Are there really infinite transcendental numbers?

[lpetrich]

Also transcendental: the Euler  Beta function B(a,b) for a, b, and a+b non-integer rational numbers:

\( \displaystyle{ \Beta(z_1,z_2) = \int_0^1 t^{z_1} (1-t)^{z_2} \, dt = \frac{\Gamma(z_1)\Gamma(z_2)}{\Gamma(z_1+z_2)} } \)

Some more partial results are for the Riemann zeta function and its relatives, the Dirichlet eta, lambda, and beta functions:
\( \displaystyle{ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} } \)
\( \displaystyle{ \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} } \)
\( \displaystyle{ \lambda(s) = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} } \)
\( \displaystyle{ \beta(s) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^s} } \)

Interrelationships:
\( \lambda(s) = (1 - 2^{-s}) \zeta(s) \text{ ... } \eta(s) = (1 - 2^{1-s}) \zeta(s) \)
None between the Dirichlet beta function and the others, however.

The Riemann zeta function has special values
\( \displaystyle{ \zeta(2n) = \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n} }{ 2(2n)! } } \)
for n a positive integer and B the Bernoulli numbers, all rational numbers. This is obviously transcendental.

But for zeta(2n+1), it is known that zeta(3) is irrational, and not known whether any of the others is irreational. However, an infinite number of the zeta(2n+1) are irrational, and at least one of zeta(5), zeta(7), ..., zeta(21).

The Dirichlet beta function has special values
\( \displaystyle{ \beta(2n+1) = \frac{(-1)^n E_{2n} (\pi/2)^{2n+1} }{ 2(2n)! } } \)
for n a nonnegative integer and E the Euler numbers, all integers. This is obviously transcendental.

But for beta(2n), it is not known whether any of them are irrational.

 

[lpetrich]

One last series sum, the Euler-Mascheroni constant

\( \displaystyle{ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \log n \right) = 0.5772156649 \dots} \)

It is not known whether it is transcendental, or even irrational, though some partial results have been found.

 

[SLD]

One last series sum, the Euler-Mascheroni constant

\( \displaystyle{ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \log n \right) = 0.5772156649 \dots} \)

It is not known whether it is transcendental, or even irrational, though some partial results have been found.

Wait, how can that even converge?

 

[SLD]

One last series sum, the Euler-Mascheroni constant

\( \displaystyle{ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \log n \right) = 0.5772156649 \dots} \)

It is not known whether it is transcendental, or even irrational, though some partial results have been found.

So I asked Chat GPT how you would go about proving that it was irrational and here is the response:

Determining whether Euler’s constant \gamma (also known as the Euler-Mascheroni constant) is irrational is an open problem in mathematics. However, there are certain approaches and insights that mathematicians have used to investigate this question. Here’s an overview of the steps and methods that might be involved:

Understanding Euler’s Constant

Euler’s constant ( \gamma ) is defined as the limiting difference between the harmonic series and the natural logarithm:
\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln n \right)

Properties and Known Results

1. Numerical Approximation: The numerical value of ( \gamma ) is approximately 0.57721. However, numerical approximation alone does not provide insight into the rationality of ( \gamma ).
2. Continued Fractions: One approach to understanding the rationality of a number is through its continued fraction expansion. For ( \gamma ), the continued fraction does not show any obvious periodicity, which is a necessary (but not sufficient) condition for rationality.
3. Series and Integral Representations: ( \gamma ) has various series and integral representations. Examining these can sometimes provide insights into the nature of the constant. For example:
\gamma = \int_1^\infty \left( \frac{1}{\lfloor x \rfloor} - \frac{1}{x} \right) dx

Analytical Approaches

1. Transcendence Theory: Techniques from transcendence theory, such as those used to prove the irrationality of ( e ) and ( \pi ), might be adapted to ( \gamma ). This often involves constructing auxiliary functions that capture certain properties of ( \gamma ).
2. Approximation by Rationals: One way to demonstrate irrationality is to show that ( \gamma ) cannot be approximated “too well” by rational numbers. This approach involves analyzing the approximations to ( \gamma ) and showing that they fail to satisfy the conditions required by the Rational Approximation Theorem (a generalization of the theorem that proves numbers like ( \pi ) and e are irrational).

