Some Consequences of Schanuel’s Conjecture

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Outline The Conjecture The Problems Main Problem

Some Consequences of Schanuel’s Conjecture

March 19, 2008

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The Conjecture

Conjecture and Corollaries The Problems

Set up

Related consequences of Schanuel’s Conjecture Main Problem

Main Result Corollaries The Proof

Some Consequences of Schanuel’s Conjecture

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Conjecture and Corollaries

Conjecture (Schanuel): Let x ₁ , . . . , x _n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e ^x

¹

, . . . , e ^x

ⁿ

) is at least n.

Corollaries:

Algebraic independence of π and e over Q .

π, log π, log log π, . . . are algebraically independent over Q .

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Conjecture and Corollaries

Conjecture (Schanuel): Let x ₁ , . . . , x _n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e ^x

¹

, . . . , e ^x

ⁿ

) is at least n.

Corollaries:

Algebraic independence of π and e over Q .

π, log π, log log π, . . . are algebraically independent over Q.

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Set up

Related consequences of Schanuel’s Conjecture

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Definitions

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L _n .

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Set up

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L. More generally:

E and L are linearly disjoint over Q .

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Set up

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L. More generally:

E and L are linearly disjoint over Q .

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Set up

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L. More generally:

E and L are linearly disjoint over Q .

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Set up

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L.

More generally:

E and L are linearly disjoint over Q .

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Set up

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L.

More generally:

E and L are linearly disjoint over Q .

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Set up

An enlightening example

Proposition: Schanuel’s Conjecture implies π 6∈ E .

Proof: by induction on n, π 6∈ E n . Base case: π 6∈ E 0 = Q is clear.

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Set up

An enlightening example

Proposition: Schanuel’s Conjecture implies π 6∈ E . Proof: by induction on n, π 6∈ E n .

Base case: π 6∈ E 0 = Q is clear.

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Set up

An enlightening example

Proposition: Schanuel’s Conjecture implies π 6∈ E . Proof: by induction on n, π 6∈ E n .

Base case: π 6∈ E 0 = Q is clear.

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite

A n−1 ⊆ E n−1 .

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite A n−1 ⊆ E n−1 .

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite

A n−1 ⊆ E n−1 .

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ).

.. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite A n−1 ⊆ E n−1 .

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ).

.. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite

A n−1 ⊆ E n−1 .

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Set up

Key Construction

Induction step:

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ).

.. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

Therefore π is algebraic over Q (exp(A n−1 )) for some finite A n−1 ⊆ E n−1 .

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Set up

Key Construction

Following similarly:

I A n−1 is algebraic over Q(exp(A n−2 )) for some finite A n−2 ⊆ E n−2 .

I A n−2 is algebraic over Q (exp(A n−3 )) for some finite A n−3 ⊆ E n−3 .

.. .

I A 1 is algebraic over Q(exp(A 0 )) for some finite A 0 ⊆ E 0 = Q.

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Set up

Key Construction

Following similarly:

I A n−1 is algebraic over Q(exp(A n−2 )) for some finite A n−2 ⊆ E n−2 .

I A n−2 is algebraic over Q (exp(A n−3 )) for some finite A n−3 ⊆ E n−3 .

.. .

I A 1 is algebraic over Q(exp(A 0 )) for some finite A 0 ⊆ E 0 = Q.

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Set up

Key Construction

Following similarly:

I A n−1 is algebraic over Q(exp(A n−2 )) for some finite A n−2 ⊆ E n−2 .

I A n−2 is algebraic over Q (exp(A n−3 )) for some finite A n−3 ⊆ E n−3 .

.. .

I A 1 is algebraic over Q(exp(A 0 )) for some finite A 0 ⊆ E 0 = Q.

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Set up

End of proof

I Set A = S

m≤n−1 A m .

I Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

I {i π} ∪ B are Q -linearly independent.

I By Schanuel’s Conjecture trdeg _Q Q(i π, B, exp(B )) ≥ |B| + 1.

