Transcendental numbers are numbers that are not the solution to any algebraic equation.

But what about $x-\pi=0$? I am guessing that it's not algebraic but I don't know why not. Polynomials are over a field, so I am guessing that $\mathbb{R}$ is implied when not specified. And since $\pi \in \mathbb{R}$, what is the problem?

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    $\begingroup$ an algebraic number is the root of a polynomial with rational coefficients (note $\pi \notin\mathbb{Q}$) $\endgroup$ – Deven Ware Dec 21 '11 at 18:46

To quote Wikipedia "In mathematics, a transcendental number is a number (possibly a complex number) that is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients." so the field is $\mathbb{Q}$ and $\pi$ is not included.

  • $\begingroup$ I apologize, I thought the coefficients were from the field! Thanks $\endgroup$ – user12205 Dec 21 '11 at 18:49
  • $\begingroup$ @Jos: In general, given two fields $E \subseteq F$ an element of $F$ is algebraic over $E$ iff it is a root of non-constant polynomial equation with coefficients from $E$. "Algebraic numbers" is a special case of this definition for $E=\mathbb Q$ and $F=\mathbb C$. However, every element $a \in E$ is algebraic over $E$, since it satisfies $x-a = 0$. $\endgroup$ – sdcvvc Dec 21 '11 at 19:34

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