Current Status

As of now, it is neither proved nor disproved whether \gamma is irrational. There have been significant results regarding the transcendence and irrationality of other constants, but \gamma remains elusive.

Key Points

1. Direct Proof: No direct proof exists yet for the irrationality of \gamma .
2. Numerical and Analytical Methods: Various methods have been attempted, including those involving series expansions, integral representations, and approximations.
3. Open Problem: This is an active area of research in number theory and mathematical analysis.

Conclusion

To determine whether Euler’s constant is irrational, one would ideally need to develop new techniques or significantly advance existing methods in transcendence theory and Diophantine approximation. While substantial progress has been made in understanding the properties of \gamma , the ultimate proof (or disproof) of its irrationality remains a major open question in mathematics.

 

[lpetrich]

I'll use LaTeX on the gamma-function content of one of my earlier posts.

\( \Gamma(1) = 1 ,\ \Gamma(\frac12) = \sqrt{\pi} \)

• Transcendental and algebraically independent of pi: \( \Gamma(\frac16) , \Gamma(\frac14) ,\ \Gamma(\frac13) ,\ \Gamma(\frac23) ,\ \Gamma(\frac34) ,\ \Gamma(\frac56) \)
• Transcendental: \( \frac{1}{\pi} \Gamma(\frac14)^2 ,\ \frac{1}{\pi} \Gamma(\frac13)^2 \)
• Not known to be irrational, let alone transcendental: \( \Gamma(\frac15) \)

From reflection:
\( \Gamma(\frac13) \Gamma(\frac23) = \frac23 \pi \sqrt{3} \)
\( \Gamma(\frac14) \Gamma(\frac34) = \pi \sqrt{2} \)
\( \Gamma(\frac16) \Gamma(\frac56) = 2\pi \)
From doubling:
\( \Gamma(\frac16) \Gamma(\frac23) = 2^{2/3} \sqrt{\pi} \Gamma(\frac13) \)
\( \Gamma(\frac13) \Gamma(\frac56) = 2^{1/3} \sqrt{\pi} \Gamma(\frac23) \)

Thus,
\( \Gamma(\frac16) = 2^{-1/3} \sqrt{3/\pi} \Gamma(\frac13)^2 \)

 

[lpetrich]

 Liouville number - a kind of transcendental number. It is a real number x with the property that for every positive integer n, there is some p and q that satisfy
\( \displaystyle{ 0 < \left| x - \frac{p}{q} \right| < \frac{1}{q^n} } \)
That means that Liouville numbers are almost rational numbers.

But they are not only irrational, they are transcendental. However, e and pi are not Liouville numbers.

Liouville numbers can be constructed from real numbers between 0 and 1. For base b, the original number's digits are a1, a2, a3, ...:
\( \displaystyle{ x_0 = \sum_{n=1}^\infty \frac{a_n}{b^n} } \)
One can make a number using powers n! instead of n of the base:
\( x_L = L(b,x_0) = \displaystyle{ \sum_{n=1}^\infty \frac{a_n}{b^{n!}} } \)
This number can be shown to be a Liouville number. I will call it a factorial Liouville number.

One of them is Liouville's constant: \( \displaystyle{ \sum_{n=1}^\infty \frac{1}{10^{n!}} } \)

How many of them?

{Real numbers in (0,1]} ~ {Factorial Liouville numbers} < {Liouville numbers} < {Real numbers}

C = |Factorial Liouville numbers| <= |Liouville numbers| <= C

Thus, |Liouville numbers| = C

Some factorial Liouville numbers are computable: the ones derived from computable numbers. Liouville's constant, for instant. Reversing the derivation gives the converse: every computable factorial Liouville number is derived from a computable number. All this is also true for definable numbers instead of computable ones.

 

[lpetrich]

AJUR Vol 13 Issue 3 08.25.16.pdf - AJUR Vol 13 Issue 3 08.25.16 pp.41-44.pdf - American Journal of Undergraduate Research - On Closure Properties of Irrational and Transcendental Numbers under Addition and Multiplication

Irrational numbers? It's almost too easy to show that they are not closed under either operation.

Transcendental numbers?

The paper cites this theorem: if x and y are transcendental numbers, then at least one of x + y and x*y is transcendental.

