Thread: Purely Transcendental

Given an element $\displaystyle x\in \mathbb{R}$ having a property that no non-constant polynomial in $\displaystyle \mathbb{Q}$ has $\displaystyle x$ as a zero, we call it "transcendental". For example, $\displaystyle \pi \in \mathbb{R}$ is an example.

However, if we allow "infinite polynomials", i.e. power series then it is no longer transcendental. Because $\displaystyle \pi$ is a zero of $\displaystyle f(x)=x - \frac{x^3}{3!}+\frac{x^5}{5!}-...$.


My question is: Is there an element in $\displaystyle \mathbb{R}$ which is purely transcendental? Meaning it has no rational power series even.

What do you mean by "rational power series"? Rational coefficients?

Originally Posted by Rebesques

What do you mean by "rational power series"? Rational coefficients?

Yes

Like $\displaystyle e^{1/\pi}=\sum_n \frac{1}{n!\pi^n}$ ?

Originally Posted by Rebesques

Like $\displaystyle e^{1/\pi}=\sum_n \frac{1}{n!\pi^n}$ ?

He is interested in those numbers, if such exist, which are not the root of any
rational power series:

$\displaystyle f(x)=\sum_{n=0}^{\infty} a_n x^n,\ a_n \in \bold{Q}$

RonL

Originally Posted by CaptainBlack

He is interested in those numbers, if such exist, which are not the root of any
rational power series:

$\displaystyle f(x)=\sum_{n=0}^{\infty} a_n x^n,\ n \in \bold{Q}$

RonL

Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as $\displaystyle \sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\displaystyle \{ a_n \}$. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

-Dan

Originally Posted by topsquark

Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as $\displaystyle \sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\displaystyle \{ a_n \}$. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

-Dan

$\displaystyle x=\sum_{n = 0}^{\infty}a_n(10)^n$ does not guarantee that

$\displaystyle
\sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}
$

for some $\displaystyle a_0, a_1, .. \in \bold{Q}$

RonL

I wonder where he comes up with those

Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental.


The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.


"Quod erat demonstratum"

Originally Posted by Rebesques

The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.

How do we know that given any $\displaystyle r\in \mathbb{R}$ there exists $\displaystyle q_i\in \mathbb{Q}$ so that $\displaystyle r=q_0+q_1x+q_2x^2+...$, i.e. is spams?

Originally Posted by Rebesques

I wonder where he comes up with those

Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental.


The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.


"Quod erat demonstratum"

It is obvious that the set of numbers which are roots of some rational power series has the same cardinality as $\displaystyle \bold{R}$, but that does not prove tht the set exhausts $\displaystyle \bold{R}$.

It is also obvious tha the set is dense in $\displaystyle \bold{R}$, as it inherits this propety from the set of algebraic numbers which it contains.

RonL

Originally Posted by CaptainBlack

$\displaystyle x=\sum_{n = 0}^{\infty}a_n(10)^n$ does not guarantee that

$\displaystyle
\sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}
$

for some $\displaystyle a_0, a_1, .. \in \bold{Q}$

RonL

Ah well. I guess when I'm off, I'm way off.

-Dan

How do we know (...) spans?

It is also obvious tha the set is dense in , as it inherits this propety from the set of algebraic numbers which it contains.

You are both right, I mistakingly took it for a maximal set.