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Daniel Strecker's Blog - daniel-strecker.com
published: 2022-08-01
last change: 2022-08-07

Transcendental Numbers

Just some notes.

Relation of Number Sets

The illustration works for both, ℝ or β„‚. Depending on the choice, 𝔸 is either the real algebraic numbers or the complex algebraic numbers, and likewise the same is true for 𝐈 and 𝐓. Not affected by this choice is β„š, which is always the set of just the quotients of integers β„€, which are also not affected.

β„€: integer numbers
β„š: rational numbers, i.e. quotients
ℝ: real numbers
β„‚: complex numbers
𝔸: algebraic numbers
𝐈 := ℝ\β„š (or β„‚\β„š): irrational numbers (not a ring)
𝐓 := ℝ\𝔸 (or β„‚\𝔸): transcendental numbers (not a ring)

Note that:
β„š ⊊ 𝔸 ⊊ ℝ
𝐓 ⊊ 𝐈 ⊊ ℝ
𝔸 and 𝐈 intersect, but neither is a subset of the other.

There are numbers which are both irrational and algebraic, but some algebraic numbers are not irrational and some irrational numbers are not algebraic.
Examples:
√(2) ∈ 𝔸 and √(2) ∈ 𝐈     (√(2) βˆ‰ 𝐓)
1 ∈ 𝔸, but 1 βˆ‰ 𝐈
Ο€ ∈ 𝐈, but Ο€ βˆ‰ 𝔸     (Ο€ ∈ 𝐓)


Lemma 0:
logba is irrational for a,b ∈ β„• with a β‰  b, b β‰  1.

Proof.
Let a, b be integers with a β‰  b, b β‰  1. Then βˆ€i,j ∈ β„• we have aⁱ β‰  bΚ² by the fundamental theorem of arithmetic.
And then
aⁱ β‰  bΚ²
⇔ (aⁱ)1/i β‰  (bΚ²)1/i
⇔ ai/i β‰  bj/i
⇔ aΒΉ β‰  bj/i
⇔ logba β‰  logb(bj/i)
⇔ logba β‰  j/i   (βˆ€i,j ∈ β„•)
⇔ βˆ„i,j ∈ β„•: logba = j/i
⇔ logba βˆ‰ β„š . ∎


Gelfond-Schneider Theorem

Theorem 1 (Gelfond-Schneider theorem):
If b ∈ π”Έβˆ–{0,1}, p ∈ πˆβˆ©π”Έ, then bα΅– is transcendental.

Corollary 1.1a (Contrapositive A of the Gelfond-Schneider theorem):
If tα΅– is algebraic, p ∈ πˆβˆ©π”Έ, and t βˆ‰ {0,1}, then t is transcendental.

Proof.
Let tα΅– algebraic, p ∈ πˆβˆ©π”Έ, and t βˆ‰ {0,1}. Then tα΅– is not transcendental. With the contrapositive of theorem 1 it follows that t βˆ‰ π”Έβˆ–{0,1} or p βˆ‰ πˆβˆ©π”Έ. Because we know that by the hypothesis p ∈ πˆβˆ©π”Έ, it follows that t βˆ‰ π”Έβˆ–{0,1}. We also know that t βˆ‰ {0,1}, so t βˆ‰ π”Έβˆ–{0,1} βˆͺ {0,1} β‡’ t βˆ‰ 𝔸 β‡’ t ∈ 𝐓. ∎

Corollary 1.1b (Contrapositive B of the Gelfond-Schneider theorem):
If bα΅– is algebraic and b ∈ 𝔸\{0,1}, then p βˆ‰ πˆβˆ©π”Έ.

Proof.
Let bα΅– algebraic and b ∈ π”Έβˆ–{0,1}. Then bα΅– is not transcendental. With the contrapositive of theorem 1 it follows that b βˆ‰ π”Έβˆ–{0,1} or p βˆ‰ πˆβˆ©π”Έ. Because we know by the hypotheis that b ∈ π”Έβˆ–{0,1}, it follows that p βˆ‰ πˆβˆ©π”Έ. ∎

Corollary 1.2:
logba is transcendental for a,b ∈ β„• with a β‰  b, b > 1.

Proof.
Let a,b ∈ β„• with a β‰  b, b > 1. Then blogba = a is algebraic. With blogba algebraic and algebraic b βˆ‰ {0,1} it follows with corollary 1.1b that logba βˆ‰ πˆβˆ©π”Έ. For a proof by contradiction, assume logba ∈ 𝔸. From lemma 0 we know that logba ∈ 𝐈. With logba ∈ 𝐈, ∈ 𝔸 it follows that logba ∈ πˆβˆ©π”Έ, which is a contradiction, because we already know that logba βˆ‰ πˆβˆ©π”Έ, so the assumption is false and it follows that logba βˆ‰ 𝔸 β‡’ logba ∈ 𝐓. ∎

Corollary 1.3:
e^Ο€ is transcendental, because
e^Ο€
= e^(-Ο€ Β· -1)
= e^(-Ο€ Β· iΒ²)
= e^(iΟ€ Β· -i)
= (e^(iΟ€))⁻ⁱ
= (-1)⁻ⁱ, which is transcendental by the Gelfond-Schneider theorem, because -1 is algebraic and -i is irrational.


Hermite-Lindemann Theorem (aka. Lindemann-Weierstraß Theorem)

Theorem 2 (Hermite-Lindemann theorem, aka. Lindemann-Weierstraß theorem):
if α₁, ..., Ξ±β‚™ are algebraic numbers that are linearly independent over the rational numbers β„š, then eα₁, ..., eΞ±β‚™ are algebraically independent over β„š.

Corollary 2.1 (Corollary to the Hermite-Lindemann theorem):
If 𝛼 ∈ 𝔸\{0}, then e𝛼 is transcendental.

Proof.
Let 𝛼 ∈ 𝔸\{0}. Then {𝛼} is a linearly independent set over the rationals. By the Hermite-Lindemann theorem, {e𝛼} is an algebraically independent set, or in other words e𝛼 is transcendental. ∎

Corollary 2.2:
e is transcendental, because 1 is algebraic, so by corollary 2.1 eΒΉ is transcendental.

Corollary 2.3 (Contrapositive to corollary 2.1):
If eα΅— is algebraic and t β‰  0, then t is transcendental.

Proof:
Let t β‰  0 and for a proof by contradiction, assume that t is not transcendental, i.e. t is algebraic, then by corollary 2.1 eα΅— is transcendental. That's a contradiction, so the assumption cannot be true, i.e. t is transcendental. ∎

Corollary 2.4:
The natural logarithm of any algebraic number 𝛼 βˆ‰ {0, 1} is transcendental, because elog(𝛼) = 𝛼 is algebraic, so by corollary 2.3, the exponent is transcendental.

That means that all of log(2), log(3), log(4), log(5/7), and log(-7/45) are transcendental.