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Is there some 'simple' example of an algebraic function (say of one variable) which is not rational and is not irrational?

peter.petrov
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    What do you cann an irrational function? – Bernard Oct 02 '19 at 23:02
  • @Bernard It is a long story... Sorry... I now understand the confusion. I mentioned 'my' definition here. https://math.stackexchange.com/questions/3378624/are-irrational-functions-superset-of-the-rational-ones?noredirect=1# – peter.petrov Oct 02 '19 at 23:28
  • The word irrational is unused for functions. $f(z)$ is a rational function = $f(z)=P(z)/Q(z)$ the quotient of two polynomials. $f(z)$ is an algebraic function = $H(z,f(z))=0$ for some non-zero two variables polynomial $H(x,y) \in \Bbb{C}[x,y]$. Every rational function is algebraic, both are fields. Transcendental function = (analytic) function which is not algebraic (for example trigonometric functions). – reuns Oct 03 '19 at 02:02

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Algebraic function means an algebraic function over a field. An algebraic function is a solution of a polynomial equation wose coefficients are in the specified field. If no field is stated, the field of the rational numbers is meant. Therefore algebraic equations are the algebraic equations over the rational numbers.
Rational functions and root functions have a function term that is a radical expression. That are the explicit algebraic functions.
But just as there are algebraic numbers that cannot be represented by radicals, there are algebraic functions (the implicit algebraic functions) whose function term cannot be represented by a radical, see e. g. the algebraic function $f$ that is determined by the irreducible algebraic equation

$$f(x)^5-f(x)+1=0.$$

see e.g. Is a polynomial equation of degree $\ge 5$ not solvable by any way?

IV_
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