Let R = {(x, y) in N times N : log_e(x + y) l | vedclass

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(A) The relation $R$ is defined by $\log_e(x + y) \leq 2$, which implies $x + y \leq e^2$. Since $e \approx 2.718$, $e^2 \approx 7.389$. Thus, $x + y

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$1$

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(A) The relation $R$ is defined by $\log_e(x + y) \leq 2$, which implies $x + y \leq e^2$. Since $e \approx 2.718$, $e^2 \approx 7.389$. Thus, $x + y \leq 7$ for $x, y \in N$.The pairs $(x, y)$ in $R$ are: $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (5,1), (5,2), (6,1)$.$A$ relation is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.Consider $(3, 4) \in R$ and $(4, 3) \in R$. For transitivity, $(3, 3)$ must be in $R$, which is true. Consider $(4, 4) 
otin R$. If we have $(3, 4) \in R$ and $(4, 3) \in R$, we need $(3, 3) \in R$ (present). If we have $(4, 3) \in R$ and $(3, 4) \in R$, we need $(4, 4) \in R$. Since $4+4=8 > 7$, $(4, 4) 
otin R$. Thus, we must add $(4, 4)$ to $R$. Similarly, checking other combinations, we find that adding $(4, 4)$ is necessary. After verification, the minimum number of elements to add is $1$.

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

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