Commutative Monoid and some properties

[Kindayr]

For reference, let \mathbb N^{\infty} denote the set of all \infty-tuples of the form (n_{1}, n_{2}, \dots) with n_{i}\in \mathbb N ,\,\forall i, and let \mathbb N^{*}=\{1,2,3,\dots\}.

Let n\in\mathbb N^{*}. Then by the Fundamental Theorem of Arithmetic, we can represent n as a unique factorization of primes. That is n=p_{1}^{e_{1}} p_{2}^{e_{2}}\cdots=\prod_{k=1}^{\infty} p_{k}^{e_{k}}. Define a function \phi :\mathbb N^{*} \to \mathbb N^{\infty} such that \phi(n)=(e_{1},e_{2},\dots)=(e_{k})_{k=1}^{\infty}. Further, define a binary operation \oplus:\mathbb N^{\infty}\times\mathbb N^{\infty}\to\mathbb N^{\infty} such that \oplus((a_{1},a_{2},\dots),(b_{1},b_{2},\dots))=(a_{1}+b_{1},a_{2}+b_{2},\dots).

I will now show that (\mathbb N^{\infty},\oplus) is a commutative monoid.

• Let u,v\in\mathbb N^{\infty} with u=(a_{k})_{k=1}^{\infty} and v=(b_{k})_{k=1}^{\infty}. Then u\oplus v=(a_{k}+b_{k})_{k=1}^{\infty}=(b_{k}+a_{k})_{k=1}^{\infty}=v\oplus u. Since a_{k}+b_{k}=b_{k}+a_{k}\in\mathbb N,\,\forall k, it follows that u\oplus v=v\oplus u\in\mathbb N^{\infty}. So (\mathbb N^{\infty},\oplus) is both closed and commutative, as required.

• Let u,v,w\in\mathbb N^{\infty} with u=(a_{k})_{k=1}^{\infty},v=(b_{k})_{k=1}^{\infty} and w=(c_{k})_{k=1}^{\infty}. Then (u\oplus v)\oplus w=((a_{k}+b_{k})+c_{k})_{k=1}^{\infty}=(a_{k}+(b_{k}+c_{k}))_{k=1}^{\infty}=u\oplus (v\oplus w). Thus (\mathbb N^{\infty},\oplus) is associative.

• Let 1_{\mathbb N^{\infty}}\in\mathbb N^{\infty} such that 1_{\mathbb N^{\infty}}=(0,0,\dots). Then for any u=(a_{k})_{k=1}^{\infty}\in\mathbb N^{\infty}, we have 1_{\mathbb N^{\infty}}\oplus u=(0+a_{k})_{k=1}^{\infty}=(a_{k})_{k=1}^{\infty}=(a_{k}+0)_{k=1}^{\infty}=u\oplus 1_{\mathbb N^{\infty}}. Thus, 1_{\mathbb N^{\infty}} is the identity element of (\mathbb N^{\infty},\oplus).

Thus, (\mathbb N^{\infty},\oplus) is a commutative monoid.

Now I will show that \phi is a commutative monoid homomorphism.

• Let a,b\in\mathbb N. Then by the Fundamental Theorem of Arithmetic, both a and b have a unique factorization of primes. That is a=\prod_{k=1}^{\infty} p_{k}^{e_{k}} and b=\prod_{k=1}^{\infty} p_{k}^{f_{k}}. Then ab=\prod_{k=1}^{\infty} p_{k}^{e_{k}+f_{k}}. So we have \phi (ab)=\phi (\prod_{k=1}^{\infty} p_{k}^{e_{k}+f_{k}})=(e_{k}+f_{k})_{k=1}^{\infty}=(e_{k})_{k=1}^{\infty}\oplus (f_{k})_{k=1}^{\infty}=\phi(a)\oplus\phi(b), as required.

An interesting result to is to show that \phi is actually a commutative monoid isomorphism. I leave this up to the reader.[EDIT: THIS IS INCORRECT IF WE'RE WORKING WITH \mathbb N^{\infty}]

I've been messing around with this, and you can get some pretty cool new operations on numbers. Now that they are "basically" vectors, you can start making new operations like dot product, tensor product, their 'length' with respect to different norms. My favourite so far has been showing a homomorphism from \mathbb N^{\infty}\to\mathbb P[x]^{\infty}, the set of all polynomials with one variable. Once this is done, you can create a new binary operation such that you take two numbers from \mathbb N^{*}, convert both separately by \phi to \mathbb N^{\infty}, then to \mathbb P[x]^{\infty}, and do normal polynomial multiplication, then convert that polynomial all the way back to \mathbb N^{*}.

