Transcendental Numbers and Algebraic Irrationals
From Liouville's construction to the transcendence of $e$ and $\pi$
Intermediate (Upper Undergraduate Level)
The real numbers are partitioned into algebraic numbers (roots of polynomials with rational coefficients) and transcendental numbers (those that are not). The irrational algebraic numbers are called algebraic irrationals. This chapter develops the theory of this classification from a number-theoretic perspective, ranging from Liouville's construction of transcendental numbers to the proofs of the transcendence of $e$ and $\pi$ via the Hermite–Lindemann–Weierstrass theorem. Advanced topics such as the Gelfond–Schneider theorem and Roth's theorem are deferred to the advanced section.
1. Overview and history
"Of what polynomial equation is a given number a root?"—this question lies at the heart of number theory. The study of numbers that are not roots of any polynomial — the transcendental numbers — originates in the construction problems of ancient Greek geometry and became one of the principal themes of number theory during the nineteenth and twentieth centuries.
A timeline of transcendence theory
| Year | Mathematician | Result |
|---|---|---|
| 1844 | Liouville | Construction of the first transcendental numbers (Liouville numbers) |
| 1873 | Hermite | Proof of the transcendence of $e$ |
| 1882 | Lindemann | Proof of the transcendence of $\pi$ (resolution of the squaring of the circle) |
| 1885 | Weierstrass | Generalisation of Lindemann's theorem (the L–W theorem) |
| 1900 | Hilbert | Seventh problem: transcendence of $\alpha^\beta$ |
| 1934 | Gelfond, Schneider | Resolution of Hilbert's seventh problem |
| 1955 | Roth | Optimal bound in Diophantine approximation (Fields Medal) |
| 1966 | Baker | Theorem on linear forms in logarithms (Fields Medal) |
Connection to the three classical Greek construction problems
Among the three classical Greek problems of compass-and-straightedge construction, squaring the circle (constructing a square of area equal to a given circle) is impossible because $\pi$ is transcendental: lengths constructible by compass and straightedge are necessarily algebraic. The other two problems (trisecting an angle and doubling the cube) are also impossible, but their impossibility follows from arguments about the degree of an algebraic number and does not require transcendence—the discussion of degrees in Sections 2 and 3 suffices.
2. Classification of algebraic numbers
Definition: algebraic and transcendental numbers
A complex number $\alpha$ is an algebraic number if there exists a nonzero polynomial $f(x) \in \mathbb{Q}[x]$ with rational coefficients such that $f(\alpha) = 0$. A complex number that is not algebraic is called a transcendental number.
Definition: minimal polynomial and degree
The minimal polynomial of an algebraic number $\alpha$ is the monic irreducible polynomial in $\mathbb{Q}[x]$ having $\alpha$ as a root. Its degree is called the degree of $\alpha$ and is denoted $\deg(\alpha)$.
- A rational number $r \in \mathbb{Q}$ has degree $1$ (with minimal polynomial $x - r$).
- $\sqrt{2}$ has degree $2$ (minimal polynomial $x^2 - 2$).
- $\sqrt[3]{2}$ has degree $3$ (minimal polynomial $x^3 - 2$).
Definition: algebraic integer
An algebraic number $\alpha$ is an algebraic integer if it is a root of some monic polynomial $f(x) \in \mathbb{Z}[x]$ with integer coefficients.
For example, $\sqrt{2}$ (a root of $x^2 - 2 = 0$) is an algebraic integer, whereas $1/2$ (minimal polynomial $x - 1/2$, monic but not in $\mathbb{Z}[x]$) is not.
Theorem: $\overline{\mathbb{Q}}$ is a field
The set $\overline{\mathbb{Q}}$ of algebraic numbers forms a field under the usual arithmetic operations. That is, the sum, difference, product, and quotient (except division by $0$) of algebraic numbers are again algebraic.
Proof (sum)
Let $\alpha$ be a root of $f(x) = 0$ and $\beta$ be a root of $g(y) = 0$, with $f, g \in \mathbb{Q}[x]$, $\deg f = m$, and $\deg g = n$. We show that $\gamma = \alpha + \beta$ is algebraic.
