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If $g(.)$ is a monotonically increasing function and $a <b$, is $a<g(a)<g(b)<b$?
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My question is in relation to a problem I am trying to solve here. If $g(.)$ is a monotonically increasing function and $a <b$, is it always true that $a<g(a)<g(b)<b$? Why or why not?
real-analysis monotone-functions
edited Jul 31 at 16:29
John Ma
37.5k93669
asked Jul 31 at 13:34
Lod
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up vote
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My question is in relation to a problem I am trying to solve here. If $g(.)$ is a monotonically increasing function and $a <b$, is it always true that $a<g(a)<g(b)<b$? Why or why not?
real-analysis monotone-functions
edited Jul 31 at 16:29
John Ma
37.5k93669
asked Jul 31 at 13:34
Lod
286
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question is in relation to a problem I am trying to solve here. If $g(.)$ is a monotonically increasing function and $a <b$, is it always true that $a<g(a)<g(b)<b$? Why or why not?
real-analysis monotone-functions
edited Jul 31 at 16:29
John Ma
37.5k93669
asked Jul 31 at 13:34
Lod
286
My question is in relation to a problem I am trying to solve here. If $g(.)$ is a monotonically increasing function and $a <b$, is it always true that $a<g(a)<g(b)<b$? Why or why not?
real-analysis monotone-functions
edited Jul 31 at 16:29
John Ma
37.5k93669
asked Jul 31 at 13:34
Lod
286
edited Jul 31 at 16:29
John Ma
37.5k93669
edited Jul 31 at 16:29
John Ma
37.5k93669
edited Jul 31 at 16:29
John Ma
37.5k93669
37.5k93669
asked Jul 31 at 13:34
Lod
286
asked Jul 31 at 13:34
Lod
286
asked Jul 31 at 13:34
Lod
286
286
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
In fact, it is impossible to have any function $g : mathbbR to mathbbR$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude
$$a < g(a) < g(b) < b$$
whereas from $b < c$ we conclude
$$ b < g(b) < g(c) < c. $$
Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.
(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
add a comment |Â
up vote
1
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You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.
As a counterexample, you can suppose $g(x) = fracx2$.
answered Jul 31 at 13:36
OmG
1,740617
add a comment |Â
up vote
1
down vote
This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$
More easily, the function $g(x)=2x$. Then you have
$$2<3<g(2)=4<g(3)=6$$
answered Jul 31 at 13:37
Suzet
2,183427
add a comment |Â
up vote
1
down vote
Why would it be true? Let $g(x) = 10^5634789 + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^5634789 < 10^5634789 + 1 < 1$?
... In fact, it's hard for me to imagine a function were this is always true.
You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.
Not only does this not need to be true, it can't be true.
answered Jul 31 at 18:03
fleablood
60.2k22575
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In fact, it is impossible to have any function $g : mathbbR to mathbbR$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude
$$a < g(a) < g(b) < b$$
whereas from $b < c$ we conclude
$$ b < g(b) < g(c) < c. $$
Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.
(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
add a comment |Â
up vote
1
down vote
accepted
In fact, it is impossible to have any function $g : mathbbR to mathbbR$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude
$$a < g(a) < g(b) < b$$
whereas from $b < c$ we conclude
$$ b < g(b) < g(c) < c. $$
Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.
(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In fact, it is impossible to have any function $g : mathbbR to mathbbR$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude
$$a < g(a) < g(b) < b$$
whereas from $b < c$ we conclude
$$ b < g(b) < g(c) < c. $$
Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.
(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
In fact, it is impossible to have any function $g : mathbbR to mathbbR$ which satisfies that inequality for all pairs $a<b$. To see this, suppose we have a triple $a < b < c$. Then from $a < b$ we conclude
$$a < g(a) < g(b) < b$$
whereas from $b < c$ we conclude
$$ b < g(b) < g(c) < c. $$
Now, combining $b < g(b)$ from the second and $g(b) < b$ from the first, this implies $b < b$, which is a contradiction.
(In fact, essentially the same argument works if the domain of $g$ is any linear order with at least three elements.)