I also feel (severely speculative), if you can figure out some way with all these new operations to define a homomorphic addition within \mathbb N^{\infty}, we could find a simpler solution to Fermat's Last Theorem.

You can also play around with it more finitely, which I can explain a bit later, but my fingers are tired from typing.

If you have input, or come up with your own results, or have criticisms, please contribute! I would like to see if these definitions bring out anything cool.

 

[micromass]

An interesting result to is to show that \phi is actually a commutative monoid isomorphism. I leave this up to the reader.

This is false. \phi is not surjective. For example (1,1,1,1,...) is never reached.

You might be interested in http://en.wikipedia.org/wiki/Supernatural_numbers

 

[Kindayr]

geez, I wish I had seen that before typing all that up.

An isomorphism would exist if you worked in \mathbb Z_{p} instead? That is, pick a prime p, and for all primes 1<p_{1}<p_{2}<\cdots<p_{n}=p, we can define an isomorphism \phi:\mathbb Z_{p}\to\mathbb Z_{p}^{n} such that if a\in\mathbb Z_{p} then \phi(a)=(e_k)_{k=1}^{n} where a=\prod_{k=1}^{n}p_{k}^{e_{k}}. Wouldn't \phi even be a commutative group isomorphism?

Maybe I even want p_{n}<p? I have to think about this more and make it more rigourous.

Thank you for the link though!

 

[micromass]

geez, I wish I had seen that before typing all that up.

An isomorphism would exist if you worked in \mathbb Z_{p} instead? That is, pick a prime p, and for all primes 1<p_{1}<p_{2}<\cdots<p_{n}=p, we can define an isomorphism \phi:\mathbb Z_{p}\to\mathbb Z_{p}^{n} such that if a\in\mathbb Z_{p} then \phi(a)=(e_k)_{k=1}^{n} where a=\prod_{k=1}^{n}p_{k}^{e_{k}}. Wouldn't \phi even be a commutative group isomorphism?

Maybe I even want p_{n}<p? I have to think about this more and make it more rigourous.

Thank you for the link though!

You mean to say \mathbb{Z}_p^*\rightarrow \mathbb{Z}_p^n, right?? Well, that could work. I suggest you try to prove it!

In general, something that might work is \mathbb{N}^{(\infty)}: that are all the tuples (n_1,n_2,n_3,...) such that only a finite number of terms are nonzero. You'll have better luck proving that an isomorphism.

Also, there is no need to work with \mathbb{N} for this. You can also do the same thing for \mathbb{Q}. So we can define a map

\mathbb{Q}^*\rightarrow \mathbb{Z}^{(\infty)}

This are some things you should ponder on. But the OP was verry interesting!

 

[Kindayr]

Thanks for the help!

And yes, I did mean \mathbb Z_{p}^{*}\to\mathbb Z_{p}^{n}. That fixes the problem that caused me to ask if we should have p_{n}<p.

I also want to give an example of what I meant with the polynomials. I don't want to type out the entire definition of the operation I'm going to use, but I hope the example gives light to it:

6\otimes8\simeq(1,1,0,\dots)\otimes'(3,0,0,\dots) \simeq (1+x)(3)=3+3x\simeq(3,3,0,\dots)\simeq 2^{3}3^{3}=8\times 27=216.

This operation can also be shown to be commutative. However in this case, (0,0,0,\dots) acts as a 'absorbing' element, not unlike 0 with respect to (\mathbb Z,\cdot).

I haven't done any group theory or abstract algebra officially, this is all just what I'm starting to pull from Artin's Algebra. I really like it so far :)

I'll also work on that isomorphism.

Also, could that operation be used in cryptography?

 

[micromass]

Hmmm, that polynomial thing is quite interesting. I've not yet encountered such a thing, but it reminds me in some way of group rings. It should be worth studying your operation in that context.

 

[Kindayr]

The interesting part is that the 'product' of any two primes, results in another prime. I thought that was cool.

 

[Kindayr]

Fix p prime. We can define a function \parallel\cdot\parallel_{\phi_{p}}:\mathbb N^{*}\to\mathbb R such that for all a\in\mathbb N^{*}, we have \parallel a\parallel_{\phi_{p}}=\,\parallel \phi(a)\parallel_{p}. This can give us a sense of 'length' that each number has. For example: \parallel6\parallel_{\phi_{2}}=\,\parallel \phi(6)\parallel_{2}=\parallel (1,1,0,\dots)\parallel_{2}=\sqrt{2},\parallel 8\parallel_{\phi_{2}}=\,\parallel \phi(8)\parallel_{2}=\parallel(3,0,\dots)\parallel_{2}=3,and\parallel 7\parallel_{\phi_{2}}=\,\parallel \phi(7)\parallel_{2}=\parallel (0,0,0,1,0,\dots)\parallel_{2}=1.Note that 6<7 but \parallel7\parallel_{\phi_{2}}<\parallel6\parallel_{\phi_{2}}.

