Example 1 (Reflexive):
Example 2 (Not Reflexive):
Example 3 (Symmetric):
Example 4 (Not Symmetric):
Example 5 (Transitive):
Example 6 (Not Transitive):
Example 7 (An Equivalence Relation): 'Is equal to' (=) on a set of real numbers.
Example 8 (Another Equivalence Relation): 'Has the same number of letters as' on a set of words.
Example 9 (Not an Equivalence Relation): 'Is a factor of' on a set of positive integers.
| Aspect | CBSE Board Exams | JEE Mains & Advanced |
|---|---|---|
| Core Understanding | Focus on clearly defining and verifying each of the three properties (reflexive, symmetric, transitive) for given relations. Questions are often direct and involve common mathematical or set-based relations. | Requires a deeper and more nuanced understanding. Questions can involve abstract relations, relations defined on complex sets (e.g., matrices, functions, complex numbers), or relations with multiple conditions. |
| Problem Complexity | You'll typically be asked to 'show that a given relation is an equivalence relation' by explicitly demonstrating all three properties with examples or general arguments. | May involve relations where one or two properties hold, but not all three, requiring you to identify the failure point. You might need to prove or disprove properties using formal, algebraic arguments for general elements, not just specific numbers. |
| Application | Direct application of definitions, often involving specific numerical or simple set elements. | Understanding how equivalence relations partition a set into equivalence classes is crucial and can be implicitly tested. You might encounter questions asking to describe the equivalence classes formed by a given equivalence relation. |
Welcome, future engineers and mathematicians! Today, we're diving deep into one of the most fundamental and elegant concepts in discrete mathematics: Equivalence Relations. This topic is not just important for your JEE preparation, but it forms a cornerstone for many advanced mathematical areas like abstract algebra and topology. So, let's build a strong foundation together!
Before we define an equivalence relation, let's quickly recall what a relation is. Given two non-empty sets, A and B, a relation R from A to B is simply a subset of the Cartesian product A Γ B. If R is a relation on a single set A, it means R is a subset of A Γ A. We often write 'a R b' to denote that the ordered pair (a, b) belongs to the relation R.
For a relation to be an equivalence relation, it must possess three crucial properties simultaneously. Let's examine each of these properties in detail.
A relation R on a set A is said to be reflexive if every element of A is related to itself.
Definition: A relation R on a set A is reflexive if for every element a in A, we have (a, a) in R (or a R a).
Key Insight: Think of it as "self-connection." Every element must "see" itself in the relation. If even one element in the set A does not have (a, a) in the relation, it's not reflexive.
A relation R on a set A is said to be symmetric if whenever an element 'a' is related to an element 'b', then 'b' must also be related to 'a'.
Definition: A relation R on a set A is symmetric if for all a, b in A, whenever (a, b) in R, then it must be that (b, a) in R (or if a R b, then b R a).
Key Insight: It's about "two-way street" connections. If there's a path from 'a' to 'b', there must be a path back from 'b' to 'a'.
A relation R on a set A is said to be transitive if whenever an element 'a' is related to 'b', and 'b' is related to 'c', then 'a' must also be related to 'c'.
Definition: A relation R on a set A is transitive if for all a, b, c in A, whenever (a, b) in R and (b, c) in R, then it must be that (a, c) in R (or if a R b and b R c, then a R c).
Key Insight: This is the "chain rule" or "domino effect." If a links to b, and b links to c, then a must link directly to c.
Now that we understand the three fundamental properties, we can define an equivalence relation.
Definition: A relation R on a set A is called an Equivalence Relation if it is simultaneously:
Intuition & Analogy: An equivalence relation essentially groups together elements of a set that are "alike" or "equivalent" in some specified way. Imagine sorting a collection of objects (say, geometric shapes) into different boxes. If you sort them by "same color," then any red shape is "equivalent" to any other red shape. Any blue shape is equivalent to any other blue shape, and so on.
This "same color" relation would form an equivalence relation.
Let A be any non-empty set. Consider the relation R defined as a R b iff a = b.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Let L be the set of all lines in a plane. Consider the relation R defined as l_1 R l_2 iff l_1 ext{ is parallel to } l_2 (l_1 || l_2).
Therefore, the relation "is parallel to" is an equivalence relation.
Let Z be the set of integers. Let n be a fixed positive integer. Consider the relation R on Z defined as a R b iff a - b ext{ is divisible by } n (or a equiv b pmod{n}).
This means a - b = kn for some integer k.
Since R is reflexive, symmetric, and transitive, congruence modulo n is an equivalence relation. This is a very important concept in number theory and abstract algebra.
One of the most profound consequences of an equivalence relation is that it partitions the original set into disjoint subsets, called equivalence classes.
Definition: Let R be an equivalence relation on a set A. For any element a in A, the equivalence class of a, denoted by [a] or ar{a}, is the set of all elements in A that are related to a.
[a] = { x in A mid x R a }
Key Properties of Equivalence Classes:
This is a crucial property. If [a] cap [b]
eq emptyset, it means there exists some c such that c in [a] and c in [b]. This implies c R a and c R b. By symmetry, a R c. By transitivity, a R c and c R b implies a R b. Similarly, b R a. This means all elements related to 'a' are also related to 'b', and vice versa, making their equivalence classes identical.
Let A = Z (integers) and the relation R be a equiv b pmod{3}. We know this is an equivalence relation.
Let's find the equivalence classes:
What about [3]?
[3] = { x in Z mid x equiv 3 pmod{3} }
Since x - 3 = 3k implies x = 3k + 3 = 3(k+1), this is the same as [0]. So [3] = [0].
Similarly, [4] = [1], [5] = [2], and so on.
The distinct equivalence classes are [0], [1], [2]. Notice that:
Thus, the relation partitions the set of integers Z into three distinct equivalence classes based on their remainder when divided by 3. These classes are often referred to as Z_3 (the set of integers modulo 3).
