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If $\alpha_1,\ldots,\alpha_n$ are linearly independent algebraic numbers over $\mathbb{Q}$ , then $e^{\alpha_1},\ldots,e^{\alpha_n}$ are algebraically independent over $\mathbb{Q}$ .
An equivalent version of the theorem states that if $\alpha_1,\ldots,\alpha_n$ are distinct algebraic numbers over $\mathbb{Q}$ , then $e^{\alpha_1},\ldots,e^{\alpha_n}$ are linearly independent over $\mathbb{Q}$ .
Some immediate consequences of this theorem:
- If $\alpha$ is a non-zero algebraic number over $\mathbb{Q}$ , then $e^{\alpha}$ is transcendental over $\mathbb{Q}$ .
- $e$ is transcendental over $\mathbb{Q}$ .
- $\pi$ is transcendental over $\mathbb{Q}$ . As a result, it is impossible to ``square the circle''!
It is easy to see that $\pi$ is transcendental over $\mathbb{Q}(e)$ iff $e$ is transcendental over $\mathbb{Q}(\pi)$ iff $\pi$ and $e$ are algebraically independent. However, whether $\pi$ and $e$ are algebraically independent is still an open question today.
Schanuel's conjecture is a generalization of the Lindemann-Weierstrass theorem. If Schanuel's conjecture were proven to be true, then the algebraic independence of $e$ and $\pi$ over $\mathbb{Q}$ can be shown.
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