Question 36

The relation $R$ on $\mathbb{N} \times \mathbb{N}$ is defined by:

1. Check Reflexivity ($P$):
   For any pair $(a, b) \in \mathbb{N} \times \mathbb{N}$, we compare $(a, b)$ with itself. Since $a \le a$ is true (and $b \le b$ is true), the condition $(a \le a) \lor (b \le b)$ is trivially satisfied. Thus, $(a, b)R(a, b)$ holds for all $(a, b)$. Therefore, $R$ is reflexive, and proposition $P$ is TRUE.

2. Check Transitivity ($Q$):
   $R$ is transitive if $(a, b)R(c, d)$ and $(c, d)R(e, f)$ together imply $(a, b)R(e, f)$.

Consider the counterexample with elements in $\mathbb{N}$:
Let $(a, b) = (5, 2)$, $(c, d) = (3, 4)$, and $(e, f) = (4, 1)$:

• $(5, 2)R(3, 4)$: $5 \le 3$ is false, but $2 \le 4$ is true. Thus, $(5, 2)R(3, 4)$ holds.
• $(3, 4)R(4, 1)$: $3 \le 4$ is true (and $4 \le 1$ is false). Thus, $(3, 4)R(4, 1)$ holds.
• Now test $(a, b)R(e, f)$, which is $(5, 2)R(4, 1)$:
  Here $a \le e$ means $5 \le 4$ (false), and $b \le f$ means $2 \le 1$ (false).
  Neither condition holds, so $(5, 2)R(4, 1)$ is FALSE!

Since transitivity fails, proposition $Q$ is FALSE.

Therefore, $P$ is true and $Q$ is false, corresponding to option (B).