Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number/Lemma

Lemma
Let:
 * $z_1 = \ln \ln 2$
 * $z_2 = 1 + \ln \ln 2$
 * $z_3 = \ln 2$
 * $z_4 = e \ln 2$

Let Schanuel's Conjecture be true.

Then $z_1$, $z_2$, $z_3$ and $z_4$ are linearly independent over the rational numbers $\Q$.

Proof
Assume the truth of Schanuel's Conjecture.

$2$ is algebraic.

Hence, by the Corollary of the weaker Hermite-Lindemann-Weierstrass theorem, $\ln 2$ is transcendental.

Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.

Equivalently, by Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers, they are linearly independent over the integers $\Z$.

Let $a, b \in \Z$ such that:
 * $a z_1 + b z_3 = 0$

Substituting:
 * $a \ln \ln 2 + b \ln 2 = 0$

Applying exponential function to both sides:
 * $\left({\ln 2}\right)^a 2^b = 1$

Since $\ln 2$ is transcendental, the above equation only has solution when $a = b = 0$.

Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, e^{z_1}, e^{z_3}}\right)$ has transcendence degree at least $2$ over the rational numbers $\Q$.

That is, the extension field $\Q \left({\ln \ln 2, \ln 2, \ln 2, 2}\right)$ has transcendence degree at least $2$ over $\Q$.

However, $2$ is algebraic.

Therefore, $\ln \ln 2$ and $\ln 2$ must be algebraically independent over $\Q$.

That is, $z_1$, $z_3$ must be algebraically independent over rational numbers $\Q$.

It follows that $z_1$, $z_3$, and $1$ are linearly independent over the rational numbers $\Q$.

By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, 1, e^{z_1}, e^{z_3}, e}\right)$ has transcendence degree at least $3$ over the rational numbers $\Q$.

That is, the extension field $\Q \left({\ln \ln 2, \ln 2, 1, \ln 2, 2, e}\right)$ has transcendence degree at least $3$ over $\Q$.

However, $1$ and $2$ are algebraic.

Therefore, $\ln \ln 2$, $\ln 2$, $e$ must be algebraically independent over $\Q$.

That is, $z_1$, $z_3$, and $e$ must be algebraically independent over rational numbers $\Q$.

It follows that $z_1$, $z_3$, $e z_3$ must be algebraically independent over rational numbers $\Q$.

That is, $z_1$, $z_3$, $z_4$ must be algebraically independent over $\Q$.

Therefore, $z_1$, $1$, $z_3$, $z_4$ must be linearly independent over $\Q$.

Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$ must be linearly independent over $\Q$.

That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$ are linearly independent over the rational numbers $\Q$.

[[Category:2]]