Equivalence relations are more than just a set of definitions; they are a powerful tool for structuring and understanding complex sets by grouping "similar" elements. Mastering this concept will serve you well in your mathematical journey!
Memorizing the definitions and properties of equivalence relations is crucial for quickly solving problems in both board exams and JEE Main. Here are some effective mnemonics and shortcuts to help you remember the three key properties:
An equivalence relation is a binary relation on a set that satisfies three fundamental properties: Reflexivity, Symmetry, and Transitivity. Think of the acronym RST to remember these three pillars:
Let's break down each property with a memorable shortcut:
A relation R on a set A is reflexive if for every element 'a' in A, (a, a) β R.
A relation R on a set A is symmetric if whenever (a, b) β R, then (b, a) β R.
A relation R on a set A is transitive if whenever (a, b) β R and (b, c) β R, then (a, c) β R.
For JEE problems, speed is key. When checking if a given relation is an equivalence relation, mentally (or quickly on scratch paper) create an 'RST' checklist:
| Property | Mental Check | Condition |
|---|---|---|
| Reflexive | Self-Reflection | For all a β A, is (a,a) β R? |
| Symmetric | Two-Way Street | If (a,b) β R, is (b,a) β R? |
| Transitive | Chain Reaction | If (a,b) β R and (b,c) β R, is (a,c) β R? |
Always verify all three properties. If even one fails, the relation is not an equivalence relation. Master these shortcuts, and you'll find equivalence relation problems much easier to tackle!
Equivalence relations are fundamental in higher mathematics and a frequently tested concept in JEE Main. Mastering the quick checks for its three defining properties is key to solving problems efficiently.
A relation R on a set A is an equivalence relation if it is:
All three conditions must be satisfied for a relation to be an equivalence relation.
Reflexive Property: Every element relates to itself.
Symmetric Property: If (a) relates to (b), then (b) relates to (a).
Transitive Property: If (a) relates to (b), and (b) relates to (c), then (a) relates to (c).
If R is an equivalence relation on set A, it partitions A into disjoint subsets called equivalence classes. Each element belongs to exactly one equivalence class.
The equivalence class of (a in A) is denoted by ([a]) or (ar{a}), and contains all elements (x in A) such that ((x, a) in R).
| Aspect | CBSE (Board Exams) | JEE Main |
|---|---|---|
| Problem Type | Usually proving a given relation is/isn't equivalence. Detailed steps required. | Often identify properties (multiple choice), or determine missing condition for equivalence. Speed is critical. |
| Solution | Elaborate mathematical proofs with 'For all...' and 'Thus R is...' statements. | Quick mental checks or counterexamples. Focus on finding the property that fails. |
Keep these rapid checks in mind to ace equivalence relation problems swiftly in JEE Main!
An equivalence relation is a special type of binary relation that essentially formalizes the idea of "sameness," "likeness," or "belonging to the same category" within a given set. When elements are related by an equivalence relation, we consider them to be "equivalent" in some specific sense.
For a relation 'R' on a set 'A' to be an equivalence relation, it must satisfy three fundamental properties, which make intuitive sense for anything we call "equivalent":
Intuitively, every object is "like itself." If we say 'a' is equivalent to 'b', it's natural to assume 'a' is equivalent to 'a'.
Mathematically: For every element a ∈ A, (a, a) ∈ R.
If 'a' is "like" 'b', then 'b' must also be "like" 'a'. There's no one-way equivalence.
Mathematically: If (a, b) ∈ R, then (b, a) ∈ R.
If 'a' is "like" 'b', and 'b' is "like" 'c', then it logically follows that 'a' must also be "like" 'c'. This property allows us to extend the idea of equivalence across multiple elements.
Mathematically: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
The most profound intuitive understanding of an equivalence relation is its ability to partition a set into disjoint (non-overlapping) subsets, called equivalence classes.
Each element within an equivalence class is equivalent to every other element in that same class, and no element in one class is equivalent to any element in another class. Think of it like sorting items into distinct bins where all items in a bin share a common characteristic.
Let's consider the set of integers Z and a relation R defined as: a R b if and only if a - b is divisible by 3 (i.e., a ≡ b (mod 3)). Let's intuitively check the properties:
This relation is indeed an equivalence relation. It partitions the integers into three distinct equivalence classes:
JEE & CBSE Focus: For both exams, the ability to verify these three properties for a given relation on various sets (numbers, lines, functions, etc.) is crucial. JEE might present more abstract definitions of relations or require proving properties for a set of functions, while CBSE generally focuses on numerical examples. The core idea of "partitioning" is universally important.
Mastering equivalence relations isn't just about definitions; it's about understanding how they naturally categorize and organize elements within a set based on shared characteristics.
While often seen as an abstract mathematical concept, equivalence relations are fundamental to how we categorize, classify, and organize information in the real world. They provide a powerful framework for grouping objects that share common properties, simplifying complex systems into manageable sets of "equivalent" elements.
The core idea of an equivalence relationβreflexivity, symmetry, and transitivityβis mirrored in many everyday scenarios, helping us define meaningful partitions and classifications.
Perhaps the most intuitive real-world application of equivalence relations comes from modular arithmetic. Consider the concept of time on a 12-hour clock:
Let's verify the properties:
This relation partitions all possible hours into 12 distinct equivalence classes (e.g., all times that show '3 o'clock'). The same principle applies to days of the week (modulo 7), determining future dates, or repeating patterns in digital displays.
In computer science and everyday organization, we constantly group items based on shared attributes. For example, consider a set of files on your computer:
This is an equivalence relation:
This relation naturally groups all files into equivalence classes based on their type, which is incredibly useful for organization, searching, and managing data.
In geometry, the concepts of congruence and similarity are prime examples of equivalence relations.
Understanding equivalence relations provides a powerful mathematical lens through which to view and structure the world around us, from the abstract realms of mathematics to concrete daily applications. For JEE, while direct questions on real-world applications are rare, appreciating these connections can deepen your conceptual grasp and problem-solving intuition.
