Mixing
Is my proof correct
$begingroup$
Just posted a similar question, but my proof had ambiguity in it so I rewrote it just so I can check if my reasoning is correct...
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ in the same interval as $x$ such that $|y - a|<|x - a|$ implies that $|f(y) - L| < |L| < epsilon$.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$ for which $|z -a| < |y - a|$ such that $|f(z) - L| < |y - L| < |L| < epsilon$, but this is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification epsilon-delta
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
$endgroup$
add a comment |
$begingroup$
Just posted a similar question, but my proof had ambiguity in it so I rewrote it just so I can check if my reasoning is correct...
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ in the same interval as $x$ such that $|y - a|<|x - a|$ implies that $|f(y) - L| < |L| < epsilon$.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$ for which $|z -a| < |y - a|$ such that $|f(z) - L| < |y - L| < |L| < epsilon$, but this is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification epsilon-delta
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
$endgroup$
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11
add a comment |
$begingroup$
Just posted a similar question, but my proof had ambiguity in it so I rewrote it just so I can check if my reasoning is correct...
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ in the same interval as $x$ such that $|y - a|<|x - a|$ implies that $|f(y) - L| < |L| < epsilon$.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$ for which $|z -a| < |y - a|$ such that $|f(z) - L| < |y - L| < |L| < epsilon$, but this is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification epsilon-delta
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
$endgroup$
Just posted a similar question, but my proof had ambiguity in it so I rewrote it just so I can check if my reasoning is correct...
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ in the same interval as $x$ such that $|y - a|<|x - a|$ implies that $|f(y) - L| < |L| < epsilon$.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$ for which $|z -a| < |y - a|$ such that $|f(z) - L| < |y - L| < |L| < epsilon$, but this is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification epsilon-delta
calculus limits functions proof-verification epsilon-delta
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
edited Jan 24 at 14:25
José Carlos Santos
168k22132236
168k22132236
asked Jan 24 at 14:06
ArielKArielK
386
asked Jan 24 at 14:06
ArielKArielK
386
asked Jan 24 at 14:06
ArielKArielK
386
386
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11
add a comment |
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is still not correct. You wrote “there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $delta$-neighborhood”, $delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
By the definition of limit, for every $epsilon>0$ there is a $delta$-neighborhood of $a$ such that if $delta>|x-a|>0$, then $|f(x)-L|<epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.
We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<epsilon$.
Again, this is not true because of what I said above. What you can say is that if $|y-a|<delta$ then $|f(y)-L|<epsilon$.
Also notice that you have not used the fact that $a neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.
What you now have is an irrational number $x$ in a $delta$-neighborhood of $a$, which tells you $|L|<epsilon$. You also have a rational number $y$ in the same $delta$-neighborhood of $a$, which tells you $|y-L|<epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a neq 0$.
To make this intuition rigorous, notice that for every $epsilon>0$, we can find such a $delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<epsilon$. The fact that $|L|<epsilon$ for every positive $epsilon$ implies $L=0$.
Now we can use $L=0$ in the other inequality. For every $epsilon>0$, we can find a $delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<epsilon$, i.e., $|y|<epsilon$. Then $|a|=|a-y+y| leq |a-y|+|y|<delta+epsilon$. I claim we can always choose $delta$ so that it is smaller than $epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2epsilon$. But this is true for every positive $epsilon$, so $a=0$. There is the contradiction.
answered Jan 24 at 14:35
kccukccu
10.6k11229
$endgroup$
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%2f3085913%2fis-my-proof-correct%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
$begingroup$
It is still not correct. You wrote “there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $delta$-neighborhood”, $delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
It is still not correct. You wrote “there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $delta$-neighborhood”, $delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
It is still not correct. You wrote “there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $delta$-neighborhood”, $delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
It is still not correct. You wrote “there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $delta$-neighborhood”, $delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 14:24
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
add a comment |
add a comment |
$begingroup$
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
By the definition of limit, for every $epsilon>0$ there is a $delta$-neighborhood of $a$ such that if $delta>|x-a|>0$, then $|f(x)-L|<epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.
We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<epsilon$.
Again, this is not true because of what I said above. What you can say is that if $|y-a|<delta$ then $|f(y)-L|<epsilon$.
Also notice that you have not used the fact that $a neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.
What you now have is an irrational number $x$ in a $delta$-neighborhood of $a$, which tells you $|L|<epsilon$. You also have a rational number $y$ in the same $delta$-neighborhood of $a$, which tells you $|y-L|<epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a neq 0$.
To make this intuition rigorous, notice that for every $epsilon>0$, we can find such a $delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<epsilon$. The fact that $|L|<epsilon$ for every positive $epsilon$ implies $L=0$.
