JEE Main 2026 — Mathematics Algebra

Given the relation $R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : \log_e(x + y) \leq 2\}$R={(x,y)∈N×N:loge​(x+y)≤2}.

This condition simplifies to:

$$x + y \leq e^2$$x+y≤e2

Since $e \approx 2.718$e≈2.718, we have $e^2 \approx 7.389$e2≈7.389.

As $x, y \in \mathbb{N}$x,y∈N, the sum $x + y$x+y must be a natural number. Therefore, the condition becomes:

$$x + y \leq 7$$x+y≤7

Let's find the number of elements in $R$R by listing the possible pairs $(x, y)$(x,y):

For $x = 1$x=1, $y \in \{1, 2, 3, 4, 5, 6\}$y∈{1,2,3,4,5,6} (6 elements)

For $x = 2$x=2, $y \in \{1, 2, 3, 4, 5\}$y∈{1,2,3,4,5} (5 elements)

For $x = 3$x=3, $y \in \{1, 2, 3, 4\}$y∈{1,2,3,4} (4 elements)

For $x = 4$x=4, $y \in \{1, 2, 3\}$y∈{1,2,3} (3 elements)

For $x = 5$x=5, $y \in \{1, 2\}$y∈{1,2} (2 elements)

For $x = 6$x=6, $y \in \{1\}$y∈{1} (1 element)

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

To make $R$R a transitive relation, we must ensure that if $(x, y) \in R$(x,y)∈R and $(y, z) \in R$(y,z)∈R, then $(x, z) \in R$(x,z)∈R.

Notice that $(x, 1) \in R$(x,1)∈R for all $x \in \{1, 2, 3, 4, 5, 6\}$x∈{1,2,3,4,5,6} and $(1, z) \in R$(1,z)∈R for all $z \in \{1, 2, 3, 4, 5, 6\}$z∈{1,2,3,4,5,6}.

For the relation to be transitive, it must contain $(x, z)$(x,z) for all possible combinations of $x$x and $z$z from the set $A = \{1, 2, 3, 4, 5, 6\}$A={1,2,3,4,5,6}.

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

The minimum number of elements required to be added to $R$R to make it transitive is the difference between the number of elements in the transitive closure and the number of elements already in $R$R:

Minimum elements to add $= 36 - 21 = 15$=36−21=15.

Answer: $15$15