Ch 1  ·  Q–
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Chapter 1 Exercise 1.1 Solutions

Relations and Functions

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

Class 12 Mathematics NCERT Class 12 Maths Relations and Functions NCERT Chapter 1 Relations and Functions Exercise 1.1 NCERT Maths Solutions Class 12 Mathematics CBSE Class 12 Reflexive Relation Symmetric Relation Transitive Relation Equivalence Relation Equivalence Classes Properties of Relations Relation Problems Proof Based Questions Step by Step Solutions Board Exam Preparation JEE Main Maths CUET Mathematics
16 Questions
35–50 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Determine whether each of the following relations are reflexive, symmetric and transitive:
  1. Relation R in the set \( A=\{1,2,3,\ldots,14\} \) defined by \( R=\{(x,y):3x-y=0\} \)
  2. Relation R in the set of natural numbers \( \mathbb{N} \) defined by \[ R=\{(x,y):y=x+5,\;x<4\} \]
  3. Relation R in the set \( A=\{1,2,3,4,5,6\} \) defined by \( R=\{(x,y):y\text{ is divisible by }x\} \)
  4. Relation R in the set of integers \( \mathbb{Z} \) defined by \( R=\{(x,y):x-y\text{ is an integer}\} \)
  5. Relation R in the set of all human beings in a town at a particular time given by:
    1. \(x\) and \(y\) work at the same place.
    2. \(x\) and \(y\) live in the same locality.
    3. \(x\) is exactly 7 cm taller than \(y\).
    4. \(x\) is wife of \(y\).
    5. \(x\) is father of \(y\).
📘 Concept & Theory
Concept / Theory
›
Before solving the question, recall the three fundamental properties of a relation.
1. Reflexive Relation
A relation \(R\) on a set \(A\) is called reflexive if every element is related to itself.
Mathematically,
\[(a,a)\in R\]
2. Symmetric Relation
A relation is symmetric if whenever one element is related to another, the reverse relation is also true.
That is, whenever
\[(a,b)\in R \implies (b,a)\in R\]
3. Transitive Relation
A relation is transitive if whenever one element is related to a second, and the second is related to a third, then the first is related to the third.
Mathematically,
\[(a,b)\in R \land (b,c)\in R \implies (a,c)\in R\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the relation explicitly whenever possible.

  2. Check whether every element is related to itself (Reflexive).

  3. Check whether reversing every ordered pair still belongs to the relation (Symmetric).

  4. Check whether chaining two related pairs always produces another related pair (Transitive).

  5. State the final conclusion clearly for each relation.

✏️ Solution
Solution - Q1. (i) Relation R in the set \(A=\{1,2,3,\ldots,14\}\) defined by \(R=\{(x,y):3x-y=0\}\)
›
Step-by-step Solution  ·  24 steps
  1. (i) Relation
  2. \[R=\{(x,y):3x-y=0\}\]
  3. Since
    \[3x-y=0\]
  4. therefore
    \[y=3x\]
  5. Now consider all values of \(x\) belonging to
    \[A=\{1,2,\ldots,14\}.\]
  6. The value of \(y\) must also belong to the set.
    \[x=1,\quad y=3\]
  7. Hence
    \[(1,3)\in R\]
  8. \[x=2,\quad y=6\]
    Hence
    \[(2,6)\in R\]
  9. Similarly,
    \[x=3,\;y=9\]
    \[x=4,\;y=12\]
  10. For
    \[x=5,\;y=15\]
    which is not in the set \(A\).
  11. \[R=\{(1,3),(2,6),(3,9),(4,12)\}.\]
  12. Checking Reflexive Property
  13. A reflexive relation must contain
    \[(1,1),(2,2),\ldots,(14,14).\]
  14. None of these satisfy
    \[y=3x\]
  15. Therefore, the relation is not reflexive.
  16. Checking Symmetric Property
  17. Since
    \[(1,3)\in R\]
    but
    \[(3,1)\notin R\]
  18. Therefore, the relation is not symmetric.
  19. Checking Transitive Property
  20. A relation is transitive if whenever
  21. \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
    then
    \[(a,c)\in R\]
  22. Consider the ordered pairs
    \[(1,3)\in R\]
    and
    \[(3,9)\in R\]
  23. Since both pairs belong to the relation, transitivity requires
    \[(1,9)\in R.\]
  24. Now check whether \((1,9)\in R\):
  25. belongs to the relation:
    \[3(1)-9=-6\neq0\]
  26. Hence
    \[(1,9)\notin R\]
  27. Thus, there exist two ordered pairs in the relation whose composition does not belong to the relation.
  28. Therefore, the relation is not transitive.
💡 Answer
Answer (i)
›
  • Reflexive : No
  • Symmetric : No
  • Transitive : No
✏️ Solution
Solution - Q1. (ii) Relation R in the set of natural numbers \(\mathbb{N}\) defined by \(R=\{(x,y):y=x+5,\;x<4\}\)
›
Step-by-step Solution  ·  19 steps
  1. (ii) Relation
  2. \[R=\{(x,y):y=x+5,\;x<4\}\]
  3. Since \(x<4\) and \(x\) is a natural number,
    \[x=1,2,3\]
  4. Therefore,
    \[(1,6),(2,7),(3,8)\]
    belong to the relation.
  5. Hence
    \[R=\{(1,6),(2,7),(3,8)\}\]
  6. Checking Reflexive Property
  7. For reflexivity,
    \[(1,1)\]
    must belong to the relation.
  8. But
    \[1\neq1+5\]
  9. Therefore,
    \[(1,1)\notin R\]
  10. Hence the relation is not reflexive.
  11. Checking Symmetric Property
  12. Since
    \[(1,6)\in R\]
  13. symmetry requires
    \[(6,1)\in R\]
  14. But
    \[1\neq6+5\]
  15. Hence
    \[(6,1)\notin R\]
  16. Therefore, the relation is not symmetric.
  17. Checking Transitive Property
  18. To violate transitivity, there must exist pairs
    \[(a,b)\]
    and
    \[(b,c)\]
  19. The second elements are
    \[6,7,8,\]
  20. whereas the first elements are only
    \[1,2,3\]
  21. No chain exists.
  22. Hence the condition for transitivity is automatically satisfied.
  23. Therefore, the relation is transitive.
💡 Answer
Answer (ii)
›
  • Reflexive : No
  • Symmetric : No
  • Transitive : Yes
✏️ Solution
Solution - Q1. (iii) Relation R in the set \(A=\{1,2,3,4,5,6\}\) defined by\(R=\{(x,y):y\text{ is divisible by }x\}\)
›
Step-by-step Solution  ·  21 steps
  1. (iii) Relation
  2. \[R=\{(x,y):y\text{ is divisible by }x\}\]
  3. List all ordered pairs. For \(x=1\)
  4. every element is divisible by 1.
    \[(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\]
  5. For \(x=2\)
    \[(2,2),(2,4),(2,6)\]
  6. For \(x=3\)
    \[(3,3),(3,6)\]
  7. For \(x=4\)
    \[(4,4)\]
  8. For \(x=5\)
    \[(5,5)\]
  9. For \(x=6\)
    \[(6,6)\]
  10. Checking Reflexive Property
  11. Every number divides itself.
    \[(a,a)\in R\]
    for every element.
  12. Therefore, the relation is reflexive
  13. Checking Symmetric Property
  14. Since
    \[(2,4)\in R,\]
  15. symmetry requires
    \[(4,2)\in R\]
  16. But $2$ is not divisible by $4$
  17. Therefore, the relation is not symmetric.
  18. Checking Transitive Property
  19. Suppose
    \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
  20. Then
    \[b=ax\]
    for some integer \(x\), and
  21. \[c=by\]
    for some integer \(y\)
  22. Substituting
    \[c=(ax)y=a(xy)\]
  23. Hence $a$ divides $c$
  24. Therefore,
    \[(a,c)\in R\]
  25. Thus the relation is transitive
💡 Answer
Answer (iii)
›
  • Reflexive : Yes
  • Symmetric : No
  • Transitive : Yes
✏️ Solution
Solution - Q1. (iv) Relation R in the set of integers \(\mathbb{Z}\) defined by \(R=\{(x,y):x-y\text{ is an integer}\}\)
›
Step-by-step Solution  ·  9 steps
  1. (iv) Relation
  2. \[R=\{(x,y):x-y\text{ is an integer}\}\]
  3. Since both $x$ and $y$ are integers, their difference is always an integer.
  4. Hence every ordered pair belongs to the relation.
  5. Checking Reflexive Property
  6. \[x-x=0\]
    which (Zero) is an integer.
  7. Therefore, the relation is reflexive.
  8. Checking Symmetric Property
  9. If
    \[x-y\]
    is an integer, then
    \[y-x=-(x-y)\]
    which is also an integer.
  10. Therefore, the relation is symmetric.
  11. Checking Transitive Property
  12. If
    \[x-y\]
    and
    \[y-z\]
    are integers, then
    \[(x-y)+(y-z)=x-z\]
    which is also an integer
  13. Hence the relation is transitive
💡 Answer
Answer (iv)
›
  • Reflexive : Yes
  • Symmetric : Yes
  • Transitive : Yes
✏️ Solution
Solution - Q1. (v)-Relation R in the set of all human beings in a town at a particular time given by: (a) \(x\) and \(y\) work at the same place.
›
Step-by-step Solution  ·  3 steps
  1. (a) Same Place of Work
  2. Reflexive: Every person works at the same place as himself or herself. → Yes
  3. Symmetric: If A works with B, then B works with A. → Yes
  4. Transitive: If A works at the same place as B, and B works at the same place as C, then A also works at the same place as C. → Yes
💡 Answer
Answer (v)-(a)
›
  • Reflexive : Yes
  • Symmetric : Yes
  • Transitive : Yes
✏️ Solution
Solution - Q1. (v)-Relation R in the set of all human beings in a town at a particular time given by: (b) \(x\) and \(y\) live in the same locality.
›
Step-by-step Solution  ·  3 steps
  1. (b) Same Locality
  2. Reflexive: Every person lives in the same locality as himself or herself. → Yes
  3. Symmetric: If A lives in the same locality as B, then B lives in the same locality as A. → Yes
  4. Transitive: If A and B live in the same locality, and B and C also live in the same locality, then A and C live in that locality. → Yes
💡 Answer
Answer (v)-(b)
›
  • Reflexive : Yes
  • Symmetric : Yes
  • Transitive : Yes
✏️ Solution
Solution - Q1. (v)-Relation R in the set of all human beings in a town at a particular time given by: (c) \(x\) is exactly 7 cm taller than \(y\).
›
Step-by-step Solution  ·  3 steps
  1. (c) Exactly 7 cm Taller
  2. Reflexive: Nobody is exactly 7 cm taller than himself or herself. → No
  3. Symmetric: If A is 7 cm taller than B, then B is 7 cm shorter than A. → Hence symmetry fails.
  4. Transitive: If A is 7 cm taller than B, and B is 7 cm taller than C, then A is 14 cm taller than C. → Therefore, transitivity fails.
💡 Answer
Answer (v)-(c)
›
  • Reflexive : No
  • Symmetric : No
  • Transitive : No
✏️ Solution
Solution - Q1. (v)-Relation R in the set of all human beings in a town at a particular time given by: (d) \(x\) is wife of \(y\).
›
Step-by-step Solution  ·  3 steps
  1. (d) Wife of
  2. Reflexive: No person is wife of oneself. → No
  3. Symmetric: If A is wife of B, then B is husband of A, not wife of A. → No
  4. Transitive: If A is wife of B, and B is wife of C, such a situation cannot occur.
    The transitive condition is not satisfied as a meaningful chain cannot exist. → No
💡 Answer
Answer (v)-(d)
›
  • Reflexive : No
  • Symmetric : No
  • Transitive : No
✏️ Solution
Solution - Q1. (v)-Relation R in the set of all human beings in a town at a particular time given by: (e) \(x\) is father of \(y\).
›
Step-by-step Solution  ·  3 steps
  1. (e) Father of
  2. Reflexive: No person is father of oneself. → No
  3. Symmetric: If A is father of B, then B cannot be father of A. → No
  4. Transitive: If A is father of B and B is father of C, then A is grandfather of C, not father of C. → Hence transitivity does not hold.
💡 Answer
Answer (v)-(e)
›
  • Reflexive : No
  • Symmetric : No
  • Transitive : No
🎯 Exam Significance
Exam Significance
›
  • This question develops the ability to identify the three fundamental properties of relations, one of the most frequently tested concepts in CBSE Board examinations.
  • JEE Main, NDA, CUET, and other entrance examinations regularly ask conceptual questions based on reflexive, symmetric and transitive relations.
  • Understanding these properties forms the foundation for equivalence relations, partial order relations, and function theory studied in higher mathematics.
  • Students should avoid memorizing answers and instead verify every property independently.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Always test reflexive, symmetric and transitive properties separately.

