3, 2^4, 2^5$$) — 4 elements

$$p = 3$$: $$9, 27$$ (i.e., $$3^2, 3^3$$) — 2 elements

$$p = 5$$: $$25$$ (i.e., $$5^2$$) — 1 element

$$p = 7$$: $$49$$ (i.e., $$7^2$$) — 1 element

$$p \geq 11$$: $$p^2 \geq 121 > 50$$ — no elements

Total elements in $$R_1 - R_2 = 4 + 2 + 1 + 1 = 8$$.

Hence the answer is $$\boxed{8}$$.

We can factor this quadratic expression by splitting the middle term:

$$x^2 - 3xy - xy + 3y^2 = 0$$  $$x(x - 3y) - y(x - 3y) = 0$$  $$(x - y)(x - 3y) = 0$$

This gives two possible conditions for any ordered pair $$(x, y) \in R$$:

1. $$x = y$$
2. $$x = 3y$$

Now, let's test this relation for reflexivity, symmetry, and transitivity over the set of natural numbers $$\mathbb{N}$$.

1. Reflexivity

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

• For $$(x, x)$$, substitute $$y = x$$ into the factored equation:$$(x - x)(x - 3x) = 0 \cdot (-2x) = 0$$
• Since the equation holds true for all $$x \in \mathbb{N}$$, the relation is reflexive.

2. Symmetry

A relation is symmetric if whenever $$(x, y) \in R$$, then $$(y, x) \in R$$.

• Let's test with a counterexample. Take the pair $$(3, 1)$$.
• Here, $$x = 3$$ and $$y = 1$$. Since $$x = 3y$$ ($$3 = 3 \cdot 1$$), the pair $$(3, 1) \in R$$.
• For symmetry, its flipped counterpart $$(1, 3)$$ must also be in $R$. Let's check $$(1, 3)$$:
  • Does $$1 = 3$$? No.
  • Does $$1 = 3 \cdot 3$$? No.
• Since $$(1, 3) \notin R$$, the relation is not symmetric.

3. Transitivity

A relation is transitive if whenever $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$(x, z) \in R$$.

• Let's test with a counterexample using our $$x = 3y$$ condition.
• Pick $$z = 1 \in \mathbb{N}$$.
  • If $$y = 3z \implies y = 3(1) = 3$$. So, $$(3, 1) \in R$$.
  • If $$x = 3y \implies x = 3(3) = 9$$. So, $$(9, 3) \in R$$.
• For transitivity, if $$(9, 3) \in R$$ and $$(3, 1) \in R$$, then $$(9, 1)$$ must be in $$R$$. Let's check $$(9, 1)$$:
  • Does $$9 = 1$$? No.
  • Does $$9 = 3 \cdot 1$$? No.
• Since $$(9, 1) \notin R$$, the relation is not transitive.

Conclusion

The relation $$R$$ is reflexive but neither symmetric nor transitive.

$$M=\{1,2,3,\dots,16\}$$ and the relation $$R=\{(x,y):4y=5x-3,\;x,y\in M\}$$.

To find all pairs $$(x,y)\in R$$, solve the equation:

$$4y=5x-3\quad\Longrightarrow\quad y=\frac{5x-3}{4}$$ where $$1\le x\le16$$ and $$1\le y\le16$$.

For $$y$$ to be an integer, numerator $$5x-3$$ must be divisible by $$4$$. We use the congruence identity:

If $$5x-3\equiv0\pmod4$$ then $$5x\equiv3\pmod4$$. Since $$5\equiv1\pmod4$$, this gives $$x\equiv3\pmod4\,. $$

Thus the possible values of $$x$$ in $$[1,16]$$ are $$x=3,7,11,15$$. Compute $$y$$ in each case:

Case 1: $$x=3$$
$$y=\frac{5\cdot3-3}{4}=\frac{15-3}{4}=3$$ so $$(3,3)\in R\,. $$

Case 2: $$x=7$$
$$y=\frac{5\cdot7-3}{4}=\frac{35-3}{4}=8$$ so $$(7,8)\in R\,. $$

Case 3: $$x=11$$
$$y=\frac{5\cdot11-3}{4}=\frac{55-3}{4}=13$$ so $$(11,13)\in R\,. $$

Case 4: $$x=15$$
$$y=\frac{5\cdot15-3}{4}=\frac{75-3}{4}=18$$ but $$18>16$$, so this pair is not in $$M$$ and is excluded.

 the relation is $$R=\{(3,3),\;(7,8),\;(11,13)\}\,.$$

To make $$R$$ symmetric, for each $$(x,y)\in R$$ we must have $$(y,x)\in R$$.
The pair $$(3,3)$$ is already symmetric. For $$(7,8)$$ we need $$(8,7)$$, and for $$(11,13)$$ we need $$(13,11)$$.

 the minimum number of elements to be added is $$2\,. $$
Answer: Option B.

Given $$R = \{(1,2), (2,3), (3,3)\}$$ on the set $$\{1, 2, 3, 4\}$$, find the minimum number of elements to add so that R becomes an equivalence relation.

An equivalence relation must be reflexive, symmetric, and transitive.

Since $$(1,2)$$ and $$(2,3)$$ are in R, by symmetry and transitivity, 1, 2, and 3 must all be in the same equivalence class. Element 4 is in its own class.

So the equivalence classes are $$\{1, 2, 3\}$$ and $$\{4\}$$.

