Mixing
Validity of Proof for 'Possibility of Subtraction' from Apostol 1
$begingroup$
I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why
Theorem 2, Possibility of Subtraction: Given $a text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $ 0-a$ is written as $-a$ and called the negative of a.
Proof from Apostol - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.
I will also use theorem 1, previously proved
Theorem 1
$ text{if } a + b = a + c text{ then } b = c $
My attempt
$$ begin{align}
text{Given } a, b text{ let } &quad a + x = b &quad text{ start }tag{1}label{eq1}\
text{there is exactly one } x text{ such that } &quad a + x = b &quadtext{from theorem 1.1, show in 2.1}tag{2}label{eq2} \
text{ proof of (2) } \
text{ consider more than one x that satisfies (2) } &quad a + x = b,; a + x' = b &quadtext{ start }tag{2.1}label{eq2.1}\
text{ then } &quad a + x = a + x' &quadtext{ }tag{2.2}label{eq2.2}\
text{ therefore x and x' are the same and (2) is proven } &quad x = x' &quadtext{ direct result of theorem 1.1 }tag{2.3}label{eq2.3}\
end{align}$$
Is this valid? If not, why?
My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.
Appreciate the input!
real-analysis proof-verification proof-writing real-numbers
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
$endgroup$
add a comment |
$begingroup$
I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why
Theorem 2, Possibility of Subtraction: Given $a text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $ 0-a$ is written as $-a$ and called the negative of a.
Proof from Apostol - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.
I will also use theorem 1, previously proved
Theorem 1
$ text{if } a + b = a + c text{ then } b = c $
My attempt
$$ begin{align}
text{Given } a, b text{ let } &quad a + x = b &quad text{ start }tag{1}label{eq1}\
text{there is exactly one } x text{ such that } &quad a + x = b &quadtext{from theorem 1.1, show in 2.1}tag{2}label{eq2} \
text{ proof of (2) } \
text{ consider more than one x that satisfies (2) } &quad a + x = b,; a + x' = b &quadtext{ start }tag{2.1}label{eq2.1}\
text{ then } &quad a + x = a + x' &quadtext{ }tag{2.2}label{eq2.2}\
text{ therefore x and x' are the same and (2) is proven } &quad x = x' &quadtext{ direct result of theorem 1.1 }tag{2.3}label{eq2.3}\
end{align}$$
Is this valid? If not, why?
My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.
Appreciate the input!
real-analysis proof-verification proof-writing real-numbers
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
$endgroup$
add a comment |
$begingroup$
I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why
Theorem 2, Possibility of Subtraction: Given $a text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $ 0-a$ is written as $-a$ and called the negative of a.
Proof from Apostol - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.
I will also use theorem 1, previously proved
Theorem 1
$ text{if } a + b = a + c text{ then } b = c $
My attempt
$$ begin{align}
text{Given } a, b text{ let } &quad a + x = b &quad text{ start }tag{1}label{eq1}\
text{there is exactly one } x text{ such that } &quad a + x = b &quadtext{from theorem 1.1, show in 2.1}tag{2}label{eq2} \
text{ proof of (2) } \
text{ consider more than one x that satisfies (2) } &quad a + x = b,; a + x' = b &quadtext{ start }tag{2.1}label{eq2.1}\
text{ then } &quad a + x = a + x' &quadtext{ }tag{2.2}label{eq2.2}\
text{ therefore x and x' are the same and (2) is proven } &quad x = x' &quadtext{ direct result of theorem 1.1 }tag{2.3}label{eq2.3}\
end{align}$$
Is this valid? If not, why?
My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.
Appreciate the input!
real-analysis proof-verification proof-writing real-numbers
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
$endgroup$
I attempted a proof before reading the solution Apostol provides. I don't think it is valid but I am trying to determine why
Theorem 2, Possibility of Subtraction: Given $a text{ and } b$, there is exactly one x such that $ a + x = b $ This is denoted by $b - a$. $ 0-a$ is written as $-a$ and called the negative of a.
Proof from Apostol - Given $a$ and $b$, choose $y$ so that $a + y = 0$ and let $x = y + b$. Then $a + x = a + (y + b) = (a + y) + b = 0 + b = b$. Therefore there is at least one x such that $a + x = b$. But by Theorem 1.1 there is at most one such x. Hence there is exactly one.
