Discrete Mathematics previous year-GATE 2025 Equivalence relation Let R and S be any two equivalence.

Discrete Mathematics previous year-GATE 2025 Equivalence relation Let R and S be any two equivalence.



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It looks like you are looking for Discrete Mathematics – Equivalence Relations – Previous Year GATE Questions related to GATE 2025. Here’s a conceptual breakdown along with a sample problem:

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 Equivalence Relation – Definition

A relation R on a set S is called an Equivalence Relation if it satisfies the following three properties:

  1. Reflexive: aRaaRa for all a∈Sa \in S.
  2. Symmetric: If aRbaRb, then bRabRa for all a,b∈Sa, b \in S.
  3. Transitive: If aRbaRb and bRcbRc, then aRcaRc for all a,b,c∈Sa, b, c \in S.

 Example GATE Question on Equivalence Relations

Question:
Let R and S be two equivalence relations on a set AA. Consider the relation R∩SR \cap S (intersection of R and S). Is R∩SR \cap S also an equivalence relation?

Solution Approach:

  • Reflexive: Since both RR and SS are reflexive, (a,a)(a, a) belongs to both RR and SS, so it must belong to R∩SR \cap S.
  • Symmetric: If (a,b)∈R∩S(a, b) \in R \cap S, then (a,b)∈R(a, b) \in R and (a,b)∈S(a, b) \in S. Since both RR and SS are symmetric, (b,a)∈R(b, a) \in R and (b,a)∈S(b, a) \in S, so (b,a)∈R∩S(b, a) \in R \cap S.
  • Transitive: If (a,b)∈R∩S(a, b) \in R \cap S and (b,c)∈R∩S(b, c) \in R \cap S, then (a,b)(a, b) and (b,c)(b, c) belong to both RR and SS. Since both RR and SS are transitive, (a,c)∈R(a, c) \in R and (a,c)∈S(a, c) \in S, so (a,c)∈R∩S(a, c) \in R \cap S.
  • Conclusion: Since R∩SR \cap S satisfies all three properties, it is also an equivalence relation.
  • Answer: Yes, R∩SR \cap S is always an equivalence relation.

Would you like more GATE previous year questions with solutions or a short trick to solve equivalence relation problems faster?

Discrete Mathematics previous year-GATE 2025 Equivalence relation Let R and S be any two equivalence.

Discrete Mathematics for Computer Science

In the context of GATE 2025 and Discrete Mathematics, understanding how operations on equivalence relations behave is crucial. Let’s delve into the properties of the intersection and union of two equivalence relations.


✅ Intersection of Two Equivalence Relations (R ∩ S)

Statement: If RR and SS are equivalence relations on a set AA, then their intersection R∩SR \cap S is also an equivalence relation.

Explanation:

An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. Let’s verify these for R∩SR \cap S:

  1. Reflexivity:

    • Since both RR and SS are reflexive, for every a∈Aa \in A, (a,a)∈R(a, a) \in R and (a,a)∈S(a, a) \in S.

    • Therefore, (a,a)∈R∩S(a, a) \in R \cap S, ensuring reflexivity.

  2. Symmetry:

    • If (a,b)∈R∩S(a, b) \in R \cap S, then (a,b)∈R(a, b) \in R and (a,b)∈S(a, b) \in S.

    • Since both RR and SS are symmetric, (b,a)∈R(b, a) \in R and (b,a)∈S(b, a) \in S.

    • Thus, (b,a)∈R∩S(b, a) \in R \cap S, confirming symmetry.

  3. Transitivity:

    • If (a,b)∈R∩S(a, b) \in R \cap S and (b,c)∈R∩S(b, c) \in R \cap S, then (a,b),(b,c)∈R(a, b), (b, c) \in R and (a,b),(b,c)∈S(a, b), (b, c) \in S.

    • Since both RR and SS are transitive, (a,c)∈R(a, c) \in R and (a,c)∈S(a, c) \in S.

    • Therefore, (a,c)∈R∩S(a, c) \in R \cap S, establishing transitivity.

Conclusion: The intersection R∩SR \cap S retains all three properties, making it an equivalence relation.


❌ Union of Two Equivalence Relations (R ∪ S)

Statement: The union R∪SR \cup S of two equivalence relations RR and SS on a set AA is not necessarily an equivalence relation.

Explanation:

While the union may preserve reflexivity and symmetry under certain conditions, transitivity is not guaranteed. Let’s illustrate this with an example:

Example:

Let A={1,2,3}A = \{1, 2, 3\}

Define:

  • R={(1,1),(2,2),(3,3),(1,2),(2,1)}R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}

  • S={(1,1),(2,2),(3,3),(2,3),(3,2)}S = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}

Then:

  • R∪S={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}R \cup S = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}

Now, check for transitivity:

  • (1,2)∈R∪S(1,2) \in R \cup S and (2,3)∈R∪S(2,3) \in R \cup S, but (1,3)∉R∪S(1,3) \notin R \cup S

Conclusion: Since transitivity fails, R∪SR \cup S is not an equivalence relation.


📌 Summary Table

Operation Resulting Relation Equivalence Relation?
R∩SR \cap S Intersection ✅ Always
R∪SR \cup S Union ❌ Not Always

📚 Further Reading and Practice

For a deeper understanding and practice problems related to equivalence relations in GATE, consider the following resources:

  • GATEOverflow Discussion on Equivalence Relations: A detailed discussion on GATE CSE 2005 Question 42, which explores the properties of intersection and union of equivalence relations.

  • BYJU’S Explanation on Equivalence Relations: Comprehensive notes on the definition, properties, and examples of equivalence relations.

  • YouTube Lecture on Equivalence Relations: A video lecture discussing equivalence relations with examples, beneficial for visual learners.

Feel free to explore these materials to reinforce your understanding of equivalence relations and their properties.

Discrete Mathematics previous year-GATE 2025 Equivalence relation Let R and S be any two equivalence.

Discrete Mathematics and Its Applications, Eighth Edition



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