Proof: let x + y = a and x*y = b and solve for x and y:

x = (1/2)*(a - sqrt(a^2 - 4*b))
y = (1/2)*(a + sqrt(a^2 - 4*b))
and the reverse: x has + and y has -.

If both a and b are algebraic, then x and y must also be algebraic, since the algebraic numbers are algebraically closed. So for x and y to be transcendental, at least one of a and b must also be transcendental.

 

[lpetrich]

real analysis - Do the Liouville Numbers form a field? - Mathematics Stack Exchange

Absolutely not, because an algebraic field needs an additive identity and a multiplicative one: 0 and 1 for the rational numbers and their supersets. Neither is a Liouville number.

Also, every real number can be expressed as the sum of two Liouville numbers and every nonzero real number as the product of two Liouville numbers. Noting
Representations of Real Numbers as Sums and Products of Liouville numbersf - Paul Erdős - Michigan Math. Journal 9, pp.59--60, 1962.

Since the reciprocal of a Liouville number is also a Liouville number, every nonzero real number is thus the solution of a linear equation whose coefficients are Liouville numbers.

This is consistent with their having cardinality C, like the real numbers.


In the almost-rational definition of Liouville numbers that I'd earlier posted on, p must be an integer and q an integer > 1.

 

[lpetrich]

Rational numbers have the property that their decimal or other place-system representations have infinitely-repeating sequences of digits that trail the first digits.

I'll show three repeats of each repeated sequence.

Decimal:
1 - 1.000... 2 - 0.5000... 3 - 0.333... 4 - 0.25000... 5 - 0.2000... 6 - 0.1666... 7 - 0.142857142857142857... 8 - 0.125000... 9 - 0.111... 10 - 0.1000...

Binary:
1 - 1.000... 10 - 0.1000... 11 - 0.010101... 100 - 0.01000... 101 - 0.001100110011... 110 - 0.0010101... 111 - 0.001001001... 1000 - 0.001000...

Ternary:
1 - 1.000... 2 - 0.111... 10 - 0.1000... 11 - 0.020202... 12 - 0.121012101210... 20 - 0.0111... 21 - 0.010212010212010212... 22 - 0.010101... 100 - 0.01000...

Quinary:
1 - 1.000... 2 - 0.222... 3 - 0.131313... 4 - 0.111... 10 - 0.1000...

Infinitely-repeating trailing digits -> rational number: easy to prove. But can one prove the converse?

 

[Swammerdami]

Infinitely-repeating trailing digits -> rational number: easy to prove. But can one prove the converse?

The key theorem one first applies for the proof is  Fermat's little theorem.

For example, 17 is coprime to 10, so Fermat's little theorem tells us that 10^16 ≡ 1 (mod 17).
In other words, 17 divides 10^16-1 evenly.
In still other words, 9999999999999999 = 17*q, where q is an integer. It happens that q = 0588235294117647.
(I've added a leading zero to give numerator and denominator the same size.)
In other w0rds 1/17 = 0588235294117647 / 9999999999999999 = 0.0588235294117647 0588235294117647 0588235294117647 ...

This is not a COMPLETE proof, but I think it points the way!

 

[lpetrich]

One can prove that converse using  Euler's theorem a.k.a. the Fermat–Euler theorem or Euler's totient theorem.

It uses  Euler's totient function - for positive integer n, phi(n) is the number of positive integers coprime or relatively prime to n.

phi(n) = n * product of (1 - 1/p) over p: all the distinct prime factors of n

The theorem itself: for a and n coprime,

a^(phi(n)) = 1 mod n

That means that a^(phi(n)) - 1 is proportional to n.

Thus, if a^q - 1 = K*n for some q, then 1/n = K/(a^q - 1) = K*(a^(-q) + a^(-2q) + a^(-3q) + ...)

If a is the number base, then we find an infinitely repeating sequence of digits with length q or some divisor of q.

For numerator N, N/n = N*K/(a^q - 1) = (N*K)*(a^(-q) + a^(-2q) + a^(-3q) + ...)

(N*K) may be larger than a^q, meaning that sequences will overlap. But their sum will still repeat.

