Mixing
Absolute Value Inequality (another)
$begingroup$
Solve the following inequality
$$|a-2| > |a+4|$$
Here I separated it into cases as shown
$a<-4$
$$-(a-2) > -(a+4) implies 2-a>-a-4 implies 0>-6$$
Always true, so we get $mathbb{R} cap (-infty , -4) = (-infty ,-4)$
$-4<a<2$
$$-(a-2)>a+4 implies 2-a>a+4 implies a<-1$$
Taking interception $(-4,2)$ $cap $ $(-infty,-1)$
$a>2$
$$a-2 > a+4 implies -2>4 $$
Always false, so no solution from there. Finally, I checked end points and noticed that they do not work in this inequality. However, I do not know how to proceed further. Could you assist me?
Regards
inequality
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
$endgroup$
add a comment |
$begingroup$
Solve the following inequality
$$|a-2| > |a+4|$$
Here I separated it into cases as shown
$a<-4$
$$-(a-2) > -(a+4) implies 2-a>-a-4 implies 0>-6$$
Always true, so we get $mathbb{R} cap (-infty , -4) = (-infty ,-4)$
$-4<a<2$
$$-(a-2)>a+4 implies 2-a>a+4 implies a<-1$$
Taking interception $(-4,2)$ $cap $ $(-infty,-1)$
$a>2$
$$a-2 > a+4 implies -2>4 $$
Always false, so no solution from there. Finally, I checked end points and noticed that they do not work in this inequality. However, I do not know how to proceed further. Could you assist me?
Regards
inequality
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
$endgroup$
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
1
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19
add a comment |
$begingroup$
Solve the following inequality
$$|a-2| > |a+4|$$
Here I separated it into cases as shown
$a<-4$
$$-(a-2) > -(a+4) implies 2-a>-a-4 implies 0>-6$$
Always true, so we get $mathbb{R} cap (-infty , -4) = (-infty ,-4)$
$-4<a<2$
$$-(a-2)>a+4 implies 2-a>a+4 implies a<-1$$
Taking interception $(-4,2)$ $cap $ $(-infty,-1)$
$a>2$
$$a-2 > a+4 implies -2>4 $$
Always false, so no solution from there. Finally, I checked end points and noticed that they do not work in this inequality. However, I do not know how to proceed further. Could you assist me?
Regards
inequality
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
$endgroup$
Solve the following inequality
$$|a-2| > |a+4|$$
Here I separated it into cases as shown
$a<-4$
$$-(a-2) > -(a+4) implies 2-a>-a-4 implies 0>-6$$
Always true, so we get $mathbb{R} cap (-infty , -4) = (-infty ,-4)$
$-4<a<2$
$$-(a-2)>a+4 implies 2-a>a+4 implies a<-1$$
Taking interception $(-4,2)$ $cap $ $(-infty,-1)$
$a>2$
$$a-2 > a+4 implies -2>4 $$
Always false, so no solution from there. Finally, I checked end points and noticed that they do not work in this inequality. However, I do not know how to proceed further. Could you assist me?
Regards
inequality
inequality
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
edited Jan 13 at 13:34
Ethan Bolker
43.1k551114
43.1k551114
asked Jan 13 at 13:16
EnzoEnzo
19917
asked Jan 13 at 13:16
EnzoEnzo
19917
asked Jan 13 at 13:16
EnzoEnzo
19917
19917
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
1
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19
add a comment |
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
1
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
1
1
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
As commented, $;a=4;$ doesn't really fits in the inequality. You can now put your solution set as
$$(-infty,-4)cupleft((-4,-2)cap(-infty,-1)right)=(-infty,-4)cup(-4,-1)=(-infty,-1)setminus{-4}$$
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
$endgroup$
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
add a comment |
$begingroup$
If you interpret the absolute values in terms of distance, it is immediate:
$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$:
$$a<frac{2-4}2=-1$$
Other method:
begin{align}
|a-2|>|a+4|&iff(a-2)^2>(a+4)^2 \
&iff a^2-4a+4>a^2+8a+16\
&iff 12a+12<0iff a<-1.
end{align}
answered Jan 13 at 13:42
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
You have to distinguish the following cases:
$$xgeq 2$$ then we get $$x-2>x+4$$
$$-4le x<2$$ then we get $$x-2>-x+4$$
$$x<-4$$ then we have $$-x+2>-x-4$$
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
$endgroup$
add a comment |
$begingroup$
$|(a+1)-3|>|(a+1)+3|;$
$x:=a+1$;
$|x-(+3)| >|x-(-3)|;$
Real number line :
Distance from a point $x$ to $(+3)$ is bigger than from $x$
to $(-3)$, i.e. $x<0.$
(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).
$x= a+1<0$, $a<-1$.
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
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',
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%2f3071997%2fabsolute-value-inequality-another%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As commented, $;a=4;$ doesn't really fits in the inequality. You can now put your solution set as
$$(-infty,-4)cupleft((-4,-2)cap(-infty,-1)right)=(-infty,-4)cup(-4,-1)=(-infty,-1)setminus{-4}$$
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
$endgroup$
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
add a comment |
$begingroup$
As commented, $;a=4;$ doesn't really fits in the inequality. You can now put your solution set as
$$(-infty,-4)cupleft((-4,-2)cap(-infty,-1)right)=(-infty,-4)cup(-4,-1)=(-infty,-1)setminus{-4}$$
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
$endgroup$
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
add a comment |
$begingroup$
As commented, $;a=4;$ doesn't really fits in the inequality. You can now put your solution set as
$$(-infty,-4)cupleft((-4,-2)cap(-infty,-1)right)=(-infty,-4)cup(-4,-1)=(-infty,-1)setminus{-4}$$
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
$endgroup$
As commented, $;a=4;$ doesn't really fits in the inequality. You can now put your solution set as
$$(-infty,-4)cupleft((-4,-2)cap(-infty,-1)right)=(-infty,-4)cup(-4,-1)=(-infty,-1)setminus{-4}$$
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 13:22
DonAntonioDonAntonio
178k1494230
178k1494230
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
add a comment |
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
What does this mathematical symbol mean ?
