Question

Prime Numbers and Vectors

Answer 1

The fundamental theorem of arithmetic says that every natural number has a unique prime factorization.

Answer 2

So we can write a number as a vector of the exponents of that factorization.

Answer 3

It would be of infinite size, since there are an infinite number of prime numbers.

Answer 4

but you are saying a vector for each number or a vector for all numbers?

Answer 5

\[
2\to \langle1\rangle\\
3\to \langle 0,1\rangle \\
24 \to \langle 3,1\rangle \\
10 \to \langle 1,0,1\rangle \\
\]

Answer 6

It would be implied the remaining elements are 0

Answer 7

gotcha

Answer 8

The vector addition for these would be multiplication
\[
\langle 1,2,0\rangle+\langle 0,0,1\rangle =\langle 1,2,1\rangle
\]Means\[
18\times5= 90
\]

Answer 9

injection from N into countable infinite multiples of N. Sweet.

Answer 10

I will be stealing this btw.

Answer 11

No, I'm not trying to prove anything here. I'm just stating some stuff.

Answer 12

me to

Answer 13

:)

Answer 14

Yeah pretty fun. =)

So maybe the gcd(a,b)=gcd(b,a) is like an inner product?

Answer 15

ignore me and carry on ;)

Answer 16

Scalar multiplication is the same as exponentiation of natural numbers.\[
2 \langle 1,2\rangle= \langle 2,4\rangle
\]Means:\[
(18)^2=324
\]

Answer 17

Since vectors can get long, we would want a way of notation that doesn't depend so heavily on position. Something like: \[
\langle 1\rangle=\mathbf p_1\\
\langle 0,1\rangle=\mathbf p_2\\
\langle0,0, 1\rangle=\mathbf p_3
\]

Answer 18

\[
\langle 0\rangle = \mathbf0
\]It wouldn't be a prime number, since 1 is not prime, but is meaningful as the identity of addition.

Answer 19

Here's an idea. It also seems if we want to "rotate" a vector from one dimension to another we raise it to this power:

\[\large \langle ...,a,0,... \rangle \ \rightarrow \ \langle ...,0,a,... \rangle \\ \large p^a \rightarrow r^a \\ \large (p^a)^{\log_p r} = r^a\]

but this might need more work.

Answer 20

The inner product doesn't seem to be a meaningful operation in this system.

Answer 21

define a new one:)

Answer 22

It would be tempting to say that the magnitude of these vectors is the actual number's prime factorization, however that would imply that all vectors have a different magnitude.

Answer 23

\[
\langle 1,1,1,1,\ldots\rangle=\mathbf1
\]

Answer 24

GCD and LCM serve as maximizing and minimizing based on their components.

Answer 25

Now sure what would be the best way to notate GCD or LCM\[
\gcd(\mathbf 0,\mathbf1) =\mathbf 0\\
\mathbf 0 + \mathbf 1=\mathbf 1
\]

Answer 26

\[
\text{lcm}(\mathbf a, \mathbf b) =\frac{\mathbf a+ \mathbf b}{\gcd(\mathbf a, \mathbf b)}
\]

Answer 27

Well the GCD is a good candidate in my opinion for an inner product because we can clearly see that \[\langle a,0 \rangle \cdot \langle 0, b \rangle = 1\] which in a sense gives us orthogonality, plus it allows us to extend it to higher dimensions. The only problem is it's more of a projection since we will always get another vector out of the GCD.

Answer 28

I think GCD would play a role in the inner product, but it might not be enough on its own.

Answer 29

GCD will always decrease,

Answer 30

But I suppose that isn't too bad.

Answer 31

Subtraction corresponds to natural number division. So I could say: \[
\text{lcm}(\mathbf a,\mathbf b)=\mathbf a+\mathbf b-\gcd(\mathbf a,\mathbf b)
\]

Answer 32

For now, we write \[
\gcd(a,b) \equiv a\cdot b
\]For lack of a better inner product.

Answer 33

A fun idea for a new function is what I'm gonna call "plum" short for "prime logarithm sum"

and what it does is this: \[plum(2^33^25^0)=3+2+0=5\] so we can now do something lik e this: \[a*plum(x)=plum(x^a)\] to allow us to take the norm as \[2 *plum(3^25^3)=2^2+3^2\]

Maybe I need to fix that somehow.

Answer 34

That would be like the trace of the vector.

Answer 35

You can't distribute exponents over addition.

Answer 36

I mean, just the fact that the order would matter is problematic

Answer 37

I guess then the norm has to just be: \[plum(x^2)=norm\]

But there's no problem with distributing exponents over addition because exponents _are_ multiplication right now.

Answer 38

What about the fact that any number and its factor, with the gcd simply returns that number?

Answer 39

that factor^

Answer 40

I think the GCD actually satisfies all the criteria of being an inner product!

\[1) \ \langle uv,w \rangle = \langle u,w \rangle \langle v,w \rangle \\ 2) \ \langle u^a,v \rangle = \langle u, v \rangle ^a \\ 3) \ \langle u,v \rangle = \langle v, u \rangle \\ 4) \ \langle u,u \rangle > 1, \ and \ equal \ \iff u=1\]

Answer 41

wait I might be totally wrong now that I'm thinking about it haha.

Answer 42

I don't know if I see what you mean, like the GCD(4,2)=2 but I don't see how the order matters exactly.

Answer 43

I think GCD is more of a projection operation, which makes it seem like the inner product.

Answer 44

wouldn't the number of factors be given by 2 to the power of the sum of the exponents?

Answer 45

Hmmm when you wrote lcm out in terms of the gcd it really reminds me of this: \[\cos \theta = \frac{a \cdot b}{ab}\]

Answer 46

In this vector system, 2 is special being in the first position. Often a scalar is just a 1D vector.

Answer 47

Also, I wonder about the wedge product.

Answer 48

We could add a new function that permutes the order of the exponents on the primes, call it r(x) Then we have the determinant: \[\Large plum(\frac{2^a3^b}{r(2^c3^d)} )=ad-bc\]hen we have

Answer 49

the 2 and 3 axes represent the powers on their respective exponents. It turns out it makes a nice square. So maybe we have a "square" rule instead of a trapezoid rule =P