  2. A relation may satisfy one property without satisfying the others.

  3. Relations based on divisibility are generally reflexive and transitive but not symmetric.

  4. Relations defined on all integers often satisfy all three properties if based on integer differences.

  5. Real-life relations can also be analyzed mathematically using the same three properties.
↑ Top
1 / 16  ·  6%
Q2 →
Q2
NUMERIC3 marks
Show that the relation \(\mathbb{R}\) in the set of real numbers, defined by \(\mathbb{R}=\{(a,b):a\leq b^2\}\) is neither reflexive nor symmetric nor transitive.
📘 Concept & Theory
Concept / Theory
›

To determine the nature of a relation, we examine each of the three properties separately.

Reflexive Relation

A relation \(R\) on a set \(A\) is reflexive if every element is related to itself.

\[ (a,a)\in R \]

for every

\[ a\in A. \]

For this relation, substituting \(b=a\), we obtain

\[ a\leq a^2. \]

This inequality must hold for every real number if the relation is reflexive.

Symmetric Relation

A relation is symmetric if

\[ (a,b)\in R \]

always implies

\[ (b,a)\in R. \]

That means, whenever

\[ a\leq b^2, \]

it must also be true that

\[ b\leq a^2. \]
Transitive Relation

A relation is transitive if

\[ (a,b)\in R \]

and

\[ (b,c)\in R \]

together imply

\[ (a,c)\in R. \]

That is, from

\[ a\leq b^2 \]

and

\[ b\leq c^2, \]

we must always be able to conclude

\[ a\leq c^2. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check reflexivity by substituting \(a=b\).

  2. If one counterexample is found, the relation is not reflexive.

  3. Choose an ordered pair satisfying the relation and verify whether the reverse ordered pair also satisfies it.

  4. To test transitivity, construct two ordered pairs belonging to the relation but whose composition does not belong to the relation.

  5. State the conclusion separately for each property.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  71 steps
  1. Checking Reflexive Property
  2. For a reflexive relation, every ordered pair
    \[(a,a)\]
    must belong to \(\mathbb{R}\).
  3. Substituting
    \[b=a\]
  4. the condition becomes
    \[a\leq a^2\]
    This inequality is not true for every real number.
  5. Choose
    \[a=\frac{1}{2}\]
  6. Then
    \[a^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\]
  7. Now compare the two numbers.
    \[\frac{1}{2}\leq\frac{1}{4}\]
    which is false because
    \[\frac{1}{2}>\frac{1}{4}\]
  8. Hence
    \[\left(\frac{1}{2},\frac{1}{2}\right)\notin \mathbb{R}\]
  9. Therefore, the relation is not reflexive.
  10. Checking Symmetric Property
  11. We need to determine whether
    \[(a,b)\in \mathbb{R}\]
    implies
    \[(b,a)\in \mathbb{R}\]
  12. Choose
    \[a=2,\quad b=-2\]
  13. Now verify whether
    \[(2,-2)\in \mathbb{R}\]
  14. Since
    \[(-2)^2=4\]
  15. we have
    \[2\leq4\]
    which is true
  16. Hence
    \[(2,-2)\in \mathbb{R}\]
  17. Now consider the reverse ordered pair
    \[(-2,2)\]
  18. The condition becomes
    \[-2\leq2^2\]
  19. Since
    \[-2\leq4\]
    this pair also belongs to the relation.
  20. This example satisfies symmetry, so it cannot be used to disprove it.
  21. Choose another example.
  22. Let
    \[a=-1,\qquad b=-2\]
  23. Then
    \[(-2)^2=4\]
  24. so
    \[-1\leq4\]
  25. Hence
    \[(-1,-2)\in R\]
  26. Now reverse the ordered pair.
    \[(-2,-1)\]
  27. Here the condition becomes
    \[-2\leq(-1)^2\]
  28. Since
    \[-2\leq1\]
    this example also satisfies symmetry.
  29. We therefore choose a better counterexample.
  30. Take
    \[a=2,\quad b=1\]
  31. Then
    \[2\leq1^2\]
    is false, so this pair is not useful.
  32. Now take
    \[a=1,\qquad b=\frac{1}{2}\]
  33. Then
    \[1\leq\left(\frac12\right)^2\]
    is false. Again not useful.
  34. Now choose
    \[a=\frac{1}{4},\quad b=\frac{1}{2}\]
  35. Then
    \[\frac{1}{4}\leq\left(\frac{1}{2}\right)^2=\frac{1}{4}\]
  36. which is true.
  37. Hence
    \[\left(\frac{1}{4},\frac{1}{2}\right)\in \mathbb{R}\]
  38. Reverse the pair.
    \[\left(\frac{1}{2},\frac{1}{4}\right)\]
  39. Now check
    \[\frac{1}{2}\leq\left(\frac{1}{4}\right)^2\]
  40. Since
    \[\left(\frac{1}{4}\right)^2=\frac{1}{16}\]
  41. we get
    \[\frac12\leq\frac1{16}\]
    which is false.
  42. Therefore,
    \[\left(\frac12,\frac14\right)\notin \mathbb{R}\]
  43. Hence the relation is not symmetric.
  44. Checking Transitive Property
  45. We must check whether
    \[(a,b)\in \mathbb{R}\]
    and
    \[(b,c)\in \mathbb{R}\]
    imply
    \[(a,c)\in \mathbb{R}\]
  46. Choose
    \[a=4,\quad b=9,\quad c=3\]
  47. First verify
    \[(a,b)\]
  48. Since
    \[9^2=81\]
  49. we obtain
    \[4\leq81\]
  50. Hence
    \[(4,9)\in \mathbb{R}\]
  51. Now verify
    \[(b,c)\]
  52. Since
    \[3^2=9\]
  53. we obtain
    \[9\leq9\]
  54. Hence
    \[(9,3)\in \mathbb{R}\]
  55. Now check
    \[(a,c)\]
  56. We have
    \[4\leq9\]
  57. Hence
    \[(4,3)\in \mathbb{R}\]
  58. this is true, so we need another example.
  59. Choose instead
    \[a=16,\quad b=25,\quad c=4\]
  60. First,
    \[16\leq25^2=625\]
  61. so
    \[(16,25)\in \mathbb{R}\]
  62. Secondly,
    \[25\leq4^2=16\]
    is false. This choice does not work.
  63. Now choose
    \[a=4,\quad b=2,\quad c=\frac32\]
  64. Check the first pair
    \[4\leq2^2=4\]
    which is true
  65. Hence
    \[(4,2)\in \mathbb{R}\]
  66. Check the second pair
    \[2\leq\left(\frac32\right)^2=\frac94\]
  67. Since
    \[2\leq\frac94\]
    this is true
  68. Hence
    \[\left(2,\frac32\right)\in \mathbb{R}\]
  69. Now check the third pair
    \[4\leq\left(\frac32\right)^2=\frac94\]
  70. Since
    \[4>\frac94\]
    this inequality is false.
  71. Therefore,
    \[\left(4,\frac32\right)\notin \mathbb{R}\]
  72. Thus
    \[(a,b)\in \mathbb{R}\]
    and
    \[(b,c)\in \mathbb{R}\]
  73. but
    \[(a,c)\notin \mathbb{R}\]
  74. Hence the relation is not transitive.
🧩 Final Conclusion
Answer
›
Property Result Reason
Reflexive No \(\left(\frac12,\frac12\right)\notin R\)
Symmetric No \(\left(\frac14,\frac12\right)\in R\) but \(\left(\frac12,\frac14\right)\notin R\)
Transitive No \((4,2)\in R,\ \left(2,\frac32\right)\in R,\ \left(4,\frac32\right)\notin R\)
Hence, the given relation is neither reflexive nor symmetric nor transitive.
🎯 Exam Significance
Exam Significance
›
  • This is one of the most important proof-based questions from the chapter "Relations and Functions" for CBSE Class 12 Board Examination.
  • Competitive examinations such as JEE Main, NDA, CUET, BITSAT and other university entrance tests frequently ask students to identify relation properties using counterexamples.
  • The key idea is that a single counterexample is sufficient to prove that a relation is not reflexive, not symmetric or not transitive.
  • Students should always justify each property independently instead of giving only the final answer.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. Reflexivity is tested by substituting the same element in both coordinates.

