# Transcendental Numbers

A **transcendental number** is a number that is not a root of any polynomial with integer coefficients. They are the opposite of algebraic numbers, which are numbers that are roots of some integer polynomial. \(e\) and \(\pi\) are the most well-known transcendental numbers.

That is, numbers like \(0, 1, \sqrt 2,\) and \(\sqrt[3]{\frac12}\) are all algebraic numbers, as they are the roots, or the 0's, of the polynomials

\[f(x)=x^2,\quad f(x)=x^2-1,\quad f(x)=x^2-2,\quad f(x)=2x^3 - 1,\]

respectively. For instance, for \(f(x)=x^2-2\), \(\sqrt 2\) is a root, because \(f\big(\sqrt 2\big) = \big(\sqrt 2\big)^2-2 = 0\). For these algebraic numbers \(a\), there exists some polynomial with integer coefficients where \(f(a) = 0\). For transcendental numbers \(t\), however, there is no \(f(t)\) where \(f(t)=0\). For instance \(2^{\sqrt2}\) is transcendental, so \(f\big({\small 2^{\sqrt2}}\big)\neq 0\) for all possible \(f(x)\) with integer coefficients..

Transcendental numbers are useful in the study of straightedge-and-compass constructions, particularly in proving the impossibility of squaring the circle (i.e. it proves that it is impossible to construct a square with area equal to the area of any given circle, including \(1\pi\), using only a straightedge and a compass, which proves that \(\pi\) is transcendental).

Surprisingly, *almost all* real numbers are transcendental, meaning that a randomly chosen real number will be transcendental with probability 1 (with respect to cardinality). Nonetheless, only a few numbers have been proven transcendental (such as \(\pi\) and \(e\)), and the vast majority remain unknowns (such as \(\pi e\)).

Additionally, an important conjecture in transcendental theory indirectly implies that Euler's identity, the famous equation

\[e^{\pi i}+1=0,\]

is the *only* nontrivial relation between \(e, \pi,\) and \(i\). If true, this is quite a remarkable result.

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## History

Leibniz was the first to use the term "transcendental" in a 1682 paper, where he proved the function \(f(x)=\sin x\) is not an **algebraic function** of \(x\), which roughly means that there are no integer polynomials \(p_0(x), p_1(x), \ldots\) such that

\[p_0(x)(\sin x)^0+p_1(x)(\sin x)^1+p_2(x)(\sin x)^2+\cdots=0\] for all \(x\).

However, it was not until 1748 that the term was applied to transcendental *numbers* when Euler conjectured that if \(\log_ab\) is irrational, it is also transcendental. However, this conjecture went unproved for over a century.

Nearly 100 years later, Liouville demonstrated that transcendental numbers existed, using a constructive proof involving continued fractions. However, his proof was only strong enough to demonstrate specifically crafted numbers (known as Liouville numbers) were transcendental, and in particular was not strong enough to detect the transcendentality of \(e\).

Some time later, Cantor demonstrated a deeper result: not only do transcendental numbers exist, but they also comprise almost *all* real numbers. This was a consequence of a deeper result regarding the uncountability of the real numbers, which also has deep applications to the study of infinities.

In 1955, the Thue-Siegel-Roth theorem (often shortened to Roth's theorem) improved significantly upon Liouville's work, allowing for the discovery of many other transcendental numbers, most notably the **Champernowne constant** \(0.1234567891011\ldots\). This work earned Roth the 1958 Fields Medal.

The field received a new boost with the discovery of the **Lindemann-Weierstraß** theorem, which finally put to rest the question of squaring the circle, showing it was impossible to construct a square with area equal to a given circle using only a straightedge and compass. In particular, the theorem proved that \(\pi\) is transcendental, and it was previously established that only algebraic lengths can be made using a straightedge and compass, so the problem was answered in the negative.

The latest major result was the **Gelfand-Schneider** theorem, which showed that \(a^b\) is transcendental for any algebraic \(a,b\) \((\)where \(a\) is not \(0,1\) and \(b\) is irrational\().\) This answers Hilbert's seventh problem affirmatively, from the famous 23 problems Hilbert proposed at the turn of the century.

## Liouville Numbers

Liouville's approach, intuitively speaking, was to argue that algebraic numbers cannot be approximated particularly well by rational numbers, so any number that *can* be approximated well by rational numbers must be transcendental. More formally, if \(\alpha\) is an algebraic number with degree \(d\) \((\)meaning the polynomial with the smallest degree that has \(\alpha\) as a root has degree \(d),\) then the inequality

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{d+\epsilon}}\]

has only finitely many solutions \(\frac{p}{q}\), where \(\epsilon\) is any positive number.

