Proof Verification: $epsilon(sigma)=epsilon(sigma^-1) , , forallsigmainS_n$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Determine sign of a permutation, calculate number of elements in the subgroup of permutations with sign = 1Equations over permutationsSign of Composition of PermutationsInversions and Multiplicativity of the Sign of a PermutationSign of permutation. Confusing exampleSign of composition of transpositionsHow to show that $sgn(sigma) = prod_1 leq i < j leq n fraci-jsigma(i) - sigma(j)$ ?The sign function is a homomorphismProve that sgn is a homomorphism from $S_nto1,-1$.The number of inversions in a permutation is equal to the number of its inverse permutation.
What was the first language to use conditional keywords?
How do living politicians protect their readily obtainable signatures from misuse?
How to install press fit bottom bracket into new frame
Putting class ranking in CV, but against dept guidelines
Why should I vote and accept answers?
Can a new player join a group only when a new campaign starts?
A term for a woman complaining about things/begging in a cute/childish way
Converted a Scalar function to a TVF function for parallel execution-Still running in Serial mode
Effects on objects due to a brief relocation of massive amounts of mass
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
What is the meaning of 'breadth' in breadth first search?
Chinese Seal on silk painting - what does it mean?
Significance of Cersei's obsession with elephants?
How fail-safe is nr as stop bytes?
Did Deadpool rescue all of the X-Force?
Drawing without replacement: why is the order of draw irrelevant?
SF book about people trapped in a series of worlds they imagine
Most bit efficient text communication method?
What is this clumpy 20-30cm high yellow-flowered plant?
Do wooden building fires get hotter than 600°C?
Is it possible for SQL statements to execute concurrently within a single session in SQL Server?
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
Is there any word for a place full of confusion?
Why does the remaining Rebel fleet at the end of Rogue One seem dramatically larger than the one in A New Hope?
Proof Verification: $epsilon(sigma)=epsilon(sigma^-1) , , forallsigmainS_n$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Determine sign of a permutation, calculate number of elements in the subgroup of permutations with sign = 1Equations over permutationsSign of Composition of PermutationsInversions and Multiplicativity of the Sign of a PermutationSign of permutation. Confusing exampleSign of composition of transpositionsHow to show that $sgn(sigma) = prod_1 leq i < j leq n fraci-jsigma(i) - sigma(j)$ ?The sign function is a homomorphismProve that sgn is a homomorphism from $S_nto1,-1$.The number of inversions in a permutation is equal to the number of its inverse permutation.
$begingroup$
Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $epsilon=sgn$
Proof:
Since the mapping $ , , epsilon:S_nrightarrowpm1$ is a group homomorphism, I'll use the fact that
beginequation
epsilon(sigmatau)=epsilon(sigma)cdotepsilon(tau)
endequation
And let $tau=sigma^-1$ which will give us $epsilon(sigmasigma^-1)=epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $epsilon(sigma)cdotepsilon(sigma^-1)=1$.
Now since $epsilon(sigma)=(-1)^textnumber of inversions of , sigma$ and $epsilon(sigma^-1)=(-1)^textnumber of inversions of , sigma^-1$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
beginalign
&(-1)^ncdot(-1)^m=1\
&(-1)^n+m=(-1)^2\
&n=m-2
endalign
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.
How does this look?
abstract-algebra permutations
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
$endgroup$
add a comment |
$begingroup$
Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $epsilon=sgn$
Proof:
Since the mapping $ , , epsilon:S_nrightarrowpm1$ is a group homomorphism, I'll use the fact that
beginequation
epsilon(sigmatau)=epsilon(sigma)cdotepsilon(tau)
endequation
And let $tau=sigma^-1$ which will give us $epsilon(sigmasigma^-1)=epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $epsilon(sigma)cdotepsilon(sigma^-1)=1$.
Now since $epsilon(sigma)=(-1)^textnumber of inversions of , sigma$ and $epsilon(sigma^-1)=(-1)^textnumber of inversions of , sigma^-1$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
beginalign
&(-1)^ncdot(-1)^m=1\
&(-1)^n+m=(-1)^2\
&n=m-2
endalign
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.