Analogies are powerful tools that simplify complex mathematical concepts by relating them to familiar real-world scenarios. For equivalence relations, these comparisons help build an intuitive understanding of the properties involved and their implications, particularly the concept of partitioning a set into disjoint subsets (equivalence classes).
An equivalence relation R on a set A must satisfy three fundamental properties:
Consider a set of people and the relation R defined as "is in the same city as".
This relation partitions the set of people into groups where everyone in a group lives in the same city. For example, all people living in Mumbai form one group, all people living in Delhi form another, and so on. These groups are precisely the equivalence classes.
Consider a set of students and the relation R defined as "has the same birthday as".
This relation partitions the students into groups based on their birth date. All students born on January 1st form one group, all born on January 2nd form another, and so forth.
Consider a set of lines in a plane and the relation R defined as "is parallel to".
This relation groups lines by their direction. All horizontal lines form one equivalence class, all vertical lines another, and lines with the same slope (e.g., slope 1) form yet another.
Notice how each of these analogies naturally leads to the idea of "grouping" or "categorizing" elements within the set. This grouping is the essence of equivalence classes. An equivalence relation on a set A partitions A into disjoint, non-empty subsets (equivalence classes) such that every element belongs to exactly one class.
Understanding these analogies will solidify your grasp of equivalence relations, a concept fundamental for both CBSE board exams and competitive exams like JEE Main. Master these basics to tackle more complex relation and function problems with confidence.
To effectively grasp the concept of Equivalence Relations, it's crucial to have a strong foundation in a few fundamental concepts from Set Theory and basic Relations. These prerequisites are not just theoretical definitions; they are the building blocks upon which equivalence relations are defined and understood.
Here are the essential concepts you must be familiar with:
JEE & CBSE Focus: While CBSE emphasizes the definitions, JEE often tests your ability to quickly identify these components from complex relation descriptions or properties.
An equivalence relation is specifically defined by the presence of three key properties. Therefore, a thorough understanding of these individual types of relations is absolutely critical.
A relation $R$ on a set $A$ is reflexive if every element of $A$ is related to itself. That is, for every $a in A$, $(a, a) in R$.
Example: "is equal to" on a set of numbers (e.g., $5=5$).
A relation $R$ on a set $A$ is symmetric if whenever an element $a$ is related to an element $b$, then $b$ is also related to $a$. That is, for every $a, b in A$, if $(a, b) in R$, then $(b, a) in R$.
Example: "is a sibling of" (If A is a sibling of B, then B is a sibling of A).
A relation $R$ on a set $A$ is transitive if whenever $a$ is related to $b$ and $b$ is related to $c$, then $a$ is also related to $c$. That is, for every $a, b, c in A$, if $(a, b) in R$ and $(b, c) in R$, then $(a, c) in R$.
Example: "is less than" (If $a < b$ and $b < c$, then $a < c$).
Tip for Success: Before moving on to equivalence relations, ensure you can define, identify, and provide examples for reflexive, symmetric, and transitive relations. Practice applying these definitions to various relation types given in set-builder form.
Mastering these prerequisites will make understanding and solving problems related to equivalence relations significantly easier, forming a solid base for advanced topics.
Verifying if a given relation is an equivalence relation is a fundamental skill in this unit. However, several common pitfalls can lead to incorrect conclusions in exams. Being aware of these traps can significantly improve accuracy and save time.
Here are some common exam traps related to Equivalence Relations:
By being mindful of these common traps, you can approach equivalence relation problems with greater confidence and accuracy in your exams. Practice rigorous verification for each property!
Mastering these points will ensure you can confidently tackle problems related to equivalence relations in your exams. Always remember to check all three properties thoroughly when asked to prove a relation is an equivalence relation.
Determining whether a given relation is an equivalence relation is a fundamental problem in this topic. The approach involves systematically checking if the relation satisfies three crucial properties: reflexivity, symmetry, and transitivity. If all three hold, the relation is an equivalence relation. If even one fails, it is not.
Only if the relation satisfies all three properties (reflexivity, symmetry, and transitivity) can it be declared an equivalence relation. If even one property fails, the relation is not an equivalence relation.
Consider the set A = {all lines in a plane} and the relation R = {(L1, L2) : L1 is parallel to L2}.
Since R satisfies all three properties, R is an equivalence relation.
Mastering these steps ensures you can confidently tackle any problem on equivalence relations. Keep practicing!
Welcome to the "CBSE Focus Areas" for Equivalence Relations! This section will guide you through the aspects of equivalence relations that are most frequently tested and crucial for your CBSE board exams.
In CBSE, understanding the fundamental definition of an equivalence relation is paramount. A relation R on a set A is called an Equivalence Relation if it satisfies the following three properties:
a in set A, (a, a) β R. (Every element is related to itself).(a, b) β R, then (b, a) β R for all a, b β A. (If a is related to b, then b is related to a).(a, b) β R and (b, c) β R, then (a, c) β R for all a, b, c β A. (If a is related to b and b is related to c, then a is related to c).For CBSE exams, you will typically be given a relation and asked to prove whether it is an equivalence relation. This involves systematically checking each of these three properties.
Board exam questions on equivalence relations usually follow a predictable pattern. Your solution should clearly demonstrate the verification of each property:
a β A. Show that (a, a) satisfies the condition of the relation.(a, b) β R for arbitrary a, b β A. Using this assumption, derive that (b, a) β R.(a, b) β R and (b, c) β R for arbitrary a, b, c β A. Using these two assumptions, derive that (a, c) β R.Each step should be well-justified using the definition of the given relation. Avoid skipping steps, as partial marks are often awarded for correctly showing individual properties.
A crucial concept often paired with equivalence relations in CBSE is Equivalence Classes. If R is an equivalence relation on a set A, then for any element a β A, the set of all elements in A that are related to a is called the equivalence class of a, denoted by [a] or ¯a.
Mathematically, [a] = {x β A | (x, a) β R}.
CBSE Focus: You'll often be asked to find the equivalence classes for a given element or to list all distinct equivalence classes that partition the set. Remember that distinct equivalence classes are always disjoint and their union is the entire set A.