Substituting $\alpha = \gamma - \beta$ into $f$ gives $f(\gamma - \beta) = 0$. Regarding this as a polynomial in $\beta$, we see that $\beta$ is a root of $h(y) = f(\gamma - y)$. Since $\beta$ is also a root of $g(y)$, it is a common root of $h(y)$ and $g(y)$, and hence the resultant
$$R(\gamma) = \mathrm{Res}_y(f(\gamma - y),\; g(y))$$is a polynomial in $\gamma$ satisfying $R(\gamma) = 0$. Since $\mathrm{Res}_y$ is the resultant with respect to $y$, $R$ is a polynomial in $\gamma$ over $\mathbb{Q}[\gamma]$ of degree at most $mn$, and $R \not\equiv 0$ (because $f, g$ are nonzero). Hence $\gamma = \alpha + \beta$ is a root of $R(\gamma) = 0$ and is therefore algebraic.
The product case is similar: setting $\gamma = \alpha\beta$ and substituting $\alpha = \gamma/\beta$ into $f$ shows that $\gamma$ is algebraic via $\mathrm{Res}_y(\beta^m f(\gamma/\beta),\; g(\beta)) = 0$. The inverse $1/\alpha$ is a root of $x^m f(1/x)$ and is therefore algebraic. Differences reduce to sums via $\alpha + (-\beta)$. $\square$
Computational aspect
For the computational representation of algebraic numbers (exact arithmetic via minimal polynomials and root isolation, arithmetic algorithms, factorisation over field extensions), see Computer Algebra, Chapter 12: Algebraic Numbers and Field Extensions. The present chapter focuses on number-theoretic properties and the theory of transcendence.
3. Algebraic irrationals and conjugate irrationals
Definition: algebraic irrational
An algebraic irrational is an algebraic number that is not rational; equivalently, an algebraic number of degree at least $2$.
Definition: conjugate irrationals
The conjugates of an algebraic number $\alpha$ are all the roots $\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_n$ of its minimal polynomial $f(x) \in \mathbb{Q}[x]$. Since $f(x)$ is irreducible over $\mathbb{Q}$, it is separable over $\mathbb{Q}$, and the $n$ conjugates are all distinct.
When $\alpha$ is an algebraic irrational, the conjugates $\alpha_2, \ldots, \alpha_n$ (other than $\alpha$ itself) are called the conjugate irrationals of $\alpha$.
Examples of conjugate irrationals
- $\sqrt{2}$: minimal polynomial $x^2 - 2$; conjugate $-\sqrt{2}$.
- $\sqrt[3]{2}$: minimal polynomial $x^3 - 2$; conjugates $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$ (with $\omega = e^{2\pi i/3}$). An example where a real number has complex conjugates.
- Golden ratio $\varphi = \dfrac{1+\sqrt{5}}{2}$: minimal polynomial $x^2 - x - 1$; conjugate $\dfrac{1-\sqrt{5}}{2} \approx -0.618$.
- $\sqrt{2} + \sqrt{3}$: minimal polynomial $x^4 - 10x^2 + 1$; conjugates $\sqrt{2} - \sqrt{3}$, $-\sqrt{2} + \sqrt{3}$, $-\sqrt{2} - \sqrt{3}$.
Theorem: symmetric functions of conjugates
Let $\alpha$ be an algebraic number of degree $n$ with minimal polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$, and let its conjugates be $\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_n$. Then:
- Trace: $\mathrm{Tr}(\alpha) = \alpha_1 + \alpha_2 + \cdots + \alpha_n = -a_{n-1} \in \mathbb{Q}$.
- Norm: $\mathrm{N}(\alpha) = \alpha_1 \cdot \alpha_2 \cdots \alpha_n = (-1)^n a_0 \in \mathbb{Q}$.
More generally, every elementary symmetric polynomial in the conjugates lies in $\mathbb{Q}$ (Vieta's formulas).
