The relation R defined on the set A = {l, 2,3,4,5} by P = {(x, y) : |x2 —y2|

Question

The relation R defined on the set A = {l, 2,3,4,5} by P = {(x, y) : |x2 &mdash;y2| < 16} is given bya){(1,1), (2,1), (3,1), (4,1), (2,3)}b){(2,2), (3,2), (4,2), (2,4)}c){(3,3), (4,3), (5,4), (3,4)}d)none of theseCorrect answer is option 'D'. Can you explain this answer?

Answer

Answer

The relation R defined on the set A ={1,2,3,4,5} by P = {(x, y) : |x^2 - y| = 2} is not an equivalence relation.

To determine if a relation is an equivalence relation, we need to check three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity: Does every element of A relate to itself?
In this case, we would need to check if (1,1), (2,2), (3,3), (4,4), and (5,5) are in P. However, if we substitute these values into the equation |x^2 - y| = 2, we get:
|1^2 - 1| = 0, which is not equal to 2.
Therefore, the relation P is not reflexive.

2. Symmetry: If (x, y) is in P, is (y, x) also in P?
Let's consider an example: (1,3) is in P because |1^2 - 3| = 2. However, if we check if (3,1) is in P, we get |3^2 - 1| = 8, which is not equal to 2.
Therefore, the relation P is not symmetric.

3. Transitivity: If (x, y) and (y, z) are in P, is (x, z) also in P?
Let's consider an example: (1,3) is in P because |1^2 - 3| = 2. (3,5) is also in P because |3^2 - 5| = 2. However, if we check if (1,5) is in P, we get |1^2 - 5| = 4, which is not equal to 2.
Therefore, the relation P is not transitive.

Since the relation P does not satisfy all three properties of an equivalence relation, it is not an equivalence relation.

Answer

The ordered pair in options a and b satisfy the given relation.Then, options a and b are correct answer.

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