Consider the relation R on the set of integers Z defined by (a, b) β R if a - b is divisible by 3 (i.e., a β‘ b (mod 3)).
a β Z, a - a = 0, which is divisible by 3. So (a, a) β R.(a, b) β R, then a - b = 3k for some integer k. Then b - a = -3k = 3(-k), which is also divisible by 3. So (b, a) β R.(a, b) β R and (b, c) β R, then a - b = 3kβ and b - c = 3kβ for integers kβ, kβ. Adding these, (a - b) + (b - c) = 3kβ + 3kβ, which simplifies to a - c = 3(kβ + kβ). This implies a - c is divisible by 3. So (a, c) β R.Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
Equivalence Classes for this example:
[0] = {..., -6, -3, 0, 3, 6, ...} (all integers divisible by 3)[1] = {..., -5, -2, 1, 4, 7, ...} (all integers leaving a remainder of 1 when divided by 3)[2] = {..., -4, -1, 2, 5, 8, ...} (all integers leaving a remainder of 2 when divided by 3)These three equivalence classes partition the set of integers Z.
For CBSE, the emphasis is on:
For JEE Main, while the definitions are the same, questions might be more abstract, involve properties of equivalence classes, or combine concepts with functions. However, a strong understanding of the CBSE-level proofs is foundational for JEE as well.
Mastering these direct applications and clear proofs will ensure you score well on equivalence relation questions in your CBSE board exams. Keep practicing!
Understanding Equivalence Relations is a foundational concept in set theory, often tested in JEE Main for its direct application of logical reasoning and properties. While seemingly straightforward, JEE questions often introduce subtle complexities, requiring a thorough grasp of definitions.
An equivalence relation 'R' on a non-empty set 'A' must satisfy three crucial properties:
JEE Main questions on equivalence relations generally fall into these categories:
Mastering equivalence relations for JEE involves not just knowing the definitions but also developing the agility to apply them to various contexts, especially those involving number theory or abstract algebra. Practice systematically checking all three properties for any given relation.
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aRa for all a in the set.aRb, then bRa must hold.aRb and bRc, then aRc must hold.aRb if a - b is a multiple of 3. A student might quickly verify reflexivity (a - a = 0, a multiple of 3) and symmetry (if a - b is a multiple of 3, then b - a = -(a - b) is also a multiple of 3). They might then assume transitivity holds without rigorous proof, just because the first two properties 'feel right' for an equivalence relation.aRb if a - b is a multiple of 3: a β Z, a - a = 0, which is a multiple of 3. So, aRa holds.aRb, then a - b = 3k for some integer k. Then b - a = -3k = 3(-k), which is also a multiple of 3. So, bRa holds.aRb and bRc, then a - b = 3kβ and b - c = 3kβ for some integers kβ, kβ. Adding these equations: (a - b) + (b - c) = 3kβ + 3kβ, which simplifies to a - c = 3(kβ + kβ). Since kβ + kβ is an integer, a - c is a multiple of 3. So, aRc holds.Relation: R on integers Z defined by R = {(a,b) | a - b is divisible by 3}.
Student's Mistake:
Assume (a,b) β R, so a - b = 3k for some integer k.
To check if (b,a) β R, we need b - a to be divisible by 3.
A student might erroneously think: "Since b - a = -(a - b) = -3k, and -3k is a negative number (if k is positive), it doesn't directly look like '3 times an integer' in the positive sense, so maybe it's not divisible by 3." This overlooks that 'divisible by 3' includes negative multiples.
Relation: R on integers Z defined by R = {(a,b) | a - b is divisible by 3}.
Correct Check for Symmetry:
Consider a relation R on the set of integers Z defined by `R = {(a, b) | a - b` is a positive integer}.
Incorrect Symmetry Check: A student might reason: 'If `a - b` is positive, then `b - a` is just the negative of that, so it's somehow related. Therefore, R is symmetric.' This skips the crucial check if `b - a` *also* fits the condition of being a 'positive integer'. For example, if `(5, 2) β R` because `5 - 2 = 3` (a positive integer), the student might wrongly assume `(2, 5) β R`.
For `R = {(a, b) | a - b` is a positive integer}` on Z:
Thus, the relation R is not symmetric. This demonstrates that reversing the elements changes the 'sign' of the difference, invalidating the original condition.
Relation: R on real numbers, aRb if |a - b| < 1.
Student's Error: Transitivity is *assumed* ('if a is close to b and b close to c, then a close to c'). This approximation incorrectly concludes R is an equivalence relation.
For R: aRb if |a - b| < 1:
Hence, R is not transitive, and not an equivalence relation.
a R b if a - b is an even number.a - b is even and b - c is even, then a - c must also be even because adding two even numbers always gives an even number. So, it's transitive."(a-b) + (b-c) = a-c and substituting 2k for even numbers) required for a rigorous proof in CBSE 12th exams. It's an intuitive understanding without formal backing.*a R b if a - b is even.a, a - a = 0. Since 0 is an even number, (a, a) β R. Thus, R is reflexive.(a, b) β R. This means a - b is an even number. So, a - b = 2k for some integer k. Then, b - a = -(a - b) = -2k = 2(-k). Since -k is an integer, b - a is even. Thus, (b, a) β R. Hence, R is symmetric.(a, b) β R and (b, c) β R. This means a - b is even and b - c is even. So, a - b = 2kβ and b - c = 2kβ for some integers kβ and kβ. Adding these equations: (a - b) + (b - c) = 2kβ + 2kβ. This simplifies to a - c = 2(kβ + kβ). Since kβ + kβ is an integer, a - c is even. Thus, (a, c) β R. Hence, R is transitive.aRb iff a-b is a multiple of n (or similar conditions like a-b is an even number) are not symmetric. They get confused by the negative sign when considering b-a, failing to recognize that if a number is a multiple of n, its negative is also a multiple of n. b-a = -(a-b) but then incorrectly assume `-(multiple of n)` is not a multiple of `n` due to the presence of the negative sign. aRb holds, meaning a-b = nk for some integer k. b-a. We know `b-a = -(a-b)`. b-a = -(nk) = n(-k). k is an integer, -k is also an integer. Therefore, b-a is explicitly shown to be a multiple of n, proving bRa holds. Z defined by aRb iff a-b is a multiple of 5.aRb βΉ a-b = 5k for some integer k.bRa, we need b-a to be a multiple of 5. We have b-a = -(a-b) = -5k.Z defined by aRb iff a-b is a multiple of 5.aRb βΉ a-b = 5k for some integer k.bRa, consider b-a. We know that b-a = -(a-b).a-b = 5k, we get b-a = -(5k) = 5(-k).k is an integer, -k is also an integer. Thus, b-a is a multiple of 5.bRa holds, and the relation R is symmetric.x is a multiple of n if x = nk for *any integer* k (positive, negative, or zero).b-a = -(a-b). If a-b is a multiple of n, then -(a-b) is inherently also a multiple of n.So, '1 meter' R '100 cm' is TRUE. The student recognizes the underlying physical equivalence despite different units.