Proof
Since $f(x)$ is monic (the leading coefficient is $1$), it factors over its splitting field as
$$f(x) = (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n).$$Expanding the right-hand side gives
$$f(x) = x^n - (\alpha_1 + \cdots + \alpha_n)\, x^{n-1} + \cdots + (-1)^n \alpha_1 \alpha_2 \cdots \alpha_n.$$Comparing with $f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0$ (Vieta's formulas) yields
\begin{align} \alpha_1 + \alpha_2 + \cdots + \alpha_n &= -a_{n-1} \\ \sum_{i < j} \alpha_i \alpha_j &= a_{n-2} \\ &\;\;\vdots \\ \alpha_1 \alpha_2 \cdots \alpha_n &= (-1)^n a_0. \end{align}Since $f(x) \in \mathbb{Q}[x]$, each $a_k \in \mathbb{Q}$, and hence $\mathrm{Tr} = -a_{n-1} \in \mathbb{Q}$ and $\mathrm{N} = (-1)^n a_0 \in \mathbb{Q}$. $\square$
Example: trace and norm
The minimal polynomial of $\alpha = 1 + \sqrt{5}$ is $x^2 - 2x - 4$. Its conjugate is $1 - \sqrt{5}$, and
$$\mathrm{Tr}(\alpha) = (1 + \sqrt{5}) + (1 - \sqrt{5}) = 2, \quad \mathrm{N}(\alpha) = (1 + \sqrt{5})(1 - \sqrt{5}) = -4.$$These agree with the coefficients of the minimal polynomial: $-a_1 = -(-2) = 2$ and $(-1)^2 a_0 = -4$.
4. Existence of transcendental numbers
The shortest path to the existence of transcendental numbers goes through Cantor's set-theoretic argument.
Theorem (Cantor, 1874)
The set $\overline{\mathbb{Q}}$ of algebraic numbers is countable, while $\mathbb{R}$ is uncountable, so transcendental numbers exist. Moreover, the set of transcendental numbers is itself uncountable, and in the sense of Lebesgue measure on $\mathbb{R}$ "almost every" real number is transcendental.
Proof
Step 1 (algebraic numbers are countable): Let $P_n$ denote the set of integer-coefficient polynomials of degree $n$. Each $f \in P_n$ is determined by its $n+1$ integer coefficients $(a_n, a_{n-1}, \ldots, a_0)$, so $P_n$ can be identified with $\mathbb{Z}^{n+1}$, which is countable. Each $f \in P_n$ has at most $n$ roots (by the fundamental theorem of algebra), so its root set is finite. Therefore
$$\overline{\mathbb{Q}} = \bigcup_{n=1}^{\infty} \bigcup_{f \in P_n} \{\text{roots of } f\}$$is the union of a countable union of countable sets, each indexed by a finite set, and is therefore countable.
Step 2 (transcendental numbers are uncountable): $\mathbb{R}$ is uncountable by Cantor's diagonal argument. Since $\overline{\mathbb{Q}} \cap \mathbb{R}$ (the real algebraic numbers) is countable, $\mathbb{R} \setminus \overline{\mathbb{Q}}$ (the real transcendentals) must be uncountable. In particular, transcendental numbers exist. $\square$
For a detailed discussion of countability, see Set Theory Basics: Properties of Countable Sets.
Constructive vs. non-constructive
Cantor's argument establishes the existence of transcendental numbers but does not construct a single explicit example. The first explicit construction of a transcendental number is due to Liouville (1844), discussed in the next section. Historically, Liouville's construction (1844) preceded Cantor's existence proof (1874) by thirty years.
From the viewpoint of measure theory, $\overline{\mathbb{Q}}$ is countable and hence has Lebesgue measure $0$. Therefore $\mathbb{R} \setminus \overline{\mathbb{Q}}$ (the set of transcendental numbers) has the same Lebesgue measure as $\mathbb{R}$. In this sense, "a real number chosen at random is transcendental with probability $1$". Nevertheless, proving the transcendence of a specific number is extraordinarily hard, and the transcendence of many fundamental constants remains open (see Advanced: Open Problems).
5. Liouville's theorem and the first transcendental numbers
Liouville gave a quantitative formulation of the fact that "algebraic irrationals are not well approximated by rationals", and used its contrapositive to construct transcendental numbers. This is the starting point of transcendence theory and the heart of this chapter.
