Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider the binary relation:S = {(x, y) | y ...
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Consider the binary relation:
S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}
S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}
Q. The reflexive transitive closure of S is
- a){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}
- b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}
- c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}
- d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0...
Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
Thus, option (B) matches the final set S.
Please comment below if you find anything wrong in the above post.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
Thus, option (B) matches the final set S.
Please comment below if you find anything wrong in the above post.
Most Upvoted Answer
Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0...
Reflexive Transitive Closure of a Binary Relation:
The reflexive transitive closure of a binary relation S is the smallest relation that contains S and is reflexive and transitive. In other words, it is the relation obtained by adding all missing reflexive and transitive pairs to the original relation S.
Given Relation:
S = {(x, y) | y = x 1 and x, y {0, 1, 2, ...}}
Step 1: Reflexive Closure:
To obtain the reflexive closure of S, we need to add all missing reflexive pairs. A pair (a, a) is reflexive if it satisfies the condition y = x + 1. Therefore, we need to add all pairs (a, a+1) to S.
The reflexive closure of S will be:
S' = {(x, y) | y = x + 1 and x, y {0, 1, 2, ...}}
Step 2: Transitive Closure:
To obtain the transitive closure of S', we need to add all missing transitive pairs. A pair (a, c) is transitive if there exists an intermediate element b such that (a, b) and (b, c) are in S'.
Let's consider an example to understand this. If (0, 1) and (1, 2) are in S', then (0, 2) should also be in S' because there exists an intermediate element 1 such that (0, 1) and (1, 2) are in S'.
Therefore, we need to add all pairs (a, c) to S' such that there exists an intermediate element b such that (a, b) and (b, c) are in S'.
The transitive closure of S' will be:
S'' = {(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}
Step 3: Reflexive Transitive Closure:
To obtain the reflexive transitive closure of S, we need to combine the reflexive closure and transitive closure obtained in the previous steps.
The reflexive transitive closure of S will be:
{(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}
Answer: Option B
Therefore, the correct answer is option B: {(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}.
The reflexive transitive closure of a binary relation S is the smallest relation that contains S and is reflexive and transitive. In other words, it is the relation obtained by adding all missing reflexive and transitive pairs to the original relation S.
Given Relation:
S = {(x, y) | y = x 1 and x, y {0, 1, 2, ...}}
Step 1: Reflexive Closure:
To obtain the reflexive closure of S, we need to add all missing reflexive pairs. A pair (a, a) is reflexive if it satisfies the condition y = x + 1. Therefore, we need to add all pairs (a, a+1) to S.
The reflexive closure of S will be:
S' = {(x, y) | y = x + 1 and x, y {0, 1, 2, ...}}
Step 2: Transitive Closure:
To obtain the transitive closure of S', we need to add all missing transitive pairs. A pair (a, c) is transitive if there exists an intermediate element b such that (a, b) and (b, c) are in S'.
Let's consider an example to understand this. If (0, 1) and (1, 2) are in S', then (0, 2) should also be in S' because there exists an intermediate element 1 such that (0, 1) and (1, 2) are in S'.
Therefore, we need to add all pairs (a, c) to S' such that there exists an intermediate element b such that (a, b) and (b, c) are in S'.
The transitive closure of S' will be:
S'' = {(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}
Step 3: Reflexive Transitive Closure:
To obtain the reflexive transitive closure of S, we need to combine the reflexive closure and transitive closure obtained in the previous steps.
The reflexive transitive closure of S will be:
{(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}
Answer: Option B
Therefore, the correct answer is option B: {(x, y) | y = x + 1 and x, y {0, 1, 2, ... }}.
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Community Answer
Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0...
Reflexive closure of a relation R on set S is the smallest reflexive relation which contains R.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
Thus, option (B) matches the final set S.
Please comment below if you find anything wrong in the above post.
If S = {(0, 1), (1, 2)} , we make it reflexive by taking its union with set {(0, 0), (1, 1), (2, 2)}. Thus, reflexive closure of S = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)}.
Now transitive closure is defined as smallest transitive relation which contains S.
We check where does it violate property of transitivity then add appropriate pair. We have (0, 1) and (1, 2) but not (0, 2). So, S = {(0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2)} now.
Thus, option (B) matches the final set S.
Please comment below if you find anything wrong in the above post.
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Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer?
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Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer?.
Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer?.
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Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Consider the binary relation:S = {(x, y) | y = x+1 and x, y ∈ {0, 1, 2, ...}}Q.The reflexive transitive closure of S isa){(x, y) | y > x and x, y ∈ {0, 1, 2, ... }}b){(x, y) | y ≥ x and x, y ∈ {0, 1, 2, ... }}c){(x, y) | y < x and x, y ∈ {0, 1, 2, ... }}d){(x, y) | y ≤ x and x, y ∈ {0, 1, 2, ... }}Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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