# Talk:Irrationality of Logarithm

"neither $\nexists n \in \N: a = b^n$ nor $\nexists n \in \N: a^n = b$" ... is there a double negative in there? --prime mover (talk) 18:32, 26 August 2014 (UTC)

Also, it's unclear why the conclusions follow from the above conditions, because the starting conditions for each section (e.g $a \perp b$) do not on the face of it appear to be equivalent to the conditions given. --prime mover (talk) 18:36, 26 August 2014 (UTC)

The neither nor statement is standard English grammar. I will fix the conditions later today, thanks for pointing that out --Ybab321 (talk) 11:12, 27 August 2014 (UTC)
It means "Neither not exists a natural number such that ..., nor not exists a natural number that ...", which means "exists a natural number that ... and exists a natural number that ..."
$\left({\neg p}\right) \downarrow \left({\neg q}\right)$ is truthtabled thus:
$\begin{array}{|cc|cc||c|} \hline p & q & \neg p & \neg q & p \downarrow q \\ \hline F & F & T & T & F \\ F & T & T & F & F \\ T & F & F & T & F \\ T & T & F & F & T \\ \hline \end{array}$
... where $\downarrow$ denotes Definition:Logical NOR.
In this context:
$p$ denotes "there exists a natural number $n$ such that $a = b^n$
$q$ denotes "there exists a natural number $n$ such that $a^n = b$
Sure that's what you meant? --prime mover (talk) 18:13, 27 August 2014 (UTC)
Haha, I see. Pardon my ignorance. --Ybab321 (talk) 20:48, 2 September 2014 (UTC)