Algebraic and Transcendental Numbers

By Grace Conway and Greg Bischoff

Introduction

The real numbers can be divided into many different categories to help facilitate different discussions of real numbers for different contexts. Usually in class we divide the real line into the rational and irrational numbers, but we don’t have to do it that way. We can discuss the real numbers by splitting the reals into what are known as the algebraic and transcendental numbers. The algebraic numbers are numbers that can be written as the root to a non-zero polynomial with rational coefficients (1). We define a polynomial as:

$a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ where $n \in \mathbb{N}, a_i \in \mathbb{Q}$

Transcendentals are numbers that are not algebraic; numbers that cannot be written as a root to a non-zero polynomial with rational coefficients (1).

Are all rational numbers also algebraic?

We may want to consider the relationship between rational and algebraic numbers since these are two ways to categorize the real numbers. As it turns out, all rational numbers are algebraic numbers. This fact follows trivially from our definition of a polynomial using only two integer terms—a binomial. By manipulating a binomial with a negative second term, we will find its root may be any desired rational.

$qx - p = 0$

$qx = p$

$x = \frac{p}{q}$ ( $p \in \mathbb{Z}, q \in \mathbb{Z} - \{0\}$ )

The root of this polynomial is the rational $p/q$ , thus the rationals are algebraic.

Are all irrational numbers also transcendental?

One may think that all irrational numbers are transcendental through the inversion of the previous point, but this is not the case. This can be seen by an example of $\sqrt{2}$ . Recall from the reading that we have previously shown that $\sqrt{2}$ is irrational. The $\sqrt{2}$ is the solution to the equation:

$x^2 = 2$

Where that solution is a root of the following polynomial:

$x^2 - 2$

What is the cardinality of the algebraic and transcendental numbers?

Now, that we know that all rational numbers are algebraic numbers, we might be curious about how many algebraic numbers exist. The cardinality of the algebraic numbers is countable infinity (3). If you are interested in the formal proof, please look here, but we will provide the highlights for you.

The proof uses a counting argument through a mapping from the polynomials of degree $n$ (from the set denoted $P^n$ ) to the natural numbers and the union of all of these polynomials, which makes the collection of all polynomials countable. Then, each polynomial is of order $n$ and has n roots. Since $n$ is countable, and the number of polynomials is countable, the number of roots of polynomials with integer coefficients in total is countably infinite.

Knowing that the set of algebraic numbers is countable, where does that leave the transcendental numbers? Since the real numbers are uncountable and are the union of the set of algebraic numbers and transcendental numbers, the set of transcendental numbers is also uncountable. This follows from the fact that the union of a countable set and an uncountable set is uncountably infinite.

How do I prove that a number is transcendental?

While there are uncountably infinitely many transcendental numbers, some famous transcendental numbers include $e$ , $\pi$ , and non-zero logarithms (1 and 2). In 1873, $e$ was the first number proven to be transcendental by Charles Hermite (2). In 1882, $\pi$ was proven to be a transcendental number by Ferdinand von Lindemann (2). A proof by contradiction is usually the way to prove a number is transcendental. The original proofs by Charles Hermite and Ferdinand von Lindemann are very heavily computational. If you are interested in seeing the original proofs, you can take a look here. We spent a lot of time trying to wrap our heads around these lengthy proofs, so we think it is easier to focus on how mathematicians have been able to simplify them. Over the years, mathematicians have defined and proven smaller theorems that are necessary in the larger proofs. An example of this is the Lindemann-Weierstrass Theorem.

The Lindemann-Weierstrass Theorem says that for any two sets of distinct algebraic numbers, $a_1,\cdots , a_n$ and $\alpha_1,\cdots , \alpha_n$ , $a_1e^{\alpha_1}+\cdots +a_ne^{\alpha_n} \neq 0$ (4). Using the Lindemann-Weierstrass Theorem, we are able to prove that a number is transcendental.

Example from here

Suppose $\pi$ is an algebraic number. By definition, $i\pi$ is also algebraic. Let $n=2, a_1=a_2=1, \alpha_1=i\pi$ , and $\alpha_2=0$ . Then, $1e^{i\pi}+1e^0=0$ . This is a contradiction of Lindemann-Weierstrass Theorem. Thus, $\pi$ is not algebraic, so it must be transcendental.

References

Math 320: Real Analysis

Algebraic and Transcendental Numbers

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