# How do primes come out of Peano arithmetic?

1. Aug 12, 2011

### Kindayr

Let $(N, s(n), 0)$ be a Peano space. That is, $N=\{1,2,3,\dots \}$ is a set in which http://en.wikipedia.org/wiki/Peano_arithmetic" [Broken] can be used.

We can then define:
• $0=\varnothing, 1=\{0\}, 2=\{0,1\},\dots \implies n=\{0,1,2,\dots ,n-2,n-1\}$
• $s(a)=a\cup \{a\}\implies s(a)=a+1$

From here we can define both addition and multiplication. I was wondering how the properties of primes come to be. That is, what makes $19=\{0,1,2,\dots ,18\}$ prime and $4=\{0,1,2,3\}$ not prime.

I've never really studied Number Theory, so I'm not strong in it at all.

(If you've noticed, I really like Peano spaces)

Last edited by a moderator: May 5, 2017
2. Aug 12, 2011

### micromass

Well, first define addition the usual recursive way:

n+0=0
n+s(m)=s(n+m)

Then define multiplication the usual recursive way

n*0=0
n*s(m)=n*m+n

Then we define that "n divides m" if there is a number x such that n*x=m. We write n|m.

Then we define a number p (which is nonzero and not one) to be prime if p|(a*b) implies that p|a or p|b.

3. Aug 13, 2011

### Kindayr

I guess I failed to explain what I was really wishing to ask.

I was just wondering how so much power and structure comes out of the prime numbers with respect to multiplication.

Maybe my question is more metamathematical than I thought it would be when I was thinking of it earlier today.

Nonetheless, thanks micromass for your help! :)

Also, could one consider the set of primes $\{2,3,5,\dots\}$ as a 'basis' for the natural numbers with respect to multiplication? Is that what the fundamental theorem of arithmetic is basically saying?

4. Aug 14, 2011

### ramsey2879

You could easily say that the prime numbers are a basis for all natural numbers > 1 since every natural number greater than 1 is a product of primes. But what then is the basis of 1? Also the fundamental theorem of arithematic is more stronger than that since it says that there is "only one way" to express a number greater than 1 as a product of prime(s) (you don't count P(1)*P(2) as different from P(2)*P(1) etc. also P(1) is considered to be simply the product a prime, i.e. P(1)).

5. Aug 15, 2011

### Kindayr

I was just fooling around in my head with this idea as the primes as a basis.

Let $a\in\mathbb N$ and let $p,p_{1},p_{2},p_{3},\dots,p_{n}$ be all the primes such that $1<p_{1}<p_{2}<p_{3}<\cdots<p_{n}=p\leq a$. By the fundamental theorem of arithmetic, we know that there exists a unique prime factorization of $a$. That is, $a=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{n}^{e_{n}}$. So we can represent $a$ with the $n$-tuple $a\simeq (e_{1},e_{2},\dots,e_{n})\in\mathbb N^{n}$. Note that $1\simeq (0,0,\dots,0)\in\mathbb N^{n}$.

Define the binary operation $\oplus : \mathbb N^{n}\times \mathbb N^{n} \to \mathbb N^{n}$ such that $\oplus ((e_{1},e_{2},\dots,e_{n}),(f_{1},f_{2},\dots,f_{n}))=(e_{1}+f_{1},e_{2}+f_{2},\dots,e_{n}+f_{n})$. Note that this corresponds to normal multiplication in $\mathbb N$. We could also define equivalence classes in $\mathbb N^{n}$ that correspond to congruence in $\mathbb Z_{p}$.

I think this pretty cool hahaha, because then with respect to some norm, we could assign 'lengths' to each of this $n$-tuples. So you could find which numbers are related with respect to length (notice the primes are all of length 1, and are orthogonal to one another).

I just thought this was cool, because I think this forms a monoid at least. The once we have a monoid, we could extend it as a Grothendiek group, I think. I just though some cool things could be done with it.

I think my definitions are a little sloppy, and could be fixed, but I hope I'm getting what I'm trying to convey over. I can make it more rigourous later in the day as I just woke up heh.

Last edited: Aug 15, 2011