Let A = {1, 2, 3} and R = {(1, 2), (2, 3)}.
Student's thought process: 'I need to check (1, 3). (1, 3) is not in R, so R is not transitive.'
Error: While the conclusion is correct, the underlying 'calculation understanding' is incomplete. The student correctly identified the needed (a,c) but might not have systematically identified all (a,b) and (b,c) chains or fully articulated why (1,3) is the ONLY pair to check. If there were other chains, they might be missed. This informal approach risks errors in more complex problems.
Let A = {1, 2, 3} and R = {(1, 2), (2, 3), (1, 3)}.
Systematic Verification:
1. Identify a pair (a, b) from R: Let (a, b) = (1, 2).
2. Identify pairs (b, c) from R where 'b' is 2: Only (2, 3) is present.
3. Form the chain: (1, 2) β R and (2, 3) β R.
4. Deduce (a, c): This implies we need (1, 3) to be in R.
5. Check R: We see that (1, 3) β R.
6. Since this is the only chain of the form (a,b) and (b,c), and the condition holds, R is transitive.
The correct equivalence classes, based on parity, are:
Let S = {1 meter, 100 centimeters, 2 meters}
Relation R defined on S such that (x, y) β R if x = y (numerically, as interpreted by the student).
Student's incorrect thought process:
To check symmetry, consider (1 meter, 100 centimeters). If this pair is in R, then (100 centimeters, 1 meter) must also be in R.
The student sees '1' and '100'. Since 1 β 100, they conclude that (1 meter, 100 centimeters) β R. If they mistakenly assume the relation meant strict numerical equality without unit conversion, they might conclude R is not an equivalence relation because a pair like (1m, 100cm) fails the check, or they might incorrectly identify elements for partitioning.
Let S = {1 meter, 100 centimeters, 2 meters}
Relation R defined on S such that (x, y) β R if x and y represent the same physical length.
Correct approach:
First, it's helpful to realize that '100 centimeters' is physically equivalent to '1 meter'. So, the distinct physical lengths in the set are {1 meter, 2 meters}.
Thus, R is an equivalence relation. The critical step here is understanding 'same physical length' implies unit consistency before comparison, not just numerical value.
To avoid this mistake, adhere strictly to the definitions and perform careful algebraic manipulation for each property:
Let R be a relation on the set of integers Z defined by `(a, b) β R` if `a - 2b` is an even integer.
Student's Incorrect Check for Symmetry:
"If `a - 2b` is an even integer, then `b - 2a` must also be an even integer because it's just a rearrangement of terms. So, R is symmetric."
(This statement is a common logical leap, ignoring the critical effect of the coefficient '2' on the variables 'a' and 'b' when swapped.)
Let R be a relation on the set of integers Z defined by `(a, b) β R` if `a - 2b` is an even integer.
Correct Check for Symmetry:
Assume `(a, b) β R`, which means `a - 2b = 2k` for some integer `k` (since it's an even integer).
We need to check if `(b, a) β R`, i.e., if `b - 2a` is an even integer.
From `a - 2b = 2k`, we can express `a` as `a = 2k + 2b`.
Now substitute this expression for `a` into `b - 2a`:
`b - 2a = b - 2(2k + 2b)`
` = b - 4k - 4b`
` = -3b - 4k`
For this expression `(-3b - 4k)` to be an even integer, `-3b` must be even (since `-4k` is always even). This implies `b` must be an even integer. However, the initial condition `a - 2b = 2k` does not require `b` to be even.
Counterexample:
Let `a=4` and `b=1`. Then `a - 2b = 4 - 2(1) = 2`, which is an even integer. So, `(4,1) β R`.
Now, check if `(1,4) β R`. We need `b - 2a` to be even: `1 - 2(4) = 1 - 8 = -7`, which is an odd integer. So, `(1,4) β R`.
Since `(4,1) β R` but `(1,4) β R`, the relation R is NOT symmetric.
Wrong Thought Process:
Correct Approach:
Conclusion: Not an equivalence relation (fails symmetry).
To correctly analyze an equivalence relation involving physical quantities, always:
Consider a set of length measurements, S = { (1, m), (0.01, km) }. A student defines a relation R on S where ((v1, u1), (v2, u2)) β R if the physical lengths are equal.
To check if ((1, m), (0.01, km)) β R, the student incorrectly assumes: "1 km = 100 m".
Based on this error, they calculate: 0.01 km = 0.01 * 100 m = 1 m.
They then falsely conclude that ((1, m), (0.01, km)) β R because 1m = 1m. This leads to an incorrect understanding of the equivalence classes.
