Let A={1,2,3, …, 20}. Let R₁ and R₂ two relation on A such that
R₁={(a, b): b is divisible by a}
R₂={(a, b): a is an integral multiple of b}.
Then, number of elements in R₁-R₂ is equal to _____________.

Solution

Maths sir · 4 steps

Count the pairs where a divides b, then remove the diagonal, the only pairs that also sit in R₂.

1. Read both relations. R₁ has (a, b) when a divides b. R₂ has (a, b) when b divides a. A pair is in both only if a | b and b | a, which for positive integers forces a = b.
2. Count R1. For a fixed a, the choices of b are the multiples of a up to 20. There are ⌊ 20/a ⌋ of them.

   Value of a	Multiples of a in A	Pairs
   1	1, 2, …, 20	20
   2	2, 4, …, 20	10
   3	3, 6, …, 18	6
   4	4, 8, 12, 16, 20	5
   5	5, 10, 15, 20	4
   6	6, 12, 18	3
   7 to 10	a and 2a	2 × 4 = 8
   11 to 20	only a itself	1 × 10 = 10

3. Remove the overlap. Adding the last column gives n(R₁) = 66. The overlap R₁ ∩ R₂ is {(1,1), (2,2), …, (20,20)}, which has 20 pairs.

   n(R₁ - R₂) = n(R₁) - n(R₁ ∩ R₂) = 66 - 20

4. So. The number of elements in R₁ - R₂ is 46.

The common slip: 66. Counting all pairs with a | b and stopping there gives 66. The 20 diagonal pairs (a, a) also lie in R₂ and must be removed.

The idea to keep. For R₁ - R₂, count R₁ and subtract only what it shares with R₂. If a | b and b | a, then a = b, so the overlap is just the diagonal.

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