Let A={1,2,3, … … … …, 100}. Let R be a relation on A defined by (x, y) ∈ R if and only if 2 x=3 y. Let R₁ be a symmetric relation on A such that R ⊂ R₁ and the number of elements in R₁ is n. Then, the minimum value of n is _________.

Solution

Maths sir · 5 steps

List the pairs in R, then add each pair's mirror image. That is the smallest symmetric relation containing R.

1. Pairs in R. From 2x = 3y, x must be a multiple of 3 and y a multiple of 2. So x = 3k, y = 2k, and every pair in R has the form (3k, 2k).
2. Count them. We need 3k ≤ 100, so k = 1, 2, …, 33. Then y = 2k ≤ 66 also stays in A. So R has 33 pairs: (3,2), (6,4), …, (99,66).
3. Make it symmetric. A symmetric relation containing (3k, 2k) must also contain (2k, 3k). The smallest such R₁ is R plus these 33 reversed pairs, and nothing else.
4. No overlap. No reversed pair is already in R. A pair (2k, 3k) in R would need 2(2k) = 3(3k), which gives k = 0. No pair has x = y either, so every pair has a distinct partner.
5. So. nₘᵢₙ = 33 + 33 = 66.

The common slip: 33. Counting only the pairs of R and not adding their reverses gives 33. A symmetric R₁ must also contain each (2k, 3k).

The idea to keep. The smallest symmetric relation containing R is R together with all its reversed pairs. Count R, double it, then subtract the pairs that are their own reverse, such as (a, a).

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