Using the same set S = { (1, m), (0.01, km) } and relation R:
This approach correctly identifies that (1, m) and (0.01, km) are not equivalent, preventing errors in defining equivalence classes or verifying properties of the relation.
a R b if |a - b| ≤ 2.a=1, b=2, then |1-2|=1 ≤ 2 (1 R 2).b=2, c=3, then |2-3|=1 ≤ 2 (2 R 3).|1-3|=2 ≤ 2 (1 R 3), the student might conclude it's transitive because these examples work, without looking for a counterexample or proving it generally.a R b if |a - b| ≤ 2, let's rigorously check transitivity:a R b and b R c. This means |a - b| ≤ 2 and |b - c| ≤ 2.a R c, i.e., |a - c| ≤ 2.a = 1, b = 3, c = 5.|a - b| = |1 - 3| = 2 ≤ 2 (so 1 R 3 holds)|b - c| = |3 - 5| = 2 ≤ 2 (so 3 R 5 holds)a R c:|a - c| = |1 - 5| = 4.4 ¬≤ 2, 1 R 5 does not hold. Thus, the relation is not transitive, and therefore, not an equivalence relation.Students often fail to grasp the fundamental principle that an equivalence relation on a set always partitions the set into disjoint, non-empty equivalence classes, and conversely, any partition of a set defines an equivalence relation. They might propose 'partitions' with overlapping subsets or incorrectly assume any arbitrary division of a set defines an equivalence relation. This leads to errors in constructing or identifying equivalence relations and their corresponding partitions, a crucial concept for JEE Advanced.
Let A = {1, 2, 3, 4}. A student proposes a 'partition' as P = {{1,2}, {2,3}, {4}}. This is not a valid partition because the subsets {1,2} and {2,3} overlap (element 2 is common). Therefore, this cannot correspond to an equivalence relation.
Consider the set A = {1, 2, 3, 4}. Let R be a relation on A defined as (a,b) β R if 'a' and 'b' have the same parity. The relation is R = {(1,1), (2,2), (3,3), (4,4), (1,3), (3,1), (2,4), (4,2)}.
This is an equivalence relation. The equivalence classes are:
The distinct equivalence classes are {1,3} and {2,4}. These form a valid partition of A since {1,3} βͺ {2,4} = A and {1,3} β© {2,4} = β .
500 cmΒ² to mΒ². They incorrectly multiply 500 by (1/100), getting 5 mΒ². Or, when calculating power, they use Force in Newtons (SI) and velocity in cm/s (CGS) directly, leading to a dimensionally inconsistent result.500 cmΒ² to mΒ²: 1 m = 100 cm 1 mΒ² = (100 cm)Β² = 10β΄ cmΒ² 500 cmΒ² = 500 cmΒ² * (1 mΒ² / 10β΄ cmΒ²) = 0.05 mΒ².F = 10 N and velocity v = 20 cm/s, for Power P = F * v, first convert v to SI units: v = 20 cm/s = 0.2 m/s. Then P = 10 N * 0.2 m/s = 2 Watts.(a, a) ∈ R for every single element ‘a’ in A.(a, b) ∈ R, then (b, a) ∈ R for all elements ‘a, b’ in A.(a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R for all elements ‘a, b, c’ in A.a in set A, (a,a) must be in relation R.(a,b) is in R, then (b,a) must also be in R.(a,b) is in R and (b,c) is in R, then (a,c) must also be in R.(a,b) satisfies a condition (e.g., a-b is an even integer), (b,a) (i.e., b-a) does not, simply due to a sign change, thereby failing to establish symmetry. Similar errors occur in transitive checks through incorrect algebraic manipulation. (a,b) β R, carefully check if (b,a) β R. If a-b satisfies property P, verify if b-a = -(a-b) also satisfies P. For many common properties (e.g., 'is even,' 'is a multiple of n'), if x satisfies P, then -x also satisfies P.(a,b) β R and (b,c) β R, use precise algebraic steps to deduce if (a,c) β R, paying attention to sign changes and variable interactions.R on integers Z: a R b if a - b is an even integer.a - b = 2k, then b - a = -2k. Since -2k is negative, it's not an even integer. So, R is not symmetric.' (Mistake: Even numbers can be negative.)R on integers Z: a R b if a - b is an even integer.a R b, then a - b = 2k for some integer k. For b R a, we check b - a = -(a - b) = -(2k) = -2k. Since k is an integer, -k is also an integer, so -2k is an even integer. Thus, b R a, and R is symmetric.'Scenario: Let set A = {1, 2, 3} and a relation R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}.
Student's Hasty Check:
Conclusion (Wrong): R is an equivalence relation.
Rigorous Check of the same R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)} on A = {1, 2, 3}:
Conclusion (Correct): Since the transitive property fails for the pair (1,2) and (2,3), R is NOT an equivalence relation.
A student's wrong approach:
Conclusion: Since the relation R is not symmetric, it is not an equivalence relation.
Remember, an equivalence relation is a fundamental concept for JEE and CBSE, requiring absolute precision in verification.
Students often misinterpret the logical implication (P → Q) fundamental to defining symmetric and transitive properties. They might incorrectly conclude a relation is not symmetric or transitive when the 'if' part of the condition (P) is not met for any relevant elements, rather than understanding that the implication is vacuously true in such cases. This leads to errors in determining if a relation is an equivalence relation.
For a relation R on set A:
Let A = {1, 2, 3}. R = {(1,1), (2,2), (3,3), (1,2)}.
Student's reasoning for Transitivity: "Since (1,2) ∈ R, we need to check for (2,c) ∈ R. There are no pairs (2,c) other than (2,2). This means we cannot form a chain (a,b), (b,c) where b=2 and c ≠ 2. Therefore, R is not transitive."
This is incorrect. The absence of such chains does not mean it's not transitive; it means the condition for checking transitivity is not met, so it holds vacuously.
Let A = {1, 2, 3}. R = {(1,1), (2,2), (3,3), (1,2)}.
Observation: R is reflexive and transitive, but not symmetric. Hence, it's not an equivalence relation.