  2. Symmetry is disproved by finding one ordered pair whose reverse does not belong to the relation.

  3. Transitivity is disproved by constructing two related pairs whose composition fails.

  4. One valid counterexample is enough to conclude that a property does not hold.
← Q1
2 / 16  ·  13%
Q3 →
Q3
NUMERIC3 marks
Check whether the relation \(R=\{(a,b):b=a+1\}\) defined on the set \(A=\{1,2,3,4,5,6\}\) is reflexive, symmetric or transitive.
📘 Concept & Theory
Concept / Theory
›

To determine the nature of a relation, we examine each property independently.

Reflexive Relation

A relation is reflexive if every element is related to itself.

\[(a,a)\in R\]

for every

\[a\in A.\]
Symmetric Relation

A relation is symmetric if

\[(a,b)\in R\]

implies

\[(b,a)\in R.\]
Transitive Relation

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

together imply

\[(a,c)\in R.\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write all ordered pairs satisfying the given condition.

  2. Check whether every element is related to itself.

  3. Reverse an ordered pair to verify symmetry.

  4. Find two consecutive ordered pairs and verify whether their composition belongs to the relation.

  5. State the conclusion separately for reflexive, symmetric and transitive properties.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  36 steps
  1. Write the Relation Explicitly
  2. he relation is defined by
    \[b=a+1\]
  3. Now consider every element of the set
    \[A=\{1,2,3,4,5,6\}\]
  4. For $a=1$
  5. we obtain $b=2$
  6. Hence
    \[(1,2)\in \mathbb{R}\]
  7. For $a=2$ we obtain $b=3$
  8. Hence
    \[(2,3)\in \mathbb{R}\]
  9. For $a=3$ we obtain $b=4$
  10. Hence
    \[(3,4)\in \mathbb{R}\]
  11. For $a=4$ we obtain $b=5$
  12. Hence
    \[(4,5)\in \mathbb{R}\]
  13. For $a=5$ we obtain $b=6$
  14. Hence
    \[(5,6)\in \mathbb{R}\]
  15. For $a=6$ we obtain $b=7$
  16. which is not an element of the given set.
  17. Therefore, the relation is
    \[R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\]
  18. Check Reflexive Property
  19. A relation is reflexive if every ordered pair \(a,a\)
  20. belongs to the relation.
  21. For example, \(1,1\) should belong to \(R\)
  22. But according to the definition,
    \[1=1+1\]
    which is false.
  23. Similarly, none of the ordered pairs
    \[(2,2),(3,3),(4,4),(5,5),(6,6)\]
    belongs to the relation.
  24. Therefore, the relation is not reflexive.
  25. Check Symmetric Property
  26. Take the ordered pair \((1,2)\in R\)
  27. If the relation were symmetric, then
    \[2,1\]
    must also belong to \(R\)
  28. Now verify the defining condition.
  29. For \(2,1\) we must have
    \[1=2+1\]
    which is false.
  30. Hence
    \[(2,1)\notin R.\]
  31. Therefore, the relation is not symmetric
  32. Check Transitive Property
  33. Choose two ordered pairs from the relation.
    \[(1,2)\in R\]
    and
    \[(2,3)\in R\]
  34. If the relation were transitive, then
    \[(1,3)\]
    must also belong to \(R\)
  35. Now verify.
    \[(1,3)\]
  36. the defining condition requires
    \[3=1+1\]
  37. Since
    \[3\neq2\]
    the condition is false.
  38. Hence
    \[(1,3)\notin R\]
  39. Thus, we have
    \[(1,2)\in R\]
    \[(2,3)\in R\]
    but
    \[(1,3)\notin R\]
  40. Therefore, the relation is not transitive
💡 Answer
Final Answer
›
Property Result Reason
Reflexive No No ordered pair of the form \((a,a)\) belongs to the relation.
Symmetric No \((1,2)\in R\), but \((2,1)\notin R\).
Transitive No \((1,2)\in R\) and \((2,3)\in R\), but \((1,3)\notin R\).

Hence, the given relation is neither reflexive, nor symmetric, nor transitive.

🎯 Exam Significance
Exam Significance
›
  • This is a standard proof-based question frequently asked in CBSE Board examinations.
  • JEE Main, NDA, CUET, BITSAT and other entrance examinations often test relation properties using similar successor-type relations.
  • Students should always list the ordered pairs before checking the three properties.
  • Learning to find suitable counterexamples is essential for solving relation-property questions quickly in competitive examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Write the relation explicitly before testing its properties.

  2. For reflexivity, every element must be related to itself.

  3. For symmetry, reversing every ordered pair must produce another valid ordered pair.

  4. For transitivity, two consecutive ordered pairs must produce a third valid ordered pair.

  5. A single counterexample is sufficient to prove that a property does not hold.
← Q2
3 / 16  ·  19%
Q4 →
Q4
NUMERIC3 marks
Show that the relation \(R=\{(a,b):a\leq b\}\) defined on the set of real numbers \(\mathbb{R}\) is reflexive and transitive but not symmetric.
📘 Concept & Theory
Concept / Theory
›

Relations defined using inequalities are among the most common examples in mathematics. To determine their nature, each property should be verified separately.

Reflexive Relation

A relation \(R\) on a set \(A\) is called reflexive if every element is related to itself.

\[(a,a)\in R\]

for every

\[a\in A.\]
Symmetric Relation

A relation is called symmetric if

\[(a,b)\in R\]

always implies

\[(b,a)\in R.\]
Transitive Relation

A relation is called transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

together imply

\[(a,c)\in R.\]

The given relation is based on the usual order relation "\(\leq\)" on the real numbers.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexivity by substituting \(a=b\).

  2. Check symmetry by taking a suitable counterexample.

  3. Use the transitive property of inequalities to prove transitivity.

  4. State the conclusion for each property separately.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  22 steps
  1. Checking Reflexive Property
  2. For every real number \(a\), we have
    \[a\leq a\]
  3. This statement is always true because every real number is equal to itself.
    \[(a,a)\in R\]
  4. for every
    \[a\in\mathbb{R}\]
  5. Hence, the relation is reflexive.
  6. Checking Symmetric Property
  7. To test symmetry, consider the ordered pair
    \[(2,5)\]
  8. Since
    \[2\leq5\]
  9. we have
    \[(2,5)\in R\]
  10. If the relation were symmetric, then
    \[(5,2)\]
    must also belong to \(R\)
  11. Now verify the condition.
    \[5\leq2\]
    This statement is false
  12. Therefore,
    \[(5,2)\notin R\]
  13. Since one counterexample is sufficient, the relation is not symmetric.
  14. Checking Transitive Property
  15. Suppose
    \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
  16. Then, by definition of the relation,
    \[a\leq b\]
    and
    \[b\leq c\]
  17. By the transitive law of inequalities,
    \[a\leq c\]
  18. Hence,
    \[(a,c)\in R\]
  19. Therefore, the relation satisfies the transitive property.
  20. Thus, the relation is transitive.
  21. Alternative Verification Using Numerical Example
  22. Take
    \[a=2,\quad b=5,\quad c=8\]
  23. Since
    \[2\leq5\]
    and
    \[5\leq8\]
  24. therefore
    \[2\leq8\]
  25. Hence,
    \[(2,8)\in R\]
  26. This illustrates the transitive property.
💡 Answer
Final Answer
›
Property Result Reason
Reflexive Yes Every real number satisfies \(a\leq a\).
Symmetric No \((2,5)\in R\), but \((5,2)\notin R\).
Transitive Yes From \(a\leq b\) and \(b\leq c\), it follows that \(a\leq c\).

Hence, the relation is reflexive and transitive but not symmetric.

🎯 Exam Significance
Exam Significance
›
  • This is one of the most important proof-based questions from the chapter "Relations and Functions" in the CBSE Class 12 syllabus.
  • Questions based on inequalities are frequently asked in CBSE Board examinations and competitive examinations such as JEE Main, CUET, NDA and BITSAT.
  • Students should remember that the relation "\(a\leq b\)" is a classic example of a relation that is reflexive and transitive but not symmetric.
  • This relation also serves as the foundation for understanding partial order relations in higher mathematics.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Every inequality of the form \(a\leq a\) proves reflexivity.

  2. A single counterexample is enough to disprove symmetry.

  3. The transitive law of inequalities directly establishes transitivity.

  4. The relation "\(a\leq b\)" is a standard example of a partial ordering relation.

  5. Always verify each property independently before drawing the final conclusion.
← Q3
4 / 16  ·  25%
Q5 →
Q5
NUMERIC3 marks
Check whether the relation $R=\{(a,b):a\leq b^3\}$ defined on the set of real numbers \(\mathbb{R}\) is reflexive, symmetric or transitive.
📘 Concept & Theory
Concept / Theory
›
The given relation is defined using the inequality
\[a\leq b^3\]
To determine whether it is reflexive, symmetric or transitive, we verify each property independently.
Reflexive Relation

A relation \(R\) on a set \(A\) is reflexive if

\[(a,a)\in R\]

for every

\[a\in A.\]

For the present relation, this requires

\[a\leq a^3\]

for every real number \(a\).

Symmetric Relation

A relation is symmetric if

\[(a,b)\in R\]

always implies

\[(b,a)\in R.\]

Thus, whenever

\[a\leq b^3,\]

it must also be true that

\[b\leq a^3.\]
Transitive Relation

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

together imply

\[(a,c)\in R.\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Substitute \(a=b\) to test reflexivity.

