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(D) Given the relation $ho = \{(x, y) \in N 	imes N : 2x + y = 41\}$.$1$. Reflexive: For $ho$ to be reflexive, $(x, x) \in ho$ for all $x \in N

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$\rho$ is not transitive

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(D) Given the relation $ho = \{(x, y) \in N 	imes N : 2x + y = 41\}$.$1$. Reflexive: For $ho$ to be reflexive, $(x, x) \in ho$ for all $x \in N$. This implies $2x + x = 41$, so $3x = 41$, which gives $x = \frac{41}{3} 
otin N$. Thus, $ho$ is not reflexive.$2$. Symmetric: For $ho$ to be symmetric, if $(x, y) \in ho$, then $(y, x) \in ho$. If $(x, y) = (1, 39) \in ho$ (since $2(1) + 39 = 41$), then $(y, x) = (39, 1)$. Checking $(39, 1)$: $2(39) + 1 = 78 + 1 = 79 
eq 41$. Thus, $(39, 1) 
otin ho$. So, $ho$ is not symmetric.$3$. Transitive: For $ho$ to be transitive, if $(x, y) \in ho$ and $(y, z) \in ho$, then $(x, z) \in ho$. Let $(x, y) = (1, 39) \in ho$ and $(y, z) = (39, -37)$. However, $z$ must be in $N$. Since $2(39) + z = 41 \Rightarrow z = 41 - 78 = -37 
otin N$. There are no pairs $(y, z) \in N 	imes N$ such that $2y + z = 41$ for $y=39$. In fact, for any $(x, y) \in ho$, $y = 41 - 2x$. For $(y, z) \in ho$, $z = 41 - 2y = 41 - 2(41 - 2x) = 41 - 82 + 4x = 4x - 41$. For $z \in N$, $4x - 41 > 0 \Rightarrow x > 10.25$. If we take $x=11$, $y=19$, $z=3$. Here $(11, 19) \in ho$ and $(19, 3) \in ho$. But $(11, 3) 
otin ho$ because $2(11) + 3 = 25 
eq 41$. Thus, $ho$ is not transitive.

Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho = \{(x, y) \in N \times N : 2x + y = 41\}$. Then:

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