Show: Let $G$ be an archimedean ordered group, and given $0 neq g in G$, $h in G$, $exists n in mathbb{Z}$...
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Here is my attempt:
Suppose $h geq 0$, then by Archimedean property of the ordering, $exists nin mathbb{N}$ s.t $h<g^{n+1}$. Hence consider the set $S$ of all $kin mathbb{N}$ such that $h<g^{k+1}$. By well ordering principle, $S$ has a minimum, call it $n+1$. Then $g^n<h<g^{n+1}$.
Now suppose $h<0$, then $h^{-1}>0$ and assume $g^{-1}>0$ so there is $n in mathbb{N}$ s.t $g^{-(n-1)} leq h^{-1}<g^{-n}$ where $n$ is the smallest positive number satisfying this property. Then $g^{n+1} geq h>g^n. But this is not what we want. Is this the right way to prove it?
group-theory
asked Nov 22 at 14:54
mathnoob
1,102115
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show 4 more comments
up vote
0
down vote
favorite
Here is my attempt:
Suppose $h geq 0$, then by Archimedean property of the ordering, $exists nin mathbb{N}$ s.t $h<g^{n+1}$. Hence consider the set $S$ of all $kin mathbb{N}$ such that $h<g^{k+1}$. By well ordering principle, $S$ has a minimum, call it $n+1$. Then $g^n<h<g^{n+1}$.
Now suppose $h<0$, then $h^{-1}>0$ and assume $g^{-1}>0$ so there is $n in mathbb{N}$ s.t $g^{-(n-1)} leq h^{-1}<g^{-n}$ where $n$ is the smallest positive number satisfying this property. Then $g^{n+1} geq h>g^n. But this is not what we want. Is this the right way to prove it?
group-theory
asked Nov 22 at 14:54
mathnoob
1,102115
It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
But the claim in your title is false
– YCor
Nov 23 at 22:34
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here is my attempt:
Suppose $h geq 0$, then by Archimedean property of the ordering, $exists nin mathbb{N}$ s.t $h<g^{n+1}$. Hence consider the set $S$ of all $kin mathbb{N}$ such that $h<g^{k+1}$. By well ordering principle, $S$ has a minimum, call it $n+1$. Then $g^n<h<g^{n+1}$.
Now suppose $h<0$, then $h^{-1}>0$ and assume $g^{-1}>0$ so there is $n in mathbb{N}$ s.t $g^{-(n-1)} leq h^{-1}<g^{-n}$ where $n$ is the smallest positive number satisfying this property. Then $g^{n+1} geq h>g^n. But this is not what we want. Is this the right way to prove it?
group-theory
asked Nov 22 at 14:54
mathnoob
1,102115
Here is my attempt:
Suppose $h geq 0$, then by Archimedean property of the ordering, $exists nin mathbb{N}$ s.t $h<g^{n+1}$. Hence consider the set $S$ of all $kin mathbb{N}$ such that $h<g^{k+1}$. By well ordering principle, $S$ has a minimum, call it $n+1$. Then $g^n<h<g^{n+1}$.
Now suppose $h<0$, then $h^{-1}>0$ and assume $g^{-1}>0$ so there is $n in mathbb{N}$ s.t $g^{-(n-1)} leq h^{-1}<g^{-n}$ where $n$ is the smallest positive number satisfying this property. Then $g^{n+1} geq h>g^n. But this is not what we want. Is this the right way to prove it?
group-theory
group-theory
asked Nov 22 at 14:54
mathnoob
1,102115
asked Nov 22 at 14:54
mathnoob
1,102115
edited Nov 23 at 22:38
asked Nov 22 at 14:54
mathnoob
1,102115
asked Nov 22 at 14:54
mathnoob
1,102115
asked Nov 22 at 14:54
mathnoob
1,102115
1,102115
It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
But the claim in your title is false
– YCor
Nov 23 at 22:34
|
show 4 more comments
It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
But the claim in your title is false
– YCor
Nov 23 at 22:34
It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
But the claim in your title is false
– YCor
Nov 23 at 22:34
But the claim in your title is false
– YCor
Nov 23 at 22:34
|
show 4 more comments
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It's absurd if $g<0$.
– YCor
Nov 23 at 21:31
It was an exercise in the book.
– mathnoob
Nov 23 at 22:26
what's in the book is correct, not what you wrote
– YCor
Nov 23 at 22:32
Yes I know, that is why I want to know the right way to prove it. The book has no proof. Please help
– mathnoob
Nov 23 at 22:33
But the claim in your title is false
– YCor
Nov 23 at 22:34