From my previous calculation of trailing digits:

• phi: 1 1 2 2 4 . 2 6 4 6 4
• 10: 0 0 1 0 0 . 1 6 0 1 0
• 2: 0 0 2 0 4 . 2 3 0
• 3: 0 1 0 2 4 . 1 6 2 0
• 5: 0 1 2 1 0

Number: number of digits that repeat, 0 for terminating (all zeros).

Zero means that the divisor evenly divides a power of the number base, while nonzero length all evenly divide phi(denominator).

For the denominator not coprime with the base, it can be separated out as a product of a part that some power of the base will divide multiplied by a part that is coprime with the base. One can then proceed as before with the second part.

If the second part is 1, then the representation will terminate, with a finite number of nonzero digits.

Thus,

(rational number) <-> (terminating or having an infinitely repeating sequence of trailing digits)

and also

(irrational number) <-> (trailing digits do not terminate or have infinitely repeating sequences)

 

[Swammerdami]

The key theorem one first applies for the proof is  Fermat's little theorem.

Fermat's little theorem is one of the most important theorems of number theory. Among many other uses, it provides the basis for the Fermat primality test, a procedure that can find composite numbers very quickly, without actually finding the factors. (Unfortunately it is a "probabilistic" test. Some composite numbers will pass a primality test -- these are called pseudo-primes.)

Although Fermat never actually provided a proof of this theorem, it turns out there is a VERY easy and elegant proof based on necklace counting.

A simple statement of Fermat's little theorem is this

For any positive integer a and any prime p, (a^p - a) is a multiple of p.​

Consider a string of beads, with p beads altogether; the beads are identical except that a different colors are allowed.
The number of possible strings is a^p of which a have beads all of a single color. For example, if a = 2 and p = 3 there are (2³ = 8) possible strings (BBB, BBw, BwB, Bww, wBB, wBw, wwB, and www) of which 2 (BBB and www) use only a single color.

The theorem applies correctly in this very simple case: there are 6 strings that use more than one color, and 6 is indeed a multiple of 3, as the theorem says it must be. But WHY does it work? Let's count necklaces!

Take each string of beads and knot it into a loop, or "necklace." Three necklaces (BBw, BwB, and wBB) are called "equivalent" because each can be rotated into any of the others in the trio. (Bww, wBw, wwB) form a second trio. The number of ways to get equivalence via rotation must be a divisor of p (but not 1 since we've already subtracted out single-color necklaces). Reflections, or inverting the necklace are not allowed for equivalence (though that doesn't matter for this trivial (2,3) case.) The necklace-counting proof is written up at the Wiki linked above.

Is it lovely when an important math theorem has such an elegant simple proof?

 

[lpetrich]

Fermat's Little Theory is easy to prove by mathematical induction.

It is obviously true for a = 1, and for any a, one can prove it for a+1:

\( \displaystyle{ (a+1)^p - (a+1) = \sum_{k=0}^p {p \choose k} a^k - (a + 1) = \sum_{k=1}^{p-1} {p \choose k} a^k + (a^p - a) = \sum_{k=1}^{p-1} {p \choose k} a^k } \)

\( \displaystyle{ {p \choose k} = \frac{p!}{k! (p-k)!} = p \frac{(p-1!}{k! (p-k)!} } \)

For 1 <= k <= (p-1), the fractional part in the right-hand side has no factors of p, thus making this binomial coefficient proportional to p:

\( \displaystyle{ {p \choose k} \propto p \to \sum_{k=1}^{p-1} {p \choose k} a^k \propto p \to a^p - a \propto p } \)

Thus proving Fermat's Little Theorem.

 

[lpetrich]

For Euler's theorem, we can use group theory. Let's see what happens when we do modular arithmetic.

The integers Z form a ring: addition is an abelian (commutative) group, and multiplication is a commutative monoid.

Let's see which properties carry over into modular arithmetic, arithmetic modulo n.

Addition and multiplication are both commutative. They are also both associative, making addition and multiplication semigroups. Multiplication is distributive over addition. They both have identities, 0 for addition, and 1 for multiplication, making both addition and multiplication monoids.

The remaining property, inverses, is more difficult. For addition, the inverse of a is n - a, and one finds an additive group over all elements: Z(n). Addition and multiplication together thus form a ring, both originally and under modular arithmetic.

Do any elements of this ring have multiplicative inverses?