I But trdeg _Q Q (iπ, B , exp(B )) = trdeg _Q Q (iπ, B , exp(A)) = trdeg _Q Q(exp(A)) = trdeg _Q Q(exp(B)) ≤ |B |.

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Set up

End of proof

I Set A = S

m≤n−1 A m .

I Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

I {i π} ∪ B are Q -linearly independent.

I By Schanuel’s Conjecture trdeg _Q Q(i π, B, exp(B )) ≥ |B| + 1.

I But trdeg _Q Q (iπ, B , exp(B )) = trdeg _Q Q (iπ, B , exp(A)) =

trdeg _Q Q(exp(A)) = trdeg _Q Q(exp(B)) ≤ |B |.

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Set up

End of proof

I Set A = S

m≤n−1 A m .

I Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

I {i π} ∪ B are Q -linearly independent.

I By Schanuel’s Conjecture trdeg _Q Q(i π, B, exp(B )) ≥ |B| + 1.

I But trdeg _Q Q (iπ, B , exp(B )) = trdeg _Q Q (iπ, B , exp(A)) = trdeg _Q Q(exp(A)) = trdeg _Q Q(exp(B)) ≤ |B |.

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Set up

End of proof

I Set A = S

m≤n−1 A m .

I Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

I {i π} ∪ B are Q -linearly independent.

I By Schanuel’s Conjecture trdeg _Q Q(i π, B, exp(B )) ≥ |B| + 1.

I But trdeg _Q Q (iπ, B , exp(B )) = trdeg _Q Q (iπ, B , exp(A)) =

trdeg _Q Q(exp(A)) = trdeg _Q Q(exp(B)) ≤ |B |.

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Set up

End of proof

I Set A = S

m≤n−1 A m .

I Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

I {i π} ∪ B are Q -linearly independent.

I By Schanuel’s Conjecture trdeg _Q Q(i π, B, exp(B )) ≥ |B| + 1.

I But trdeg _Q Q (iπ, B, exp(B )) = trdeg _Q Q (iπ, B , exp(A)) = trdeg _Q Q(exp(A)) = trdeg _Q Q(exp(B)) ≤ |B |.

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Main Result Corollaries The Proof

Main result

We say K ₁ and K ₂ are linearly disjoint over k iff:

{x ₁ , . . . , x _n } ⊆ K ₁ linearly independent over k ⇒ linearly independent over K 2 .

Theorem: Schanuel’s Conjecture implies E and L are linearly

disjoint over Q .

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Corollaries:

I E ∩ L = Q .

I π, log π, log log π, . . . are algebraically independent over E.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L.

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Corollaries:

I E ∩ L = Q .

I π, log π, log log π, . . . are algebraically independent over E.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L.

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Corollaries:

I E ∩ L = Q .

I π, log π, log log π, . . . are algebraically independent over E.

I e , e ^e , e ^e

^e

, . . . are algebraically independent over L.

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The Proof

Let’s prove E m and L n are linearly disjoint.

Take {l _i } ⊆ L n linearly independent over Q and {e _i } ⊆ E m such that P

l _i e _i = 0.

Proceeding as before:

∃ finite A ⊆ E m−1 such that A ∪ {e _i } algebraic over Q(exp(A)).

∃ finite C ⊆ L n finite such that exp(C ) ∪ {l _i } algebraic over Q (C ). Take B ⊆ A such that exp(B) is a transcendence basis of

Q (exp(A)).

Take D ⊆ C such that D is a transcendence basis of Q(C ).

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The Proof

Let’s prove E m and L n are linearly disjoint.

Take {l _i } ⊆ L n linearly independent over Q and {e _i } ⊆ E m such that P

l _i e _i = 0.

Proceeding as before:

∃ finite A ⊆ E m−1 such that A ∪ {e _i } algebraic over Q(exp(A)).

∃ finite C ⊆ L n finite such that exp(C ) ∪ {l _i } algebraic over Q (C ).

Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

Take D ⊆ C such that D is a transcendence basis of Q(C ).

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The Proof

Let’s prove E m and L n are linearly disjoint.

Take {l _i } ⊆ L n linearly independent over Q and {e _i } ⊆ E m such that P

l _i e _i = 0.