Theorem (Liouville, 1844): an approximation barrier for algebraic irrationals
Let $\alpha$ be a real algebraic irrational of degree $n \ge 2$. Then there exists a constant $c = c(\alpha) > 0$ such that, for every rational number $p/q$ with $q > 0$,
$$\left|\alpha - \dfrac{p}{q}\right| > \dfrac{c}{q^n}.$$Proof
Let the minimal polynomial of $\alpha$ be $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x]$ with $a_n \ne 0$.
Step 1: If $|\alpha - p/q| \ge 1$, the bound is trivial provided $c \le 1$. Henceforth assume $|\alpha - p/q| < 1$.
Step 2: We show $f(p/q) \ne 0$. Because $f$ is irreducible over $\mathbb{Q}$ with $\deg f = n \ge 2$, no rational number can be a root of $f$ (a rational root would give a linear factor, contradicting irreducibility).
Step 3: We derive a lower bound for $|f(p/q)|$. Clearing denominators,
$$f\!\left(\dfrac{p}{q}\right) = \dfrac{a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_0 q^n}{q^n}.$$The numerator is a nonzero integer, so $|f(p/q)| \ge 1/q^n$.
Step 4: We apply the mean value theorem. Since $|\alpha - p/q| < 1$, the point $p/q$ lies in the interval $I = [\alpha - 1, \alpha + 1]$. Set
$$M = \max_{t \in I} |f'(t)|$$($M$ is finite because $f'$ is continuous and $I$ is compact). Since $f(\alpha) = 0$, the mean value theorem yields a point $\xi$ between $p/q$ and $\alpha$ such that
$$\left|f\!\left(\dfrac{p}{q}\right)\right| = |f'(\xi)|\left|\dfrac{p}{q} - \alpha\right| \le M \left|\dfrac{p}{q} - \alpha\right|.$$Step 5: Combining Steps 3 and 4,
$$\dfrac{1}{q^n} \le \left|f\!\left(\dfrac{p}{q}\right)\right| \le M \left|\dfrac{p}{q} - \alpha\right|,$$hence
$$\left|\alpha - \dfrac{p}{q}\right| \ge \dfrac{1}{M q^n}.$$Setting $c = \min(1, 1/M)$ gives $|\alpha - p/q| > c/q^n$ in all cases. $\square$
Definition: Liouville number
A real number $\alpha$ is a Liouville number if, for every positive integer $n$, there exist integers $p, q$ with $q \ge 2$ such that
$$0 < \left|\alpha - \dfrac{p}{q}\right| < \dfrac{1}{q^n}.$$Theorem: Liouville numbers are transcendental
Every Liouville number is transcendental.
Proof
We prove the contrapositive. If $\alpha$ is an algebraic irrational of degree $n$, then by Liouville's theorem $|\alpha - p/q| > c/q^n$ for some $c > 0$. Hence, for sufficiently large $q$, no rational $p/q$ satisfies $|\alpha - p/q| < 1/q^{n+1}$. This means $\alpha$ is not a Liouville number. $\square$
Example: the Liouville constant (the first transcendental number)
The Liouville constant
$$L = \sum_{k=1}^{\infty} 10^{-k!} = 0.1\underbrace{1}_{2!}\underbrace{0\cdots01}_{3! - 2!}\underbrace{0\cdots01}_{4! - 3!}\cdots = 0.110001000000000000000001\ldots$$is transcendental.
Proof: Set the $n$-th partial sum
$$\dfrac{p_n}{q_n} = \sum_{k=1}^{n} 10^{-k!}, \quad q_n = 10^{n!}.$$Then $p_n$ is an integer and $q_n = 10^{n!}$. Estimating the tail,
$$0 < L - \dfrac{p_n}{q_n} = \sum_{k=n+1}^{\infty} 10^{-k!} < \dfrac{2}{10^{(n+1)!}} = \dfrac{2}{q_n^{n+1}}.$$For sufficiently large $n$, $2/q_n^{n+1} < 1/q_n^n$, so $L$ is a Liouville number and hence transcendental. $\square$
Going further
Liouville's theorem shows that "an algebraic irrational of degree $n$ admits no approximation better than $1/q^n$", but this is not the optimal bound. Roth's theorem (1955) showed that for every $\varepsilon > 0$, only finitely many rationals $p/q$ satisfy $|\alpha - p/q| < 1/q^{2+\varepsilon}$. This result earned a Fields Medal. For details, see Advanced: Diophantine Approximation and Roth's Theorem.