First, an inverse must be unique: inv1(a)*a = inv2(a)*a = 1 implies inv1(a) = inv2(a) for all a. That is, multiplication must be cancellative. When will that happen?

a1*b = a2*b mod n
gives us
(a1 - a2)*b = 0 mod n
If b contains some factors of n, then a1-a2 may be nonzero mod n. But if b is coprime or relatively prime to n, then a1-a2 = 0 mod n -- cancellation.

So we consider Zx(n) -- all the elements of Z(n) coprime with n under multiplication. This is obviously closed, since if neither a and b have prime factors shared with n, then a*b will also lack shared prime factor. The multiplicative identity 1 is also in it, since it has no prime factors. So Zx(n) is a monoid.

All that remains is to show that inverses exist.

 Bézout's identity and Bézout's Identity - ProofWiki have proofs of that identity:

For integers a and b, there will be other integers x and y satisfying a*x + b*y = gcd(a,b)

Let us take a in Zx(n) and b = n: a*x + n*y = 1 since a and n are coprime.

Taking modulo n, this means that a*x = 1 and thus that x is the inverse of a. Since this proof works on every element of Zx(n), that means that every element has at least one inverse. But since multiplication is cancellative in Zx(n), that means that inverses are unique, and that Zx(n) is thus a group.

 

[lpetrich]

Now that we have established that Zx(n) is a group, we can use this result to prove Euler's theorem.

Element a will generate a cyclic group {a} with order or number of elements m, where m is the smallest positive integer that makes a^m = 1. By Lagrange's theorem, its order evenly divides the order of the full group Zx(n), which is Euler's totient function phi(n). Thus,

a^(phi(n)) = 1 mod n

 Multiplicative group of integers modulo n and Modulo Multiplication Group -- from Wolfram MathWorld

Every finite ring can be decomposed into a direct product of rings with orders p^m for distinct primes p that give the order of the original ring when multiplied together.

Likewise, Z(n) = direct product of Z(p^m) and Zx(n) = direct product of Zx(p^m)

Every finite abelian group is isomorphic to a product of Z(p^m) where the Z(p^m) can repeat and can share the same p values. Thus, Z2*Z2 and Z2*Z4 are both possible.

Zx(p^m) ~ Z(p^(m-1)*(p-1))
Zx(2) ~ identity group, Zx(4) ~ Z2, Zx(2^m) ~ Z2*Z(2^(m-2))

 

[lpetrich]

These decompositions can tell us about maximum orders of elements in Zx(n, and thus the maximum number of repeats in a place-system representation.

For powers of primes, if that prime is odd, then the maximum order is p^(m-1)*(p-1)

For 2, however, the maximum order is, for 2, 1, for 4, 2, for 2, 4, for 2^m, 2^(m-2) = 2^(m-2)*(2-1)

Thus, the maximum is, for odd n, phi(n), for even n and power of 2 <= 2, phi(n), and for even n and power of 2 > 2, phi(n/2)


Returning to transcendental numbers, here is one: Champernowne Constant -- from Wolfram MathWorld

Take positive integers represented with some base and concatenate the digits:

0.12345678910111213...

(decimal version) - it is transcendental.

 

[lpetrich]

A good Wikipedia article will have plenty of references for its assertions, and  Transcendental number is no exception. Transcendental Number -- from Wolfram MathWorld also has plenty of references.

Transcendence of Periods: The State of the Art (2006) by Michel Waldschmidt

Periods? From Periods - a kind of number that M. Kontsevich and D. Zagier call a "period".

Periods are complex numbers whose real and imaginary parts are rational numbers with rational coefficients over domains of R^n given by polynomial inequalities with rational coefficients. Periods can also be defined with algebraic numbers instead of rational numbers.

This kind of number is countable, like algebraic numbers and computable numbers, since each one has a finite number of parameters, parameters that are all rational numbers.

All algebraic numbers are periods. For example, sqrt(2) = integral of 1 over x with 2*x^2 <= 1.

Logarithms of algebraic numbers, pi, elliptic integrals of algebraic numbers, and zeta functions of integers >= 2 are all periods.

But e, 1/pi, and the Euler-Mascheroni constant are probably not periods.

It's rather scattershot what is and is not a period.

From the beta integral, for p, q positive integers, Gamma(p/q)^q is a period.