Proceeding as before:

∃ finite A ⊆ E m−1 such that A ∪ {e _i } algebraic over Q(exp(A)).

∃ finite C ⊆ L n finite such that exp(C ) ∪ {l _i } algebraic over Q (C ).

Take B ⊆ A such that exp(B) is a transcendence basis of Q (exp(A)).

Take D ⊆ C such that D is a transcendence basis of Q(C ).

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The Proof

We have B ∪ D linearly independent over Q .

By Schanuel’s Conjecture

trdeg _Q Q (B, D, exp(B), exp(D)) ≥ |B | + |D|. However

trdeg _Q Q(B, D, exp(B), exp(D)) = trdeg _Q Q(exp(B), D) ≤ |B|+|D|.

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The Proof

We have B ∪ D linearly independent over Q . By Schanuel’s Conjecture

trdeg _Q Q (B, D, exp(B), exp(D)) ≥ |B | + |D|.

However

trdeg _Q Q(B, D, exp(B), exp(D)) = trdeg _Q Q(exp(B), D) ≤ |B|+|D|.

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The Proof

We have B ∪ D linearly independent over Q . By Schanuel’s Conjecture

trdeg _Q Q (B, D, exp(B), exp(D)) ≥ |B | + |D|.

However

trdeg _Q Q (B, D, exp(B), exp(D)) = trdeg _Q Q (exp(B), D) ≤ |B|+|D|.

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The Proof

Therefore Q (exp(B)) and Q (D) are free over Q , and the same is true for Q (exp(B)) and Q (D ).

Since Q is algebraically closed, Q (exp(B )) and Q (D) are linearly

independent over Q (see Lang’s Algebra).

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Some Consequences of Schanuel’s Conjecture

Some Consequences of Schanuel’s Conjecture

March 19, 2008

The Conjecture

Conjecture and Corollaries The Problems

Set up

Related consequences of Schanuel’s Conjecture Main Problem

Main Result Corollaries The Proof

Conjecture and Corollaries

Conjecture (Schanuel): Let x 1 , . . . , x n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e x

, . . . , e x

) is at least n.

Corollaries:

Algebraic independence of π and e over Q .

π, log π, log log π, . . . are algebraically independent over Q .

Conjecture and Corollaries

Conjecture (Schanuel): Let x 1 , . . . , x n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e x

, . . . , e x

) is at least n.

Corollaries:

Algebraic independence of π and e over Q .

π, log π, log log π, . . . are algebraically independent over Q.

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Definitions

I Define E 0 to be the set of algebraic numbers E 0 = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e x : x ∈ E n−1 ).

I E = S

n≥0 E n .

I Similarly define L 0 = Q .

I Inductively, for n ≥ 1, define L n = L n−1 (log(x) : x ∈ L n−1 ).

I L = S

n≥0 L n .

Some Consequences

I π 6∈ E .

I π, log π, log log π, . . . are algebraically independent over E.

I e 6∈ L.

Conjecture (Schanuel): Let x ₁ , . . . , x _n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e ^x

, . . . , e ^x

Conjecture (Schanuel): Let x ₁ , . . . , x _n be Q -linearly independent complex numbers. Then the transcendence degree over Q of the field Q (x 1 , . . . , x n , e ^x

, . . . , e ^x

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I Define E ₀ to be the set of algebraic numbers E ₀ = Q .

I Inductively, for n ≥ 1, define E n = E n−1 (e ^x : x ∈ E n−1 ).

I Similarly define L ₀ = Q .

I Inductively, for n ≥ 1, define L _n = L n−1 (log(x) : x ∈ L n−1 ).

n≥0 L _n .

I e , e ^e , e ^e

I e , e ^e , e ^e

I e , e ^e , e ^e

I e , e ^e , e ^e

I e , e ^e , e ^e

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−1 (e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−2 )(e ^x : x ∈ E n−1 ).

I π is algebraic over E n−2 (e ^x : x ∈ E n−1 ). .. .

I π is algebraic over Q (e ^x : x ∈ E n−1 ).