If you removed all logarithms, would every real number have a unique factorization?

Question

empty · Answer

The reason I say this is because it seems like the only thing that breaks factorization is stuff like this:

$$2^{\log_2(3)} = 3$$

So it would have a nonunique prime factorization. Without logs though, you can't do this anymore.

empty · Answer

Like for instance, now we can factor something as:

$$2^\sqrt{5} \ne 3^k$$ since in order to "flip factors around" you'd have to have logs:

$$\sqrt{5} \log_3(2) = k$$

empty · Answer

Or at least that's what I think, unless someone can come up with some kinda counter example I'd be really curious to see.

empty · Answer

Or even just something to try to prove that this is really true in some more concrete sense? I dunno haa.

ganeshie8 · Answer

By factorization, do you mean power of primes and allow real numbers as exponents ?

empty · Answer

Yeah, I suppose so, whatever it takes to generalize the concept of prime factorization to work for larger than what we currently have.

ganeshie8 · Answer

$$\prod\limits_{k=1}^{n}{p_k}^{e_k}~ = ~e^{\sum\limits_{k=1}^n \ln{p_k}^{e_k}}$$

ganeshie8 · Answer

If I let $n\to\infty$, above expression can be made to converge to ANY real number by choosing the sequence $\{{p_k}^{e_k}\}$ appropriately I guess...

empty · Answer

Yeah, I agree, that's why logarithms have to go. Unfortunately it doesn't seem like a very simple matter since we can of course get anything with any infinite sequence.

At the very least it seems as though we can have all primes raised to rational exponents only, that would remain closed.

ganeshie8 · Answer

https://www.wolframalpha.com/input/?i=%5Cprod%5Climits_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+k%5E(1%2Fk%5E2)

ganeshie8 · Answer

That is one factorization of the real number 
2.553712682748209052939314574440964078667151038214817128135...

empty · Answer

Well a few things, k isn't prime, and the 'factorization' you give is not finite. Limits can give you anything you want which is why I'm saying we have to remove them and it's not completely crazy since factorizations are finite normally.


Now if you want to include infinite numbers like this, than that's another story and also kind of interesting.
$$\prod_{i=1}^\infty p_i$$

empty · Answer

So does this real number you have here a factorization in this way? Not that I know of.

empty · Answer

But I'm not really interested in getting all real numbers. Already we don't have any logarithms with is infinitely many numbers. I'm trying to throw out as few things as possible while maintaining the structure of factorizability.

ganeshie8 · Answer

$k$ can be made prime by grouping the integers

ganeshie8 · Answer

With a finite product, I don't see yet how we can have two different factorizations hmm

empty · Answer

To make the distinction more clear, let's just give them a name, if it's not too lame sounding lol.
The numbers I want, let's call them "Super integers" since they have finite factorizations in primes whereas normal integers have finite factorizations in the primes.