Purely Transcendental

**ThePerfectHacker** · August 12th 2007, 07:46 AM

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

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

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

**Rebesques** · August 12th 2007, 08:11 AM

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

**ThePerfectHacker** · August 12th 2007, 08:41 AM

Originally Posted by Rebesques

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

Yes

**Rebesques** · August 12th 2007, 08:52 AM

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

**CaptainBlack** · August 12th 2007, 11:12 AM

Originally Posted by Rebesques

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

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

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

RonL

**topsquark** · August 12th 2007, 11:41 AM

Originally Posted by CaptainBlack

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

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

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 $\sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\{ a_n \}$ . (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

**CaptainBlack** · August 12th 2007, 12:48 PM

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 $\sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\{ a_n \}$ . (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

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

$<br /> \sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}<br />$

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

RonL

**Rebesques** · August 12th 2007, 05:25 PM

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"

**ThePerfectHacker** · August 12th 2007, 07:45 PM

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

**CaptainBlack** · August 12th 2007, 09:24 PM

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 $\bold{R}$ , but that does not prove tht the set exhausts $\bold{R}$ .

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

RonL

**topsquark** · August 13th 2007, 04:26 AM

Originally Posted by CaptainBlack

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

$<br /> \sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}<br />$

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

RonL

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

-Dan

**Rebesques** · August 13th 2007, 05:14 PM

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.

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