  2. Choose one suitable ordered pair and check whether its reverse belongs to the relation.

  3. Find two ordered pairs belonging to the relation and verify whether their composition also belongs to the relation.

  4. Use counterexamples wherever necessary.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  34 steps
  1. Check Reflexive Property
  2. For reflexivity, every ordered pair
    \[(a,a)\]
  3. must belong to the relation.
  4. This requires
    \[a\leq a^3\]
  5. This inequality is not true for every real number.
  6. Take
    \[a=\frac12\]
    Then
    \[a=\frac12\]
    and
    \[a^3=\left(\frac12\right)^3=\frac18\]
  7. Now compare the two values.
    \[\frac12\leq\frac18\]
  8. which is false because
    \[\frac12>\frac18\]
  9. Hence
    \[\left(\frac12,\frac12\right)\notin R\]
  10. Therefore, the relation is not reflexive.
  11. Check Symmetric Property
  12. Take the ordered pair
    \[(1,2)\]
  13. Since
    \[2^3=8\]
    we obtain
    \[1\leq8\]
  14. Therefore,
    \[(1,2)\in R\]
  15. Now reverse the ordered pair.
    \[(2,1)\]
  16. For this pair, the condition becomes
    \[2\leq1^3\]
  17. That is,
    \[2\leq1\]
    which is false.
  18. Hence
    \[(2,1)\notin R.\]
  19. Therefore, the relation is not symmetric
  20. Check Transitive Property
  21. To verify transitivity, consider the following ordered pairs. Choose
    \[a=8,\qquad b=2,\qquad c=\frac32\]
  22. First verify
    \[(a,b)\]
  23. Since
    \[2^3=8\]
  24. we have
    \[8\leq8\]
  25. Hence
    \[(8,2)\in R\]
  26. Now verify
    \[(b,c)\]
  27. Since
    \[\left(\frac32\right)^3=\frac{27}{8}=3.375\]
  28. we obtain
    \[2\leq\frac{27}{8}\]
  29. which is true. Hence
    \[\left(2,\frac32\right)\in R\]
  30. Now examine
    \[(a,c)\]
  31. The condition becomes
    \[8\leq\left(\frac32\right)^3\]
  32. Since
    \[8>\frac{27}{8}\]
  33. the inequality is false.
  34. Therefore,
    \[\left(8,\frac32\right)\notin R\]
  35. Thus,
    \[(a,b)\in R\]
    \[(b,c)\in R\]
  36. but
    \[(a,c)\notin R\]
  37. Hence, the relation is not transitive.
💡 Answer
Final Answer
›
Property Result Reason
Reflexive No \(\left(\frac12,\frac12\right)\notin R\)
Symmetric No \((1,2)\in R\), but \((2,1)\notin R\)
Transitive No \((8,2)\in R,\ \left(2,\frac32\right)\in R,\ \left(8,\frac32\right)\notin R\)

Hence, the given relation is neither reflexive, nor symmetric, nor transitive.

🎯 Exam Significance
Exam Significance
›
  • This is an important proof-based question from the CBSE Class 12 chapter "Relations and Functions".
  • JEE Main, CUET, NDA, BITSAT and other competitive examinations frequently ask students to determine the properties of relations using counterexamples.
  • Students should remember that proving a relation is not reflexive, not symmetric or not transitive requires only one valid counterexample.
  • Questions involving powers such as \(b^2\), \(b^3\) and absolute values are commonly asked in board and entrance examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. For reflexivity, substitute the same element in both coordinates.

  2. For symmetry, reverse a valid ordered pair and verify whether it still satisfies the defining condition.

  3. For transitivity, construct two related ordered pairs and test their composition.

  4. A single counterexample is sufficient to disprove a property.

  5. Always verify each property independently before giving the final conclusion.
← Q4
5 / 16  ·  31%
Q6 →
Q6
NUMERIC3 marks
Show that the relation $R=\{(1,2),(2,1)\}$ defined on the set $A=\{1,2,3\}$ is symmetric but neither reflexive nor transitive.
📘 Concept & Theory
Concept / Theory
›

To determine the nature of a relation, we examine each of the three fundamental properties separately.

Reflexive Relation

A relation is reflexive if every element of the set is related to itself.

\[(a,a)\in R\]

for every

\[a\in A.\]
Symmetric Relation

A relation is symmetric if whenever

\[(a,b)\in R,\]

then

\[(b,a)\in R.\]
Transitive Relation

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

together imply

\[(a,c)\in R.\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given relation explicitly.

  2. Check whether every element is related to itself.

  3. Verify that every ordered pair has its reverse in the relation.

  4. Use two related ordered pairs to test the transitive property.

  5. State the conclusion for each property independently.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. Write the Relation
  2. he given relation is
    \[R=\{(1,2),(2,1)\}\]
  3. The set is
    \[A=\{1,2,3\}\]
  4. Check Reflexive Property
  5. For a relation to be reflexive, the following ordered pairs must belong to the relation:
    \[(1,1),\quad(2,2),\quad(3,3)\]
  6. However, the given relation contains only
    \[(1,2)\quad\text{and}\quad(2,1)\]
  7. he ordered pairs
    \[(1,1),\ (2,2),\ (3,3)\]
    are not present.
  8. Hence, the relation is not reflexive.
  9. Check Symmetric Property
  10. The relation contains
    \[(1,2)\]
  11. Its reverse ordered pair is
    \[(2,1)\]
  12. which also belongs to the relation.
  13. Similarly,
    \[(2,1)\in R\]
    and its reverse
    \[(1,2)\in R\]
  14. Thus, every ordered pair appears together with its reverse.
  15. Therefore, the relation is symmetric.
  16. Check Transitive Property
  17. Consider the two ordered pairs
    \[(1,2)\in R\]
    and
    \[(2,1)\in R\]
  18. If the relation were transitive, then
    \[(1,1)\]
    must also belong to the relation.
  19. But
    \[(1,1)\notin R\]
  20. Therefore, the transitive property is violated.
  21. Similarly, consider
    \[(2,1)\in R\]
    and
    \[(1,2)\in R\]
  22. Then transitivity requires
    \[(2,2)\in R\]
  23. However,
    \[(2,2)\notin R\]
  24. Hence, the relation is not transitive.
💡 Answer
Final Answer
›
Property Result Reason
Reflexive No \((1,1), (2,2)\) and \((3,3)\) do not belong to the relation.
Symmetric Yes \((1,2)\) and \((2,1)\) are both present.
Transitive No \((1,2)\) and \((2,1)\) require \((1,1)\), which is absent.

Hence, the given relation is symmetric but neither reflexive nor transitive.

🎯 Exam Significance
Exam Significance
›
  • This is a frequently asked conceptual proof in CBSE Class 12 Board examinations.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often test the understanding of relation properties through small finite sets.
  • Students should remember that symmetry only requires the reverse ordered pair, whereas transitivity requires a new ordered pair obtained by chaining two existing pairs.
  • Finite-set relations are ideal for checking properties by direct inspection.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Reflexivity requires every element to be related to itself.

  2. Symmetry requires each ordered pair to appear together with its reverse.

  3. Transitivity requires composition of two related pairs to produce another valid pair.

  4. A relation may satisfy one property without satisfying the other two.

  5. Always verify each property independently before drawing the final conclusion.
← Q5
6 / 16  ·  38%
Q7 →
Q7
NUMERIC3 marks
Show that the relation \(R=\{(x,y):x\text{ and }y\text{ have the same number of pages}\}\) defined on the set \(A\) of all the books in a college library is an equivalence relation.
📘 Concept & Theory
Concept / Theory
›

An equivalence relation is a relation that satisfies all the following three properties simultaneously.

1. Reflexive Property

A relation \(R\) on a set \(A\) is reflexive if every element is related to itself.

\[(a,a)\in R\]

for every

\[a\in A.\]
2. Symmetric Property

A relation is symmetric if

\[(a,b)\in R\]

implies

\[(b,a)\in R.\]
3. Transitive Property

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

together imply

\[(a,c)\in R.\]

If a relation possesses all three properties, then it is called an equivalence relation.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Interpret the meaning of the relation in everyday language.

  2. Verify that every book has the same number of pages as itself.

  3. Show that if one book has the same number of pages as another, then the reverse statement is also true.

  4. Show that if two books each have the same number of pages as a third book, then they must have the same number of pages as each other.

  5. Conclude that the relation satisfies all three properties.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. The relation is defined by
    \[R=\{(x,y):x\text{ and }y\text{ have the same number of pages}\}.\]
  2. We shall verify the three properties one by one.
  3. Check Reflexive Property
  4. Consider any book
    \[x\in A\]
  5. Every book obviously has the same number of pages as itself.
  6. Therefore,
    \[(x,x)\in R\]
  7. Since this is true for every book in the library, the relation is reflexive.
  8. Check Symmetric Property
  9. Suppose
    \[(x,y)\in R\]
  10. This means that book \(x\) and book \(y\) have the same number of pages
  11. If book \(x\) has the same number of pages as book \(y\), then book \(y\) also has the same number of pages as book \(x\).
  12. Therefore,
    \[(y,x)\in R.\]
  13. Hence, the relation is symmetric.
  14. Check Transitive Property
  15. Suppose
    \[(x,y)\in R\]
    and
    \[(y,z)\in R.\]
  16. Then,
    • Book \(x\) and book \(y\) have the same number of pages.
    • Book \(y\) and book \(z\) have the same number of pages.
  17. Therefore, all three books have an equal number of pages.
  18. Hence, book \(x\) and book \(z\) also have the same number of pages.
  19. Thus,
    \[(x,z)\in R\]
  20. Therefore, the relation is transitive.
💡 Answer
Final Answer
›
The given relation satisfies
  • Reflexive Property
  • Symmetric Property
  • Transitive Property

Hence, the relation is an equivalence relation.

Equivalence Classes

The relation partitions the books in the library into distinct groups.

Each equivalence class contains all books having exactly the same number of pages.

For example, if there are books containing

\[250\]

pages, then all such books belong to one equivalence class.

Similarly, books having

\[320\]

pages form another equivalence class.

No book can belong to two different equivalence classes because a book has only one fixed number of pages.

🎯 Exam Significance
Exam Significance
›
  • This is a classic proof-based question frequently asked in CBSE Class 12 Board examinations.
  • Questions involving real-life situations such as books, students, families, and geometric figures are commonly used to test the concept of equivalence relations.
  • JEE Main, CUET, NDA, BITSAT and other competitive examinations often ask students to identify whether a given practical relation is an equivalence relation.
  • Understanding equivalence relations is essential because they partition a set into disjoint equivalence classes, an important concept in higher mathematics.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. An equivalence relation must satisfy reflexive, symmetric and transitive properties simultaneously.

  2. "Having the same number of pages" naturally satisfies all three properties.

  3. Equivalence relations divide a set into mutually exclusive equivalence classes.

  4. Each equivalence class consists of elements sharing the same defining characteristic.

  5. Always verify all three properties separately before concluding that a relation is an equivalence relation.
← Q6
7 / 16  ·  44%
Q8 →
Q8
NUMERIC3 marks
Show that the relation $R=\{(a,b):|a-b|\text{ is even}\}$ defined on the set $A=\{1,2,3,4,5\}$ is an equivalence relation. Also show that all the elements of $\{1,3,5\}$ are related to each other and all the elements of $\{2,4\}$ are related to each other. But no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$
📘 Concept & Theory
Concept / Theory
›

The relation is defined using the parity (oddness or evenness) of the difference of two numbers.

