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#### QuestForInsight

##### Member

- Jul 22, 2012

- 35

**Problem:**let $\mathbb{N} = \left\{0, ~ 1, ...\right\}$ be the set of natural numbers. Prove that $(\mathbb{N},~\le)$ is a poset under the ordinary order.

**Solution:**let $x \in\mathbb{N}$, then $x \le x$ as of course $x = x$. If also $y \in\mathbb{N}$, then $x \le y$ and $y \le x$ implies $x = y$. Lastly, $x \le y$ and $y \le z$ implies $x \le z$. Therefore $(\mathbb{N}, ~ \le)$ is a poset under the ordinary order.

Is the above valid? I feel like I'm basically stating what I was asked to prove!