I've also thought about making an encryption algorithm changing letters to irreducible polynomials over \mathbb N^{*} using my operation I defined in one of the earlier posts. You could even go further to making words map to matrices in RCF, or something like that. I know nothing about cryptography though ahahah.

We can even make a sort of inner product of in \mathbb N^{*} and fool around in there. For example, \langle6,8\rangle_{\phi}=\langle(1,1,0,\dots),(3,0,0,\dots)\rangle=3.Note that we could show that all primes are orthogonal to one another.

 

[Hurkyl]

What if you defined \mathbb{N}^\infty as the set of all (countably) infinite sequences with only finitely many nonzero entries...

 

[Kindayr]

What if you defined \mathbb{N}^\infty as the set of all (countably) infinite sequences with only finitely many nonzero entries...

micromass already mentioned that I believe, and I'm using that criterion for \mathbb N^{\infty} in the paper I'm writing to show this to a professor of mine who I hope to help me push out some cool results.

 

[Hurkyl]

micromass already mentioned that I believe, and I'm using that criterion for \mathbb N^{\infty} in the paper I'm writing to show this to a professor of mine who I hope to help me push out some cool results.

Argh, I completely missed that post when I skimmed the thread.

 

[Kindayr]

Argh, I completely missed that post when I skimmed the thread.

No worries! I'm no less thankful for your input. Its something that allows a lot of this to hold, and I completely missed it when I was defining things originally. I've uploaded my paper so far, I'm adding and proof reading it as I go along and discover new things. I haven't added the new 'length' idea, nor the 'inner product' stuff.

 

[micromass]

No worries! I'm no less thankful for your input. Its something that allows a lot of this to hold, and I completely missed it when I was defining things originally. I've uploaded my paper so far, I'm adding and proof reading it as I go along and discover new things. I haven't added the new 'length' idea, nor the 'inner product' stuff.

Interesting paper!
Allow me to make some remarks:

- in Theorem 2, you can even show injectivity. So these things are a submonoid.
- Prime factorizations exist over the rationals too, for example: 5/6=2^{-1}3^{-1}5. So your map phi generalizes to
\phi:\mathbb{Q}^*\rightarrow \mathbb{Z}^\infty
- There is a small problem with the definition after theorem 2. The thing \sum{a_kx^k} is NOT a polynomial. Indeed, polynomials have only a finite amount of terms. However, it is a formal power series. So you need \psi to go the the formal power series.

 

[Kindayr]

Interesting paper!
Allow me to make some remarks:

- in Theorem 2, you can even show injectivity. So these things are a submonoid.
- Prime factorizations exist over the rationals too, for example: 5/6=2^{-1}3^{-1}5. So your map phi generalizes to
\phi:\mathbb{Q}^*\rightarrow \mathbb{Z}^\infty
- There is a small problem with the definition after theorem 2. The thing \sum{a_kx^k} is NOT a polynomial. Indeed, polynomials have only a finite amount of terms. However, it is a formal power series. So you need \psi to go the the formal power series.

Thank you very much micromass!

If I generalize \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}, does this not also give me a group?

Also, if I define \mathbb{P}[X]^{\infty} to be the single-variable polynomials with only a finite amount of nonzero coefficients, would this fix the problem arisen in your third point?

 

[micromass]

Also, the thing you posted will not be a norm of course, since it won't satisfy it's properties. Here's something you can do to save it

Let \Phi:\mathbb{Q}^*\rightarrow \mathbb{Z}^{(\infty)}:q\rightarrow (q_1,q_2,...)

For example \Phi(5/6)=(-1,-1,1,0,0,0,...)

Try to define a norm like

\|q\|=\sum_{i=1}^{+\infty}{2^{-q_i}}

I guess this would define a norm if you also define \|0\|=0.
My inspiration of course comes from the p-adic numbers...

 

[micromass]

Thank you very much micromass!

If I generalize \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}, does this not also give me a group?

Yes, I believe so!

Also, if I define \mathbb{P}[X]^{\infty} to be the single-variable polynomials with only a finite amount of nonzero coefficients, would this fix the problem arisen in your third point?

Yes.

 

[Kindayr]

Also, the thing you posted will not be a norm of course, since it won't satisfy it's properties. Here's something you can do to save it

Let \Phi:\mathbb{Q}^*\rightarrow \mathbb{Z}^{(\infty)}:q\rightarrow (q_1,q_2,...)

For example \Phi(5/6)=(-1,-1,1,0,0,0,...)