Recall that:

  • The difference of two odd numbers is always even.
  • The difference of two even numbers is always even.
  • The difference of one odd number and one even number is always odd.

These facts help us determine whether two elements are related.

Equivalence Relation

A relation is called an equivalence relation if it is

  • Reflexive
  • Symmetric
  • Transitive
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexivity by computing \(|a-a|\).

  2. Use the property \(|a-b|=|b-a|\) to prove symmetry.

  3. Use parity (odd/even nature) of numbers to prove transitivity.

  4. Identify the equivalence classes.

  5. Verify that elements belonging to different classes are not related.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  32 steps
  1. The given relation is
    \[R=\{(a,b):|a-b|\text{ is even}\}.\]
  2. Check Reflexive Property
  3. Take any element
    \[a\in A\]
  4. Then
    \[|a-a|=0\]
  5. Since $0$ is an even integer,
    \[(a,a)\in R\]
    This is true for every element of the set.
  6. Hence, the relation is reflexive.
  7. Check Symmetric Property
  8. Suppose
    \[(a,b)\in R\]
  9. Then |a-b| is even.
  10. Since
    \[|a-b|=|b-a|\]
  11. it follows that
    \[|b-a|\]
    is also even.
  12. Hence,
    \[(b,a)\in R\]
  13. Therefore, the relation is symmetric.
  14. Check Transitive Property
  15. Suppose
    \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
  16. Then
    \[|a-b|\]
    and
    \[|b-c|\]
    are even integers.
  17. This means that
    • \(a\) and \(b\) have the same parity (both odd or both even).
    • \(b\) and \(c\) also have the same parity.
  18. Therefore, \(a\) and \(c\) must also have the same parity.
  19. Hence,
    \[|a-c|\]
    is even.
  20. Thus,
    \[(a,c)\in R\]
  21. Therefore, the relation is transitive.
  22. Conclusion
  23. Since the relation is
    • Reflexive
    • Symmetric
    • Transitive
  24. it is an equivalence relation.
  25. Showing that all elements of \(\{1,3,5\}\) are related
  26. Consider every pair.
    Pair \(|a-b|\) Even? Related?
    \((1,3)\) \(|1-3|=2\) Yes Yes
    \((1,5)\) \(|1-5|=4\) Yes Yes
    \((3,5)\) \(|3-5|=2\) Yes Yes
  27. Hence, every element of \(\{1,3,5\}\) is related to every other element of the same set.
  28. Showing that all elements of \(\{2,4\}\) are related
  29. Consider the pair
    \[(2,4)\]
  30. Then
    \[|2-4|=2\]
    which is even.
  31. Therefore,
    \[(2,4)\in R\]
  32. Also, by symmetry,
    \[(4,2)\in R\]
  33. Further,
    \[(2,2)\in R\]
    and
    \[(4,4)\in R\]
    by reflexivity.
  34. Hence, every element of $\{2,4\}$ is related to every other element of the same set.
  35. Showing that no element of \(\{1,3,5\}\) is related to any element of \(\{2,4\}\)
  36. Consider all possible pairs.
    Pair \(|a-b|\) Even? Related?
    \((1,2)\) 1 No No
    \((1,4)\) 3 No No
    \((3,2)\) 1 No No
    \((3,4)\) 1 No No
    \((5,2)\) 3 No No
    \((5,4)\) 1 No No
  37. Thus, no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$
  38. Equivalence Classes
  39. The relation divides the set
    \[A=\{1,2,3,4,5\}\]
  40. into the following equivalence classes.
    \[[1]=[3]=[5]=\{1,3,5\}\]
    \[[2]=[4]=\{2,4\}\]
    These two classes are disjoint and together cover the entire set.
💡 Answer
Final Answer
›
Property Result
Reflexive Yes
Symmetric Yes
Transitive Yes
Type of Relation Equivalence Relation
Equivalence Classes \(\{1,3,5\}\) and \(\{2,4\}\)

Hence, the given relation is an equivalence relation having two equivalence classes:

\[\boxed{\{1,3,5\}\quad\text{and}\quad\{2,4\}}\]
🎯 Exam Significance
Exam Significance
›
  • This is one of the most important proof-based questions from the chapter "Relations and Functions" in the CBSE Class 12 syllabus.
  • Questions based on parity (odd-even property) are frequently asked in Board examinations and competitive examinations such as JEE Main, CUET and NDA.
  • This problem illustrates how an equivalence relation partitions a set into disjoint equivalence classes.
  • Students should remember that members of different equivalence classes can never be related.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. The difference of two odd numbers or two even numbers is always even.

  2. The difference of an odd number and an even number is always odd.

  3. The given relation satisfies reflexive, symmetric and transitive properties.

  4. Equivalence classes contain elements having the same parity.

  5. An equivalence relation partitions a set into mutually exclusive classes.
← Q7
8 / 16  ·  50%
Q9 →
Q9
NUMERIC3 marks
Show that each of the following relations defined on the set $A=\{x\in\mathbb{Z}:0\leq x\leq12\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.
  1. \[R=\{(a,b):|a-b|\text{ is a multiple of }4\}\]
  2. \[R=\{(a,b):a=b\}\]
📘 Concept & Theory
Concept / Theory
›

A relation is called an equivalence relation if it satisfies all three properties:

  • Reflexive
  • Symmetric
  • Transitive

An equivalence relation partitions a set into mutually disjoint subsets called equivalence classes.

The equivalence class of an element \(a\) is denoted by

\[[a]=\{x\in A:(x,a)\in R\}.\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexive, symmetric and transitive properties separately.

  2. Conclude that the relation is an equivalence relation.

  3. Determine all elements satisfying the given condition with the element 1.

  4. Write the equivalence class containing 1.

✏️ Solution
Solution (i) \(R=\{(a,b):|a-b|\text{ is a multiple of }4\}\)
›
Step-by-step Solution  ·  20 steps
  1. Given Relation
  2. R=\{(a,b):|a-b|\text{ is a multiple of }4\}
  3. Check Reflexive Property
  4. Take any element
    \[a\in A\]
  5. Then
    \[|a-a|=0\]
  6. Since
    \[0=4\times0\]
    zero is a multiple of 4
  7. Therefore,
    \[(a,a)\in R\]
  8. Hence, the relation is reflexive.
  9. Check Symmetric Property
  10. Suppose
    \[(a,b)\in R\]
  11. Then
    \[|a-b|\]
    is a multiple of 4.
  12. Since
    \[|a-b|=|b-a|\]
  13. it follows that
    \[|b-a|\]
    is also a multiple of 4.
  14. Hence,
    \[(b,a)\in R\]
  15. Therefore, the relation is symmetric.
  16. Check Transitive Property
  17. Suppose
    \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
  18. Then
    \[a-b=4m\tag{1}\]
    and
    \[b-c=4n\tag{2}\]
    for some integers $m,\;n.$
  19. Adding these equations (1) and (2),
    \[(a-b)+(b-c)=4m+4n\]
  20. Therefore,
    \[a-c=4(m+n)\]
  21. Hence, $|a-c|$ is also a multiple of 4.
  22. Thus,
    \[(a,c)\in R\]
  23. Therefore, the relation is transitive.
  24. Conclusion

  25. Since the relation is reflexive, symmetric and transitive, it is an equivalence relation.

    Elements Related to 1

    An element \(x\) is related to 1 if

    \[|x-1|\]

    is a multiple of 4.

    Checking every element of

    \[A=\{0,1,2,\ldots,12\},\]
    \(x\) \(|x-1|\) Multiple of 4?
    01No
    10Yes
    21No
    32No
    43No
    54Yes
    65No
    76No
    87No
    98Yes
    109No
    1110No
    1211No

    Hence, the equivalence class containing 1 is

    \[[1]=\{1,5,9\}.\]
✏️ Solution
Solution (ii) \(R=\{(a,b):a=b\}\)
›
Step-by-step Solution  ·  16 steps
  1. Given Relation
  2. \[R=\{(a,b):a=b\}.\]
    This is the equality relation.
  3. Check Reflexive Property
  4. For every element
    \[a\in A\]
  5. we have $a=a$
  6. Hence,
    \[(a,a)\in R\]
  7. Therefore, the relation is reflexive.
  8. Check Symmetric Property
  9. Suppose
    \[(a,b)\in R.\]
  10. Then
    \[a=b\]
  11. Therefore,
    \[b=a\]
  12. Hence,
    \[(b,a)\in R\]
  13. Thus, the relation is symmetric.
  14. Check Transitive Property
  15. Suppose
    \[(a,b)\in R\]
    and
    \[(b,c)\in R\]
  16. Then
    \[a=b\]
    and
    \[b=c\]
  17. Therefore,
    \[a=c\]
  18. Hence,
    \[(a,c)\in R\]
  19. Thus, the relation is transitive.
  20. Conclusion

  21. The equality relation satisfies all three properties.

    Therefore, it is an equivalence relation.

    Elements Related to 1

    An element is related to 1 only if

    \[a=1.\]

    Hence, the equivalence class containing 1 is

    \[[1]=\{1\}.\]
💡 Answer
Final Answer
›
Relation Reflexive Symmetric Transitive Equivalence Relation Elements Related to 1
\(|a-b|\) is a multiple of 4 Yes Yes Yes Yes \(\{1,5,9\}\)
\(a=b\) Yes Yes Yes Yes \(\{1\}\)
🎯 Exam Significance
Exam Significance
›
  • This question is frequently asked in CBSE Class 12 Board examinations as a proof-based question on equivalence relations.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often require students to identify equivalence classes formed by modular arithmetic or equality relations.
  • The first relation demonstrates the concept of congruence modulo 4, which is widely used in Number Theory and higher mathematics.
  • The second relation represents the identity (equality) relation, the simplest example of an equivalence relation.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Every equality relation is an equivalence relation.

  2. Relations based on divisibility or congruence often produce equivalence classes.

  3. The equivalence class of an element contains all elements related to it.

  4. Equivalence classes partition a set into disjoint subsets.

  5. Always verify reflexive, symmetric and transitive properties separately before concluding.
← Q8
9 / 16  ·  56%
Q10 →
Q10
NUMERIC3 marks
Give an example of a relation which is
  1. Symmetric but neither r`eflexive nor transitive.
  2. Transitive but neither reflexive nor symmetric.
  3. Reflexive and symmetric but not transitive.
  4. Reflexive and transitive but not symmetric.
  5. Symmetric and transitive but not reflexive.
📘 Concept & Theory
Concept / Theory
›

Before constructing examples, recall the definitions of the three basic properties of relations.

