Help in understanding why the arithmetic derivative is well-defined.

I'm reading into the arithmetic derivative at the moment and I just don't get why it is well-defined. For reference, Ufnarovski tried explaining it here on page 2. I would like to understand the not explicitly stated induction. Thanks in advance!

Solution 1:

To be perfectly frank, any notion is well defined so long as there are no contradictions. Here, we are really just defining a special function from $\mathbb{N}$ to $\mathbb{N}\cup\{0\}$. So long as it is defined on all inputs then we are technically done.

There is a small wrinkle that we also want this function to satisfy some properties, but that isn't a real issue; once you have the function written down you just need to verify that the properties hold. In this case, we need to verify two properties:

1) $p'=1$ for all primes $p$

2) $(ab)'=a'b+ab'$ for any $a,b\in\mathbb{N}$

That property 1 is true is obvious from the definition.

That property 2 is true is less obvious, but the paper you linked goes through the small bit of math necessary to show that the Leibniz rule holds.

So what have we shown? That the function $(\cdot)':\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ as defined satisfies the two properties we wanted. What is brushed over very lightly is the author's comments that state this function is the only way we can define the notion of arithmetic derivative that we want. But this issue has nothing to do with the fact that our derivative is well-defined. Uniqueness and being well-defined are two separate ideas.

But let's talk about uniqueness anyway, since that is a part of your post. Suppose you have a two functions $f$ and $g$ that are arithmetic derivatives. What are $f(1)$ and $g(1)$? By the fact that $f$ and $g$ satisfy the Leibniz rule we know that $f(1)=g(1)=0$; the paper even shows the derivation of this fact. Let this sink in; any arithmetic derivative must send $1$ to $0$. Now let's try and do some (strong) induction:

Assume that for all natural numbers $1\leq k\leq n$ we have shown that $f(k)=g(k)$. What can we say about $n+1$? If it is prime, then by our first property we see that $f(n+1)=g(n+1)=1$. If it is composite then we may write it as the product of two naturals $a$ and $b$, both of which are less than $n$. Well what is $f(ab)$? By the Leibniz rule and by our inductive hypothesis, we know the following: $$f(n+1)=f(ab)=f(a)b+af(b)=g(a)b+ag(b)=g(ab)=g(n+1)$$

Thus it must by the case that $f$ and $g$ are the same function, and the arithmetic derivative is unique.

Solution 2:

The expression marked (1) is a bit contrived, but it is well-defined because the author implicitly assumes that $n_i>0$ for all $i$ in the expression $n=\prod_{i=1}^kp_i^{n_i}$. This determines the $p_i$ and $n_i$ uniquely, because (positive) integers have unique factorizations. Rewriting expression (1) as $$n'=n\sum_{i=1}^k\frac{n_i}{p_i}=\sum_{i=1}n_i\frac{n}{p_i},$$ shows that $n'$ is an integer, because the $p_i$ divide $n$ and the $n_i$ are integers by definition.