EDIT: I am updating this to illustrate my proof thus far. I would appreciate any critique.

Question details: "Use the well ordering principle to prove that every integer $n\ge2$ has a prime factor.

One way to prove this for a given integer $n\ge2$ is to apply the well-ordering principle to the set

$X=\{d\in{\Bbb Z}:d\ge2 \land d\ |\ n\}$,

the set of all factors $d$ of $n$ such that $d\ge2$.

(a) Prove that $X$ is not empty.

(b) Prove that if $p$ is the minimal element of $X$, then $p$ must be a prime number.

(c) Finish the proof of the theorem."

(a) Trivial. By definition of $X$, $n\in X$ if $n\ge2$. Thus, $X$ is nonempty given any integer $n\ge2$.

(b) Suppose $p$ is the minimal element of $X$. Then, it is a divisor of $n$, and it is also $\ge2$. Consider some integer $q\ge2$ that is a divisor of $p$. Then, $q\in X$ by definition of $X$, since any divisor of $p$ is also a divisor of $n$ (and $q\ge2$). Since $p$ is the minimal element of $X$, then any other element of $X$ $\le p$. Therefore, $q\le p$. Furthermore, $q$ is a divisor of $p$, so $q\ge p$. Then, $q=p$. Thus, $p$ only has two positive integer factors - itself, and 1. Therefore, $p$ is prime.

(c) Suppose, for the sake of contradiction, that there exists a set $C$, of integers $\ge2$ that cannot be factored into a product of two prime numbers. By the well-ordering principle, there is some $m\in C$ such that $m$ is the minimal element of $C$. Because $m\in C$ and therefore cannot be prime, it must be the product of two integers $s,t$ such that $2\le s,t \lt m$. Therefore, $s,t\notin C$ because they are smaller than $m$, which is the minimal element of $C$.

Since $s,t\notin C$, they can be factored into the products of primes, $a_1,a_2,b_1$ and $b_2$ such that $s=a_1*a_2$ and $t=b_1*b_2$. Since $s,t$ are divisors of $m$,

$m=s*t=(a_1*a_2)(b_1*b_2)$, the product of primes.

This is a contradiction, because $m\in C$. Thus, set $C$ cannot exist. As a consequence, we have proved that every integer $n\ge2$ has a prime factor.

[The End]

Have I got it write? If not, please point out any flaws. If I do have it correct, what I would especially appreciate is any input on the proof-writing itself - is it nonsensical or inefficient? Thanks!

EDIT: I am updating this to illustrate my proof thus far. I would appreciate any critique.

Question details: "Use the well ordering principle to prove that every integer $n\ge2$ has a prime factor.

One way to prove this for a given integer $n\ge2$ is to apply the well-ordering principle to the set

$X=\{d\in{\Bbb Z}:d\ge2 \land d\ |\ n\}$,

the set of all factors $d$ of $n$ such that $d\ge2$.

(a) Prove that $X$ is not empty.

(b) Prove that if $p$ is the minimal element of $X$, then $p$ must be a prime number.

(c) Finish the proof of the theorem."

(a) Trivial. By definition of $X$, $n\in X$ if $n\ge2$. Thus, $X$ is nonempty given any integer $n\ge2$.

(b) Suppose $p$ is the minimal element of $X$. Then, it is a divisor of $n$, and it is also $\ge2$. Consider some integer $q\ge2$ that is a divisor of $p$. Then, $q\in X$ by definition of $X$, since any divisor of $p$ is also a divisor of $n$ (and $q\ge2$). Since $p$ is the minimal element of $X$, then any other element of $X$ $\le p$. Therefore, $q\le p$. Furthermore, $q$ is a divisor of $p$, so $q\ge p$. Then, $q=p$. Thus, $p$ only has two positive integer factors - itself, and 1. Therefore, $p$ is prime.

(c) Suppose, for the sake of contradiction, that there exists a set $C$, of integers $\ge2$ that cannot be factored into a product of two prime numbers. By the well-ordering principle, there is some $m\in C$ such that $m$ is the minimal element of $C$. Because $m\in C$ and therefore cannot be prime, it must be the product of two integers $s,t$ such that $2\le s,t \lt m$. Therefore, $s,t\notin C$ because they are smaller than $m$, which is the minimal element of $C$.

