Functions Question 471
Question: Let R={$(x,y):x,y\in N $ and $ x^{2}-4xy+3y^{2}=0$}, Where N is the set of all natural numbers. Then the relation R is:
Options:
A) Reflexive and symmetric.
B) Symmetric and transitive.
C) Reflexive and symmetric.
D) Reflexive but neither symmetric nor transitive.
Show Answer
Answer:
Correct Answer: A
Solution:
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First, we need to find the pairs (x,y) that satisfy the equation $x^2 - 4xy + 3y^2 = 0$ where x and y are natural numbers.
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This equation can be factored as: (x - 3y)(x - y) = 0
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For this to be true, either x - 3y = 0 or x - y = 0
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Case 1: If x - 3y = 0, then x = 3y. The smallest natural number solution is (3,1). Case 2: If x - y = 0, then x = y. This gives us solutions like (1,1), (2,2), (3,3), etc.
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Now, let’s check each property:
Reflexive: A relation R on a set A is reflexive if (a,a) ∈ R for every a ∈ A. This is true for our relation as (1,1), (2,2), (3,3), etc. are all in the set R.
Symmetric: A relation R on a set A is symmetric if (a,b) ∈ R implies (b,a) ∈ R for all a,b ∈ A.
- This is not true for our relation. For example, (3,1) ∈ R, but (1,3) ∉ R.
Transitive: A relation R on a set A is transitive if (a,b) ∈ R and (b,c) ∈ R imply (a,c) ∈ R for all a,b,c ∈ A.
- This is not true for our relation. For example, (3,1) ∈ R and (1,1) ∈ R, but (3,1) ∉ R.
- Therefore, the relation R is reflexive, but neither symmetric nor transitive.
The correct answer is A) Reflexive but neither symmetric nor transitive.
BETA