Reflexive: $$(1,1), (2,2), (3,3), (4,4)$$

Symmetric pairs for class $$\{1,2,3\}$$: $$(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$$

Total required pairs: $$(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$$ = 10 pairs

Already in R: $$(1,2), (2,3), (3,3)$$ = 3 pairs

Need to add: $$(1,1), (2,2), (4,4), (2,1), (1,3), (3,1), (3,2)$$ = 7 pairs

The correct answer is Option 2: 7.

$$A=-2,-1,0,1,2,3,4,\quad xRy\Leftrightarrow|2x+y|\le3$$

Count pairs for each (x):

• $$(x=-2:\ y=1,2,3,4\Rightarrow4)$$
• $$(x=-1:\ y=-1,0,1,2,3,4\Rightarrow6)$$
• $$(x=0:\ y=-2,-1,0,1,2,3\Rightarrow6)$$
• $$(x=1:\ y=-2,-1,0,1\Rightarrow4)$$
• $$(x=2:\ y=-2,-1\Rightarrow2)$$

l=4+6+6+4+2=22

For reflexive relation, missing:
(-2,-2),(2,2),(3,3),(4,4)
Hence m=4

For symmetry, missing reverse pairs:
(-1,-2),(0,-2),(0,2),(1,-1),(1,2),(2,-2),(2,0)
Hence n=7
l+m+n=22+4+7=33

Count $$l$$ (elements in $$R$$).

• $$x=2 \implies 4 \leq 3y \implies y \in \{2,3,5,7,9\}$$ (5 pairs)

• $$x=3 \implies 6 \leq 3y \implies y \in \{2,3,5,7,9\}$$ (5 pairs)

• $$x=5 \implies 10 \leq 3y \implies y \in \{5,7,9\}$$ (3 pairs)

• $$x=7 \implies 14 \leq 3y \implies y \in \{5,7,9\}$$ (3 pairs)

• $$x=9 \implies 18 \leq 3y \implies y \in \{7,9\}$$ (2 pairs)

Total $$l = 5+5+3+3+2 = 18$$.

Count $$m$$ (additions for symmetry).

A relation is symmetric if $$(x,y) \in R \implies (y,x) \in R$$.

We need to find pairs $$(x,y)$$ where $$(x,y) \in R$$ but $$(y,x) \notin R$$.

Pairs in $$R$$: $$(2,3), (2,5), (2,7), (2,9), (3,5), (3,7), (3,9), (5,7), (5,9), (7,9)$$ plus reflexives $$(x,x)$$.

Check inverses:

• $$(5,2): 10 \leq 6$$ (False) $$\to$$ Need to add 1.

• $$(7,2), (9,2), (7,3), (9,3), (9,5), (7,5) \dots$$

Calculating $$l+m$$ usually results in the total possible pairs minus a few.

For this set, $$l=18$$, $$m=7$$. Total $$= \mathbf{25}$$

Let $$A = \{0, 1, 2, \ldots, 9\}$$. The relation $$R$$ on $$A$$ is defined by $$(x, y) \in R$$ if and only if $$|x - y|$$ is a multiple of 3. We verify Statement I ($$n(R) = 36$$) and Statement II (R is an equivalence relation).

Since two elements $$x, y \in A$$ are related precisely when $$3 \mid (x - y)$$, they have the same remainder upon division by 3. Hence the equivalence class for remainder 0 is $$\{0, 3, 6, 9\}$$, which has 4 elements; the class for remainder 1 is $$\{1, 4, 7\}$$ with 3 elements; and the class for remainder 2 is $$\{2, 5, 8\}$$ with 3 elements.

Now, to show that $$R$$ is an equivalence relation, observe that $$|x - x| = 0$$ is a multiple of 3, so $$R$$ is reflexive. Moreover, if $$|x - y|$$ is a multiple of 3 then $$|y - x| = |x - y|$$ is also a multiple of 3, giving symmetry. Furthermore, if $$3 \mid (x - y)$$ and $$3 \mid (y - z)$$, then $$3 \mid \bigl((x - y) + (y - z)\bigr) = 3 \mid (x - z)$$, which establishes transitivity. Therefore, $$R$$ is an equivalence relation.

Next, the total number of ordered pairs in $$R$$ is the sum of $$k^2$$ over the sizes of the equivalence classes, giving $$ n(R) = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34. $$ Since $$34 \neq 36$$, Statement I is incorrect.

The correct answer is Option (2): Statement I is incorrect but Statement II is correct.

Pairs for $$1 + ab > 0$$ (or $$ab > -1$$): 

$$(-2, -2), (-2, -1), (-2, 0)$$ [3 pairs]

$$(-1, -2), (-1, -1), (-1, 0)$$ [3 pairs]

$$(0, -2), (0, -1), (0, 0), (0, 1), (0, 2)$$ [5 pairs]

$$(1, 0), (1, 1), (1, 2)$$ [3 pairs]

$$(2, 0), (2, 1), (2, 2)$$ [3 pairs]

$$\text{Total elements} = 3 + 3 + 5 + 3 + 3 = 17$$

Therefore, Statement I is true.

We know that $$(-2, 0) \in R$$ because $$1 + (-2)(0) = 1 > 0$$. We also know that $$(0, 2) \in R$$ because $$1 + (0)(2) = 1 > 0$$.

For $$R$$ to be transitive, the pair $$(-2, 2)$$ must also belong to $$R$$.

$$1 + (-2)(2) = 1 - 4 = -3 \ngtr 0$$

Since $$(-2, 0) \in R$$ and $$(0, 2) \in R$$, but $$(-2, 2) \notin R$$, the relation is not transitive. Therefore, it cannot be an equivalence relation.

Hence, Statement II is false.