Inverse of set operations
$begingroup$
$A cup B equiv C $ then what is $A$ in terms of $B,C$?
I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.
From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.
Complement is easy as it is it's own inverse $(A^C)^C=A$
not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.
when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?
elementary-set-theory
$endgroup$
add a comment |
$begingroup$
$A cup B equiv C $ then what is $A$ in terms of $B,C$?
I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.
From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.
Complement is easy as it is it's own inverse $(A^C)^C=A$
not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.
when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?
elementary-set-theory
$endgroup$
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago
add a comment |
$begingroup$
$A cup B equiv C $ then what is $A$ in terms of $B,C$?
I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.
From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.
Complement is easy as it is it's own inverse $(A^C)^C=A$
not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.
when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?
elementary-set-theory
$endgroup$
$A cup B equiv C $ then what is $A$ in terms of $B,C$?
I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.
From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.
Complement is easy as it is it's own inverse $(A^C)^C=A$
not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.
when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?
elementary-set-theory
elementary-set-theory
edited 4 hours ago
Arjang
asked 5 hours ago
ArjangArjang
5,59262363
5,59262363
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago
add a comment |
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".
On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.
If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$
What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).
$endgroup$
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
It's not possible.
Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.
1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)
2) ${x not in C, x in B} = emptyset$ (as $B subset C$)
3) ${x in C, x not in B} = Csetminus B$.
4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).
Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.
Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.
[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.
[1 and 4] and [1 and 2 and 4] = $C^c cup B$
[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
$endgroup$
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%2f3080006%2finverse-of-set-operations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".
On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.
If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$
What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).
$endgroup$
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".
On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.
If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$
What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).
$endgroup$
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".
On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.
If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$
What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).
$endgroup$
Inverse operators for union, intersection, or set difference are impossible in general.
Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".
On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.
If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.
This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$
What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.
(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).
edited 3 hours ago
answered 4 hours ago
Henning MakholmHenning Makholm
239k17304541
239k17304541
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
Nice counter example!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
$begingroup$
So if $C=Atriangle B$ then $Ctriangle B=(Atriangle B)setminus B cup Bsetminus(Atriangle B)=[(Asetminus B cup Bsetminus A)setminus B ]cup [Bsetminus(Asetminus B cup Bsetminus A]=(Asetminus B)cup Bsetminus(Bsetminus A) = (Asetminus B)cup (Bcap A) = A$. .... nifty!
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
It's not possible.
Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.
1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)
2) ${x not in C, x in B} = emptyset$ (as $B subset C$)
3) ${x in C, x not in B} = Csetminus B$.
4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).
Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.
Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.
[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.
[1 and 4] and [1 and 2 and 4] = $C^c cup B$
[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
$endgroup$
add a comment |
$begingroup$
It's not possible.
Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.
1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)
2) ${x not in C, x in B} = emptyset$ (as $B subset C$)
3) ${x in C, x not in B} = Csetminus B$.
4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).
Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.
Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.
[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.
[1 and 4] and [1 and 2 and 4] = $C^c cup B$
[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
$endgroup$
add a comment |
$begingroup$
It's not possible.
Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.
1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)
2) ${x not in C, x in B} = emptyset$ (as $B subset C$)
3) ${x in C, x not in B} = Csetminus B$.
4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).
Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.
Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.
[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.
[1 and 4] and [1 and 2 and 4] = $C^c cup B$
[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
$endgroup$
It's not possible.
Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.
1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)
2) ${x not in C, x in B} = emptyset$ (as $B subset C$)
3) ${x in C, x not in B} = Csetminus B$.
4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).
Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.
Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.
====
FWIW
We can list all possible sets possible to describe they are
The four above:
1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.
[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.
[1 and 4] and [1 and 2 and 4] = $C^c cup B$
[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$
[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.
edited 3 hours ago
answered 4 hours ago
fleabloodfleablood
69.1k22685
69.1k22685
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%2f3080006%2finverse-of-set-operations%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
$begingroup$
What does $equiv$ mean in this context? Why aren't you saying they are equal?
$endgroup$
– fleablood
4 hours ago