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
answered Jul 31 at 17:35
Daniel Schepler
6,6681513
6,6681513
add a comment |Â
add a comment |Â
up vote
1
down vote
You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.
As a counterexample, you can suppose $g(x) = fracx2$.
answered Jul 31 at 13:36
OmG
1,740617
add a comment |Â
up vote
1
down vote
You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.
As a counterexample, you can suppose $g(x) = fracx2$.
answered Jul 31 at 13:36
OmG
1,740617
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.
As a counterexample, you can suppose $g(x) = fracx2$.
answered Jul 31 at 13:36
OmG
1,740617
You can't say $a < g(a)$ and also $g(b)$. But you can say $g(a) < g(b)$ as it is monotinocally increasing.
As a counterexample, you can suppose $g(x) = fracx2$.
answered Jul 31 at 13:36
OmG
1,740617
answered Jul 31 at 13:36
OmG
1,740617
answered Jul 31 at 13:36
OmG
1,740617
answered Jul 31 at 13:36
OmG
1,740617
1,740617
add a comment |Â
add a comment |Â
up vote
1
down vote
This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$
More easily, the function $g(x)=2x$. Then you have
$$2<3<g(2)=4<g(3)=6$$
answered Jul 31 at 13:37
Suzet
2,183427
add a comment |Â
up vote
1
down vote
This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$
More easily, the function $g(x)=2x$. Then you have
$$2<3<g(2)=4<g(3)=6$$
answered Jul 31 at 13:37
Suzet
2,183427
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$
More easily, the function $g(x)=2x$. Then you have
$$2<3<g(2)=4<g(3)=6$$
answered Jul 31 at 13:37
Suzet
2,183427
This is not true. For instance, the exponential function satisfy the inequalities $$2<3<e^2<e^3$$
More easily, the function $g(x)=2x$. Then you have
$$2<3<g(2)=4<g(3)=6$$
answered Jul 31 at 13:37
Suzet
2,183427
answered Jul 31 at 13:37
Suzet
2,183427
answered Jul 31 at 13:37
Suzet
2,183427
answered Jul 31 at 13:37
Suzet
2,183427
2,183427
add a comment |Â
add a comment |Â
up vote
1
down vote
Why would it be true? Let $g(x) = 10^5634789 + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^5634789 < 10^5634789 + 1 < 1$?
... In fact, it's hard for me to imagine a function were this is always true.
You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.
Not only does this not need to be true, it can't be true.
answered Jul 31 at 18:03
fleablood
60.2k22575
add a comment |Â
up vote
1
down vote
Why would it be true? Let $g(x) = 10^5634789 + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^5634789 < 10^5634789 + 1 < 1$?
... In fact, it's hard for me to imagine a function were this is always true.
You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.
Not only does this not need to be true, it can't be true.
answered Jul 31 at 18:03
fleablood
60.2k22575
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Why would it be true? Let $g(x) = 10^5634789 + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^5634789 < 10^5634789 + 1 < 1$?
... In fact, it's hard for me to imagine a function were this is always true.
You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.
Not only does this not need to be true, it can't be true.
answered Jul 31 at 18:03
fleablood
60.2k22575
Why would it be true? Let $g(x) = 10^5634789 + x$. The $g'(x) = 1 > 0$ so it is monotonically increasing. Let $0 < 1$. Does it follow that $0 < 10^5634789 < 10^5634789 + 1 < 1$?
... In fact, it's hard for me to imagine a function were this is always true.
You are essentially saying that for any $a < b$ then $a < g(a)$ but $b < g(b)$. Well then what about $a < b < c$? If it's true that $a < g(a) < g(b) < b$ then it can not be true that $b < g(b) < g(c) < c$ because $g(b) < b$.
Not only does this not need to be true, it can't be true.
answered Jul 31 at 18:03
fleablood
60.2k22575
answered Jul 31 at 18:03
fleablood
60.2k22575
answered Jul 31 at 18:03
fleablood
60.2k22575
answered Jul 31 at 18:03
fleablood
60.2k22575
60.2k22575
add a comment |Â
add a comment |Â
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