Proof: If $a\in\mathbb{N}$ is well-ordered, since $a\ge 1$ and $1$ is the minimum element of $\mathbb{N}$, then $a/2\ge 1/2$. Furthermore, since $1/2$ is the minimum element of $a/2$, all positive rational numbers in the form $a/2$ is well-ordered.
Is my Proof Correct? If not, what alternative can be used.
On
Did you know that $\mathbb{N}$ is well ordered? Assuming this, let $A=\left\{\dfrac{a}{2} \mid a\in\mathbb{N}\right\}$. Take $B\subseteq A$ such that $B\neq\emptyset$. Define $C=\{ 2z\mid z\in B\}$. Clearly $C\neq\emptyset$ since $B\neq\emptyset$. Moreover, if we take $z\in B$ then there exist $a\in\mathbb{N}$ such that $z=\dfrac{a}{2}$. Then $2z=a$. Thus $C\subseteq\mathbb{N}$ and therefore there exist $\min(C)$. Let $w=\min(C)$. Can you prove from here that $\dfrac{w}{2}=\min(B)$?
A set is well ordered if every nonempty subset has a least element.
Your set is
$$X = \mathbb N/2 = \{\frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}, \frac{5}{2}, ...\}.$$
You want to show that every nonempty subset $A$ of $X$ has a least element.
Already, you know that $\mathbb N$ is well ordered.
Isn't there a natural association between subsets of $\mathbb N$ and subsets of $\mathbb N/2$? Namely, if you take subset of $\mathbb N/2$, and you double each element in the set, then you get a subset of $\mathbb N$. Use this idea to prove that $A$ has a smallest element.