6. The Hermite–Lindemann–Weierstrass theorem
Liouville's method can construct transcendental numbers, but it cannot establish the transcendence of "naturally occurring" constants such as $e$ and $\pi$. That problem was resolved by Hermite (1873) and Lindemann (1882), and the result was generalised by Weierstrass (1885).
Theorem (Hermite, 1873)
Euler's number $e$ is transcendental.
Proof (a modern reformulation of Hermite's method)
Argue by contradiction. Assume $e$ is algebraic, so that
$$a_0 + a_1 e + a_2 e^2 + \cdots + a_n e^n = 0, \quad a_i \in \mathbb{Z},\; a_0 \ne 0.$$Choose a sufficiently large prime $p$ and construct the auxiliary polynomial
$$f(t) = \dfrac{t^{p-1}\bigl((t-1)(t-2)\cdots(t-n)\bigr)^p}{(p-1)!}.$$For each $k = 0, 1, \ldots, n$, define the integral
$$I_k = \int_0^k e^{k-t} f(t)\, dt.$$Iterated integration by parts gives
$$I_k = e^k \sum_{j=0}^{np+p-1} f^{(j)}(0) - \sum_{j=0}^{np+p-1} f^{(j)}(k),$$where $f^{(j)}$ denotes the $j$-th derivative of $f$. Set $S = \sum_{k=0}^n a_k I_k$.
Estimate 1 ($S$ is a nonzero integer): The construction of $f(t)$ gives the following:
- $f^{(j)}(0) = 0$ for $j < p-1$, since $t = 0$ is a root of multiplicity $p - 1$ of $f(t) \cdot (p-1)!$.
- $f^{(p-1)}(0) = \bigl((-1)(-2)\cdots(-n)\bigr)^p = (-1)^{np}(n!)^p$.
- $f^{(j)}(0) \equiv 0 \pmod{p}$ for $j \ge p$.
- For each integer $k = 1, \ldots, n$, $f^{(j)}(k) \equiv 0 \pmod{p}$, because $k$ is a root of multiplicity $p$ of $(t-k)^p$.
Substituting these,
$$S = \sum_{k=0}^n a_k \left(e^k \sum_j f^{(j)}(0) - \sum_j f^{(j)}(k)\right).$$Using the hypothesis $\sum a_k e^k = 0$ kills part of the $e^k$ terms, and after simplification
$$S \equiv a_0 \cdot (-1)^{np} (n!)^p \pmod{p}.$$Choosing a prime $p > n$ with $p > |a_0|$ ensures $p \nmid a_0(n!)^p$, so $S \not\equiv 0 \pmod{p}$, and in particular $S \ne 0$.
Estimate 2 ($|S|$ is small): On the other hand, estimating $|I_k| \le \int_0^k e^{k-t} |f(t)|\, dt$ directly,
$$|I_k| \le e^n \cdot n \cdot \dfrac{n^{p-1} \cdot n^{np}}{(p-1)!} = e^n \cdot \dfrac{n^{np+p}}{(p-1)!}.$$As $p \to \infty$, $(p-1)!$ grows far faster than $n^{np+p}$ (Stirling's formula), so for sufficiently large $p$
$$|S| \le \sum_{k=0}^n |a_k| \cdot |I_k| < 1.$$But $S$ is a nonzero integer, so $|S| \ge 1$. Contradiction. $\square$
Key idea of the proof
The heart of Hermite's proof is the construction of the auxiliary polynomial $f(t)$ parameterised by a prime $p$. The mod-$p$ behaviour of $f^{(j)}(0)$ and $f^{(j)}(k)$ enables the integrality argument ($S \ne 0$), while the rapid growth of $(p-1)!$ supplies the analytic estimate ($|S| < 1$). The dichotomy "$S$ is an integer with $|S| < 1$, hence a contradiction" was later extended by Lindemann to prove the transcendence of $\pi$.
