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Maths Question 1 – JEE-MAIN 2026

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Consider the relation R on the set {−2,−1,0,1,2} defined by (a,b)∈R if and only if 1+ab>0. Then, among the statements : I. The number of elements in R is 17 II. R is an equivalence relation

A relation R on a set S is a subset of S×S. An equivalence relation must satisfy reflexivity, symmetry, and transitivity.

Step 1: Determine the elements of the relation R (Statement I)✦ Active

The set is S={−2,−1,0,1,2}. The relation is (a,b)∈R if 1+ab>0, which means ab>−1. We list all pairs (a,b) from S×S satisfying ab>−1:

For a=−2: ab∈{4,2,0,−2,−4}. Pairs satisfying ab>−1: (−2,−2),(−2,−1),(−2,0) (3 pairs).

For a=−1: ab∈{2,1,0,−1,−2}. Pairs satisfying ab>−1: (−1,−2),(−1,−1),(−1,0) (3 pairs).

For a=0: ab∈{0,0,0,0,0}. Pairs satisfying ab>−1: (0,−2),(0,−1),(0,0),(0,1),(0,2) (5 pairs).

For a=1: ab∈{−2,−1,0,1,2}. Pairs satisfying ab>−1: (1,0),(1,1),(1,2) (3 pairs).

For a=2: ab∈{−4,−2,0,2,4}. Pairs satisfying ab>−1: (2,0),(2,1),(2,2) (3 pairs).

The total number of elements in R is 3+3+5+3+3=17. Thus, Statement I is true.

Step 2: Check for reflexivity and symmetry (Statement II)○ Expand

**Reflexivity:** For any a∈S, we need (a,a)∈R, which means 1+a2>0. Since a∈{−2,−1,0,1,2}, a2∈{0,1,4}. In all cases, 1+a2 is 1,2, or 5, which are all greater than 0. So, R is reflexive.

**Symmetry:** If (a,b)∈R, then 1+ab>0. Since ab=ba, it follows that 1+ba>0, so (b,a)∈R. Thus, R is symmetric.

💡 Teacher's Secret Hint

Remember that ab=ba for real numbers, which simplifies the symmetry check.

Step 3: Check for transitivity (Statement II)○ Expand

**Transitivity:** If (a,b)∈R and (b,c)∈R, then (a,c)∈R. This means if 1+ab>0 and 1+bc>0, then 1+ac>0. Let's test with a counterexample:

Let a=−2, b=0, c=1 from the set S.

1. For (a,b)=(−2,0): 1+(−2)(0)=1>0. So (−2,0)∈R.

2. For (b,c)=(0,1): 1+(0)(1)=1>0. So (0,1)∈R.

3. Now check (a,c)=(−2,1): 1+(−2)(1)=1−2=−1. Since −1gtr0, (−2,1)∉R.

Since transitivity fails, R is not an equivalence relation. Thus, Statement II is false.

Based on the analysis, only Statement I is true.

💡 Teacher's Secret Hint

To disprove transitivity, you only need one counterexample. Look for elements that might cause the product ac to be negative or zero when ab and bc are positive.

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