Showing that $ U_x=left{a^p:pinmathbb{Q},pxright} $ are contiguous classes for $ a>1 $
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I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.
Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]
I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.
Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like
Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.
Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?
real-analysis exponential-function proof-explanation
asked Dec 30 '18 at 14:56
marco21marco21
308211
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add a comment |
$begingroup$
I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.
Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]
I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.
Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like
Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.
Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?
real-analysis exponential-function proof-explanation
asked Dec 30 '18 at 14:56
marco21marco21
308211
$endgroup$
add a comment |
$begingroup$
I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.
Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]
I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.
Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like
Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.
Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?
real-analysis exponential-function proof-explanation
asked Dec 30 '18 at 14:56
marco21marco21
308211
$endgroup$
I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.
Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]
I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.
Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like
Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.
Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?
real-analysis exponential-function proof-explanation
real-analysis exponential-function proof-explanation
asked Dec 30 '18 at 14:56
marco21marco21
308211
asked Dec 30 '18 at 14:56
marco21marco21
308211
edited Dec 30 '18 at 17:21
marco21
asked Dec 30 '18 at 14:56
marco21marco21
308211
asked Dec 30 '18 at 14:56
marco21marco21
308211
asked Dec 30 '18 at 14:56
marco21marco21
308211
308211
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1 Answer
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1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.
2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
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$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
add a comment |
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1 Answer
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$begingroup$
1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.
2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
$endgroup$
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
add a comment |
$begingroup$
1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.
2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
$endgroup$
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
add a comment |
$begingroup$
1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.
2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
$endgroup$
1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.
2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
edited Dec 30 '18 at 18:30
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
answered Dec 30 '18 at 18:06
JAskgaardJAskgaard
1467
1467
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
add a comment |
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
$endgroup$
– marco21
Dec 30 '18 at 18:21
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
$begingroup$
Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
$endgroup$
– JAskgaard
Dec 30 '18 at 18:33
add a comment |
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