Reflexive Relation

A relation \(R\) on a set \(A\) is reflexive if

\[(a,a)\in R\]

for every

\[a\in A.\]
Symmetric Relation

A relation is symmetric if

\[(a,b)\in R\]

implies

\[(b,a)\in R.\]
Transitive Relation

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

imply

\[(a,c)\in R.\]

To answer this question, we construct suitable examples satisfying only the required properties.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Select a small finite set.

  2. Construct a relation satisfying the required property.

  3. Verify reflexive, symmetric and transitive properties one by one.

  4. Explain why the remaining properties fail.

✏️ Solution
Solution (i) Symmetric but Neither Reflexive nor Transitive
›
Step-by-step Solution  ·  6 steps
  1. Take
    \[A=\{1,2,3\}\]
  2. and define
    \[R=\{(1,2),(2,1)\}\]
  3. Verification
  4. Reflexive: No, because \((1,1)\), \((2,2)\) and \((3,3)\) are absent.
  5. Symmetric: Yes, because together with \((1,2)\), the pair \((2,1)\) is also present.
  6. Transitive: No, because
    \[(1,2),(2,1)\in R\]
    but
    \[(1,1)\notin R.\]
  7. Hence, this relation is symmetric but neither reflexive nor transitive.
✏️ Solution
(ii) Transitive but Neither Reflexive nor Symmetric
›
Step-by-step Solution  ·  6 steps
  1. Take
    \[A=\{1,2,3\}\]
  2. and define
    \[R=\{(1,2)\}\]
  3. Verification
  4. Reflexive: No, because \((1,1)\), \((2,2)\) and \((3,3)\) are absent.
  5. Symmetric: No, because \((2,1)\notin R\).
  6. Transitive: Yes, because there are no two ordered pairs of the form
    \[(a,b)\]
    and
    \[(b,c)\]
    whose composition can violate transitivity.

  7. Hence, this relation is transitive but neither reflexive nor symmetric.

✏️ Solution
(iii) Reflexive and Symmetric but Not Transitive
›
Step-by-step Solution  ·  6 steps
  1. Take
    \[A=\{1,2,3\}\]
  2. and define
    \[R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}\]
  3. Verification
  4. Reflexive: Yes, because every element is related to itself.
  5. Symmetric: Yes, because every ordered pair has its reverse.
  6. Transitive: No, because
    \[(1,2),(2,3)\in R\]
    but
    \[(1,3)\notin R\]
  7. Hence, this relation is reflexive and symmetric but not transitive.
✏️ Solution
(iv) Reflexive and Transitive but Not Symmetric
›
Step-by-step Solution  ·  6 steps
  1. Take
    \[A=\{1,2,3\}\]
  2. and define
    \[R=\{(1,1),(2,2),(3,3),(1,2)\}\]
  3. Verification
  4. Reflexive: Yes, because all diagonal pairs are present.
  5. Symmetric: No, because
    \[(2,1)\notin R\]
  6. Transitive: Yes, because every possible chain satisfies the transitive condition.
  7. Hence, this relation is reflexive and transitive but not symmetric.
✏️ Solution
(v) Symmetric and Transitive but Not Reflexive
›
Step-by-step Solution  ·  6 steps
  1. Take
    \[A=\{1,2,3\}\]
  2. and define
    \[R=\{(1,1)\}\]
  3. Verification
  4. Reflexive: No, because
    \[(2,2)\]
    and
    \[(3,3)\]
    are absent.
  5. Symmetric: Yes, because
    \[(1,1)\]
    is its own reverse.
  6. Transitive: Yes, because
    \[(1,1)\]
    combined with itself again gives
    \[(1,1).\]
  7. Hence, this relation is symmetric and transitive but not reflexive.
💡 Answer
Final Answer
›
Part Example of Relation Properties
(i) \(\{(1,2),(2,1)\}\) Symmetric only
(ii) \(\{(1,2)\}\) Transitive only
(iii) \(\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}\) Reflexive and Symmetric
(iv) \(\{(1,1),(2,2),(3,3),(1,2)\}\) Reflexive and Transitive
(v) \(\{(1,1)\}\) Symmetric and Transitive
🎯 Exam Significance
Exam Significance
›
  • This is one of the most important conceptual questions from the chapter "Relations and Functions".
  • CBSE Board examinations frequently ask students to construct examples satisfying specified combinations of relation properties.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often include similar conceptual questions.
  • Students should learn to verify every property separately instead of memorizing examples.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. A relation may satisfy one, two or all three fundamental properties.

  2. Reflexive, symmetric and transitive properties are independent of one another.

  3. Small finite sets provide the easiest way to construct examples.

  4. Counterexamples are sufficient to show that a property does not hold.

  5. Always test each property individually before drawing the final conclusion.
← Q9
10 / 16  ·  63%
Q11 →
Q11
NUMERIC3 marks
Show that the relation \(R=\{(P,Q):\text{distance of }P\text{ from the origin is the same as the distance of }Q\text{ from the origin}\}\) defined on the set \(A\) of all points in a plane is an equivalence relation. Further, show that the set of all points related to a point \(P\neq(0,0)\) is the circle passing through \(P\) with origin as centre.
📘 Concept & Theory
Concept / Theory
›

Let the coordinates of two points be

\[P(x_1,y_1)\]

and

\[Q(x_2,y_2).\]

The distance of a point from the origin is given by the distance formula

\[OP=\sqrt{x_1^2+y_1^2}\]

and

\[OQ=\sqrt{x_2^2+y_2^2}.\]

According to the given relation,

\[(P,Q)\in R\]

if and only if

\[OP=OQ.\]

A relation is called an equivalence relation if it is

  • Reflexive
  • Symmetric
  • Transitive
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Express the relation in terms of distances from the origin.

  2. Verify reflexive, symmetric and transitive properties one by one.

  3. Determine all points related to a fixed point \(P\).

  4. Recognize the geometric figure formed by these points.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  35 steps
  1. The relation is
    \[(P,Q)\in R\]
  2. if
    \[OP=OQ\]
  3. Check Reflexive Property
  4. Take any point
    \[P\in A\]
  5. The distance of the point \(P\) from the origin is equal to itself.
    \[OP=OP\]
  6. Therefore,
    \[(P,P)\in R\]
  7. Since this is true for every point in the plane, the relation is reflexive.
  8. Check Symmetric Property
  9. Suppose
    \[(P,Q)\in R\]
  10. Then
    \[OP=OQ\]
  11. Since equality is symmetric,
    \[OQ=OP\]
  12. Therefore,
    \[(Q,P)\in R\]
  13. Hence, the relation is symmetric.
  14. Check Transitive Property
  15. Suppose
    \[(P,Q)\in R\]
    and
    \[(Q,S)\in R\]
  16. Then
    \[OP=OQ\]
    and
    \[OQ=OS\]
  17. Using the transitive property of equality,
    \[OP=OS\]
  18. Therefore,
    \[(P,S)\in R\]
  19. Hence, the relation is transitive.
  20. Conclusion

  21. Since the relation satisfies

    • Reflexive Property
    • Symmetric Property
    • Transitive Property

    the given relation is an equivalence relation.

  22. Finding the Set of All Points Related to a Fixed Point
  23. Let
    \[P(a,b),\qquad P\neq(0,0)\]
  24. The distance of \(P\) from the origin is
    \[OP=\sqrt{a^2+b^2}\]
  25. Now let \(Q(x,y)\) be any point related to \(P\).
  26. Then, by the definition of the relation,
    \[OQ=OP\]
  27. Substituting the distance formula,
  28. \[\sqrt{x^2+y^2}=\sqrt{a^2+b^2}.\]
  29. Squaring both sides,
    \[x^2+y^2=a^2+b^2\]
  30. Here,\(a^2+b^2\) is a fixed positive constant because
    \[P\neq(0,0)\]
  31. Therefore,
    \[x^2+y^2=r^2\]
    where
    \[r=\sqrt{a^2+b^2}\]
    This is the standard equation of a circle having Centre at the origin
  32. Radius
    \[r=\sqrt{a^2+b^2}\]
  33. Since the point \(P(a,b)\) satisfies this equation, the circle passes through \(P\).
  34. Hence, the set of all points related to \(P\) is precisely the circle with centre at the origin passing through \(P\).
  35. Equivalence Class of a Point
  36. If
    \[P(a,b)\neq(0,0)\]
    then its equivalence class is
    \[[P]=\{(x,y):x^2+y^2=a^2+b^2\}.\]
  37. This represents the entire circle centred at the origin passing through \(P\).
  38. Special Case
  39. If
    \[P=(0,0)\]
    then
    \[OP=0\]
  40. The only point whose distance from the origin is zero is the origin itself.
  41. Therefore,
    \[[(0,0)]=\{(0,0)\}\]
💡 Answer
Final Answer
›
Property Result Reason
Reflexive Yes \(OP=OP\)
Symmetric Yes \(OP=OQ\Rightarrow OQ=OP\)
Transitive Yes \(OP=OQ\) and \(OQ=OS\Rightarrow OP=OS\)
Type of Relation Equivalence Relation
Equivalence Class of \(P(a,b)\) \(\{(x,y):x^2+y^2=a^2+b^2\}\)

Hence, the given relation is an equivalence relation, and the set of all points related to any point

\[P\neq(0,0)\]

is the circle having the origin as centre and passing through \(P\).

🎯 Exam Significance
Exam Significance
›
  • This is one of the most important proof-based questions from Chapter 1 "Relations and Functions" for the CBSE Class 12 Board Examination.
  • It combines concepts from Coordinate Geometry and Relations, making it valuable for JEE Main, CUET, NDA, BITSAT and other entrance examinations.
  • Students should recognize that "equal distance from a fixed point" naturally describes a circle.
  • The question illustrates how an equivalence relation partitions the plane into concentric circles centred at the origin.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. This is one of the most important proof-based questions from Chapter 1 "Relations and Functions" for the CBSE Class 12 Board Examination.

  2. It combines concepts from Coordinate Geometry and Relations, making it valuable for JEE Main, CUET, NDA, BITSAT and other entrance examinations.

  3. Students should recognize that "equal distance from a fixed point" naturally describes a circle.

  4. The question illustrates how an equivalence relation partitions the plane into concentric circles centred at the origin.
← Q10
11 / 16  ·  69%
Q12 →
Q12
NUMERIC3 marks
Show that the relation \(R=\{(T_1,T_2):T_1\text{ is similar to }T_2\}\) defined on the set \(A\) of all triangles is an equivalence relation. Consider the following three right-angled triangles:
  • \(T_1\) having sides \(3,\;4,\;5\)
  • \(T_2\) having sides \(5,\;12,\;13\)
  • \(T_3\) having sides \(6,\;8,\;10\)

Determine which triangles among \(T_1\), \(T_2\) and \(T_3\) are related.