I will also use theorem 1, previously proved
Theorem 1
$ text{if } a + b = a + c text{ then } b = c $
My attempt
$$ begin{align}
text{Given } a, b text{ let } &quad a + x = b &quad text{ start }tag{1}label{eq1}\
text{there is exactly one } x text{ such that } &quad a + x = b &quadtext{from theorem 1.1, show in 2.1}tag{2}label{eq2} \
text{ proof of (2) } \
text{ consider more than one x that satisfies (2) } &quad a + x = b,; a + x' = b &quadtext{ start }tag{2.1}label{eq2.1}\
text{ then } &quad a + x = a + x' &quadtext{ }tag{2.2}label{eq2.2}\
text{ therefore x and x' are the same and (2) is proven } &quad x = x' &quadtext{ direct result of theorem 1.1 }tag{2.3}label{eq2.3}\
end{align}$$
Is this valid? If not, why?
My thoughts - one way that I could see it being invalid is that in (2.1) I break $a + x = b$ into two cases, x and x'. The form of the resulting cases is same as the original, so then, using my logic, you would have to break each case into two more cases on to infinity.
Appreciate the input!
real-analysis proof-verification proof-writing real-numbers
real-analysis proof-verification proof-writing real-numbers
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
asked Jan 11 at 18:17
Jake KirschJake Kirsch
617
617
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.
What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.
answered Jan 11 at 19:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
add a comment |
$begingroup$
No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
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%2f3070173%2fvalidity-of-proof-for-possibility-of-subtraction-from-apostol-1%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
$begingroup$
You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.
What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.
answered Jan 11 at 19:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
add a comment |
$begingroup$
You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.
What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.
answered Jan 11 at 19:21
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
add a comment |
$begingroup$
You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.
What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.
answered Jan 11 at 19:21
user3482749user3482749
4,266919
$endgroup$
You've got half of the answer: you've attempted to show that if such an $x$ exists, then there is at most 1. Presumably, Thereom 1.1 is something of the form "if $a + b = c + b$, then $a = c$.
What you haven't done is show that $x$ exists at all. For example, every part of your proof so far is entirely and completely valid if we restrict our variables to take values only in the natural numbers. However, the final result is clearly not true in the natural numbers: there is no natural number $x$ such that $2 + x = 1$.
answered Jan 11 at 19:21
user3482749user3482749
4,266919
answered Jan 11 at 19:21
user3482749user3482749
4,266919
answered Jan 11 at 19:21
user3482749user3482749
4,266919
answered Jan 11 at 19:21
user3482749user3482749
4,266919
4,266919
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
add a comment |
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Ok I think I understand, Apostol proves it by first stating "choose y so that a + y = 0" which I think can be read as "there exists a y s.t. a + y = 0 (existence of negatives)". We don't have to do this for a or b because it's given. Then he says "let x = y + b" which i think can be read as "there exists an x s.t. x = y + b (valid because of the addition operation that we're assuming real numbers have)". So the key point here is that he states there must be an x that exists if the givens and the axioms hold true, and from that, there must also be a y that exists?
$endgroup$
– Jake Kirsch
Jan 11 at 20:52
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Apostol's $y$ exists by some presumably-previous result (or axiom maybe?), which is what you're calling "existence of negatives". He then picks his $x$ as something that you know exists ($y + b$), then checks that it satisfies the required equation, and that it's the only one that does, though he abreviates the latter down to essentially nothing: your attempt gives that latter step in full.
$endgroup$
– user3482749
Jan 11 at 20:55
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
$begingroup$
Yes, he states several axioms of the real number system, as that is the starting point for the way he develops calculus for the reader. The existence of negatives is one of those, I believe it's also known as the additive inverse. I think it's making sens for me now, thank you for the help!
$endgroup$
– Jake Kirsch
Jan 11 at 20:59
add a comment |
$begingroup$
No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
No, this is not correct. First, you write “given $a,b$, let $a+x=b$”. This doesn't make sense. Besides, you don't explain why such a $x$ exists. And then you only prove that if such a $x$ exists, then it is unique. But, again, you don't prove that it exists.
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 11 at 18:24
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
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%2f3070173%2fvalidity-of-proof-for-possibility-of-subtraction-from-apostol-1%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
