In my textbook, it says that
an irrational algebraic function is a function in which the independent variables appear under a radical sign or in a power with a rational number for its exponent.
I understand the variable should be under the square root.
However, I wasn't sure about it being to the power of a rational number.
One example would be $x^{\frac{1}{2}}$. However, $x^2$ would work because the power is a rational number, but isn't it considered a quadratic?
Could someone please explain this to me?
Thank you very much.
Best Answer
An Algebraic Function is the root of a polynomial equation. Thus a solution to $F(x)^5+xF(x)^2+x=0$ is an algebraic function.
A Rational Function is the quotient of two polynomials. This $$G(x)=\frac {x^5+x^2+1}{x^2-2}$$ is a rational function.
It's clear that every rational function is algebraic (since $G(x)$ satisfies $q(x)G(x)-p(x)=0$).
I've never heard the term used before, but it seems fair to say that an Irrational Algebraic Function is any Algebraic Function which isn't rational.
For example: $H(x)=\sqrt x$ is an irrational algebraic function. It's algebraic because it satisfies the polynomial $H(x)^2-x=0$. It's not rational because there are no two polynomials $p(x),q(x)$ such that $H(x) = \frac {p(x)}{q(x)}$ (if there were, we'd have $p(x)=q(x)\times H(x)$ but then the degree of $p(x)$ would not be an integer).
Not every Irrational Algebraic function has such a simple form, however. This follows from Abel's theorem on the Unsolvability of the general quintic equation.