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DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Im working on #2. Just a quick question which may turn out to be more complicated, but on showing the inverse is in H
'log(n^-1)' if n is an integer is n inverse (n^-1) also an integer? i know log(n^-1) wont always be an integer but 'n' inverse would be, wouldn't it?

Shadow · **Posted:** Sun, 19 Feb 2012 02:42:56 UTC

DgrayMan wrote:

Im working on #2. Just a quick question which may turn out to be more complicated, but on showing the inverse is in H
'log(n^-1)' if n is an integer is n inverse (n^-1) also an integer? i know log(n^-1) wont always be an integer but 'n' inverse would be, wouldn't it?

The elements of

are real numbers

such that

for some $n\in\mathbb{N}$ . Now, since the original, underlying group is $(\mathbb{R},+)$ we need to consider $x^{-1}=-x$ and as is is true that $e^{-x}$ is also a positive integer, and the answer should be obvious: not unless n=1