Let $$R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : \log_e(x + y) \le 2\}$$. Then the minimum number of elements, required to be added in R to make it a transitive relation, is __________.

$$\ln(x+y) \le 2 \implies x + y \le e^2$$

Since $$x, y \in \mathbb{N}$$, their sum must be an integer: $$x + y \le 7$$

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

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

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

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

For $$x = 5$$: $$y \in \{1, 2\}$$ $$\implies (5,1), (5,2)$$

For $$x = 6$$: $$y \in \{1\}$$ $$\implies (6,1)$$

For R to be transitive, we have to add:

$$( 6 , 2 ) , ( 6 , 3 ) , ( 6 , 4 ) , ( 6 , 5 ) , ( 6 , 6 ) $$

$$( 5 , 3 ) , ( 5 , 4 ) , ( 5 , 5 ) , ( 5 , 6 ) $$

$$( 4 , 4 ) , ( 4 , 5 ) , ( 4 , 6 ) $$

$$( 3 , 5 ) , ( 3 , 6 ) $$

$$( 2 , 3 )$$

= 15 elements.