$endgroup$
– Enzo
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
Set difference. Sometimes it is also used $;A-B;$ ... A simple minus sign.
$endgroup$
– DonAntonio
Jan 13 at 13:23
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
The right answer seems to be $(-infty, -1)$ according to my textbook. Do we get it from your final interval?:
$endgroup$
– Enzo
Jan 13 at 13:24
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
$begingroup$
That is correct, @Enzo . Your solution didn't even take into account $;a=-4;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way
$endgroup$
– DonAntonio
Jan 13 at 13:34
add a comment |
$begingroup$
If you interpret the absolute values in terms of distance, it is immediate:
$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$:
$$a<frac{2-4}2=-1$$
Other method:
begin{align}
|a-2|>|a+4|&iff(a-2)^2>(a+4)^2 \
&iff a^2-4a+4>a^2+8a+16\
&iff 12a+12<0iff a<-1.
end{align}
answered Jan 13 at 13:42
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
If you interpret the absolute values in terms of distance, it is immediate:
$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$:
$$a<frac{2-4}2=-1$$
Other method:
begin{align}
|a-2|>|a+4|&iff(a-2)^2>(a+4)^2 \
&iff a^2-4a+4>a^2+8a+16\
&iff 12a+12<0iff a<-1.
end{align}
answered Jan 13 at 13:42
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
If you interpret the absolute values in terms of distance, it is immediate:
$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$:
$$a<frac{2-4}2=-1$$
Other method:
begin{align}
|a-2|>|a+4|&iff(a-2)^2>(a+4)^2 \
&iff a^2-4a+4>a^2+8a+16\
&iff 12a+12<0iff a<-1.
end{align}
answered Jan 13 at 13:42
BernardBernard
121k740116
$endgroup$
If you interpret the absolute values in terms of distance, it is immediate:
$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$:
$$a<frac{2-4}2=-1$$
Other method:
begin{align}
|a-2|>|a+4|&iff(a-2)^2>(a+4)^2 \
&iff a^2-4a+4>a^2+8a+16\
&iff 12a+12<0iff a<-1.
end{align}
answered Jan 13 at 13:42
BernardBernard
121k740116
edited Jan 13 at 13:51
answered Jan 13 at 13:42
BernardBernard
121k740116
answered Jan 13 at 13:42
BernardBernard
121k740116
answered Jan 13 at 13:42
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
You have to distinguish the following cases:
$$xgeq 2$$ then we get $$x-2>x+4$$
$$-4le x<2$$ then we get $$x-2>-x+4$$
$$x<-4$$ then we have $$-x+2>-x-4$$
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
$endgroup$
add a comment |
$begingroup$
You have to distinguish the following cases:
$$xgeq 2$$ then we get $$x-2>x+4$$
$$-4le x<2$$ then we get $$x-2>-x+4$$
$$x<-4$$ then we have $$-x+2>-x-4$$
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
$endgroup$
add a comment |
$begingroup$
You have to distinguish the following cases:
$$xgeq 2$$ then we get $$x-2>x+4$$
$$-4le x<2$$ then we get $$x-2>-x+4$$
$$x<-4$$ then we have $$-x+2>-x-4$$
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
$endgroup$
You have to distinguish the following cases:
$$xgeq 2$$ then we get $$x-2>x+4$$
$$-4le x<2$$ then we get $$x-2>-x+4$$
$$x<-4$$ then we have $$-x+2>-x-4$$
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
answered Jan 13 at 13:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.3k42865
75.3k42865
add a comment |
add a comment |
$begingroup$
$|(a+1)-3|>|(a+1)+3|;$
$x:=a+1$;
$|x-(+3)| >|x-(-3)|;$
Real number line :
Distance from a point $x$ to $(+3)$ is bigger than from $x$
to $(-3)$, i.e. $x<0.$
(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).
$x= a+1<0$, $a<-1$.
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
$|(a+1)-3|>|(a+1)+3|;$
$x:=a+1$;
$|x-(+3)| >|x-(-3)|;$
Real number line :
Distance from a point $x$ to $(+3)$ is bigger than from $x$
to $(-3)$, i.e. $x<0.$
(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).
$x= a+1<0$, $a<-1$.
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
$|(a+1)-3|>|(a+1)+3|;$
$x:=a+1$;
$|x-(+3)| >|x-(-3)|;$
Real number line :
Distance from a point $x$ to $(+3)$ is bigger than from $x$
to $(-3)$, i.e. $x<0.$
(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).
$x= a+1<0$, $a<-1$.
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
$|(a+1)-3|>|(a+1)+3|;$
$x:=a+1$;
$|x-(+3)| >|x-(-3)|;$
Real number line :
Distance from a point $x$ to $(+3)$ is bigger than from $x$
to $(-3)$, i.e. $x<0.$
(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).
$x= a+1<0$, $a<-1$.
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
answered Jan 13 at 16:14
Peter SzilasPeter Szilas
11.3k2822
11.3k2822
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%2f3071997%2fabsolute-value-inequality-another%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
$begingroup$
With $;a=4;$ your inequality is $$2=|4-2|>|4+4|=8...text{you still think this is true?}$$
$endgroup$
– DonAntonio
Jan 13 at 13:18
1
$begingroup$
@DonAntonio Not really, thanks for pointing out.
$endgroup$
– Enzo
Jan 13 at 13:19