📘 Concept & Theory
Concept / Theory
›

Two triangles are said to be similar if

  • their corresponding angles are equal, and
  • their corresponding sides are proportional.

If the corresponding sides satisfy

\[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},\]

then the two triangles are similar.

An equivalence relation is a relation that is

  • Reflexive
  • Symmetric
  • Transitive
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexive, symmetric and transitive properties using the definition of similar triangles.

  2. Compare the corresponding sides of the given triangles.

  3. Check whether the ratios of corresponding sides are equal.

  4. Identify the triangles that are related.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  35 steps
  1. The given relation is
    \[R=\{(T_1,T_2):T_1\text{ is similar to }T_2\}\]
  2. Check Reflexive Property
  3. Every triangle is similar to itself because
    • all corresponding angles are equal, and
    • all corresponding sides are in the ratio
    \[1:1\]
  4. Hence,
    \[(T,T)\in R\]
  5. for every triangle \(T\)
  6. Therefore, the relation is reflexive
  7. Check Symmetric Property
  8. Suppose
    \[(T_1,T_2)\in R\]
  9. This means that \(T_1\) is similar to \(T_2\)
  10. Then the corresponding sides satisfy
    \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.\]
  11. Taking reciprocals,
    \[\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}.\]
  12. Hence, \(T_2\) is also similar to \(T_1\)
  13. Therefore,
    \[(T_2,T_1)\in R\]
  14. Thus, the relation is symmetric.
  15. Check Transitive Property
  16. Suppose
    \[(T_1,T_2)\in R\]
    and
    \[(T_2,T_3)\in R\]
  17. Then
    \[T_1 \sim T_2\]
    and
    \[T_2\sim T_1\]
  18. Hence, the corresponding sides satisfy

    \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\]
    and
    \[\frac{a_2}{a_3}=\frac{b_2}{b_3}=\frac{c_2}{c_3}\]
  19. Multiplying corresponding ratios, we obtain
    \[\frac{a_1}{a_3}=\frac{b_1}{b_3}=\frac{c_1}{c_3}.\]
  20. Therefore,
    \[T_1 \sim T_3\]
  21. Hence,
    \[(T_1,T_3)\in R\]
  22. Thus, the relation is transitive.
  23. Conclusion
  24. he relation satisfies
    • Reflexive Property
    • Symmetric Property
    • Transitive Property
  25. Therefore, it is an equivalence relation.
  26. Determining Which Triangles are Related
  27. Compare \(T_1\) and \(T_2\)
  28. The corresponding side ratios are
    \[\frac35,\qquad\frac4{12}=\frac13,\qquad\frac5{13}.\]
  29. Since these ratios are not equal, $T_1 \text{ and} T_2$ are not similar
  30. Therefore,
    \[(T_1,T_2)\notin R\]
  31. Compare \(T_1\) and \(T_3\)
  32. he corresponding side ratios are
    \[\frac36=\frac12,\]
    \[\frac48=\frac12,\]
    \[\frac5{10}=\frac12.\]
  33. Since all three ratios are equal, $T_1 \text{ and} T_3$ are similar
  34. Therefore,
    \[(T_1,T_3)\in R\]
  35. Compare \(T_2\) and \(T_3\)
  36. The corresponding side ratios are
    \[\frac56,\]
    \[\frac{12}8=\frac32,\]
    \[\frac{13}{10}.\]
    These ratios are not equal.
  37. Hence, $T_2 \text{ and} T_3$ are not similar
  38. Therefore,
    \[(T_2,T_3)\notin R\]
💡 Answer
Final Answer
›
Pair of Triangles Similar? Related?
\(T_1,\;T_2\) No No
\(T_1,\;T_3\) Yes Yes
\(T_2,\;T_3\) No No

Hence, among the three given triangles, only

\[\boxed{T_1\text{ and }T_3}\]

are related.

🎯 Exam Significance
Exam Significance
›
  • This is an important proof-based question that combines the concepts of relations and triangle similarity.
  • CBSE Board examinations frequently ask students to identify whether a geometric relation is an equivalence relation.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often test similarity using proportional sides.
  • Students should remember that similarity naturally satisfies reflexive, symmetric and transitive properties.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Similarity of triangles is an equivalence relation.

  2. To verify similarity, compare the ratios of corresponding sides.

  3. Equal ratios imply similarity.

  4. Different ratios imply the triangles are not related.

  5. Among the given triangles, only \(T_1\) and \(T_3\) belong to the same equivalence class.
← Q11
12 / 16  ·  75%
Q13 →
Q13
NUMERIC3 marks
Show that the relation R defined in the set A of all polygons as $R = \{(P_1, P_2) :P_1 \text{ and } P_2 \text{ have same number of sides}\}$, is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle $T$ with sides 3, 4 and 5?
📘 Concept & Theory
Concept / Theory
›

A polygon is a closed plane figure formed by a finite number of line segments.

The defining characteristic of the given relation is the number of sides of a polygon.

Two polygons are related if and only if they possess exactly the same number of sides.

A relation is called an equivalence relation if it satisfies all three properties:

  • Reflexive
  • Symmetric
  • Transitive
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexive property by considering any polygon.

  2. Use the symmetry of equality to prove the symmetric property.

  3. Use equality of the number of sides to establish transitivity.

  4. Determine the equivalence class of the given triangle.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  26 steps
  1. The relation is defined as
    \[(P_1,P_2)\in R\]
  2. if and only if \(P_1\) and \(P_2\) have the same number of sides.
  3. Check Reflexive Property
  4. Consider any polygon
    \[P\in A\]
  5. Every polygon has exactly the same number of sides as itself.
  6. Hence,
    \[(P,P)\in R\]
  7. This is true for every polygon in the set.
  8. Therefore, the relation is reflexive.
  9. Check Symmetric Property
  10. Suppose
    \[(P_1,P_2)\in R\]
  11. Then \(P_1\) and \(P_2\) have the same number of sides.
  12. Therefore, \(P_2\) also has the same number of sides as \(P_1\).
  13. Hence,
  14. Therefore, the relation is symmetric.
  15. Check Symmetric Property
  16. Check Transitive Property
  17. Suppose
    \[(P_1,P_2)\in R\]
    and
    \[(P_2,P_3)\in R\]
  18. Then
    • \(P_1\) and \(P_2\) have the same number of sides.
    • \(P_2\) and \(P_3\) have the same number of sides.
  19. Therefore, \(P_1\) and \(P_3\) also have the same number of sides.
  20. Hence,
    \[(P_1,P_3)\in R\]
  21. Thus, the relation is transitive.
  22. Conclusion
  23. The given relation satisfies
    • Reflexive Property
    • Symmetric Property
    • Transitive Property
  24. Hence, it is an equivalence relation.
  25. Finding the Set of All Elements Related to the Triangle \(T\)
  26. The given triangle \(T\) has 3 sides
  27. According to the definition of the relation, every polygon related to \(T\) must also have exactly 3 sides
  28. Therefore, the equivalence class containing \(T\) consists of all triangles.
  29. Hence,
    \[[T]=\{\text{all triangles}\}.\]
  30. This includes every type of triangle such as
    • Equilateral triangles
    • Isosceles triangles
    • Scalene triangles
    • Right-angled triangles
    • Acute-angled triangles
    • Obtuse-angled triangles
  31. Every triangle has exactly three sides, so each of them is related to the given triangle.
💡 Answer
Final Answer
›
Property Result Reason
Reflexive Yes Every polygon has the same number of sides as itself.
Symmetric Yes If one polygon has the same number of sides as another, then the reverse is also true.
Transitive Yes If two polygons each have the same number of sides as a third polygon, then they have the same number of sides as each other.
Type of Relation Equivalence Relation
Equivalence Class of Triangle \(T\) \(\{\text{All triangles}\}\)

Hence, the relation is an equivalence relation, and the set of all elements related to the given triangle is the set of all triangles.

🎯 Exam Significance
Exam Significance
›
  • This is a standard proof-based question from the chapter "Relations and Functions" frequently asked in CBSE Class 12 Board examinations.
  • It demonstrates how a common geometric property can define an equivalence relation.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often ask students to identify equivalence classes formed by geometric objects.
  • The question highlights that an equivalence relation partitions a set into groups having the same defining characteristic.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Having the same number of sides naturally satisfies reflexive, symmetric and transitive properties.

  2. Every equivalence relation partitions a set into disjoint equivalence classes.

  3. The equivalence class of a triangle consists of all triangles.

  4. Polygons having different numbers of sides belong to different equivalence classes.

  5. Always verify the three properties separately before concluding that a relation is an equivalence relation.
← Q12
13 / 16  ·  81%
Q14 →
Q14
NUMERIC3 marks
Let \(L\) be the set of all lines in the XY-plane and let $R=\{(L_1,L_2):L_1\text{ is parallel to }L_2\}$ Show that \(R\) is an equivalence relation. Find the set of all lines related to the line $y=2x+4$
📘 Concept & Theory
Concept / Theory
›

Two lines in a plane are parallel if they have the same slope.

The slope-intercept form of a line is

\[y=mx+c,\]

where

  • \(m\) is the slope of the line.
  • \(c\) is the y-intercept.

Two non-vertical lines

\[y=m_1x+c_1\]

and

\[y=m_2x+c_2\]

are parallel if and only if

\[m_1=m_2.\]

An equivalence relation is a relation that is

  • Reflexive
  • Symmetric
  • Transitive
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify reflexive property using the fact that every line is parallel to itself.

