Talk:Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals

Your objections
Re your first objection, I am failing to see how this definition cannot apply.

Let there be rational numbers $a$ and $b$ such that $a \times 1 + b \times \left({2 i \pi}\right) = 0$.

Equating real parts and imaginary parts would give us $a = 0$ and $b = 0$.

Therefore, $1$ and $2 i \pi$ are linearly independent, just as what is described in the definition.

--kc_kennylau (talk) 10:06, 24 December 2016 (EST)


 * For a start, Definition:Linearly Independent/Set is not the page linked to on the page in question, you've just used Definition:Linearly Independent.


 * For another thing, it needs to be stated, in order to bring it into context, what the group $G$ is and what the ring $R$ is.


 * And for a third thing, it needs to be demonstrated that the numbers in question actually are independent, which I see you have now done above. But, as you should understand by now, putting part of the proof on the talk page is inadequate. --prime mover (talk) 10:18, 24 December 2016 (EST)


 * Should I create a new sub-definition? --kc_kennylau (talk) 10:25, 24 December 2016 (EST)


 * If I were doing it, I would (a) change the page to put the correct link in place, (b) by stating which is the group and which the scalar ring is in this context, indicate how $\left\{ {1, 2 \pi i}\right\}$ is a subset of the unitary $G$-module formed as required, and (c) that they form a linearly independent set. --prime mover (talk) 10:31, 24 December 2016 (EST)


 * My opinion is that rather than stating many times that $G$ is $\C$ and $R$ is $\Q$, I would create a new sub-definition just for the complex numbers... --kc_kennylau (talk) 10:36, 24 December 2016 (EST)


 * Are you sure it's not that $G$ is $\Q$ and $R$ is $\C$? It's just that I'm not sure myself. It would be good to see this backed up with some resources. This is all stuff which we have not properly covered, and it needs to be done thoroughly. --prime mover (talk) 11:50, 24 December 2016 (EST)