Why we get inequality form an equation?
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In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$varLambda leq frac1by^n$ (see equation 7 on page 5)
But we get it from an equation. As I understand , it should be $varLambda = frac1by^n$.
How $varLambda$ could be less than $frac1by^n$?
If it is $varLambda leq frac1by^n$ then why not $varLambda geq frac1by^n$?
real-analysis inequality proof-explanation diophantine-equations diophantine-approximation
asked Sep 5 at 6:26
Mike SQ
265216
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up vote
2
down vote
favorite
In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$varLambda leq frac1by^n$ (see equation 7 on page 5)
But we get it from an equation. As I understand , it should be $varLambda = frac1by^n$.
How $varLambda$ could be less than $frac1by^n$?
If it is $varLambda leq frac1by^n$ then why not $varLambda geq frac1by^n$?
real-analysis inequality proof-explanation diophantine-equations diophantine-approximation
asked Sep 5 at 6:26
Mike SQ
265216
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$varLambda leq frac1by^n$ (see equation 7 on page 5)
But we get it from an equation. As I understand , it should be $varLambda = frac1by^n$.
How $varLambda$ could be less than $frac1by^n$?
If it is $varLambda leq frac1by^n$ then why not $varLambda geq frac1by^n$?
real-analysis inequality proof-explanation diophantine-equations diophantine-approximation
asked Sep 5 at 6:26
Mike SQ
265216
In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$varLambda leq frac1by^n$ (see equation 7 on page 5)
But we get it from an equation. As I understand , it should be $varLambda = frac1by^n$.
How $varLambda$ could be less than $frac1by^n$?
If it is $varLambda leq frac1by^n$ then why not $varLambda geq frac1by^n$?
real-analysis inequality proof-explanation diophantine-equations diophantine-approximation
real-analysis inequality proof-explanation diophantine-equations diophantine-approximation
asked Sep 5 at 6:26
Mike SQ
265216
asked Sep 5 at 6:26
Mike SQ
265216
asked Sep 5 at 6:26
Mike SQ
265216
asked Sep 5 at 6:26
Mike SQ
265216
asked Sep 5 at 6:26
Mike SQ
265216
265216
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
0
down vote
accepted
$$a=b$$ implies both $$ale b$$ and $$age b.$$
It's the author's choice to weaken the comparison for the requirements of the exposition.
Quiz:
Are these propositions true ? ($land$ is and, $lor$ is or)
$a=bimplies ale bland age b$ ?
$a=bimplies ale blor age b$ ?
$a=bimplies a< blor a=blor a> b$ ?
- $a=bimplies a< bland a=bland a> b$ ?
answered Sep 5 at 6:33
Yves Daoust
115k666209
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
add a comment |Â
up vote
0
down vote
From $$tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$
$$left(1+frac1bright)left(frac xyright)^n-1=frac1by^n.$$
This certainly implies (as long as $b,y>0$) that also
$$left|left(1+frac1bright)left(frac xyright)^n-1right|le frac1by^n.$$
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$$a=b$$ implies both $$ale b$$ and $$age b.$$
It's the author's choice to weaken the comparison for the requirements of the exposition.
Quiz:
Are these propositions true ? ($land$ is and, $lor$ is or)
$a=bimplies ale bland age b$ ?
$a=bimplies ale blor age b$ ?
$a=bimplies a< blor a=blor a> b$ ?
- $a=bimplies a< bland a=bland a> b$ ?
answered Sep 5 at 6:33
Yves Daoust
115k666209
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
add a comment |Â
up vote
0
down vote
accepted
$$a=b$$ implies both $$ale b$$ and $$age b.$$
It's the author's choice to weaken the comparison for the requirements of the exposition.
Quiz:
Are these propositions true ? ($land$ is and, $lor$ is or)
$a=bimplies ale bland age b$ ?
$a=bimplies ale blor age b$ ?
$a=bimplies a< blor a=blor a> b$ ?
- $a=bimplies a< bland a=bland a> b$ ?
answered Sep 5 at 6:33
Yves Daoust
115k666209
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$$a=b$$ implies both $$ale b$$ and $$age b.$$
It's the author's choice to weaken the comparison for the requirements of the exposition.
Quiz:
Are these propositions true ? ($land$ is and, $lor$ is or)
$a=bimplies ale bland age b$ ?
$a=bimplies ale blor age b$ ?
$a=bimplies a< blor a=blor a> b$ ?
- $a=bimplies a< bland a=bland a> b$ ?
answered Sep 5 at 6:33
Yves Daoust
115k666209
$$a=b$$ implies both $$ale b$$ and $$age b.$$
It's the author's choice to weaken the comparison for the requirements of the exposition.
Quiz:
Are these propositions true ? ($land$ is and, $lor$ is or)
$a=bimplies ale bland age b$ ?
$a=bimplies ale blor age b$ ?
$a=bimplies a< blor a=blor a> b$ ?
- $a=bimplies a< bland a=bland a> b$ ?
answered Sep 5 at 6:33
Yves Daoust
115k666209
edited Sep 5 at 7:38
answered Sep 5 at 6:33
Yves Daoust
115k666209
answered Sep 5 at 6:33
Yves Daoust
115k666209
answered Sep 5 at 6:33
Yves Daoust
115k666209
115k666209
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
add a comment |Â
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
Seriously! I am surprised!!
– Mike SQ
Sep 5 at 6:36
1
1
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
@MikeSQ: don't forget that $ale b$ means "$a=b$ or $a<b$", whichever is true.
– Yves Daoust
Sep 5 at 7:35
add a comment |Â
up vote
0
down vote
From $$tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$
$$left(1+frac1bright)left(frac xyright)^n-1=frac1by^n.$$
This certainly implies (as long as $b,y>0$) that also
$$left|left(1+frac1bright)left(frac xyright)^n-1right|le frac1by^n.$$
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
add a comment |Â
up vote
0
down vote
From $$tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$
$$left(1+frac1bright)left(frac xyright)^n-1=frac1by^n.$$
This certainly implies (as long as $b,y>0$) that also
$$left|left(1+frac1bright)left(frac xyright)^n-1right|le frac1by^n.$$
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From $$tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$
$$left(1+frac1bright)left(frac xyright)^n-1=frac1by^n.$$
This certainly implies (as long as $b,y>0$) that also
$$left|left(1+frac1bright)left(frac xyright)^n-1right|le frac1by^n.$$
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
From $$tag6(b+1)x^n-by^n=1$$ we find after dividing by $by^n$
$$left(1+frac1bright)left(frac xyright)^n-1=frac1by^n.$$
This certainly implies (as long as $b,y>0$) that also
$$left|left(1+frac1bright)left(frac xyright)^n-1right|le frac1by^n.$$
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
answered Sep 5 at 6:35
Hagen von Eitzen
267k21259482
267k21259482
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
add a comment |Â
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
If you don't mind I will ask the same question again as it is yet to be clear to me! Are you saying $$left|left(1+frac1bright)left(frac xyright)^n-1right|geq frac1by^n.$$ is not possible? Do you disagree with Daoust's answer above?
– Mike SQ
Sep 5 at 6:56
add a comment |Â
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