# The Arithmetic Derivative (Part 2)

__Our first pattern__

So we've taken the usual rules for differentiation and applied them to natural numbers:

### D(ab) = aD(b) + D(a)b

with initial conditions

### D(0) = 0

### D(1) = 0

### D(p) = 1 for all primes p

We can use the following code to generate a graph of natural numbers and their derivatives:

k = 10000

N = list(range(k+1))

DN = [0]*(k+1)

for n in range(2,k+1):

if is_prime(n):

DN[n] = 1

else:

a = factor(n)[0][0]

b = n/a

DN[n] = a*DN[b] + DN[a]*b

list_plot(DN,color='#1E5090')

*(n, D(n))*

*(n, D(n))*

There are a few patterns we can see after writing out a few examples in the last post. The derivative of a prime is 1, but what if a number is a product of two primes?

### D(pq) = pD(q) + D(p)q = p + q

Cool! So the derivative of the product of two primes is just the sum of those two primes. If ** p** and

**are equal, then this turns into**

*q*### D(p^2) = 2p

Is this a pattern forming? Let's find out!

### D(p^3) = pD(p^2) + p^2 = p(2p) + p^2 = 3p^2

### D(p^4) = pD(p^3) + p^3 = p(3p^2) + p^3 = 4p^3

I believe we have enough data points to form a conjecture - that is, a guess at a theorem!

It seems that if ** p** is a prime, then the derivative of its powers are given by

### D(p^k) = kp^(k-1)

Let's go ahead and prove this! We will do so using the __principle of induction__** . **So we first want to establish our base case: for

**, we have**

*k = 0**D(p^0) = D(1) = 0*

*D(p^0) = D(1) = 0*

and

*kp^(k-1) = 0*,

*kp^(k-1) = 0*

so the equality holds.

Now we assume this holds true for some fixed ** k** and show it holds true for

**. Notice that if this is true, then the fact that the equality holds for**

*k + 1***implies it holds for**

*k = 0***, which implies it holds for**

*k = 1***, and so on, for all natural numbers**

*k = 2***. That's induction!**

*k*If we assume

*D(p^k) = kp^(k-1)*,

*D(p^k) = kp^(k-1)*

then

### D(p^(k+1)) = D(p*p^k) = pD(p^k) + D(p)p^k =

### = p(kp^(k-1)) + p^k = kp^k + p^k = (k+1)p^k

Therefore the equality holds for ** k+1** and we're done with our induction!

__More patterns ahead!__

So far we know the following things about our arithmetic derivative:

### D(p) = 1

### D(pq) = p + q

### D(p^k) = kp^(k-1)

Let's try to generalize in the direction of adding prime factors to our number. So

### D(p) = 1

### D(p1p2) = p1 + p2

### D(p1p2p3) = p1D(p2p3) + p2p3 = p1(p2 + p3) + p2p3 = p1p2 + p1p3 + p2p3

### D(p1p2p3p4) = p1p2p3 + p1p2p4 + p1p3p4 + p2p3p4

So what's the pattern here? If ** n = p1p2...pk** is the product of

**distinct primes, then**

*k***is a sum of**

*D(n)***terms, where each term is**

*k***with a single prime factor missing. In symbols,**

*p1p2...pk*where the term ** pi** with the hat over it indicates we leave it out. For example,

This lovely characterization lets us prove one of the earliest results of __Number Theory__: the infinitude of primes! __Euclid__ proved in 300BC that there are infinitely many prime numbers. His original reasoning is actually very similar to the reasoning we'll give using the derivative!

__The infinitude of primes__

Suppose that we only have ** k** distinct primes -

**. Then we'll show that this leads to a**

*p1, p2, p3, ..., pk***, and therefore is an example of a**

*contradiction***.**

__proof by contradiction__Multiply them all together to get another number ** n = p1p2p3...pk** and look at its derivative

**. We just found that**

*D(n)*If the only primes are ** p1, p2, ..., pk**, then

**must be divisible by one of these, since EVERY natural number greater than 1 is divisible by some prime. But the term**

*D(n)*has no ** p1** term, so it's not divisible by

**. So we know**

*p1***is not divisible by**

*D(n)***.**

*p1*Similarly, the term

is NOT divisible by ** p2**, so we know that

**is not divisible by**

*D(n)*

*p2.*Continuing with this logic, the term

is not divisible by ** pi**, so we know

**is not divisible by**

*D(n)*

*pi.*In other words: D(n) IS NOT DIVISIBLE BY ANY PRIME!

This is our contradiction. So ** p1, p2, ..., pk** cannot be a complete list of primes. Since

**was general, this means we can have no FINITE set of primes at all.**

*k*

*Thus, infinite primes!***Thank you for reading!**

**Jonathan Gerhard**

**J****Math****G**

**J**

**Math**

**G**