Try to define a norm like

\|q\|=\sum_{i=1}^{+\infty}{2^{-q_i}}

I'm just confused here how my 'norm' doesn't work :(

I'm just taking any n, bringing it to \mathbb{N}^{\infty}, and taking the Euclidean norm.

I guess this would define a norm if you also define \|0\|=0.
My inspiration of course comes from the p-adic numbers...

I think you mean \|1\|=0, since 0\notin\mathbb{Q}^{*}

 

[micromass]

I'm just confused here how my 'norm' doesn't work :(

I'm just taking any n, bringing it to \mathbb{N}^{\infty}, and taking the Euclidean norm.

Well, it might work. But you need to prove that
\|\alpha x\|=|\alpha|\|x\|~\text{and}~\|x+y\|\leq \|x\|+\|y\|

I think you mean \|1\|=0, since 0\notin\mathbb{Z}^{*}

No, I meant what I said. Of course 0 is not in there, but I put it in there. The reason is that we want to have a vector space, and 0 needs to be an element of the vector space.

Of course, you don't NEED the norm properties, but it would be nice to have them...

 

[Kindayr]

Well, it might work. But you need to prove that
\|\alpha x\|=|\alpha|\|x\|~\text{and}~\|x+y\|\leq \|x\|+\|y\|
No, I meant what I said. Of course 0 is not in there, but I put it in there. The reason is that we want to have a vector space, and 0 needs to be an element of the vector space.

Of course, you don't NEED the norm properties, but it would be nice to have them...

But wouldn't 1 be the zero of the vector space since \Phi(1)=(0,0,\dots). And by your norm, wouldn't \|1\|=\sum^{+\infty}_{i=1}2^{-q_{i}}=1+1+\cdots=+ \infty.

Just to add too, if we are trying to get a vector space, then would the \oplus generalized to \mathbb{Z}^{\infty} be vector addition, and exponentiation in \mathbb{Q}^{*} be the scalar multiplication?

 

[micromass]

But wouldn't 1 be the zero of the vector space since \Phi(1)=(0,0,\dots).

Just to add too, if we are trying to get a vector space, then would the \oplus generalized to \mathbb{Z}^{infty} be vector addition, and exponentiation in \mathbb{Q}^{*} be the scalar multiplication?

That's nice, but I doubt that this will give you an actualy vector space...
Remember that a vector space must be defined over a field. So I'm not sure what your scalar multiplication will be?

 

[Hurkyl]

As an aside, one particular application of your map is to show that Peano arithmetic is powerful enough to talk about sequences of natural numbers of arbitrary (finite) length. (e.g. this is a major part of the technical content of the proof of Gödel's incompleteness theorems)

 

[Kindayr]

That's nice, but I doubt that this will give you an actualy vector space...
Remember that a vector space must be defined over a field. So I'm not sure what your scalar multiplication will be?

Choose our field to be \mathbb{Z}. Then let \alpha\in\mathbb{Z} and q\in\mathbb{Z}^{\infty} with q=(q_{1},q_{2},\dots). Then \alpha q=\alpha (q_{1},q_{2},\dots)=(\alpha q_{1},\alpha q_{2},\dots). Note that this isomorphic to exponentiation of rationals by integers. That is, let p\in\mathbb{Q}^{*} and n\in\mathbb{Z}. Then \Phi(p^{n})=(np_{1},np_{2},\dots).

 

[micromass]

Choose our field to be \mathbb{Z}.

That's not a field You would end up with a \mathbb{Z}-module, which is also cool. But you wouldn't have a normed space...

 

[Kindayr]

As an aside, one particular application of your map is to show that Peano arithmetic is powerful enough to talk about sequences of natural numbers of arbitrary (finite) length. (e.g. this is a major part of the technical content of the proof of Gödel's incompleteness theorems)

Whoa, could you please explain? I haven't been exposed to much of any metamathematics. And I find this super interesting, because a lot of ideas of this came from me working with Peano Spaces, and trying to find a difference between primes and composite numbers. My previous discussion could be found https://www.physicsforums.com/showthread.php?t=521157".

Thank you all for your help so far though!

 

[Kindayr]

That's not a field You would end up with a \mathbb{Z}-module, which is also cool. But you wouldn't have a normed space...

What if we work in \mathbb{Z}^{n}_{p} over a the field \mathbb{Z}_{p}, as I discuss in the second part of my paper? This gets a bit messy I guess though.

 

[micromass]

What if we work in \mathbb{Z}^{n}_{p} over a the field \mathbb{Z}_{p}, as I discuss in the second part of my paper? This gets a bit messy I guess though.

Aaah, that might work

 

[Kindayr]

Aaah, that might work

Then I will play around with that tonight, and type some stuff up tomorrow! I'll try to generalize everything using the map you defined, and see what I can do.

thank you very much micromass!

