JEE Main 2026 — Sets and Relations Question with Solution

Given the relation $R=\{(x,y)\in \mathbb{N}\times \mathbb{N}:{\log }_e(x+y)\le 2\}$.

This condition simplifies to:
$x+y\le e^2$

Since $e\approx 2.718$, we have $e^2\approx 7.389$.
As $x,y\in \mathbb{N}$, the sum $x+y$ must be a natural number. Therefore, the condition becomes:
$x+y\le 7$

Let's find the number of elements in $R$ by listing the possible pairs $(x,y)$:
For $x=1$, $y\in \{1,2,3,4,5,6\}$ (6 elements)
For $x=2$, $y\in \{1,2,3,4,5\}$ (5 elements)
For $x=3$, $y\in \{1,2,3,4\}$ (4 elements)
For $x=4$, $y\in \{1,2,3\}$ (3 elements)
For $x=5$, $y\in \{1,2\}$ (2 elements)
For $x=6$, $y\in \{1\}$ (1 element)

Total number of elements in $R=6+5+4+3+2+1=21$.

To make $R$ a transitive relation, we must ensure that if $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$.
Notice that $(x,1)\in R$ for all $x\in \{1,2,3,4,5,6\}$ and $(1,z)\in R$ for all $z\in \{1,2,3,4,5,6\}$.
For the relation to be transitive, it must contain $(x,z)$ for all possible combinations of $x$ and $z$ from the set $A=\{1,2,3,4,5,6\}$.

Thus, the transitive closure of $R$ will be the Cartesian product $A\times A$, which contains $6\times 6=36$ elements.

The minimum number of elements required to be added to $R$ to make it transitive is the difference between the number of elements in the transitive closure and the number of elements already in $R$:
Minimum elements to add $=36-21=15$.

Answer: $15$