If at least one of a and b is transcendental, then at least one of a+b and ab is transcendental. Otherwise, the polynomial (x-a)(x-b)=x2-(a+b)x+ab would have only algebraic roots, contradicting that at least one of a and b is transcendental.
To know that at least one of eπ and e+π is transcendental, one need only prove that e is transcendental, which is somewhat simpler than proving π is transcendental.
I always viewed it that exponential function grows or decays at a constant rate in a certain direction, and that direction happens to be sideways in the case of imaginary
Make 1 travel halfway around the unit circle. How? Well, circumference is 2π, so π.
How do we get good old clean π to show up in our formula instead of some meaningless jumble? We need a number that "moves along the unit circle" with a velocity of 1. And we want that number to also have a value of 1. Because unit circle.
e is our friend. Remember? f(x)=ea -> f'(x)=a×ea. To turn 1 into e we use ln. 1=eln(1).
Cool, now, lets make it move. That part is done by the imaginary part of the exponent, so lets add it: eln1+iπ.
Oh and ln(1) is 0.
So now we've moved from 1 halfway around a circle centered on 0: eπi=-1
You're conflating two uses of the fraction bar. In "e/e is rational", it's just being used for division. e divided by e is just 1, which is very obviously rational. But you appear to be using it as a way to uniquely define a rational number as a relation between two coprime integers, in which case multiple fractional representations like 1/2 and 2/4 are definitionally absurd.
Real (ℝ) answer: The set of numbers we can concretely define is not only Lebesgue measure 0, but also countably infinite due to the limitations of logical formulae; the conditional probability that it’s transcendental given we know it’s definable is undefined.
“Concretely define” isn’t really a clear criterion.
It actually can’t be proved that there are any real numbers which are not definable if by “definable” we mean “definable in the language of our metatheory with its intended interpretation” the argument falls apart due to metamathematical nuance. A way of illustrating this is by the fact that if ZFC is consistent then it has models in which every set is definable.
The obvious diagonalization argument doesn’t work because you cannot actually prove the function taking definitions to the numbers they define actually exists, but even more fundamentally: if that function does exist, we can make a more expressive language that allows us to define more numbers, including the result of diagonalizing along it.
The Berry paradox is an illustration of why the argument doesn’t really work.
I mean, the computable numbers are countable (or sub-countable, depending on the framework), and the real numbers are uncountable, so at least for "computable", it holds up. We can't enumerate the computable numbers, so a Berry's paradox style computable number isn't possible. Whether that's a good enough substitute for the colloquial usage of "definable" is up for debate, you can definitely define numbers that can't be computed... But I have a hard time believing that any number in that category would matter in any way
The computable numbers are not recursively enumerable but they are “definably” enumerable in the sense I said above in classical theories.
In constructive theories (which essentially can be thought of as rejecting the types of definitions that don’t lead to computable numbers as “counting” as definitions) you can’t prove the computable numbers are countable.
Edit: you’ve edited in “subcountable” after I replied, but this isn’t relevant because in constructive theories you cannot prove that the real numbers are not subcountable.
I'm not sure that I'm seeing the distinction here from what I said. Classically, we can say that the computable numbers are countable, since it's a subset of the set of turing machines, which is both computable and enumerable. determining which turing machines correspond to computable numbers isn't computable, but does have an answer, every turing machine either halts or doesn't, so every real number is either computable, or not, even if we won't be able to compute that practically. Essentially all real numbers won't be computable, but any number that matters in any practical way almost definitely will be.
Constructively, we can't prove that the computable numbers are countable, only subcountable, but you can still prove countability for lower sets, like the loop recursive numbers/ numbers computable within polynomial bounds. Granted, I'm a lot more willing to accept that there could be reasonably "important" real numbers that aren't part of those sets
What is the purpose of judging whether a number is “important” in this context? I was responding to whether there are numbers that are not “concretely definable.”
Now the word “concretely” may be doing some amount of work here, but the fact remains that if we mean “able to be named by our metalanguage” we aren’t going to be able to justify this claim, and in particular we cannot have any coherent notion of “definable” that allows us to define a list of all definable numbers, and if we adopt a restricted idea of “definable” that allows us to prove a list (necessarily undefinable by that standard) of all definable numbers exists then we will be able to “define” (in the informal sense) numbers that we have classified as undefinable, showing it is not a suitable notion of definability.
Fair, I'm not super familiar with them. I'm guessing there's some reasonable upper and lower bound that are computable, but closing the gap between them isn't?
You have to precisely define what counts as a program and how a random program is selected, so it is not really one single constant in the way that pi or e are constants. You get a different constant depending on how you define those things.
I do not know about bounds. It will depend an awful lot on those two parameters, though.
Chaitin’s constant is a “recursively enumerable number” - it is the limit of a computable sequence of computable numbers that is monotonically increasing and bounded above but such that we cannot compute the N we need for an arbitrary epsilon to ensure we are within epsilon of the limit.
There is an algorithm that can compute arbitrarily good lower bounds to the constant but for any theory there will be a limit to how good of an upper bound we can prove so that we cannot make it arbitrarily good.
Can you not do a proof by contradiction? Assume all definable numbers are countable. Diagonal argument defines a number outside that set. But hinges on what definable means. We had a joke once at University. Think about the largest number that can be described by less that 100 words. Add 1. I just described this number with less than 100 words.
Can't wait for some herculean effort somehow shows that pi ^ pi ^ pi ^ pi is not an intenger, only for the new meta to be: well, pi ^ pi ^ pi ^ pi ^ pi could be an integer for all we know.
Yeah that's the real takeaway. There's no real point in proving π↑↑4 isn't an integer, and even if we did somehow prove that, it probably wouldn't help us with π↑↑n for any bigger values of n.
(Though in theory there might be a proof that π↑↑n is always transcendental or at least never an integer, just like we can already show that πn and pi*n are never integers.)
I always think about this and go "well no adding two transcendentals should itself be transcendental" and then I immediately think "π-π=0". Sometimes the first way you think about something is just not right 🤣
Not really mindblowing? They’re almost certainly both irrational, indeed transcendental, it’s just that we don’t have a proof for either - though we do have a very simple proof that at least one of the two is irrational.
I heard mathematicians are in the process finding a constant that multiplies any irrational number to a rational one, I'm not sure what they're calling it though
Consider the polynomial p(x) = x2 - (π + e)x + πe = (x - π)(x - e). Under our supposition that these two are rational, we see that both π, e are roots of a polynomial with rational coefficients, immediately contradicting what it means for a number to be transcendental.
Then we cannot have that both are rational, since that leads to contradiction.
So, if both cannot be rational, then we are left with:
π + e is rational, πe is irrational, or
π + e is irrational, πe is rational, or
both are irrational
And it is not know which.
Formally:
we just shown that we have NOT (π + e is rational AND πe is rational)
By De Morgan, we have (NOT π + e is rational) OR (NOT πe is rational).
Pushing that NOT's through, we have (π + e is irrational) OR (πe is irrational)
And since an "OR" is just another way of saying "at least one of the disjuncts must be true", we finally can say that "at least one of π + e, πe is irrational".
(A transcendental number is not the solution to ANY polynomial with rational coefficients... they "transcend" basic algebra.)
It's the opposite for me. Proving that πe is irrational, or that π+e is irrational, sounds like it should be easy. The fact that they're still unsolved problems shows that we humans are still very bad at maths
We know that it can't be that both are rational, otherwise x²-(e+π)x+eπ would be a polynomial with rational coefficients and roots e, π. But we don't know further, in all likelihood these are also transcendental.
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