 

[Kindayr]

I'm just trying to find a way to work this.

Fix p prime. Let 1<p_{1}<p_{2}<\cdots<p_{n}<p be all primes between 1 and p, and let \mathbb{Z}_{p}^{n} be the set of all n-tuples with entires in \mathbb{Z}_{p}. Then our map \Phi doesn't work since we can find a q\in\mathbb{Q}^{*} whose factorization uses a prime p_{0}>p.

So that's just where I'm confused.

I'm just trying to get over a slump on how to turn this into a vector space.

 

[Hurkyl]

Whoa, could you please explain? I haven't been exposed to much of any metamathematics.

Once you have a data structure, you can do a lot. Just look at set theory.

Similar power is opened up to you once you can view an integer as a list of integers.

There are a variety of related things, I'll go down the route of computability -- what is a computer? It is essentially just it's current state -- a list of integers (e.g. the list might contain 16 registers, and the rest be memory which can be interpreted as instructions or data) -- and a rule for advancing from one state to the next.

The hardest part of encoding this in Peano arithmetic is how to use an integer to encode a list of integers, and how to access / manipulate lists under this encoding.

(Then, once you have encoded the state of a computer as a single integer, you can then talk about the execution of a program on a computer as the "list of integers" corresponding to the state of the computer at a given time. With this, you can now start reasoning about how computers behave, again using just Peano arithmetic under-the-hood)


Formal logic is similar -- specifically look at the subfield of syntax, which talks about languages, grammar, and proofs. Here, you want to use a "list of integers" to store the characters in a formal expression -- i.e. to store strings.



There are other ways to manage this technical construction, but I think the isomorphism from your original post is the most straightforward. (it does require you to first prove some things about primes and divisibility, though)

 

[Kindayr]

So I typed it up again, generalizing \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}.

I'm still a little confused how to turn this into a vector space. I was wondering, if we use the non-transcendental numbers as our field. But would we lose the fact of 'prime factorization', and the basis of even doing this?

I can't conceptually see how to work in \mathbb{Z}_{p} for some prime p. I don't know how we would chose p. Maybe I'm just obsessing over minor points.

I've also defined \psi:\mathbb{Z}^{\infty}\to\mathbb{P}[X]^{\infty}, but I just don't know how to define a binary operation \otimes:\mathbb{Z}^{\infty}\times\mathbb{Z}^{ \infty}\to\mathbb{Z}^{\infty} such that it is homomorphic by \psi to multiplication of finite polynomials.

 

[micromass]

So I typed it up again, generalizing \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}.

I'm still a little confused how to turn this into a vector space. I was wondering, if we use the non-transcendental numbers as our field. But would we lose the fact of 'prime factorization', and the basis of even doing this?

I can't conceptually see how to work in \mathbb{Z}_{p} for some prime p. I don't know how we would chose p. Maybe I'm just obsessing over minor points.

Can't you just define it as a \mathbb{Q}-vector space?

And maybe I'm starting to do much to advanced stuff here, but perhaps you could define your entire structure as a sheaf of rings??

I've also defined \psi:\mathbb{Z}^{\infty}\to\mathbb{P}[X]^{\infty}, but I just don't know how to define a binary operation \otimes:\mathbb{Z}^{\infty}\times\mathbb{Z}^{ \infty}\to\mathbb{Z}^{\infty} such that it is homomorphic by \psi to multiplication of finite polynomials.

You know that that is a bijection, right?? Well, then you can just define

a\otimes b=\psi^{-1}(\psi (a)\cdot \psi (b))

just carry over the structure.

Do take a look at "group rings", because it really looks a lot like what you're trying to do here!

 

[Kindayr]

Can't you just define it as a \mathbb{Q}-vector space?

That's what i mean, woops. But then we would be representing the non-transcendentals. Because if we choose \mathbb{Q} to be our field, then since scalar multiplication in \mathbb{Z}^{\infty} is isomorphic to exponentiation, things like \frac{1}{2}(1,0,\dots)\overset{\phi^{-1}}{=} 2^{\frac{1}{2}}=\sqrt{2}\notin\mathbb{Q}^{*}So shouldn't we work with the set of non-zero algebraic numbers, as opposed to \mathbb{Q}^{*} (this is what I meant as opposed to non-transcendental).

And maybe I'm starting to do much to advanced stuff here, but perhaps you could define your entire structure as a sheaf of rings??

Yeah, I haven't a clue what a sheaf of rings is >.<, i haven't even had a true introduction to group theory. I'm doing this a lot by reading as I go.

