Show $F_n$ has a least element and it is prime.

Question

let $F_n$ be the set of positive factors of n greater than 1 and $n\in \mathbb{N}$.

Show $F_n$ has a least element and it is prime.

Answer 1

Hint: Use the well ordering principle.

You'll have that $F_n$ has a least element (say $a$).

Now, it all boils down to show that $a$ is prime.

Suppose, on contrary, $a$ isn't prime. Then $\exists b<a$ such that $b|a$ which contradicts the minimality of $a$.

Answer 2

let $F_n=\{x:x|n ~and~x>0\}$

by WOA $F_n$ has a least element be $q$.

Suppose $q$ isn't a prime.

Which means it can be factored in to a prime $q=p_1k $ by doing this repeatedly the least element of $F_n$ should be a prime.

Answer 3

Note that $F_n$ isn't empty ($n\mid n \Rightarrow n \in F_n \Rightarrow |F_n| > 0$) and has the least element. This is due to the fact that $(\forall x \in F_n)(1 < x \leq n \wedge x \in \mathbb{Z})$ and set of integers is countable, so $F_n$ must be finite. You can also invoke to well-ordering principle.

Assume that $m = \min(F_n)$ and $m$ isn't prime. Then $\left(\exists~d \in \mathbb{Z}\right)\left(d < m \wedge d | m\right)$.

Using $\left(\forall a,b,c\in \mathbb{Z}\right)\left( (a\mid b \wedge b \mid c) \Longrightarrow a \mid c\right)$ we obtain $d \in F_n$.

$$\left(d \mid m \wedge m \mid n \right)\Longrightarrow(d \mid n) \Longrightarrow d \in F_n$$

$\left(d < m \wedge d\in F_n \right)$ is contrary to the assumption $m = \min(F_n)$, so $m$ is prime.