(a, b) belongs to the relation R, then (b, a) must also belong to R for every pair (a, b) in R. If even one pair (a, b) exists for which (b, a) is not in R, the relation is not symmetric.(a, b) belongs to R and (b, c) belongs to R, then (a, c) must also belong to R for every such triplet (a, b, c). If you find even one instance where (a,b) ∈ R, (b,c) ∈ R but (a,c) ∉ R, the relation is not transitive.R on set A = {1, 2, 3, 4} defined as R = {(a, b) | a divides b}. A student might check for transitivity: (2,4) ∈ R and (4,8) is not possible here. Let's take (1,2) ∈ R and (2,4) ∈ R, then (1,4) ∈ R. They might conclude it's transitive just from this specific example, without a general proof or careful consideration of all valid (a,b) and (b,c) pairs.R on set A = {1, 2, 3} defined as R = {(1,1), (2,2), (3,3), (1,2), (2,1), (1,3)}.(x,y) ∈ R...).(a,b) and (b,c) in the relation, or incorrectly assuming transitivity holds simply because no such pairs exist, without proper justification. They might check only obvious or a few specific cases, leading to an incorrect conclusion about the entire relation. (a,b) β R AND (b,c) β R, and 'Q' is (a,c) β R. Students often rush, fail to systematically identify all such 'P' conditions, or forget that if 'P' is false (i.e., no such b exists or no such (a,b) and (b,c) pairs exist), the implication is vacuously true for that specific triplet (a,b,c), but this does not automatically mean the relation is transitive overall without thorough examination. For JEE Advanced, this can be a subtle trap in multi-correct questions. a, b, c in the set, if the ordered pairs (a,b) and (b,c) both belong to the relation R, then the ordered pair (a,c) must also belong to R. (a,b) β R.(a,b), identify all pairs (b,c) β R where the second element of the first pair matches the first element of the second.(a,c) is present in R. If even one such (a,c) is missing, the relation is not transitive.b can be found to 'connect' a and c through (a,b) and (b,c), then that specific triplet (a,b,c) vacuously satisfies the condition, but the entire relation needs to be checked exhaustively.Let set A = {1, 2, 3} and R = {(1, 2), (2, 1)}.
Student's thought process: 'I see (1,2) and (2,1). What about (1,1)? It's not there. Oh, but there's no (1,2) and (2,3) so no need to check (1,3). It must be transitive!'
Incorrect conclusion: R is transitive.
Let set A = {1, 2, 3} and R = {(1, 2), (2, 1)}.
Correct check:
(a,b) = (1,2) β R.(b,c) = (2,1) β R.a=1, b=2, c=1. According to transitivity, (a,c) = (1,1) must be in R.(1,1) β R.Correct conclusion: R is NOT transitive.
(a,b) and (b,c) that could potentially form a chain.(a,b) β R, (b,c) β R, but (a,c) β R.(a,b) and (b,c) pairs exist for a given b, then transitivity holds for that specific a,b,c. However, this must be considered for all possible a,b,c in the set, not just one instance.(a,a) β R.(b,a) β R.(a,c) β R.Mistake: This is a specific instance, not a general proof. Further, $le$ is not symmetric (e.g., $1 le 2$ but $2
otle 1$), so it's not an equivalence relation. All three properties must be *generally* proven.
All properties hold generally. Thus, R is an equivalence relation.
a ∈ A, it must be true that (a, a) ∈ R.(a, b) ∈ R for any a, b ∈ A, then it must also be true that (b, a) ∈ R.(a, b) ∈ R and (b, c) ∈ R for any a, b, c ∈ A, then it must also be true that (a, c) ∈ R.a - b). They often incorrectly conclude that because b - a = -(a - b), the condition for (b, a) is not met, especially for conditions like 'multiple of n' or 'even/odd'. This leads to an incorrect assessment of symmetry, making the entire classification of the relation erroneous. b - a = -(a - b) and link it back to the original condition.x is a multiple of n, then -x is also a multiple of n).a R b implies a condition for (a - b), explicitly show how b - a = -(a - b) still satisfies that condition. This means recognizing that if X meets a property (e.g., 'is a multiple of n'), then -X often also meets it (e.g., - (k * n) = (-k) * n). Relation R on Z: a R b if (a - b) is divisible by 5.
Student's Incorrect Symmetry Check:
"Assume a R b, so a - b = 5k for some integer k.
For b R a, we need to check if b - a is divisible by 5. We have b - a = -(a - b) = -5k.
Since -5k is negative, it's not divisible by 5. Therefore, R is not symmetric."
Relation R on Z: a R b if (a - b) is divisible by 5.
Correct Symmetry Check:
"Assume a R b. By definition, a - b = 5k for some integer k.
To check for b R a, we need to determine if b - a is divisible by 5.
We know that b - a = -(a - b).
Substituting the expression for (a - b), we get b - a = -(5k) = 5(-k).
Since k is an integer, -k is also an integer.
Therefore, b - a is a multiple of 5 (i.e., divisible by 5), which means b R a.
Hence, R is symmetric."
(a - b), explicitly demonstrate how (b - a) = -(a - b) satisfies the given condition.x is divisible by n if x = kn for any integer k (positive, negative, or zero).When an equivalence relation is defined on a set of elements that are physical quantities, the first and most critical step is to ensure all elements are expressed in a single, consistent system of units (e.g., all lengths in meters, all masses in kilograms). Only after this conversion can the relation's criteria be accurately applied to check for reflexivity, symmetry, and transitivity.
CBSE/JEE Tip: While less common in standard CBSE equivalence relation problems (which often use integers or abstract sets), JEE Advanced questions might implicitly test this by combining concepts from different subjects or using real-world scenarios.
Consider a set A = { '1 meter', '100 cm', '200 cm' } and a relation R defined as (x, y) β R if 'x and y represent the same length'.
Wrong Approach: Checking for reflexivity, a student might incorrectly think ( '1 meter', '100 cm' ) is not related to itself if they interpret 'same length' as 'same numerical value without unit conversion'. They might compare '1' and '100' directly, concluding 1 β 100, thus leading to a false negative for symmetry or even reflexivity if the elements were listed as '1m' and '100cm'.