  2. Verify symmetry using the definition of parallel lines.

  3. Verify transitivity using equality of slopes.

  4. Determine the slope of the given line.

  5. Write the equation of every line having the same slope.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  25 steps
  1. The relation is defined as
    \[(L_1,L_2)\in R\]
    if and only if \(L_1\) is parallel to \(L_2\).
  2. Check Reflexive Property
  3. Consider any line
    \[L\in A.\]
  4. Every line is parallel to itself because it has exactly the same slope as itself.
  5. Therefore,
    \[(L,L)\in R\]
  6. >Hence, the relation is reflexive.
  7. Check Symmetric Property
  8. Suppose
    \[(L_1,L_2)\in R\]
  9. This means
    \[L_1\parallel L_2\]
  10. If one line is parallel to another, then the second line is also parallel to the first.
  11. Hence,
    \[(L_2,L_1)\in R\]
  12. Therefore, the relation is symmetric
  13. Check Transitive Property
  14. Suppose
    \[(L_1,L_2)\in R\]
    and
    \[(L_2,L_3)\in R.\]
  15. Then
    • \(L_1\) and \(L_2\) have the same slope.
    • \(L_2\) and \(L_3\) have the same slope.
  16. Therefore,
    \[\text{slope of }L_1=\text{slope of }L_3.\]
  17. Hence,
    \[L_1\parallel L_3\]
  18. Thus,
    \[(L_1,L_3)\in R\]
  19. Therefore, the relation is transitive.
  20. Conclusion
  21. Since the relation is
    • Reflexive
    • Symmetric
    • Transitive
  22. the relation is an equivalence relation
  23. Finding the Set of All Lines Related to $y=2x+4$
  24. The given line is already in slope-intercept form.
    \[y=2x+4\]
  25. Its slope is $m=2$
  26. Every line parallel to this line must have the same slope.
  27. Therefore, the equation of every line related to the given line is
    \[y=2x+c\]
    where $c is any real number
  28. The value of \(c\) changes the position of the line but does not change its slope.
  29. Hence, all such lines are parallel to
    \[y=2x+4\]
  30. Therefore, the equivalence class of the given line is
    \[[y=2x+4]=\{\,y=2x+c:\;c\in\mathbb{R}\,\}\]
💡 Answer
Final Conclusion
›
Property Result Reason
Reflexive Yes Every line is parallel to itself.
Symmetric Yes If \(L_1\parallel L_2\), then \(L_2\parallel L_1\).
Transitive Yes If \(L_1\parallel L_2\) and \(L_2\parallel L_3\), then \(L_1\parallel L_3\).
Type of Relation Equivalence Relation
Equivalence Class of \(y=2x+4\) \(\{\,y=2x+c:\;c\in\mathbb{R}\,\}\)

Hence, the relation "is parallel to" is an equivalence relation, and every line having slope 2 belongs to the equivalence class of the line

\[\boxed{y=2x+4.}\]
🎯 Exam Significance
Exam Significance
›
  • This is an important proof-based question from Chapter 1 "Relations and Functions" that combines coordinate geometry with relation properties.
  • CBSE Board examinations frequently ask students to identify equivalence relations defined using geometric concepts.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often require students to recognize that parallel lines have equal slopes.
  • The question demonstrates how an equivalence relation partitions the set of all lines into families of parallel lines.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. The relation "is parallel to" satisfies reflexive, symmetric and transitive properties.

  2. Parallel lines have equal slopes.

  3. Changing the y-intercept changes the position of the line but not its slope.

  4. All lines of the form \(\,y=2x+c\,\) are parallel to \(\,y=2x+4\,\).

  5. Each family of parallel lines forms one equivalence class.
← Q13
14 / 16  ·  88%
Q15 →
Q15
NUMERIC3 marks
Let $A=\{1,2,3,4\}$ and $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ Choose the correct answer.

(A) \(R\) is reflexive and symmetric but not transitive.

(B) \(R\) is reflexive and transitive but not symmetric.

(C) \(R\) is symmetric and transitive but not reflexive.

(D) \(R\) is an equivalence relation.

📘 Concept & Theory
Concept / Theory
›

To identify the correct option, each property of the relation must be checked separately.

Reflexive Relation

A relation is reflexive if

\[(a,a)\in R\]

for every element of the set.

Symmetric Relation

A relation is symmetric if

\[(a,b)\in R\]

implies

\[(b,a)\in R.\]
Transitive Relation

A relation is transitive if

\[(a,b)\in R\]

and

\[(b,c)\in R\]

imply

\[(a,c)\in R.\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check whether every diagonal pair is present.

  2. Verify whether every ordered pair has its reverse.

  3. Test all possible chains of ordered pairs for transitivity.

  4. Select the option matching the verified properties.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  28 steps
  1. The given relation is
    \[R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}\]
  2. Check Reflexive Property
  3. The set is
    \[A=\{1,2,3,4\}\]
  4. For reflexivity, the following ordered pairs must be present
    \[(1,1),\quad(2,2),\quad(3,3),\quad(4,4)\]
  5. All four ordered pairs are present in the relation.
  6. Therefore, the relation is reflexive.
  7. Check Symmetric Property
  8. Consider the ordered pair
    \[(1,2)\in R.\]
  9. If the relation were symmetric, then \((2,1)\) must also belong to the relation.
  10. But
    \[(2,1)\notin R\]
  11. Hence, the relation is not symmetric.
  12. Check Transitive Property
  13. We examine all possible chains.
  14. Chain 1
  15. \[(1,2)\in R,\qquad(2,2)\in R\]
  16. Therefore, transitivity requires
    \[(1,2)\in R\]
    which is already present.
  17. Chain 2
  18. \[(1,1)\in R,\qquad(1,2)\in R\]
  19. Therefore,
    \[(1,2)\in R\]
    which is present.
  20. Chain 3
  21. \[(1,1)\in R,\qquad(1,3)\in R\]
  22. Therefore,
    \[(1,3)\in R\]
    which is present.
  23. Chain 4
  24. \[(1,3)\in R,\qquad(3,3)\in R\]
  25. Therefore,
    \[(1,3)\in R\]
    which is present.
  26. Chain 5
  27. \[(1,3)\in R,\qquad(3,2)\in R\]
  28. Therefore, transitivity requires
    \[(1,2)\in R\]
    which is present.
  29. Chain 6
  30. \[(3,3)\in R,\qquad(3,2)\in R\]
  31. Therefore,
    \[(3,2)\in R\]
    which is present.
  32. Chain 7
  33. \[(3,2)\in R,\qquad(2,2)\in R.\]
  34. Therefore,
    \[(3,2)\in R\]
    which is present.
  35. Chain 8
  36. \[(4,4)\in R,\qquad(4,4)\in R\]
  37. Therefore,
    \[(4,4)\in R\]
    which is present.
  38. No chain violates the transitive property.
  39. Hence, the relation is transitive.
💡 Answer
Final Conclusion
›
Property Result
Reflexive Yes
Symmetric No
Transitive Yes

Therefore, the correct option is

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

🎯 Exam Significance
Exam Significance
›
  • This is a frequently asked multiple-choice question in CBSE Board examinations.
  • Students should verify each property systematically instead of guessing from the ordered pairs.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT often include similar conceptual MCQs.
  • The safest strategy is to check reflexivity first, then symmetry, and finally transitivity.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Reflexivity requires every diagonal pair to be present.

  2. A single missing reverse ordered pair is enough to prove that a relation is not symmetric.

  3. For transitivity, every possible chain must satisfy the required conclusion.

  4. The given relation satisfies reflexive and transitive properties only.

  5. The correct answer is Option (B).
← Q14
15 / 16  ·  94%
Q16 →
Q16
NUMERIC3 marks
Let $R=\{(a,b):a=b-2,\;b>6\}$ be a relation in the set of natural numbers \(\mathbb{N}\). Choose the correct answer.

(A) \((2,4)\in R\)

(B) \((3,8)\in R\)

(C) \((6,8)\in R\)

(D) \((8,7)\in R\)

📘 Concept & Theory
Concept / Theory
›

An ordered pair \((a,b)\) belongs to the relation if and only if it satisfies both of the following conditions simultaneously:

  • \[a=b-2\]
  • \[b>6.\]

    • If even one of these conditions is not satisfied, then the ordered pair does not belong to the relation.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check whether \(a=b-2\).

  2. Verify whether \(b>6\).

  3. Accept only the ordered pair satisfying both conditions.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. The relation is
    \[R=\{(a,b):a=b-2,\;b>6\}\]
  2. We examine each option one by one.
  3. Option (A)
  4. Given — pair:
    \[(2,4)\]
  5. Check the first condition.
    \[b-2=4-2=2\]
  6. Thus,
    \[a=b-2\]
    is satisfied.
  7. Now check the second condition.
    \[4>6\]
  8. which is false.
  9. Therefore,
    \[(2,4)\notin R\]
  10. Option (B)
  11. Given — pair:
    \[(3,8)\]
  12. Check the first condition.
    \[b-2=8-2=6\]
  13. But
    \[a=3\neq6\]
    Hence, the first condition fails.
  14. Therefore,
    \[(3,8)\notin R\]
  15. Option (C)
  16. Given — pair:
    \[(6,8)\]
  17. Check the first condition.
    \[b-2=8-2=6\]
  18. Thus,
    \[a=b-2\]
  19. Now check the second condition.
    \[8>6\]
  20. which is true. Hence, both conditions are satisfied.
  21. Therefore,
    \[(6,8)\in R\]
  22. Option (D)
  23. Given — pair:
    \[(8,7)\]
  24. Check the first condition.
    \[b-2=7-2=5\]
  25. But
    \[a=8\neq5\]
  26. Hence, the first condition is not satisfied.
  27. Therefore,
    \[(8,7)\notin R\]
💡 Answer
Final Conclusion
›
Option \(a=b-2\) \(b>6\) Belongs to \(R\)?
(A) \((2,4)\) Yes No No
(B) \((3,8)\) No Yes No
(C) \((6,8)\) Yes Yes Yes
(D) \((8,7)\) No Yes No

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

🎯 Exam Significance
Exam Significance
›
  • This is a common CBSE multiple-choice question based on the definition of a relation.
  • Students should always verify every condition given in the definition before selecting an option.
  • Competitive examinations such as JEE Main, CUET, NDA and BITSAT frequently ask similar questions involving ordered pairs.
  • Remember that all conditions connected by "and" must hold simultaneously.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. A relation may contain multiple conditions.

  2. An ordered pair belongs to the relation only when every condition is satisfied.

  3. Check algebraic conditions first, followed by domain restrictions.

  4. Do not stop after verifying only one condition.

  5. The correct answer is Option (C).
← Q15
16 / 16  ·  100%
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NCERT Class 12 Maths Ch 1 Exercise 1.1 Solutions
NCERT Class 12 Maths Ch 1 Exercise 1.1 Solutions — Complete Notes & Solutions · academia-aeternum.com
Relations form the foundation of many important concepts in higher mathematics, and understanding their properties is essential for mastering Class 12 Mathematics. The solutions to Exercise 1.1 of NCERT Chapter 1, Relations and Functions, are designed to help students develop a clear conceptual understanding of reflexive, symmetric, transitive and equivalence relations. Each solution begins with the underlying theory and key concepts, followed by a systematic roadmap that explains the approach…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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