How does this look?
abstract-algebra permutations
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
$endgroup$
add a comment |
$begingroup$
Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $epsilon=sgn$
Proof:
Since the mapping $ , , epsilon:S_nrightarrowpm1$ is a group homomorphism, I'll use the fact that
beginequation
epsilon(sigmatau)=epsilon(sigma)cdotepsilon(tau)
endequation
And let $tau=sigma^-1$ which will give us $epsilon(sigmasigma^-1)=epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $epsilon(sigma)cdotepsilon(sigma^-1)=1$.
Now since $epsilon(sigma)=(-1)^textnumber of inversions of , sigma$ and $epsilon(sigma^-1)=(-1)^textnumber of inversions of , sigma^-1$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
beginalign
&(-1)^ncdot(-1)^m=1\
&(-1)^n+m=(-1)^2\
&n=m-2
endalign
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.
How does this look?
abstract-algebra permutations
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
$endgroup$
Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $epsilon=sgn$
Proof:
Since the mapping $ , , epsilon:S_nrightarrowpm1$ is a group homomorphism, I'll use the fact that
beginequation
epsilon(sigmatau)=epsilon(sigma)cdotepsilon(tau)
endequation
And let $tau=sigma^-1$ which will give us $epsilon(sigmasigma^-1)=epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $epsilon(sigma)cdotepsilon(sigma^-1)=1$.
Now since $epsilon(sigma)=(-1)^textnumber of inversions of , sigma$ and $epsilon(sigma^-1)=(-1)^textnumber of inversions of , sigma^-1$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
beginalign
&(-1)^ncdot(-1)^m=1\
&(-1)^n+m=(-1)^2\
&n=m-2
endalign
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.
How does this look?
abstract-algebra permutations
abstract-algebra permutations
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
asked Mar 27 at 17:36
kareem bokaikareem bokai
868
868
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:48
cspruncsprun
2,839211
$endgroup$
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
add a comment |
$begingroup$
When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.
answered Mar 27 at 17:48
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.
There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
$endgroup$
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%2f3164833%2fproof-verification-epsilon-sigma-epsilon-sigma-1-forall-sigm%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:48
cspruncsprun
2,839211
$endgroup$
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
add a comment |
$begingroup$
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:48
cspruncsprun
2,839211
$endgroup$
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
add a comment |
$begingroup$
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:48
cspruncsprun
2,839211
$endgroup$
You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)
You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:48
cspruncsprun
2,839211
answered Mar 27 at 17:48
cspruncsprun
2,839211
answered Mar 27 at 17:48
cspruncsprun
2,839211
answered Mar 27 at 17:48
cspruncsprun
2,839211
2,839211
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
add a comment |
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
I see what you mean with the first remark, although am still not sure how you'd have $epsilon(sigma)=epsilon(sigma^-1)=1$ or $-1$ from the first statement
$endgroup$
– kareem bokai
Mar 27 at 18:01
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Both $epsilon(sigma)$ and $epsilon(sigma^-1)$ are either $1$ or $-1$. If they are to multiply to $1$, you cannot have them be unequal.
$endgroup$
– csprun
Mar 27 at 18:02
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
$begingroup$
Ah right ofcourse.. Thanks for the feedback :)
$endgroup$
– kareem bokai
Mar 27 at 18:06
add a comment |
$begingroup$
When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.
answered Mar 27 at 17:48
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.
answered Mar 27 at 17:48
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.
answered Mar 27 at 17:48
jmerryjmerry
17k11633
$endgroup$
When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.
In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.
The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.
answered Mar 27 at 17:48
jmerryjmerry
17k11633
answered Mar 27 at 17:48
jmerryjmerry
17k11633
answered Mar 27 at 17:48
jmerryjmerry
17k11633
answered Mar 27 at 17:48
jmerryjmerry
17k11633
17k11633
add a comment |
add a comment |
$begingroup$
You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.
There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
$endgroup$
add a comment |
$begingroup$
You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.
There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
$endgroup$
add a comment |
$begingroup$
You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.
There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
$endgroup$
You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.
There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
answered Mar 27 at 17:49
Torsten SchoenebergTorsten Schoeneberg
4,6342834
4,6342834
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%2f3164833%2fproof-verification-epsilon-sigma-epsilon-sigma-1-forall-sigm%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