You know that that is a bijection, right?? Well, then you can just define

a\otimes b=\psi^{-1}(\psi (a)\cdot \psi (b))

just carry over the structure.

Perfect! Can't believe I didn't think of that, thank you very much!

Do take a look at "group rings", because it really looks a lot like what you're trying to do here!

No I haven't yet, totally forgot you mentioned that. I'll definitely take a look now before I continue :)

 

[micromass]

That's what i mean, woops. But then we would be representing the non-transcendentals. Because if we choose \mathbb{Q} to be our field, then since scalar multiplication in \mathbb{Z}^{\infty} is isomorphic to exponentiation, things like \frac{1}{2}(1,0,\dots)\overset{\phi^{-1}}{=} 2^{\frac{1}{2}}=\sqrt{2}\notin\mathbb{Q}^{*}

Well, defining it like that isn't exactly what I mean

You know that \mathbb{Q} is isomorphic to \mathbb{Z}^{\infty} (if we adjoin a certain element 0 to the last set). So define scalar multiplication by

\alpha x=\varphi(\varphi^{-1}(\alpha)\varphi^{-1}(x)). So just "lift" the scalar multiplication in \mathbb{Q} to a new one in \mathbb{Z}^\infty.

Yeah, I haven't a clue what a sheaf of rings is >.<, i haven't even had a true introduction to group theory. I'm doing this a lot by reading as I go.

Well, I think that researching things because you need it in this problem is the only true way to learn something. So all power to you!
But sheafs is too advanced It would work though.

 

[Kindayr]

Well, defining it like that isn't exactly what I mean

You know that \mathbb{Q} is isomorphic to \mathbb{Z}^{\infty} (if we adjoin a certain element 0 to the last set). So define scalar multiplication by

\alpha x=\varphi(\varphi^{-1}(\alpha)\varphi^{-1}(x)). So just "lift" the scalar multiplication in \mathbb{Q} to a new one in \mathbb{Z}^\infty.

I though I've already defined the scalar multiplication in \mathbb{Q} in \mathbb{Z}^{\infty} through defining \phi(a)=\phi(\prod^{\infty}_{k=1}p_{k}^{e_{k}})=(e_{1},e_{2},\dots)=(e_{k})^{\infty}_{k=1}. And since \phi is an isomorphism (I now show this properly in my paper), multiplication of numbers in \mathbb{Q}^{*} is isomorphic to the 'addition' i defined (\oplus) in \mathbb{Z}^{\infty}.

So there our vector addition is defined as multiplication in \mathbb{Q}^{*} (or algebraic numbers, to generalize).

Then it makes sense that our scalar multiplication (from \mathbb{Q}) would be isomorphic to exponentiation. Since it can be shown, for example, that

a^{2} \overset{\phi}{=} 2 (e_{k})^{\infty} _{k=1} = (e_{k}) ^{\infty} _{k=1} \oplus (e_{k}) ^{\infty}_{k=1}.

I guess I'm actually generalizing even more instead of \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}, but to \phi:\mathbb{A}^{*}\to\mathbb{Q}^{\infty}, where \mathbb{A}^{*} is the set of all non-zero algebraic numbers.

Am I making sense or am I just crazy?

 

[micromass]

I though I've already defined the scalar multiplication in \mathbb{Q} in \mathbb{Z}^{\infty} through defining \phi(a)=\phi(\prod^{\infty}_{k=1}p_{k}^{e_{k}})=(e_{1},e_{2},\dots)=(e_{k})^{\infty}_{k=1}. And since \phi is an isomorphism (I now show this properly in my paper), multiplication of numbers in \mathbb{Q}^{*} is isomorphic to the 'addition' i defined (\oplus) in \mathbb{Z}^{\infty}.

So there our vector addition is defined as multiplication in \mathbb{Q}^{*} (or algebraic numbers, to generalize).

Then it makes sense that our scalar multiplication (from \mathbb{Q}) would be isomorphic to exponentiation. Since it can be shown, for example, that a^{2} \: \overset{\phi}{=} 2 (e_{k})^{\infty} _{k=1} = (e_{k}) ^{\infty) _{k=1} \oplus (e_{k}) ^{\infty )_{k=1}.

I guess I'm actually generalizing even more instead of \phi:\mathbb{Q}^{*}\to\mathbb{Z}^{\infty}, but to \phi:\mathbb{A}^{*}\to\mathbb{Q}^{\infty}, where \mathbb{A}^{*} is the set of all non-zero algebraic numbers.

Am I making sense or am I just crazy?

Hmm, I'm not sure how you would generalize all of this to algebraic numbers. But it does make sense what you're saying...

 

[Kindayr]

Hmm, I'm not sure how you would generalize all of this to algebraic numbers. But it does make sense what you're saying...

