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 ℂ. 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 ℚ, which is always the set of just the quotients of integers ℤ, which are also not affected.

ℤ: integer numbers
ℚ: rational numbers, i.e. quotients
ℝ: real numbers
ℂ: complex numbers
𝔸: algebraic numbers
𝐈 := ℝ\ℚ (or ℂ\ℚ): irrational numbers (not a ring)
𝐓 := ℝ\𝔸 (or ℂ\𝔸): transcendental numbers (not a ring)

Note that:
ℚ ⊊ 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) ∉ 𝐓)
1 ∈ 𝔸, but 1 ∉ 𝐈
π ∈ 𝐈, but π ∉ 𝔸     (π ∈ 𝐓)

Lemma 0:
logba is irrational for a,b ∈ ℕ with a ≠ b, b ≠ 1.

Proof.
Let a, b be integers with a ≠ b, b ≠ 1. Then ∀i,j ∈ ℕ we have aⁱ ≠ bʲ by the fundamental theorem of arithmetic.
And then
aⁱ ≠ bʲ
(aⁱ)1/i ≠ (bʲ)1/i
ai/i ≠ bj/i
a¹ ≠ bj/i
logba ≠ logb(bj/i)
logba ≠ j/i   (∀i,j ∈ ℕ)
∄i,j ∈ ℕ: logba = j/i
logba ∉ ℚ . ∎

## Gelfond-Schneider Theorem

Theorem 1 (Gelfond-Schneider theorem):
If b ∈ 𝔸∖{0,1}, p ∈ 𝐈∩𝔸, then bᵖ is transcendental.

Corollary 1.1a (Contrapositive A of the Gelfond-Schneider theorem):
If tᵖ is algebraic, p ∈ 𝐈∩𝔸, and t ∉ {0,1}, then t is transcendental.

Proof.
Let tᵖ algebraic, p ∈ 𝐈∩𝔸, and t ∉ {0,1}. Then tᵖ is not transcendental. With the contrapositive of theorem 1 it follows that t ∉ 𝔸∖{0,1} or p ∉ 𝐈∩𝔸. Because we know that by the hypothesis p ∈ 𝐈∩𝔸, it follows that t ∉ 𝔸∖{0,1}. We also know that t ∉ {0,1}, so t ∉ 𝔸∖{0,1} ∪ {0,1} t ∉ 𝔸 t ∈ 𝐓. ∎

Corollary 1.1b (Contrapositive B of the Gelfond-Schneider theorem):
If bᵖ is algebraic and b ∈ 𝔸\{0,1}, then p ∉ 𝐈∩𝔸.

Proof.
Let bᵖ algebraic and b ∈ 𝔸∖{0,1}. Then bᵖ is not transcendental. With the contrapositive of theorem 1 it follows that b ∉ 𝔸∖{0,1} or p ∉ 𝐈∩𝔸. Because we know by the hypotheis that b ∈ 𝔸∖{0,1}, it follows that p ∉ 𝐈∩𝔸. ∎

Corollary 1.2:
logba is transcendental for a,b ∈ ℕ with a ≠ b, b > 1.

Proof.
Let a,b ∈ ℕ with a ≠ b, b > 1. Then blogba = a is algebraic. With blogba algebraic and algebraic b ∉ {0,1} it follows with corollary 1.1b that logba ∉ 𝐈∩𝔸. For a proof by contradiction, assume logba ∈ 𝔸. From lemma 0 we know that logba ∈ 𝐈. With logba ∈ 𝐈, ∈ 𝔸 it follows that logba ∈ 𝐈∩𝔸, which is a contradiction, because we already know that logba ∉ 𝐈∩𝔸, so the assumption is false and it follows that logba ∉ 𝔸 logba ∈ 𝐓. ∎

Corollary 1.3:
e^π is transcendental, because
e^π
= e^(-π · -1)
= e^(-π · i²)
= e^(iπ · -i)
= (e^(iπ))⁻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, ..., αₙ are algebraic numbers that are linearly independent over the rational numbers ℚ, then eα₁, ..., eαₙ are algebraically independent over ℚ.

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 e¹ is transcendental.

Corollary 2.3 (Contrapositive to corollary 2.1):
If eᵗ is algebraic and t ≠ 0, then t is transcendental.

Proof:
Let t ≠ 0 and for a proof by contradiction, assume that t is not transcendental, i.e. t is algebraic, then by corollary 2.1 eᵗ 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 ∉ {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.