Now we can use $L=0$ in the other inequality. For every $epsilon>0$, we can find a $delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<epsilon$, i.e., $|y|<epsilon$. Then $|a|=|a-y+y| leq |a-y|+|y|<delta+epsilon$. I claim we can always choose $delta$ so that it is smaller than $epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2epsilon$. But this is true for every positive $epsilon$, so $a=0$. There is the contradiction.
answered Jan 24 at 14:35
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
By the definition of limit, for every $epsilon>0$ there is a $delta$-neighborhood of $a$ such that if $delta>|x-a|>0$, then $|f(x)-L|<epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.
We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<epsilon$.
Again, this is not true because of what I said above. What you can say is that if $|y-a|<delta$ then $|f(y)-L|<epsilon$.
Also notice that you have not used the fact that $a neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.
What you now have is an irrational number $x$ in a $delta$-neighborhood of $a$, which tells you $|L|<epsilon$. You also have a rational number $y$ in the same $delta$-neighborhood of $a$, which tells you $|y-L|<epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a neq 0$.
To make this intuition rigorous, notice that for every $epsilon>0$, we can find such a $delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<epsilon$. The fact that $|L|<epsilon$ for every positive $epsilon$ implies $L=0$.
Now we can use $L=0$ in the other inequality. For every $epsilon>0$, we can find a $delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<epsilon$, i.e., $|y|<epsilon$. Then $|a|=|a-y+y| leq |a-y|+|y|<delta+epsilon$. I claim we can always choose $delta$ so that it is smaller than $epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2epsilon$. But this is true for every positive $epsilon$, so $a=0$. There is the contradiction.
answered Jan 24 at 14:35
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
By the definition of limit, for every $epsilon>0$ there is a $delta$-neighborhood of $a$ such that if $delta>|x-a|>0$, then $|f(x)-L|<epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.
We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<epsilon$.
Again, this is not true because of what I said above. What you can say is that if $|y-a|<delta$ then $|f(y)-L|<epsilon$.
Also notice that you have not used the fact that $a neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.
What you now have is an irrational number $x$ in a $delta$-neighborhood of $a$, which tells you $|L|<epsilon$. You also have a rational number $y$ in the same $delta$-neighborhood of $a$, which tells you $|y-L|<epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a neq 0$.
To make this intuition rigorous, notice that for every $epsilon>0$, we can find such a $delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<epsilon$. The fact that $|L|<epsilon$ for every positive $epsilon$ implies $L=0$.
Now we can use $L=0$ in the other inequality. For every $epsilon>0$, we can find a $delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<epsilon$, i.e., $|y|<epsilon$. Then $|a|=|a-y+y| leq |a-y|+|y|<delta+epsilon$. I claim we can always choose $delta$ so that it is smaller than $epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2epsilon$. But this is true for every positive $epsilon$, so $a=0$. There is the contradiction.
answered Jan 24 at 14:35
kccukccu
10.6k11229
$endgroup$
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.
By the definition of limit, for every $epsilon>0$ there is a $delta$-neighborhood of $a$ such that if $delta>|x-a|>0$, then $|f(x)-L|<epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.
We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<epsilon$.
Again, this is not true because of what I said above. What you can say is that if $|y-a|<delta$ then $|f(y)-L|<epsilon$.
Also notice that you have not used the fact that $a neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.
What you now have is an irrational number $x$ in a $delta$-neighborhood of $a$, which tells you $|L|<epsilon$. You also have a rational number $y$ in the same $delta$-neighborhood of $a$, which tells you $|y-L|<epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a neq 0$.
To make this intuition rigorous, notice that for every $epsilon>0$, we can find such a $delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<epsilon$. The fact that $|L|<epsilon$ for every positive $epsilon$ implies $L=0$.
Now we can use $L=0$ in the other inequality. For every $epsilon>0$, we can find a $delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<epsilon$, i.e., $|y|<epsilon$. Then $|a|=|a-y+y| leq |a-y|+|y|<delta+epsilon$. I claim we can always choose $delta$ so that it is smaller than $epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2epsilon$. But this is true for every positive $epsilon$, so $a=0$. There is the contradiction.
answered Jan 24 at 14:35
kccukccu
10.6k11229
answered Jan 24 at 14:35
kccukccu
10.6k11229
answered Jan 24 at 14:35
kccukccu
10.6k11229
answered Jan 24 at 14:35
kccukccu
10.6k11229
10.6k11229
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%2f3085913%2fis-my-proof-correct%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
$begingroup$
is the problem from spivak calculus?
$endgroup$
– Bijayan Ray
Jan 24 at 14:09
$begingroup$
It's not really a problem, he just said that I should have no trouble conving myself that this was true, and indeed i had no trouble, but i wanted to check if i can prove it in a formal way
$endgroup$
– ArielK
Jan 24 at 14:11