How many elements does relation R have on {−2,−1,0,1,2} with condition 1+ab>0?

Total ordered pairs = 25. Pairs where ab ≤ −1 are excluded: (1,−1),(−1,1),(1,−2),(−1,2),(2,−1),(−2,1),(2,−2),(−2,2) — 8 pairs. So |R| = 25 − 8 = 17.

Is the relation R defined by 1+ab>0 reflexive?

Yes. For all a in S, 1+a² ≥ 1 > 0, so (a,a) ∈ R for every element.

Yes. Since ab = ba, if 1+ab > 0 then 1+ba > 0. So (a,b) ∈ R implies (b,a) ∈ R.

No. Counterexample: (−1,0) ∈ R since 1+0=1>0, and (0,2) ∈ R since 1+0=1>0, but (−1,2) ∉ R since 1+(−2)=−1 which is not >0.

What is the correct answer for this JEE Main problem?

Only Statement I is true. R has exactly 17 elements, but R is not an equivalence relation as it fails transitivity. Answer: A.

How to quickly count excluded pairs for this relation?

For a=1: b=−1,−2 fail. For a=−1: b=1,2 fail. For a=2: b=−1,−2 fail. For a=−2: b=1,2 fail. Total 8 pairs excluded, so |R|=17.

Why does R fail to be an equivalence relation?

R fails transitivity. An equivalence relation requires all three: reflexivity, symmetry, and transitivity. Even one failure disqualifies it.

What type of relation is R if it is reflexive and symmetric but not transitive?

R is called a tolerance relation — reflexive and symmetric but not transitive.

JEERANKERS

MathRelations

Q MCQ Relations & Functions

Consider the relation $R$ on the set $\{-2,-1,0,1,2\}$ defined by $(a,b)\in R$ 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) Only I is true     B) Only II is true
C) Both I and II are true     D) Neither I nor II is true

✅ Correct Answer

A) Only I is true

✓

Solution

Count elements of R

Total ordered pairs from $S=\{-2,-1,0,1,2\}$: $5\times5=25$.

$(a,b)\notin R$ when $ab\leq-1$. Excluded pairs:

$(1,-1),\ (-1,1)$ → $ab=-1$
$(1,-2),\ (-1,2),\ (2,-1),\ (-2,1)$ → $ab=-2$
$(2,-2),\ (-2,2)$ → $ab=-4$

Total excluded $= 8$

$$|R|=25-8=\mathbf{17}\ \checkmark \quad \text{Statement I is TRUE}$$

Test: Reflexivity

For all $a\in S$: $1+a^2\geq1>0$, so $(a,a)\in R$. ✅ Reflexive.

Test: Symmetry

$ab=ba$, so $1+ab>0\Leftrightarrow1+ba>0$. ✅ Symmetric.

Test: Transitivity — Counterexample

Take $a=-1,\ b=0,\ c=2$:

$(-1,0)\in R$: $\ 1+(-1)(0)=1>0$ ✅
$(0,\ 2)\in R$: $\ 1+(0)(2)=1>0$ ✅
$(-1,2)\notin R$: $1+(-1)(2)=-1\not>0$ ❌

Transitivity fails → $R$ is not an equivalence relation.

Statement II is FALSE.

📘 Key Concept

An equivalence relation must be reflexive, symmetric, and transitive — all three. Failing even one disqualifies it. To count $|R|$, count excluded pairs and subtract from the total: faster than listing all valid pairs.

FAQs

How to quickly count elements of R?

⌄

Row by row: for $a=1$, $b=-1,-2$ fail (2 pairs); $a=-1$, $b=1,2$ fail (2); $a=2$, $b=-1,-2$ fail (2); $a=-2$, $b=1,2$ fail (2). Total $8$ excluded. $|R|=25-8=17$.

Is (0,0) ∈ R?

⌄

Yes. $1+(0)(0)=1>0$. All pairs involving $0$ are in $R$ since $0\cdot a=0$ and $1+0=1>0$.

Why does transitivity fail?

⌄

$0$ acts as a "false bridge": $-1$ relates to $0$ (product $0$) and $0$ relates to $2$ (product $0$), but $-1$ and $2$ have product $-2$, so $1+(-2)=-1<0$. They are not related.

Would R be an equivalence relation if the set were {0,1,2}?

⌄

Yes. All products $\geq0$, so $1+ab\geq1>0$ always. Every ordered pair is in $R$ — this universal relation is an equivalence relation.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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