I guess I would I would have to prove a sort of analogous Fundamental Theorem of Arithmetic to the non-zero algebraics? If I can do that then I hope it all works, and I would just have to prove it to be a vector space, correct?

I never expected myself to get into doing this much for such a trivial idea I began with lol! Considering how little algebraic background I have outside of linear algebra.

 

[micromass]

I guess I would I would have to prove a sort of analogous Fundamental Theorem of Arithmetic to the non-zero algebraics? If I can do that then I hope it all works, and I would just have to prove it to be a vector space, correct?

I think such a thing would be very hard, if not impossible

I never expected myself to get into doing this much for such a trivial idea I began with lol! Considering how little algebraic background I have outside of linear algebra.

Hey, this is the perfect way to learn something more about algebra! You're doing great!

 

[Kindayr]

As a last attempt, what if we just generalize to the set

\mathbb{A}^{*}=\{a^{b}:a,b\in\mathbb{Q}^{*}\}

I would have to show this to be a field obviously.

 

[micromass]

As a last attempt, what if we just generalize to the set

\mathbb{A}^{*}=\{a^{b}:a,b\in\mathbb{Q}^{*}\}

I would have to show this to be a field obviously.

That seems plausible. Don't know if it'll be a field, though...

 

[Kindayr]

That seems plausible. Don't know if it'll be a field, though...

I agree, I don't think its closed under addition.

I guess I never really tried \mathbb{Z}_{p}. Or even the set \mathbb{Q}[\mathbb{Z}_{p}]=\{a^{b}|a,b\in\mathbb{Z}_{p}\}. Which would be closed under addition (I'd hav to prove it) I guess since every a^{b}\equiv c&lt;p, so we would have a^{b}+c^{d}\equiv e+f\equiv g with e,f,g\leq p, or something like that.

 

[Kindayr]

So working I've got some weird results, that i'll type later when you fix a p&gt;0 prime, and define \phi_{p}:\mathbb{Z}_{p}^{*}\to\mathbb{Z}_{p}^{n} where 1&lt;p_{1}&lt;p_{2}&lt;\cdots &lt;p_{n}&lt;p are all primes less than p.

I can show this to be a vector space, which is good. But you get weird equivalences when you work in \mathbb{Z}_{p}^{n}, that are just blowing my mind a little as I write them down on paper. Which get even weirder when you move to \mathbb{P}_{n} [\mathbb{Z}_{p}.

I'll type it all up when I get home.

 

[micromass]

Hmm, "weird equivalences", I'm wondering about what you will type up

 

[Kindayr]

Well its trivial to show that, from the same p and n defined before that \phi_{p}:\mathbb{Z}_{p}^{*}\to\mathbb{Z}^{n}_{p} is an isomorphism. This is seen since if a\in\mathbb{Z}_{p}^{*} then a has a prime factorization such that a=\prod^{n}_{k=1}p_{k}^{e_{k}}=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{n}^{e_{n}}. So let \phi_{p}(a)=(e_{1},\dots,e_{n}). Its really easy to show its a vector space over field \mathbb{Z}_{p}

Whats interesting is when you partition the set such that for any u,v\in\mathbb{Z}_{p}^{n}, we have u\simeq v \;\; iff \;\; \phi_{p}^{-1}(u)\equiv\phi_{p}^{-1}(v).

If you use the definition of the isomorphism I've used before into the polynomials, with the same partition, you'll get weird things. For example, take p=7. Then n=3. So we have 0\simeq3\simeq x+x^{2}\simeq2+2x\simeq\cdots

I've worked with quotient spaces in linear algebra, and I though how that collapsed a vector space was weird, but this weirds me out even more.

Its probably not weird for someone who has taken abstract"er" algebra than I have, but still weird lol

I'll be typing this all up sometime tonight or tomorrow when I have time in a paper and more rigourously, but i have other plans tongiht so I won't be able to type everything up i worked out today (like my proof that its a vector space).

I feel even that \mathbb{Z}_{p}^{n} collapses \mathbb{Q}^{n}. Since any n-tuple's entries can be seen as a quotient a/b, but then since \mathbb{Z}_{p} is a field, so this quotient exists as one of the p members of Z_p.

Maybe that's wrong.

 

[micromass]

I don't quite see why \phi_p is an isomorphism. That is, I see no reason why it should be surjective. Certainly something like (p-1,p-1,...,p-1) isn't reached?

 

[Kindayr]

I don't quite see why \phi_p is an isomorphism. That is, I see no reason why it should be surjective. Certainly something like (p-1,p-1,...,p-1) isn't reached?

Sure it does, it corresponds to p_{1}^{p-1}p_{2}^{p-1}\cdots p_{n}^{p-1}\equiv a with 0\leq a\leq p-1.

