Let A={0,3,4,6,7,8,9,10} and R be the relation defined on A such that R={(x, y) ∈ A × A: x-y is odd positive integer or x-y=2}. The minimum number of elements that must be added to the relation R, so that it is a symmetric relation, is equal to ____________.

Solution

Maths sir · 4 steps

Every pair in R has x bigger than y, so no mirror pair exists yet: count R and that is the answer.

1. No reverse pairs. Both conditions make x - y positive, so every (x, y) in R has x > y. Its reverse (y, x) has y - x < 0, so it can never be in R. For symmetry, each pair in R needs exactly one new element added.
2. Odd positive differences. x - y is odd only when one number is odd and the other is even. A has 3 odd numbers {3, 7, 9} and 5 even numbers {0, 4, 6, 8, 10}. Each odd and even pair gives exactly one ordered pair with a positive difference, so there are 3 × 5 = 15 such pairs.
3. Difference exactly 2. Look for x and x - 2 both in A: (6, 4), (8, 6), (10, 8) and (9, 7). That gives 4 pairs. None of them repeat the first group, because a difference of 2 is even.

   |R| = 15 + 4 = 19

4. So. Adding the 19 reversed pairs makes R symmetric, and no smaller number works. The answer is 19.

The common slip: 15. Counting only the odd-difference pairs and forgetting the four pairs with x - y = 2 gives 15, not 19.

The idea to keep. When every pair in a relation has x > y, none of the reverse pairs can already be there. The number of pairs to add for symmetry is then the size of the relation itself.

Answer from the NTA final answer key. Solution by PrepWiser.