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by dead10ck 4316 days ago
I feel like I'm missing something. Isn't the derivative of a constant just defined to be 0? Why does the definition in the article restrict it to primes?
2 comments

You're right, it's not the normal definition of derivative. He's defining a function on numbers that has some similarities to the derivative (the multiplication rule is the same, and therefore so are some other properties), but isn't the derivative.

It's common in mathematics to take a word that already means something, and re-use it for a different but related concept. For example, in group and field theory, you talk about addition and multiplication, but they're not necessarily the standard definition.

> It's common in mathematics to take a word that already means something, and re-use it for a different but related concept. For example, in group and field theory, you talk about addition and multiplication, but they're not necessarily the standard definition.

That's true, but I feel like the article mislead me because it didn't give me a proper and timely explanation of what it means by "derivative".

I find I have the same problem explaining computer technologies to people.
This article is defining a new notion which is different from the derivative of a function. It's called derivative because of its inspirations and similarities to the classical idea of a derivative. It's interesting because it appears to have close relationships to important open problems in number theory, such as the Goldbach conjecture.

And it's not restricted to primes, it's just defined first for primes and then extended to all integers later.

What is confusing is that the article introduces both derivatives (of a function and of a number) as D, when both are definitely not the same thing.

EDIT: that is, the fact that they have the same name and the same symbol is what's confusing.

For better or worse that's very common. A big part of learning to read advanced math is keeping track of the "kind of thing" each variable is meant to be. If you do that, then overloading D is a convenience which illustrates similarities.

But if you're not a professional mathematician then such overloading is terrifically confusing. Actually, I'm fairly certain it still is even for professional mathematicians.

Ahh, I see. That makes more sense.

However, I'm not sure I understand how the definition extends to all numbers. Edit: nvm, I'm dumb.