Beta integral? One has to use a generalized one.

\( \displaystyle{ B(a,b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} = \int_0^t t^{a-1} (1-t)^{b-1} \,dt = \int_0^1\int_0^1 t^{a-1} u^{b-1} \delta(t+u-1) \, dt \, du } \)

\( \displaystyle{ B(a_1,a_2, \dots ,a_n) = \frac{ \prod_{k=1}^n \Gamma(a_k) }{\Gamma(\sum_{k=1}^n a_k)} = \int_0^1 \int_0^1 \cdots \int_0^1 \prod_{k=1}^n (t_k)^{a_k-1} \delta(\sum_{k=1}^n t_k - 1) \, dt_1 \, dt_2 \cdots dt_n } \)

 

[lpetrich]

For nonzero algebraic a's and algebraic b's, \( \displaystyle{ \sum_i b_i \log a_i } \) is either zero or transcendental.

For algebraic-coefficient polynomials P(z), Q(z) of complex variable z, with deg(P) < deg(Q), then

integral of P(z)/Q(z) over z over path C is either 0 or transcendental if path C is either closed or else with algebraic or infinite endpoints.

"he value of any elliptic integral of the first or the second kind with algebraic coefficients between algebraic limits is either zero or else transcendental."

For an ellipse with algebraic axis lengths a and b, its perimeter is transcendental.

Likewise for the perimeter of the lemniscate, an 8 or oo shape with equation in coordinates x, y:

(x² + y²)² = x² - y²

In polar coordinates: r = sqrt(cos(2a))

The length: ds = 2*sqrt(sec(2a)) da
= 2 * dx / sqrt(1-x^2) / sqrt(1-2x^2) where x = sin(a) and r = sqrt(1 - 2x^2)
= sqrt(2) * dy / sqrt(1-y^2) / sqrt(1-y^2/2) where x = y/sqrt(2) and r = sqrt(1 - y^2)

For algebraic-number limits, this length is a period, and that includes the total length.

 Lemniscate constant and  Lemniscate elliptic functions

 

[lpetrich]

Using the beta function, some gamma-function values are given in terms of elliptic integrals.

\( \displaystyle{ \int_0^1 \frac{dx}{\sqrt{x - x^3}} = \frac12 B(1/4,1/2) = \frac{\Gamma(1/4)^2}{2^{3/2}\pi^{1/2}} } \)

\( \displaystyle{ \int_0^1 \frac{dx}{\sqrt{1 - x^3}} = \frac13 B(1/3,1/2) = \frac{\Gamma(1/3)^3}{2^{4/3}3^{1/2}\pi} } \)

If a, b, and a+b are non-integer rational numbers, then B(a,b) is transcendental.

Gamma(1/4) and pi are transcendental and algebraically independent, as are Gamma(1/3) and pi.

Gamma(3/4) is algebraically related to Gamma(1/4) and pi, and Gamma(1/6), Gamma(2/3), and Gamma(5/6) to Gamma(1/3) and pi.

No transcendence results are known for gamma functions that are not algebraically related to 1, pi, Gamma(1/4), and Gamma(1/3). Factorials, for instance, are algebraically related to 1, and Gamma(1/2) = sqrt(pi), another algebraic relation.

Algebraically independent: every rational/algebraic-coefficient polynomial in the values is nonzero except for the trivial one.

 

[lpetrich]

Part of the proof involves Weierstrass elliptic functions, an alternate formulation of this kind of function.

• (dP/dz)^2 = 4*P^3 - g2*P - g3
• dZ/dz = - P
• Z(z + w) = Z(z) + h

At least two of these four numbers - g2, g3, w/pi, h/pi - are algebraically independent.

Gamma(1/4), and Gamma(1/3), like pi and e, are not Liouville numbers.

It's hard to go much further, but at least one of Gamma(1/5) and Gamma(2/5) is not only transcendental, but algebraically independent of pi.

Returning to 1/3 and 1/4, each set is algebraically independent:

• Gamma(1/4), pi, e^pi
• Gamma(1/3), pi, e^(pi*sqrt(3))

Conjecture: at least 3 of these numbers is algebraically independent: Gamma(1/5), Gamma(2/5), pi, e^(pi*sqrt(5))