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As a proven fact, there exists algebraic numbers that are not representable by radicals (add, subtract, multiply, divide, $n$th-root operation). My question is that is there a way to specify such number using a function.

Understanding non-solvable algebraic numbers

The above post provides some interesting ways to represnet such numbers, however none of the ways provided are functions.

For examle, if you represent $\sqrt2$ and $-\sqrt2$ in the form of a solution to a lowest-degree polynomial form, they both correspond to the same expression $x^2-2=0$.

That is, in such manner we cannot define a function $sqrt(-2)=root(1,0,-2)=1.4142...$ because $root(1,0,-2)$ can also be $-1.4142...$ so it's not a function.

However, radical expressions do correspond to composition of functions.

For instance, we can express the number $2\sqrt2 + 3$ in function form $$add(multiply(2,sqrt(2)), 3)$$ and if you gives the same input, we always end up with the same number using an algorithm defined by the function.

We can have infinite expressions too, for instance the number given by the expression $${1\over1^3}+{1\over2^3}+...=1.20205...$$

Whether the number is algebraic or transcendental, we have a definite function to express this number (in this case $Zeta(3)$), and we can use the algorithm defined by this function to calculate the decimal representation of this number.

Although it is known that some algebraic numbers cannot be represented by a finite number of radical operations, what if infinite operations are also allowed?

And my requirement does not even restrict it to radicals. Any function-like behavior is enough so we can define whatever new operator we want as long as it maps an input to only one possible output.

Also, my requirement does not need the function to be injective or surjective, we can have the number $1$ be represented by either $1$ or $\sqrt4 - 1$ or $\sum_1^\infty 0.9^n$.

So my question is, is there such a way, or a proof that there isn't such a way, or any theory/study into such a way to specify a function representation for all algebraic numbers in modern mathematics?

EDIT:

One answer mentioned a function using a triplet $a, b, P$ to fix the value of a number. Although strictly speaking it is a function with carefully specified domain, the domains of two inputs $a,b$ of the function are dependent on the other input $P$, so we do not know whether it is a function at the point inputs are given.

Intuitively, I would like a (finite or infinite) set of operators $f_1, f_2, ...$ with no-variable domain (the domain can be specified before the inputs are given), that each takes arbitrary numbers of integer inputs and can represent an infinite algorithm as well, such that all algebraic numbers can be written as a composition of $f_i$.

cr001
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1 Answers1

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The canonic way to represent exactly the real algebraic numbers is by represented $x$ by the coefficients of non-zero integer polynomial $P$ and two rational numbers $a$, $b$ such that $a < x < b$, $P(x) = 0$, and whenever $a < y < b$ and $P(y) = 0$, then $x = y$.

So essentially, we just make explicit which root of the polynomial we want by giving some suitable small rational interval around it which only contains a single root. This process defines a function from a suitable subset of $\mathbb{Q} \times \mathbb{Q} \times \mathbb{N}^*$ onto the algebraic numbers, but of the course domain of the function is a bit messy.

If you want to work in the complex numbers, replace rationals $a$, $b$ with $a < x < b$ with a complex rational $c$ and $k \in \mathbb{N}$ such that $x \in B(c,2^{-k})$, where $B(c,2^{-k})$ is the ball around $c$ with radius $2^{-k}$.

Arno
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  • What about imaginary algebraic numbers which does not have order? – cr001 Sep 18 '17 at 16:02
  • For example, even if we use the absolute value of the complex number, we are not going to be able to distinguish $i$ and $-i$. – cr001 Sep 18 '17 at 16:05
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    In general, algebraic numbers are the real roots of non-constant integer polynomials. However, point of the intervals is not about order, but about a basis of the topology. So in $\mathbb{C}$, you just use rational balls. – Arno Sep 18 '17 at 16:08
  • I understood that part. Thanks. Next question is you need to know the actual number first in order to find suitable $a$ and $b$ but the whole point is that I want to get the actual number first. How can $a$ and $b$ be specified without knowing the number itself? – cr001 Sep 18 '17 at 16:12
  • Well, that is the point of the entire thing, isn't it? Knowing an algebraic number essentially means knowing $a$, $b$, $P$. – Arno Sep 18 '17 at 16:16
  • That is what I am asking. Knowing $\sqrt2$ means we automatically get an algorithm to find $1.4142...$ but we don't know it is $1.4142...$ at first. What if the two roots are different by like $0.00001$? We have to find the actual value first in order to get $a$ and $b$. But I am asking for a function to get its value through a definite algorithm. – cr001 Sep 18 '17 at 16:19
  • If the function is defined on $a$, $b$, $P$, then there is an algorithm that computes the decimal expansion of its value. Moreover, given integer coefficients of a polynomial, you can find some bounds on how close distinct roots can be, so you can even decide whether or not some $a$, $b$, $P$ are in the domain of the function. – Arno Sep 18 '17 at 16:25
  • I got your point. What I was trying to say is that the whole mapping of $a,b,P$ is only a function sometimes. For example, {$-100,100, x^2-2=0$} is not a function. In order to know whether it is a function we need to know its value first, but if it is not a function we cannot even compute its value. – cr001 Sep 18 '17 at 16:29
  • You seem to have some misconceptions over what a function is. – Arno Sep 18 '17 at 16:33
  • What misconception? In my definition, a function is a mapping that takes an input and provides the same single output every time through an algorithm. For example $add(x, y)$ takes whatever $x$ and $y$ and every time it takes the same $x$ and $y$ it always gives the same single result. However if I define a mapping with input $a,b,P$ then $-100,100,x^2-2=0$ will not give a single result. – cr001 Sep 18 '17 at 16:36
  • What you mention is much closer to computable function than to function. In any way, what is add(apple, orange)? A function has a domain, and it doesnt make sense to apply a function to something not inside its domain. The domain of the function here is specified in my answer, and your putative input is not part of it. Hence, you cannot apply the function. – Arno Sep 18 '17 at 17:08
  • The fundamental difference of the "function" I am looking for and what you provide is that, the domain of the function you specified depends on one of the input, specifically $P$. So basically you need to find the domain first based on one of the input, and define a new function based on that domain. Then what is the actual domain of your function? It is certainly not R X R X P[R]. If we consider the domain as only P[R], then it is not a function because no function can map one polynomial to one specific number. – cr001 Sep 18 '17 at 18:40
  • What I want is a function whose each domain does not depends on value of other inputs. It is fine that the domain of the input depends on the value of the input itself, but not cross-dependency. Intuitively, I want a (finite or infinite) set of operators on integers (plus $i$) that can be "composed" in finite or infinite ways to represent all algebraic numbers without ambiguity. The operator does not need to be able to write in a "operator way", though, as we can perfectly use $Zeta(x)$ to represent infinite summation with a pattern. – cr001 Sep 18 '17 at 18:44
  • In fact, can you tell me what is the domain of your function? Then maybe I can try to figure out what exactly is going on. I tried to specify such domain but found it extremely hard. – cr001 Sep 18 '17 at 19:06