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?


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 $f_i$.


1 Answer 1


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 $2^{-k}$.

  • $\begingroup$ What about imaginary algebraic numbers which does not have order? $\endgroup$
    – cr001
    Sep 18, 2017 at 16:02
  • $\begingroup$ For example, even if we use the absolute value of the complex number, we are not going to be able to distinguish $i$ and $-i$. $\endgroup$
    – cr001
    Sep 18, 2017 at 16:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – Arno
    Sep 18, 2017 at 16:08
  • $\begingroup$ 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? $\endgroup$
    – cr001
    Sep 18, 2017 at 16:12
  • $\begingroup$ Well, that is the point of the entire thing, isn't it? Knowing an algebraic number essentially means knowing $a$, $b$, $P$. $\endgroup$
    – Arno
    Sep 18, 2017 at 16:16

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