Since $s,t\notin C$, they can be factored into the products of primes, $a_1,a_2,b_1$ and $b_2$ such that $s=a_1*a_2$ and $t=b_1*b_2$. Since $s,t$ are divisors of $m$,

$m=s*t=(a_1*a_2)(b_1*b_2)$, the product of primes.

This is a contradiction, because $m\in C$. Thus, set $C$ cannot exist. As a consequence, we have proved that every integer $n\ge2$ has a prime factor.

[The End]

Have I got it right? If not, please point out any flaws. If I do have it correct, what I would especially appreciate is any input on the proof-writing itself - is it nonsensical or inefficient? Thanks!

This problem is asked in a very specific way:

"One way to prove this for a given integer $n\ge2$ is to apply the well-ordering principle to the set

As I recall, using the well-ordering principle to prove stuff usually involves supposing a set of counter examples and seeking a contradiction.

Asking me to prove that $X$ is non-empty, and whatever (b) is asking me - is confusing me. I'm not exactly sure how to begin. This is what I have, but I don't really think it makes much sense:

(a) Theorem. $X$ is not empty.

Proof. If $X$ is nonempty, then there is at least one element in X. Consider $d=2$ and $n=4$. Then, $d\ge2$, and $d|n$. Thus, $2\in X$. Therefore, $X$ is not empty.

(b) Theorem. If $p$ is the minimal element of $X$, then $p$ is prime. Proof. The smallest element of X is $2\in X$, because it is an element of the set (reference (a)), and because $2=d\ge2$ by definition of $X$. $2$ is a prime number. Therefore, the minimal element of $X$ is prime.

Am I lost? Or am I in the ballpark? Honestly, I don't understand (b), and part (a) is sort of weird to me. $X$ is as set of factors of $n$? So is it a set of values for $d$, or for $d$ and $n$, or for $n$. I honestly dunno.

Any assistance would be appreciated!

EDIT: I am updating this to illustrate my proof thus far. I would appreciate any critique.

Question details: "Use the well ordering principle to prove that every integer $n\ge2$ has a prime factor.

One way to prove this for a given integer $n\ge2$ is to apply the well-ordering principle to the set

(a) Trivial. By definition of $X$, $n\in X$ if $n\ge2$. Thus, $X$ is nonempty given any integer $n\ge2$.

(b) Suppose $p$ is the minimal element of $X$. Then, it is a divisor of $n$, and it is also $\ge2$. Consider some integer $q\ge2$ that is a divisor of $p$. Then, $q\in X$ by definition of $X$, since any divisor of $p$ is also a divisor of $n$ (and $q\ge2$). Since $p$ is the minimal element of $X$, then any other element of $X$ $\le p$. Therefore, $q\le p$. Furthermore, $q$ is a divisor of $p$, so $q\ge p$. Then, $q=p$. Thus, $p$ only has two positive integer factors - itself, and 1. Therefore, $p$ is prime.

(c) Suppose, for the sake of contradiction, that there exists a set $C$, of integers $\ge2$ that cannot be factored into a product of two prime numbers. By the well-ordering principle, there is some $m\in C$ such that $m$ is the minimal element of $C$. Because $m\in C$ and therefore cannot be prime, it must be the product of two integers $s,t$ such that $2\le s,t \lt m$. Therefore, $s,t\notin C$ because they are smaller than $m$, which is the minimal element of $C$.

Since $s,t\notin C$, they can be factored into the products of primes, $a_1,a_2,b_1$ and $b_2$ such that $s=a_1*a_2$ and $t=b_1*b_2$. Since $s,t$ are divisors of $m$,

$m=s*t=(a_1*a_2)(b_1*b_2)$, the product of primes.

This is a contradiction, because $m\in C$. Thus, set $C$ cannot exist. As a consequence, we have proved that every integer $n\ge2$ has a prime factor.

[The End]

Have I got it write? If not, please point out any flaws. If I do have it correct, what I would especially appreciate is any input on the proof-writing itself - is it nonsensical or inefficient? Thanks!