Taking the absolute value in inequalities. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Taking the square roots in inequalitiesA system of absolute value equalitiesabsolute value of an expressionSolve a system of equation with absolute value and rotationmaximizing absolute value in linear programmingGetting rid of absolute value bars to rewrite a linear program in standard formMaximum value of a function whose sum of roots is knownProof verification of if $x_n$ is monotone then $y_n = 1over x_1 + x_2 + dots + x_n$ is monotoneInequalities with multiple conditions helpWriting Absolute Value For Equations With Inequalities
Trademark violation for app?
A term for a woman complaining about things/begging in a cute/childish way
Crossing US/Canada Border for less than 24 hours
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
What does Turing mean by this statement?
In musical terms, what properties are varied by the human voice to produce different words / syllables?
Semigroups with no morphisms between them
Karn the great creator - 'card from outside the game' in sealed
How long can equipment go unused before powering up runs the risk of damage?
What is best way to wire a ceiling receptacle in this situation?
How does Belgium enforce obligatory attendance in elections?
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
preposition before coffee
How much damage would a cupful of neutron star matter do to the Earth?
Misunderstanding of Sylow theory
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
How do I find out the mythology and history of my Fortress?
Why are my pictures showing a dark band on one edge?
What initially awakened the Balrog?
How many time has Arya actually used Needle?
Is there public access to the Meteor Crater in Arizona?
Is it possible for SQL statements to execute concurrently within a single session in SQL Server?
Can the Flaming Sphere spell be rammed into multiple Tiny creatures that are in the same 5-foot square?
What's the difference between the capability remove_users and delete_users?
Taking the absolute value in inequalities.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Taking the square roots in inequalitiesA system of absolute value equalitiesabsolute value of an expressionSolve a system of equation with absolute value and rotationmaximizing absolute value in linear programmingGetting rid of absolute value bars to rewrite a linear program in standard formMaximum value of a function whose sum of roots is knownProof verification of if $x_n$ is monotone then $y_n = 1over x_1 + x_2 + dots + x_n$ is monotoneInequalities with multiple conditions helpWriting Absolute Value For Equations With Inequalities
$begingroup$
If I've a expression: $-4<3$ and take the absolute value $|-4|<|3|implies 4<3$ which is false. So I though that maybe the inequality sign would change. But $|-2|<|3| implies 2>3$ which is also false.
My problem is that I have the inequality $-x<y$ where $x,y>0$. What will happen if I take the absolute value of it: $|-x|<|y|.$ I though it would be $0<|x|<|y|implies 0<x<y$.
But from the examples above it seems that this ain't true.
EDIT:
I have a increasing sequence $(x_1>x_2>x_3...$ etc.), $x_1,x_2,...,x_n$ where all $x's$ are positive and a constant $m=1/2$. The inequality: $-x_n<1/2$ is always valid because the $x's$ are positive. But $x_3=1$. So for all $n geq 3$, $|-x_n|>1/2$ .
Given this context helps in solving my problem in taking the absolute value?
Taking the absolute value on both side we have for all $n>3, |-x_n|>|1/2|>0 implies x_n>1/2>0$.
Is this correct?
real-analysis algebra-precalculus absolute-value
asked Apr 2 at 2:34
PintecoPinteco
827313
$endgroup$
|
show 5 more comments
$begingroup$
If I've a expression: $-4<3$ and take the absolute value $|-4|<|3|implies 4<3$ which is false. So I though that maybe the inequality sign would change. But $|-2|<|3| implies 2>3$ which is also false.
My problem is that I have the inequality $-x<y$ where $x,y>0$. What will happen if I take the absolute value of it: $|-x|<|y|.$ I though it would be $0<|x|<|y|implies 0<x<y$.
But from the examples above it seems that this ain't true.
EDIT:
I have a increasing sequence $(x_1>x_2>x_3...$ etc.), $x_1,x_2,...,x_n$ where all $x's$ are positive and a constant $m=1/2$. The inequality: $-x_n<1/2$ is always valid because the $x's$ are positive. But $x_3=1$. So for all $n geq 3$, $|-x_n|>1/2$ .
Given this context helps in solving my problem in taking the absolute value?
Taking the absolute value on both side we have for all $n>3, |-x_n|>|1/2|>0 implies x_n>1/2>0$.
Is this correct?
real-analysis algebra-precalculus absolute-value
asked Apr 2 at 2:34
PintecoPinteco
827313
$endgroup$
6
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
1
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55
|
show 5 more comments
$begingroup$
If I've a expression: $-4<3$ and take the absolute value $|-4|<|3|implies 4<3$ which is false. So I though that maybe the inequality sign would change. But $|-2|<|3| implies 2>3$ which is also false.
My problem is that I have the inequality $-x<y$ where $x,y>0$. What will happen if I take the absolute value of it: $|-x|<|y|.$ I though it would be $0<|x|<|y|implies 0<x<y$.
But from the examples above it seems that this ain't true.