Theorem (Lindemann, 1882)
The number $\pi$ is transcendental.
Proof
Argue by contradiction. Assume $\pi$ is algebraic. Since $i$ is a root of $x^2 + 1 = 0$, it is algebraic; by the field property of algebraic numbers (Section 2), $i\pi$ is also algebraic. Since $\alpha = i\pi \ne 0$ is algebraic, the special case of the Lindemann–Weierstrass theorem below forces $e^{i\pi}$ to be transcendental.
However, Euler's identity $e^{i\pi} = -1$ shows that $e^{i\pi}$ is algebraic (it is a root of $x + 1 = 0$). Contradiction.
Therefore $\pi$ is not algebraic, that is, $\pi$ is transcendental. $\square$
The proof reduces to the special case of the Lindemann–Weierstrass theorem below: "$e^\alpha$ is transcendental for every nonzero algebraic number $\alpha$". The theorem itself is proved by extending Hermite's method (used for the transcendence of $e$) to a sum of several exponentials. We state the full theorem.
Theorem (Lindemann–Weierstrass, 1885)
If $\alpha_1, \ldots, \alpha_n$ are algebraic numbers that are linearly independent over $\mathbb{Q}$, then $e^{\alpha_1}, \ldots, e^{\alpha_n}$ are algebraically independent over $\mathbb{Q}$.
In particular, $e^\alpha$ is transcendental for every nonzero algebraic number $\alpha$.
Outline of proof (special case $n = 1$)
Let $\alpha \ne 0$ be algebraic, and assume $e^\alpha$ is algebraic to derive a contradiction. If $e^\alpha$ is algebraic, collecting all its conjugates yields an algebraic relation of the form
$$\beta_0 + \beta_1 e^{\alpha_1} + \cdots + \beta_m e^{\alpha_m} = 0,$$where the $\alpha_i$ are algebraic numbers related to $\alpha$ and $\beta_i \in \overline{\mathbb{Q}}$. Following Hermite's strategy, one constructs an auxiliary polynomial $f(t)$ and evaluates the integrals $I_k = \int_0^{\alpha_k} e^{\alpha_k - t} f(t)\, dt$. Choosing the prime $p$ appropriately makes the integer part nonzero, while the absolute value of the integral tends to $0$ as $p \to \infty$, yielding a contradiction. The full statement ($n \ge 2$) uses the product structure over all conjugates and the same argument.
A complete proof requires Galois theory and the norm of algebraic integers. See Advanced, Section 3 for the details.
Corollaries of the Lindemann–Weierstrass theorem
The following are all transcendental:
- $e^\alpha$ for nonzero algebraic $\alpha$. In particular, $e = e^1$ is transcendental.
- $\ln \alpha$ for algebraic $\alpha \ne 0, 1$. If $\ln \alpha$ were algebraic, then $e^{\ln \alpha} = \alpha$ would force a contradiction with $\alpha \ne 0, 1$.
- $\sin \alpha, \cos \alpha, \tan \alpha$ for nonzero algebraic $\alpha$. This follows, for example, from $e^{i\alpha} = \cos\alpha + i\sin\alpha$.
- $\pi = -i \ln(-1)$. Since $-1$ is algebraic, $\ln(-1)$ is transcendental; and $\pi$ is a constant multiple of $\ln(-1)$, hence also transcendental.
For the significance of the Lindemann–Weierstrass theorem in computer algebra (its connection to exact arithmetic of algebraic numbers), see Computer Algebra, Chapter 12, Theorem 12.2.
7. Summary and references
Summary
- The real numbers split into algebraic numbers (roots of polynomials with rational coefficients) and transcendental numbers. The algebraic numbers are countable but the transcendentals are uncountable.
- Algebraic irrationals are algebraic numbers of degree $\ge 2$; the other roots of the same minimal polynomial are the conjugate irrationals. The trace and norm of the conjugates are rational.
- Liouville's theorem: an algebraic irrational of degree $n$ satisfies $|\alpha - p/q| > c/q^n$. By contrapositive, every Liouville number is transcendental.
- Hermite–Lindemann–Weierstrass: $e^\alpha$ is transcendental for every nonzero algebraic $\alpha$. As corollaries, $e$ and $\pi$ are transcendental.