However, this result is difficult to use in showing a number is transcendental, as it is necessary to demonstrate infinite solutions \(\frac{p}{q}\) for **every** value of \(d\). As a result, Liouville's proof was only sufficient to show specially crafted numbers were transcendental, most notably the number

\[\sum_{i=0}^{\infty}\frac{1}{10^{i!}}=0.110001000000000000000001\ldots\]

known as **Liouville's constant**. More generally, \(\displaystyle \sum_{i=0}^{\infty}\frac{a_k}{b^{k!}}\) \((\)i.e. \(0.a_1a_2000a_300000000000000000a_4\ldots_b)\) is transcendental for \(0 \leq a_k \leq b-1\), and this class of numbers is known as **Liouville numbers**.

These numbers are chosen specifically to satisfy the following property:

For any integer \(n\) and Liouville number \(L\), there exist an infinite number of pairs \((p,q)\) satisfying the inequality \[ \] \[\left|L-\frac{p}{q}\right|<\frac{1}{q^n}.\]

If proved, this implies \(L\) is transcendental, as if it were algebraic with degree \(d\) then there would only be finitely many solutions to \(\left|L-\frac{p}{q}\right|<\frac{1}{q^{d+\epsilon}}\), contradicting the fact that there are infinite solutions for any \(n\).

The proof makes use of the following lemma:

If \(\alpha\) is an algebraic number with degree \(n\), then there exists a real number \(A\) such that \(\displaystyle \left|\alpha-\frac{p}{q}\right|>\frac{A}{q^n}\) for all \(p\) and \(q.\)

Now, if \(L\) is algebraic, then there exists some \(A\) such that

\[\left|L-\frac{p}{q}\right|>\frac{A}{q^n}\]

for all \(p,q\). But from the above property, there exist integers \(a,b\) such that

\[\left|L-\frac{a}{b}\right|<\frac{1}{b^m}\]

for any \(m\). In particular, setting \(m=n+r\) gives

\[\left|L-\frac{a}{b}\right|<\frac{1}{b^n+r} = \frac{1}{b^r} \cdot \frac{1}{b^{n}} \leq \frac{1}{2^r}\frac{1}{b^{n}},\]

so choosing a big enough \(r\) such that \(\frac{1}{2^r}<A\),

\[\left|L-\frac{a}{b}\right|<\frac{1}{b^n+r} = \frac{1}{b^r} \cdot \frac{1}{b^{n}} \leq \frac{1}{2^r}\frac{1}{b^{n}}<\frac{A}{b^n}\]

for any \(n\), which contradicts the original assertion that \(L\) is algebraic. Hence, \(L\) must be transcendental. \(_\square\)

## Cantor's Proof of Transcendentality

Cantor demonstrated that transcendental numbers exist in his now-famous diagonal argument, which demonstrated that the real numbers are **uncountable**. In other words, there is no bijection between the real numbers and the natural numbers, meaning that there are "more" real numbers than there are natural numbers (despite there being an infinite number of both).

Cantor also showed that there *is* a bijection between the natural numbers and the algebraic numbers, meaning that there are "more" real numbers than there are algebraic numbers. This directly implies that there must be real numbers that are not algebraic, which means that transcendental numbers exist.

In fact, Cantor's argument is stronger than this, since it demonstrates an important result:

Almost allreal numbers are transcendental.

In this sense, the phrase "almost all" has a specific meaning: all numbers except a countable set. In particular, if a real number were chosen randomly (the term randomly is used loosely here, as it is impossible to choose a real number uniformly at random, but the intuition suffices), the probability it will be algebraic is 0.

## The Thue-Siegel-Roth Theorem

Liouville's theorem was sufficient to show transcendental numbers existed, but only specially crafted ones. Thue was the first to realize the broader application of Liouville's theorem to Diophantine equations if the exponent \(d\) in the inequality

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{d+\epsilon}}\]

could be reduced. **Thue's theorem**, in 1909, improved the bound to

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{\frac{d}{2}+1+\epsilon}}.\]

This was later improved again by Siegel, and later Dyson, who both demonstrated a bound of

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{O\left(\sqrt{d}\right)}},\]

where big O notation is used. Finally, Roth effectively settled Liouville's work and earned a Fields Medal in the process, by improving the bound to \(2+\epsilon:\)

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2+\epsilon}},\]

which is quite a remarkable improvement. This theorem is interchangeably known as the **Thue-Siegel-Roth** theorem, honoring the progression of improved bounds, and as **Roth's theorem** after its ultimate discoverer.

## Dirichlet's Approximation Theorem

It is worth noting the purpose of the \(\epsilon\) in the previous section: the statement fails to remain true when the exponent is reduced to 2. In particular,

there are infinitely many \(p,q\) satisfying \(\displaystyle \left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}}\) for any

real\(\alpha\), whether algebraic or transcendental.