EDIT:
I have a increasing sequence $(x_1>x_2>x_3...$ etc.), $x_1,x_2,...,x_n$ where all $x's$ are positive and a constant $m=1/2$. The inequality: $-x_n<1/2$ is always valid because the $x's$ are positive. But $x_3=1$. So for all $n geq 3$, $|-x_n|>1/2$ .
Given this context helps in solving my problem in taking the absolute value?
Taking the absolute value on both side we have for all $n>3, |-x_n|>|1/2|>0 implies x_n>1/2>0$.
Is this correct?
real-analysis algebra-precalculus absolute-value
asked Apr 2 at 2:34
PintecoPinteco
827313
$endgroup$
If I've a expression: $-4<3$ and take the absolute value $|-4|<|3|implies 4<3$ which is false. So I though that maybe the inequality sign would change. But $|-2|<|3| implies 2>3$ which is also false.
My problem is that I have the inequality $-x<y$ where $x,y>0$. What will happen if I take the absolute value of it: $|-x|<|y|.$ I though it would be $0<|x|<|y|implies 0<x<y$.
But from the examples above it seems that this ain't true.
EDIT:
I have a increasing sequence $(x_1>x_2>x_3...$ etc.), $x_1,x_2,...,x_n$ where all $x's$ are positive and a constant $m=1/2$. The inequality: $-x_n<1/2$ is always valid because the $x's$ are positive. But $x_3=1$. So for all $n geq 3$, $|-x_n|>1/2$ .
Given this context helps in solving my problem in taking the absolute value?
Taking the absolute value on both side we have for all $n>3, |-x_n|>|1/2|>0 implies x_n>1/2>0$.
Is this correct?
real-analysis algebra-precalculus absolute-value
real-analysis algebra-precalculus absolute-value
asked Apr 2 at 2:34
PintecoPinteco
827313
asked Apr 2 at 2:34
PintecoPinteco
827313
edited Apr 2 at 3:07
Pinteco
asked Apr 2 at 2:34
PintecoPinteco
827313
asked Apr 2 at 2:34
PintecoPinteco
827313
asked Apr 2 at 2:34
PintecoPinteco
827313
827313
6
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
1
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55
|
show 5 more comments
6
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
1
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55
6
6
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
1
1
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
No. Apparently, you are trying to remove the absolute value from both sides from the inequality. But that's would mean you apply the inverse function to both sides, but absolute value does not have an inverse.
answered Apr 4 at 12:08
user1747134user1747134
239
$endgroup$
add a comment |
Your Answer
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',
autoActivateHeartbeat: false,
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171385%2ftaking-the-absolute-value-in-inequalities%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Apparently, you are trying to remove the absolute value from both sides from the inequality. But that's would mean you apply the inverse function to both sides, but absolute value does not have an inverse.
answered Apr 4 at 12:08
user1747134user1747134
239
$endgroup$
add a comment |
$begingroup$
No. Apparently, you are trying to remove the absolute value from both sides from the inequality. But that's would mean you apply the inverse function to both sides, but absolute value does not have an inverse.
answered Apr 4 at 12:08
user1747134user1747134
239
$endgroup$
add a comment |
$begingroup$
No. Apparently, you are trying to remove the absolute value from both sides from the inequality. But that's would mean you apply the inverse function to both sides, but absolute value does not have an inverse.
answered Apr 4 at 12:08
user1747134user1747134
239
$endgroup$
No. Apparently, you are trying to remove the absolute value from both sides from the inequality. But that's would mean you apply the inverse function to both sides, but absolute value does not have an inverse.
answered Apr 4 at 12:08
user1747134user1747134
239
answered Apr 4 at 12:08
user1747134user1747134
239
answered Apr 4 at 12:08
user1747134user1747134
239
answered Apr 4 at 12:08
user1747134user1747134
239
239
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171385%2ftaking-the-absolute-value-in-inequalities%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
6
$begingroup$
Absolute value is not a monotonic function
$endgroup$
– J. W. Tanner
Apr 2 at 2:42
$begingroup$
You dont know which number has a greater distance from 0. Talking about x (or-x) and y.
$endgroup$
– randomgirl
Apr 2 at 2:47
1
$begingroup$
As you have demonstrated, this is nonsensical. $x < y$ in no way implies $|x| < |y|$. "Taking the absolute value" of both sides of an inequality is a misleading phrase since you are not justified in doing so without additional context. It may be helpful to mentally replace the absolute value function with an arbitrary function $f$ and consider what it would be to apply $f$ to both sides of an inequality.
$endgroup$
– Brian
Apr 2 at 2:47
$begingroup$
So if I assumed that $x>y>0$, for example, then I would know that $x$ is far from $0$ than $y$ and I would be able to do somthing about it? There must be more information some sort of relation between this numbers?
$endgroup$
– Pinteco
Apr 2 at 2:52
$begingroup$
I'll give some context and put in the post.
$endgroup$
– Pinteco
Apr 2 at 2:55