# A question on transcendental numbers

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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an algebraic number is the root of a polynomial with rational coefficients (note $\pi \notin\mathbb{Q}$) –  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.

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I apologize, I thought the coefficients were from the field! Thanks –  user12205 Dec 21 '11 at 18:49
@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 $x-a = 0$. –  sdcvvc Dec 21 '11 at 19:34

I think the field has been rather sloppy.

A transcendental number is a number which is:

1. Not rational (cannot be expressed by the simple division of two integers) and
2. Has some non-finite relation to the Platonic space of geometry -- the only realm (except perhaps for Nature) where the word "transcendental" applies without argument.

That means pi is transcendental, but the diagonal of the unit square is not. A controversial, but imminently defensible, point.

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'Transcendental' has a formal mathematical definition, and this isn't it. There's no formal definition of 'non-finite relation to the Platonic space of geometry'; the closest thing to that concept is the notion of 'constructible number'. –  Steven Stadnicki Jan 12 at 1:07
From wikipedia: "In mathematics, a transcendental number is a number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with rational coefficients". But what is the root of x^2-4=0? A rational. So are you arguing that transcendental numbers can be rational? –  Mark J Jan 12 at 1:28
No - I'm saying that the 'Platonic space of geometry' doesn't enter into the matter at all. How do you ascribe a geometric meaning to, e.g., the real root of $x^5+x=3$ that doesn't apply to arbitrary numbers? –  Steven Stadnicki Jan 12 at 2:38
You didn't address my point: are you or are you not saying that transcendental numbers can be rational (since you claim it's well-defined). –  Mark J Jan 12 at 3:09
I don't see where 'well-defined' has anything to do with rationality. I'm not disagreeing with your point 1; certainly transcendental numbers can't be rational. But your point 2 is completely inaccurate in its description. –  Steven Stadnicki Jan 12 at 4:16
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