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Are there really infinite transcendental numbers?

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.
 
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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.
 
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lpetrich said:
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.
Click to expand...
Wait, how can that even converge?
 
lpetrich said:
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.
Click to expand...
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.
 
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,
\( \Gamma(\frac16) = 2^{-1/3} \sqrt{3/\pi} \Gamma(\frac13)^2 \)
 
 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.
 
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.
 
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.
 
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?
 
lpetrich said:
Infinitely-repeating trailing digits -> rational number: easy to prove. But can one prove the converse?
Click to expand...

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!
 
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)
 
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