Are these two spaces isomorphic?

Question

Ok, first of all, I am sorry if all of what I am going to write turns out to actually be rubbish, or it has already been explained/discussed somewhere but i couldn't find it; but i had an idea today and i wanted to know if that is actually legitimate.

We know that every natural number has a unique prime factorization. for example: $48=2^4 \times 3$, knowing this we can unicquely identify a number with a vector that displays the exponents of all the prime numbers in its prime factorization, for example $$48:=(4,1,0,0,0,...,0)$$(I am not counting 1 as a prime)

And this vector will only represent the number $48$ and the number $48$ will only be represented by that vector so thi definition should be accurate. Now it is pretty clear that adding up vectors actually means multiplicating numbers in this representation $$14\times12=(1, 0, 0, 1, 0, ..., 0)+(2,1,0,...,0)=(3,1,0,1,0,...,0)=2^3\times3\times7= 168$$ And multiplying a vector for a numerical value would just be the same as raising the number to that value $$ 14^2=2\times(1,0,0,1,0,...,0)=(2,0,0,2,0,...,0)=2^2\times7^2=196$$ And we also have a null element which equals to one $$1=(0,0,0,0,...,0)=2^0\times3^0\times5^0...=1 $$ Therefore, having defined these two operations, shouldn't these mean that the natural Numbers are a Vector space? And furthermore, there is something that weirds me out, sice we can represent every natural numbers with a vector that has endless entries (because there are indeed an infinite amount of prime numbers) shouldn't this mean that the "vector space of natural numbers" which we defined earlier is isomorphic to $(\mathbb{N}+\{0\})^\infty$ because it seems to me that those vectors are elements of $(\mathbb{N}+\{0\})^\infty$ (I added 0 separately because i don't usually count it as a Natural Number) and what we described actually looks like a bijection.

I apologize for any spellig or math mistake i may have made in this post, and wait impatiently for your answers.

Answer 1

A vector space is defined over a field. What you have defined here is a commutative monoid endowed with an action of the natural number it is not a vector space.

Answer 2

First: +1 for discovering this way to encode the multiplicative structure of the positive integers: each is represented by a sequence of natural numbers all but finitely many of which are $0$.

You add those, by adding corresponding entries, which (as you point out) corresponds to multiplying the positive integers they represent. The sequence of all zeroes is the "$0$" element. And you know how to "multiply" any sequence by a natural number - essentially by repeated "addition".

The structure you are using isn't a vector space since it does not satisfy all the vector space axioms. For that to be true you would have to be able to multiply by an arbitrary number, like 1/2 or $\pi$. Then you could find the "negative" of any vector: the thing to add that gives you the identity. Clearly there's no such sequence in your system since the elements are the powers of the primes that the number represented uses, and those can't be negative.

To summarize: you've discovered a nice way to represent the positive integers as sequences of natural numbers that turns multiplication into addition. Your argument relies on knowing any integer is uniquely the product of primes. You don't end up with a vector space, but that doesn't make the idea less valuable. There are some places in mathematics where this is a useful way to look at things. You can find a discussion here: https://en.wikipedia.org/wiki/Supernatural_number .