Partially ordered set and a method that can quickly solve these questions?!

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I prepare for Msc Qualification exam. This is 2015 Exams, with Answer is option (2).

Set $M={2,3,4,...}$ is given. Suppose $M \times M$ is sorted in which $ (a,b) \leq (c,d)$ if and only if $c$ is Divisible on $a$ and condition $b \leq d$ is hold. Which of these option is true about minimal and maximal elemetns of partial ordered set $(M \times M, \leq)$? in following option $p$ is arbitrary prime number.

1) each pair $(p,m)$ for $m \in M$ be a minimal element and there is no maximal element.

2) each pair $(p,m)$ for $m \in M$ be a minimal element and there is maximal element.

3) each pair $(p,2)$ for $m \in M$ be a minimal element and there is no maximal element.

4) each pair $(p,m)$ for $m \in M$ be a minimal element and there is maximal element.


my challenge is via a method that can solve this problem and like it, in a quick way? is it any description or hint?

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It seems to me that option (3) is correct.

  • $\boxed{(p, m) \text{ is not minimal for }m \geq 3}$: Since $m \geq 3$, we know that $m - 1 \geq 2$, so $(p, m - 1) \in M \times M - \{(p, m)\}$. Since $p \mid p$ and $m - 1 \leq m$, we have that $(p, m - 1) \leq (p, m)$.

  • $\boxed{\text{There is no maximal element}}$: Given any candidate maximal element $(p, q) \in M \times M$, notice that $p \mid p$ and $q \leq q + 1$ so that $(p, q) \leq (p, q + 1)$. Thus, since $(p, q + 1) \in M \times M - \{(p, q)\}$, we have found a different element larger than the candidate maximal element.