It is clearer to prove a stronger result:

For any integer \(n\), there exist \(p,q\) such that \(1 \leq q \leq n\) and \(\displaystyle \left|\alpha-\frac{p}{q}\right|<\frac{1}{nq}.\)

Consider the interval

\[[0,1) = \left[0, \frac{1}{n}\right) \cup \left[\frac{1}{n}, \frac{2}{n}\right) \cup \ldots \cup \left[\frac{n-1}{n}, n\right),\]

where there are \(n\) sub-intervals. Now consider the \(n+1\) fractional parts \(\{0 \cdot \alpha\}, \{1 \cdot \alpha\}, \ldots, \{n \cdot \alpha\}\). By the pigeonhole principle, some two of these fractional parts belong to the same interval, so there exist \(i,j\) for which \(\{i \cdot \alpha\}\) and \(\{j \cdot \alpha\}\) belong to the same interval. But then \(\big|(i-j)\alpha-k\big|<\frac{1}{n}\) for some integer \(k\), since \((i-j)\alpha\) is within \(\frac{1}{n}\) of an integer \(k\). As a result,

\[\left|\alpha-\frac{k}{i-j}\right|<\frac{1}{n(i-j)},\]

precisely as desired. \(_\square\)

## Lindemann-Weierstrauss Theorem

The greatest triumph of transcendental number theory was the **Lindemann-Weierstrauss theorem**:

If \(\alpha_1, \alpha_2, \ldots, \alpha_n\) are algebraic numbers that are linearly independent over \(\mathbb{Q}\), then \(\{e^{\alpha_1}, e^{\alpha_2}, \ldots, e^{\alpha_n}\}\) are

algebraically independentover \(\mathbb{Q}\), meaning that the set is not the solution to any nontrivial polynomial with rational coefficients.

This helps to discover several more transcendental numbers, most notably \(e\) and \(\pi\):

\(e\) and \(\pi\) are both transcendental.

Note that \(\{1\}\) is a linearly independent set, so \(\{e\}\) is an algebraically independent set. This implies that \(\{e\}\) is not the solution to any nontrivial polynomial, so \(e\) is transcendental by definition. This same logic shows that \(e^{\alpha}\) is transcendental for any algebraic \(\alpha\).

Now suppose \(\pi\) were algebraic. Since \(i\) is also algebraic, this implies that \(\pi i\) is algebraic. Hence \(e^{\pi i}\) is transcendental as \(e^{\alpha}\) is transcendental for any algebraic \(\alpha\). However, \(e^{\pi i}=-1\), which is clearly not transcendental, so \(\pi\) must be transcendental as desired. \(_\square\)

## Applications and Open Problems

Surprisingly, relatively little is known about transcendental numbers in general. In fact, there is little general strategy available for determining whether a specific number is transcendental, especially when the number is unrelated to the exponential function \(e^x\) and the logarithmic function \(\ln x\). For example, while some numbers are known to be transcendental:

- \(e\)
- \(\pi\)
- Champernowne's constant \(0.1234567891011\ldots\)
- Gelfond's constant \(e^{\pi}\)
- Gelfond-Schneider constant \(\left(\sqrt{2}\right)^{\sqrt{2}}\)
- \(\ln 2\)
- \(\sin 1\) \(\big(\)which is different from \(\sin 1^\circ\big),\)

the majority are not. For instance, it is unknown whether

- \(\pi e\)
- \(\pi+e\) (though it is known that at least one of \(\pi e\) and \(\pi+e\) is transcendental)
- \(\pi^{\pi}\)
- \(e^e\)
- \(\pi^e\)
- the Euler-Mascheroni constant \(\gamma\)
- Apéry's constant \(\zeta(3) = 1.20205\ldots\) (Apéry showed in 1978 that it is irrational)

are transcendental. More specifically, while the **Gelfond-Schneider theorem** showed that \(a^b\) is transcendental for any algebraic \(a,b\) (other than the trivial cases \(a=0,1\) and \(b\) is rational), this is still a countable set of numbers. In fact, it is not known whether many of the constants on the above list are even rational.

Several conjectures have been made regarding these types of numbers, such as **Schanuel's conjecture** which states that \(\mathbb{Q}(z_1, z_2, \ldots, z_n, e^{z_1}, e^{z_2}, \ldots, e^{z_n})\) has transcendence degree at least \(n\), which generalizes the Lindemann-Weierstrauss theorem above. If proved, it would establish the nature of numbers such as \(\pi+e\) and \(e^e\). In some sense, this also implies Euler's theorem \(e^{\pi i}+1=0\) is the only relation between \(e, \pi,\) and \(i\).

Additionally, as an improvement to Roth's theorem, Lang conjectured that

\[\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^{2}\log(q)^{1+\epsilon}},\]

which is as yet unproved.

\[\large \color{skyblue}{\boxed{0.1234567891011...}} \quad \color{violet}{\boxed{\sqrt{1729}}} \quad \color{green}{\boxed{\sqrt{2}}} \quad \color{blue}{\boxed{e} } \quad \color{red}{\boxed{\pi} }\]

How many of the above numbers are transcendental numbers?

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## See Also

**Cite as:**Transcendental Numbers.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/transcedental-number/