Further topics
Advanced topics in transcendence theory—the Gelfond–Schneider theorem (transcendence of $\alpha^\beta$), Roth's theorem (the optimal bound for Diophantine approximation), Baker's theorem, Schanuel's conjecture, and others—are treated in Advanced: Developments in Transcendence Theory.
Exercises
- Find the minimal polynomial of $\alpha = \sqrt{2} + i$, and list its degree and all its conjugates.
- Show that $L_2 = \sum_{k=1}^{\infty} 2^{-k!}$ is transcendental using Liouville's theorem.
- Deduce from the Lindemann–Weierstrass theorem that $\ln 2$ is transcendental. (Hint: suppose $\ln 2$ is algebraic and consider $e^{\ln 2}$.)
Sample solutions (click to expand)
-
Setting $\alpha = \sqrt{2} + i$, we have $\alpha - i = \sqrt{2}$, so $(\alpha - i)^2 = 2$, that is, $\alpha^2 - 2i\alpha - 1 = 2$. Squaring both sides of $\alpha^2 - 3 = 2i\alpha$ yields $\alpha^4 - 6\alpha^2 + 9 = -4\alpha^2$, that is, $\alpha^4 - 2\alpha^2 + 9 = 0$. Irreducibility over $\mathbb{Q}$ can be verified by the rational root theorem and the absence of quadratic factors, so the degree is $4$ and the conjugates are $\sqrt{2} - i$, $-\sqrt{2} + i$, and $-\sqrt{2} - i$.
-
Setting $p_n/q_n = \sum_{k=1}^{n} 2^{-k!}$, we get $q_n = 2^{n!}$. The tail satisfies
$$0 < L_2 - \dfrac{p_n}{q_n} = \sum_{k=n+1}^{\infty} 2^{-k!} < \dfrac{2}{2^{(n+1)!}} = \dfrac{2}{q_n^{n+1}}.$$For sufficiently large $n$, $2/q_n^{n+1} < 1/q_n^n$, so $L_2$ is a Liouville number and therefore transcendental.
-
Assume $\ln 2$ is algebraic. Since $\ln 2 \ne 0$, the corollary of the Lindemann–Weierstrass theorem forces $e^{\ln 2}$ to be transcendental. However, $e^{\ln 2} = 2$ is an algebraic number (a root of $x - 2 = 0$). Contradiction. Hence $\ln 2$ is transcendental.
References
- Baker, A. (1975). Transcendental Number Theory. Cambridge University Press.
- Niven, I. (1956). Irrational Numbers. MAA.
- Lang, S. (1966). Introduction to Transcendental Numbers. Addison-Wesley.
- Murty, M. R., & Rath, P. (2014). Transcendental Numbers. Springer.
- Nishioka, K. (2003). What Are Transcendental Numbers? (in Japanese). Kodansha Bluebacks.
Related articles
- Developments in transcendence theory (advanced) — Gelfond–Schneider, Roth, Baker, and Schanuel's conjecture
- Computer Algebra, Chapter 12: Algebraic Numbers and Field Extensions — Computational representation of algebraic numbers and arithmetic algorithms
- Set Theory Basics: Properties of Countable Sets — Proof of countability of algebraic numbers
- Number Theory Intermediate: Table of Contents
Frequently Asked Questions
- What is a transcendental number?
- A real (or complex) number that is not a root of any polynomial with rational coefficients. The canonical examples are π and Euler's number e. Whereas the algebraic numbers form only a countable set, the transcendentals account for "almost all" real numbers (with Lebesgue measure 1).
- How was the transcendence of e proved?
- Hermite proved it in 1873 using an auxiliary polynomial and an integral construction. By choosing a sufficiently large prime p, an expression that must be a nonzero integer is forced to have absolute value less than 1, producing a contradiction.
- Why are Liouville numbers transcendental?
- An algebraic number (a root of a degree-d integer-coefficient polynomial) is known to satisfy the lower bound |α − p/q| ≥ c/q^d. A Liouville number admits rational approximations that violate this lower bound for every d, so it cannot be algebraic and must be transcendental.