Solving compound inequalities

up vote
1
down vote

favorite

Consider the following two inequalities,

$frac{a}{1-a} < b$

and

$frac{a}{1-a}< (1-b)$

Is it correct to substitute the first into the second, and write,

$b<(1-b)$

to derive $b < frac{1}{2}$ ?

EDIT:

It is also known that $0<a<0.5$ and $0<b<1$

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

add a comment |

up vote
1
down vote

favorite

Consider the following two inequalities,

$frac{a}{1-a} < b$

and

$frac{a}{1-a}< (1-b)$

Is it correct to substitute the first into the second, and write,

$b<(1-b)$

to derive $b < frac{1}{2}$ ?

EDIT:

It is also known that $0<a<0.5$ and $0<b<1$

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

add a comment |

up vote
1
down vote

favorite

Consider the following two inequalities,

$frac{a}{1-a} < b$

and

$frac{a}{1-a}< (1-b)$

Is it correct to substitute the first into the second, and write,

$b<(1-b)$

to derive $b < frac{1}{2}$ ?

EDIT:

It is also known that $0<a<0.5$ and $0<b<1$

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

Consider the following two inequalities,

$frac{a}{1-a} < b$

and

$frac{a}{1-a}< (1-b)$

Is it correct to substitute the first into the second, and write,

$b<(1-b)$

to derive $b < frac{1}{2}$ ?

EDIT:

It is also known that $0<a<0.5$ and $0<b<1$

inequality systems-of-equations

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

edited Nov 21 at 21:02

asked Nov 21 at 20:47

user3222

407

asked Nov 21 at 20:47

user3222

407

asked Nov 21 at 20:47

user3222

407

add a comment |

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

No we can't, we have three cases

$b<1-b implies b<frac12$

$$frac{a}{1-a} <b< 1-b$$

$b>1-b implies b>frac12$

$$frac{a}{1-a} <1-b<b$$

$b=1-b implies b=frac12$

$$frac{a}{1-a} <frac12$$

answered Nov 21 at 20:50

gimusi

88.2k74394

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

1

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

add a comment |

up vote
1
down vote

Surely not. We may also have $${aover 1-a}<1-b<b$$for example $$a=0.002\b={2over 3}$$

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

add a comment |

up vote
1
down vote

If you have two inequalities, you can combine to get $frac{a}{1-a}< min{ b, (1-b)}$

Inequalities which can be combined are
$$a < b\
b < c$$to get $a < c$ and you can also add and multiply, i.e.
$$a < b\
c < d$$
gives e.g.
$$ac < bd\
a+c < b+d$$

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008332%2fsolving-compound-inequalities%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

No we can't, we have three cases

$b<1-b implies b<frac12$

$$frac{a}{1-a} <b< 1-b$$

$b>1-b implies b>frac12$

$$frac{a}{1-a} <1-b<b$$

$b=1-b implies b=frac12$

$$frac{a}{1-a} <frac12$$

answered Nov 21 at 20:50

gimusi

88.2k74394

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

1

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

add a comment |

up vote
1
down vote

accepted

No we can't, we have three cases

$b<1-b implies b<frac12$

$$frac{a}{1-a} <b< 1-b$$

$b>1-b implies b>frac12$

$$frac{a}{1-a} <1-b<b$$

$b=1-b implies b=frac12$

$$frac{a}{1-a} <frac12$$

answered Nov 21 at 20:50

gimusi

88.2k74394

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

1

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

add a comment |

up vote
1
down vote

accepted

No we can't, we have three cases

$b<1-b implies b<frac12$

$$frac{a}{1-a} <b< 1-b$$

$b>1-b implies b>frac12$

$$frac{a}{1-a} <1-b<b$$

$b=1-b implies b=frac12$

$$frac{a}{1-a} <frac12$$

answered Nov 21 at 20:50

gimusi

88.2k74394

No we can't, we have three cases

$b<1-b implies b<frac12$

$$frac{a}{1-a} <b< 1-b$$

$b>1-b implies b>frac12$

$$frac{a}{1-a} <1-b<b$$

$b=1-b implies b=frac12$

$$frac{a}{1-a} <frac12$$

answered Nov 21 at 20:50

gimusi

88.2k74394

answered Nov 21 at 20:50

gimusi

88.2k74394

answered Nov 21 at 20:50

gimusi

88.2k74394

answered Nov 21 at 20:50

gimusi

88.2k74394

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

1

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

add a comment |

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

1

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

thanks for the answer. What if it is also known that $0<a<0.5 $ and $0<b<1$ ? I want to get rid of $a$ and know a bound for the value of $b$. Is that possible ?
– user3222
Nov 21 at 20:58

@user3222 In the first case we have $frac{a}{1-a} <b implies a<b-ab implies a<frac{b}{1+b}$
– gimusi
Nov 21 at 21:02

add a comment |

up vote
1
down vote

Surely not. We may also have $${aover 1-a}<1-b<b$$for example $$a=0.002\b={2over 3}$$

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

add a comment |

up vote
1
down vote

Surely not. We may also have $${aover 1-a}<1-b<b$$for example $$a=0.002\b={2over 3}$$

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

add a comment |

up vote
1
down vote

Surely not. We may also have $${aover 1-a}<1-b<b$$for example $$a=0.002\b={2over 3}$$

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

Surely not. We may also have $${aover 1-a}<1-b<b$$for example $$a=0.002\b={2over 3}$$

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

answered Nov 21 at 20:53

Mostafa Ayaz

12.8k3733

add a comment |

up vote
1
down vote

If you have two inequalities, you can combine to get $frac{a}{1-a}< min{ b, (1-b)}$

Inequalities which can be combined are
$$a < b\
b < c$$to get $a < c$ and you can also add and multiply, i.e.
$$a < b\
c < d$$
gives e.g.
$$ac < bd\
a+c < b+d$$

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

add a comment |

up vote
1
down vote

If you have two inequalities, you can combine to get $frac{a}{1-a}< min{ b, (1-b)}$

Inequalities which can be combined are
$$a < b\
b < c$$to get $a < c$ and you can also add and multiply, i.e.
$$a < b\
c < d$$
gives e.g.
$$ac < bd\
a+c < b+d$$

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

add a comment |

up vote
1
down vote

If you have two inequalities, you can combine to get $frac{a}{1-a}< min{ b, (1-b)}$

Inequalities which can be combined are
$$a < b\
b < c$$to get $a < c$ and you can also add and multiply, i.e.
$$a < b\
c < d$$
gives e.g.
$$ac < bd\
a+c < b+d$$

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

If you have two inequalities, you can combine to get $frac{a}{1-a}< min{ b, (1-b)}$

Inequalities which can be combined are
$$a < b\
b < c$$to get $a < c$ and you can also add and multiply, i.e.
$$a < b\
c < d$$
gives e.g.
$$ac < bd\
a+c < b+d$$

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

edited Nov 21 at 20:58

answered Nov 21 at 20:53

Andreas

7,5441037

answered Nov 21 at 20:53

Andreas

7,5441037

answered Nov 21 at 20:53

Andreas

7,5441037

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ytukyg