For the same set A = { '1 meter', '100 cm', '200 cm' } and relation R defined as (x, y) β R if 'x and y represent the same length'.
Correct Approach: Convert all elements to a common unit, e.g., meters:
Now, the effective set for comparison is { '1 m', '1 m', '2 m' }. When checking if ( '1 meter', '100 cm' ) β R, we convert both to meters: '1m' and '1m'. Since 1m = 1m, they are related. This correct conversion ensures that properties like reflexivity (an element is related to itself), symmetry (if xRy then yRx), and transitivity (if xRy and yRz then xRz) are accurately verified.
a in set A, check if (a, a) belongs to the relation R.(a, b) in R, check if (b, a) also belongs to R.(a, b) and (b, c) in R, check if (a, c) also belongs to R.R on the set of integers Z defined as (a, b) β R if a < b.(a, a) β R means a < a, which is false. Not reflexive. (Students often confuse < with ≤).a < b, then b < a is false. Not symmetric.a < b and b < c, then a < c. This property holds.R on A = {1, 2, 3} as R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.(1, 1), (2, 2), (3, 3) β R. Holds for all a β A.(1, 2) β R β (2, 1) β R. Holds.(1, 2), (2, 1) β R β (1, 1) β R. Also (2, 1), (1, 2) β R β (2, 2) β R. Holds.R is an equivalence relation. This demonstrates a complete and systematic verification.JEE Advanced Tip: Always look for edge cases or counterexamples, especially for transitivity, as these are common traps.
Consider the relation R defined on integers Z where (a, b) β R if a - b β₯ 0.
Incorrect Symmetry Check: A student might reason: "If a - b β₯ 0, then swapping a and b to b - a must also be β₯ 0." This leads to an erroneous conclusion that R is symmetric. For example, if (5, 3) β R because 5 - 3 = 2 β₯ 0, they might assume (3, 5) β R without checking 3 - 5.
Consider the relation R defined on integers Z where (a, b) β R if a - b β₯ 0.
Correct Symmetry Check:
Assume (a, b) β R, so a - b β₯ 0.
For R to be symmetric, if (a, b) β R, then (b, a) β R must also hold. This means we need to check if b - a β₯ 0.
However, we know that b - a = -(a - b). Since a - b β₯ 0, it follows that -(a - b) β€ 0. Thus, b - a β€ 0.
For example, let a = 5, b = 3. Then (5, 3) β R because 5 - 3 = 2 β₯ 0.
But for (3, 5), we check 3 - 5 = -2, which is NOT β₯ 0. Therefore, (3, 5) β R.
Hence, R is NOT symmetric. This relation is a partial order, not an equivalence relation.
When dealing with equivalence relations on sets whose elements involve units (e.g., lengths, masses):
Let S be a set of length measurements, e.g., S = { (5m), (500cm), (0.005km) }. Define a relation R on S such that (lβ, lβ) β R if lβ represents the same physical length as lβ. A student attempts to check transitivity:
Using the same set S and relation R: (lβ, lβ) β R if lβ represents the same physical length as lβ. To correctly verify transitivity:
a β‘ b (mod n).[a] of an element a, systematically check *every* element x in the given set A. If (x, a) satisfies the relation's condition, then x belongs to [a]. Ensure all elements related to a are included. Then, identify all *distinct* equivalence classes; remember that [a] = [b] if a and b are related. A = {1, 2, ..., 8} and a relation R on A defined as (x, y) β R if x^2 - y^2 is a multiple of 5 (i.e., x^2 β‘ y^2 (mod 5)).[1] correctly identifies {1, 4, 6}. However, by miscalculating 8^2 - 1^2 = 63 and wrongly concluding 63 is a multiple of 5, they might erroneously include 8 in [1]. This leads to the incorrect class [1] = {1, 4, 6, 8}.R on A = {1, 2, ..., 8} where x^2 β‘ y^2 (mod 5):x^2 mod 5 for all x β A:1^2β‘1, 2^2β‘4, 3^2β‘4, 4^2β‘1, 5^2β‘0, 6^2β‘1, 7^2β‘4, 8^2β‘4.x^2 mod 5 value:[1] = {1, 4, 6} (elements where x^2 β‘ 1 (mod 5))[2] = {2, 3, 7, 8} (elements where x^2 β‘ 4 (mod 5))[5] = {5} (elements where x^2 β‘ 0 (mod 5))A.x mod n or f(x) mod n for all relevant elements in the set to minimize calculation errors.To correctly verify if a relation R on a set A is an equivalence relation, one must systematically and rigorously check all three properties for all relevant elements/pairs:
Always assume a property is NOT true until you can prove it for ALL cases, or find a single counterexample to disprove it.
Consider set A = {1, 2, 3} and relation R = {(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)}.
A student might verify:
Using the same set A = {1, 2, 3} and relation R = {(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)}:
Since R is not transitive, it is not an equivalence relation.
a R b if and only if a - b is a positive even number.a R b if and only if a - b is a positive even number.| Recommendation | JEE Main Specific Tip | |
|---|---|---|
| ✅ | Master Definitions: Thoroughly understand the precise definitions of reflexive, symmetric, and transitive properties. | Memorize the 'for all' (∀) and 'there exists' (∃) conditions within each definition. |
| ✅ | Systematic Verification: Always verify each property separately and explicitly. Do not assume or combine steps. | Allocate sufficient time to check transitivity carefully; it's a common trap. |
| ✅ | General Proofs: Use arbitrary elements (e.g., x, y, z) for proofs, not specific numbers, to ensure generality. | If unsure, try to construct a counter-example to disprove a property. If you can't, proceed with a formal proof. |
| ✅ | Transitivity Checklist: For transitivity, list all possible (a,b) and (b,c) pairs, and then verify the presence of (a,c). This prevents missing critical cases. | Critical for Negative Marking: An incorrect conclusion on equivalence relations due to 'approximate' checking can lead to significant loss of marks. Be precise! |
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