I've read this:

In mathematics, a transcendental number is a (possibly complex) number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with rational coefficients.

Why does the concept of transcendental numbers was appended to polynomials with rational coefficients? Why not polynomials with real or complex coefficients? By the few mathematical readings I have done, I felt that it's not safe to work with reals or complex numbers in some mathematical concepts, I speculate that this safety (in the case of the transcendental numbers) can only be granted with rational numbers.


2 Answers 2


Any real number $a$ is the solution of a polynomial equation with real coefficients: $$ x-a=0. $$

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    $\begingroup$ Furthermore, every complex number is also a solution of a polynomial (in fact at worst quadratic) equation with real coefficients. $\endgroup$ Jun 24, 2014 at 16:13

An algebraic number can equivalently be defined as a root of a polynomial whose coefficients are all integers. So, the algebraic numbers are closely related to the integers, certainly an interesting class of objects to study.

When extending past the rational numbers, it is very natural to look at those numbers that occur as such roots before looking at the reals (at least from an algebraic point of view). The reason is that the reals require an analytic extension of the rationals, while the algebraic numbers are algebraic (i.e., not analytic).

Lastly, it makes no sense to wonder about reals or complex numbers that are roots of polynomials with real or complex coefficients: every real numbers (resp. complex number) is the root of a linear polynomial with real (resp. complex) coefficients.

  • $\begingroup$ What is the problem with the analytic extension of the rationals? $\endgroup$
    – Red Banana
    Apr 18, 2013 at 9:47
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    $\begingroup$ there is no problem with it. It's just a different sort of extension. A non-algebraic one. $\endgroup$ Apr 18, 2013 at 9:49

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