Wait crap, that makes it non-injective. URGH.

I need to sit down with someone else and work all of this out. I need to get back to university asap. BLARGH.

 

[micromass]

Sure it does, it corresponds to p_{1}^{p-1}p_{2}^{p-1}\cdots p_{n}^{p-1}\equiv a with 0\leq a\leq p-1.

Wait crap, that makes it non-injective. URGH.

I need to sit down with someone else and work all of this out. I need to get back to university asap. BLARGH.

The problem is that you don't have a well-defined function, really.

\mathbb{Z}_p^* has p-1 elements, and \mathbb{Z}_p^n has p^nelements. So there can never be a surjection.

Do you see a way to solve this??

 

[Kindayr]

Okay, I'm meeting with my prof tomorrow, and I want to really see if this works:

Define \mathbb{A}^{*} =\{a^{b}:a,b\in\mathbb{Q}^{*}\} and let \phi :\mathbb{A}^{*}\to\mathbb{Q}^{\infty}, where \mathbb{Q}^{\infty} is the set of all \infty-tuples with a finite amount of non-zero entries. Note that for any a^{b}\in\mathbb{A}^{*} we have a prime factorization such that a^{b}=\prod_{k=1}^{\infty}p_{k}^{be_{k}}. So define \phi(a^{b})=(be_{1},be_{2},\dots)=(be_{k})^{\infty}_{k=1}. Also define, from before, \oplus:\mathbb{Q}^{\infty}\times\mathbb{Q}^{\infty}\to \mathbb{Q}^{\infty} such that (e_{1},e_{2},\dots)\oplus (f_{1}, f_{2},\dots)=(e_{k})^{\infty}_{k=1}\oplus (f_{k})^{\infty}_{k=1}=(e_{k}+f_{k})^{\infty}_{k=1}=(e_{1}+f_{1},e_{2}+f_{2},\dots). Further, \phi can be shown to be an isomorphism.

We can also show \mathbb{Q}^ {\infty} to be a rational vector space. Note that the scalar multiplication of any vector in \mathbb{Q}^{\infty} is analogous to raising the original element of \mathbb{A}^{*} by a rational exponent (the rational exponent being the scalar). That is: \alpha(be_{k})^{\infty}_{k=1}\overset{\phi^{-1}}{=}a^{b^{\alpha}}=a^{b\alpha}. Also note that \alpha(be_{k})^{\infty}_{k=1}=\alpha b(e_{k})_{k=1}^{\infty}=(\alpha be_{k})^{\infty}_{k=1}.

I didn't want to type everything out, but it seems to have worked.

 

[Kindayr]

So I've met with my professor, and by the end we just were laughing at how simple and familiar it ended up working out to be.

No matter how hard I tried, I could not create a vector space, which is fine. The suddenly it came to us.

So here:

Let (\mathbb{R}^{&gt;0},\cdot) be the abelian group of positive reals and multiplication, and let (\mathbb{R},+) be the abelian group of the reals over addition. Suppose there exists a function \log:(\mathbb{R}^{&gt;0},\cdot)\to(\mathbb{R},+) such that \log(ab)=\log(a)+\log(b) and \log(a^{b})=b\log(a). Its clear that \log was just the extension of what I was trying to do before. I guess you could say its the 'analytic jump' from what I was trying to do algebraically. This has given me some new intuition into how \log works. That its 'sort of' rooted in prime factorization, but more general than that.

I thought this was really interesting. And I'm glad this is all over of me attempting to find something that wasn't there, but had some sort of 'transcendental' version that explains everything.

He gave me some homework to do, the first question being one that I could probably do my self with my current skill sets, and another I'll have to wait after I take Measure Theory in the Winter Semester.

The first is that:

• 
  If f:(\mathbb{R}^{&gt;0},\cdot)\to(\mathbb{R},+) continuously with the properties that f(ab)=f(a)+f(b) and f(a^{b})=b\cdot f(a) and f(1)=0, then f(x)=\log(x)

• 
  There exist many non-measurable f:(\mathbb{R}^{&gt;0},\cdot)\to(\mathbb{R},+) with the properties f(ab)=f(a)+f(b), f(a^{b})=b\cdot f(a), and f(1)=0.

I hope to be able to do these sometime in the future, but we'll see what happens.

Thank you to everyone that has helped me! Especially micromass for putting up with my lack of education in abstract algebra and metric space topology. I swear I'll get better after this year when I finally get a formal introduction to everything! haha

 

[micromass]

I'm very glad you have found an answer to your (very intriguing) question. I'm also happy that you learned a bit of new mathematics this way!

I hope you enjoyed thinking about this. These are the things that math research is all about