Minimum Elements Added to Make Relation Transitive

To find the minimum number of elements required to be added to $R$ to make it a transitive relation, we first determine the elements of the relation $R$.

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

Since $\log_e(x + y) \le 2$, exponentiating both sides gives: $x + y \le e^2$

Knowing that $e \approx 2.718$, we have $e^2 \approx 7.389$. Since $x, y \in \mathbb{N} = \{1, 2, 3, \dots\}$, $x + y$ must be a positive integer. Therefore, the inequality simplifies to: $x + y \le 7$

The set of all possible values for $x$ and $y$ is $S = \{1, 2, 3, 4, 5, 6\}$.

Now, let's list the elements of $R$ based on the values of $x$:

• For $x = 1$: $y \in \{1, 2, 3, 4, 5, 6\} \implies 6 \text{ pairs}$
• For $x = 2$: $y \in \{1, 2, 3, 4, 5\} \implies 5 \text{ pairs}$
• For $x = 3$: $y \in \{1, 2, 3, 4\} \implies 4 \text{ pairs}$
• For $x = 4$: $y \in \{1, 2, 3\} \implies 3 \text{ pairs}$
• For $x = 5$: $y \in \{1, 2\} \implies 2 \text{ pairs}$
• For $x = 6$: $y \in \{1\} \implies 1 \text{ pair}$

The total number of elements currently in $R$ is: $|R| = 6 + 5 + 4 + 3 + 2 + 1 = 21$

To make $R$ transitive, let $R'$ be the transitive closure of $R$:

1. For every $x \in S$, we have $x + 1 \le 7$, which means $(x, 1) \in R$.
2. For every $y \in S$, we have $1 + y \le 7$, which means $(1, y) \in R$.

By the definition of a transitive relation, if $(x, 1) \in R'$ and $(1, y) \in R'$, then $(x, y)$ must also belong to $R'$ for every $x, y \in S$.

Thus, $R'$ must contain all elements of the Cartesian product $S \times S$: $S \times S = \{1, 2, 3, 4, 5, 6\} \times \{1, 2, 3, 4, 5, 6\}$

The total number of elements in $S \times S$ is: $|S \times S| = 6 \times 6 = 36$

Since $S \times S$ is a universal relation on $S$, it is inherently transitive. Hence, the minimal transitive relation containing $R$ is $S \times S$.

Therefore, the minimum number of elements required to be added to $R$ is: $|S \times S| - |R| = 36 - 21 = 15$