I am working on some things related to arithmetic derivatives, and I was writing a program to calculate the arithmetic derivative. In working on my program I came across an assumption that I have made, but I could not prove that it was true and it was quite frustrating to me.
My hypothesis is for any $n \in \mathbb{N}$, the smallest number, $a \in \mathbb{N}/\{1\}$ such that $a \vert n$ should be prime. My beginning of a proof for this was this
Using the fundamental theorem of arithmetic, for all $n \in \mathbb{N}/\{1\}$, that either $n$ is prime, or that $n$ has a unique prime factorization. If $n$ is prime, the proof is trivial, as the prime factorization for all $p \in \mathbb{P}$ is just $p$; however, for the case of $n$ being composite, the proof became quite difficult for me.
For the case of a number $c$ such that $c \in \mathbb{N}/\{1\}$ and $c \notin \mathbb{P}$, we can say that there exists at least two numbers, $a$ and $b$ such that fulfill the following four statement
- $1 < a \le b < c$
- $a \vert c$
- $b \vert c$
- $ab = c$
My thought process thereby went into dividing $c$ into odds and evens. We can say that this theorem is true for all even numbers, $e$, rather trivially as well because for all $e \in \{2k:\mathbb{N}/\{1\}\}$ that $2\vert e$ and $2$ is also the smallest integer that could possibly fulfill the above criteria thereby making $a = 2$ and fulfilling the theorem for all even numbers.
The part that I cannot figure out however is odd numbers. My intuition is telling me this must be true, but I cannot figure out a proof of this theorem for odds and I was wondering if
- Is this theorem actually true?
- If so, how can this be proven for odds?
Best Answer
Let $n\in\Bbb N$ and let $a\in\Bbb N\setminus\{1\}$ be the smallest divisor of $n$. If $a$ is prime, you're done. If not, then $a$ is composite, and so can be written $a = bc$ for $b,c\in\Bbb N\setminus\{1\}$. But then $b,c < a$, and $b\mid n$, contradicting the assumption that $a$ was the smallest divisor.