Let A={-3,-2,-1,0,1,2,3} and R be a relation on A defined by x R y if and only if 2 x-y ∈{0,1}. Let l be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+n is equal to:

Solution

Maths sir · 5 steps

List the pairs of R by hand first, then count what is missing for reflexive and for symmetric.

1. Two lines. 2x - y ∈ {0, 1} means y = 2x or y = 2x - 1. Try every x in A and keep only the pairs where y also lies in A.

   x	y = 2x	y = 2x - 1	Pairs in R
   -3, -2, 3	-6, -4, 6 (not in A)	-7, -5, 5 (not in A)	none
   -1	-2	-3	(-1,-2), (-1,-3)
   0	0	-1	(0,0), (0,-1)
   1	2	1	(1,2), (1,1)
   2	4 (not in A)	3	(2,3)

2. Count l. R has 2 + 2 + 2 + 1 = 7 pairs, so l = 7.
3. Reflexive needs. All 7 pairs (a, a) for a ∈ A must be present. R already has (0,0) and (1,1), so m = 7 - 2 = 5.
4. Symmetric needs. Each of the 5 non-diagonal pairs (-1,-2), (-1,-3), (0,-1), (1,2), (2,3) needs its reverse, and none of those reverses is in R. So n = 5.
5. So. l + m + n = 7 + 5 + 5 = 17, which is option (a).

The idea to keep. For small sets, write out the relation pair by pair. Reflexive: add the missing (a,a) for every element. Symmetric: add the reverse of every off-diagonal pair that lacks one.

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