bijective function from a circle to a line
$begingroup$
I am just reading a book which says the square has the same size as the line. It is saying every coordinate in the square have two coordinates x and y, and we can make a bijective function this in a line if we transform the coordinates as $ z = .x_1 y_1 x_2 y_2dots $. Ex: $ x=1/2$ and $ y=1/3 $ $to z= 0.530303030dots$
Read this, I thought actually we could use this to map circle also (actually for any shape) Every point on the circe line is map like this for just one point on the line. Also they cardinality stay same. Is this correct?
If yes what happens if one of the coordinates is negative? Lat says one of the point on the circle x=2 and y = -3, So if how we organise these numbers after each other. $ 0.2(-3)00 $ Obviously it is not looking right.
geometry
asked Dec 12 '18 at 13:50
arkheinarkhein
95
$endgroup$
add a comment |
$begingroup$
I am just reading a book which says the square has the same size as the line. It is saying every coordinate in the square have two coordinates x and y, and we can make a bijective function this in a line if we transform the coordinates as $ z = .x_1 y_1 x_2 y_2dots $. Ex: $ x=1/2$ and $ y=1/3 $ $to z= 0.530303030dots$
Read this, I thought actually we could use this to map circle also (actually for any shape) Every point on the circe line is map like this for just one point on the line. Also they cardinality stay same. Is this correct?
If yes what happens if one of the coordinates is negative? Lat says one of the point on the circle x=2 and y = -3, So if how we organise these numbers after each other. $ 0.2(-3)00 $ Obviously it is not looking right.
geometry
asked Dec 12 '18 at 13:50
arkheinarkhein
95
$endgroup$
1
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52
add a comment |
$begingroup$
I am just reading a book which says the square has the same size as the line. It is saying every coordinate in the square have two coordinates x and y, and we can make a bijective function this in a line if we transform the coordinates as $ z = .x_1 y_1 x_2 y_2dots $. Ex: $ x=1/2$ and $ y=1/3 $ $to z= 0.530303030dots$
Read this, I thought actually we could use this to map circle also (actually for any shape) Every point on the circe line is map like this for just one point on the line. Also they cardinality stay same. Is this correct?
If yes what happens if one of the coordinates is negative? Lat says one of the point on the circle x=2 and y = -3, So if how we organise these numbers after each other. $ 0.2(-3)00 $ Obviously it is not looking right.
geometry
asked Dec 12 '18 at 13:50
arkheinarkhein
95
$endgroup$
I am just reading a book which says the square has the same size as the line. It is saying every coordinate in the square have two coordinates x and y, and we can make a bijective function this in a line if we transform the coordinates as $ z = .x_1 y_1 x_2 y_2dots $. Ex: $ x=1/2$ and $ y=1/3 $ $to z= 0.530303030dots$
Read this, I thought actually we could use this to map circle also (actually for any shape) Every point on the circe line is map like this for just one point on the line. Also they cardinality stay same. Is this correct?
If yes what happens if one of the coordinates is negative? Lat says one of the point on the circle x=2 and y = -3, So if how we organise these numbers after each other. $ 0.2(-3)00 $ Obviously it is not looking right.
geometry
geometry
asked Dec 12 '18 at 13:50
arkheinarkhein
95
asked Dec 12 '18 at 13:50
arkheinarkhein
95
asked Dec 12 '18 at 13:50
arkheinarkhein
95
asked Dec 12 '18 at 13:50
arkheinarkhein
95
asked Dec 12 '18 at 13:50
arkheinarkhein
95
95
1
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52
add a comment |
1
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52
1
1
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. That map is specifically constructed to work for points in the square $[0, 1]times[0, 1]$. Any other shape (including a square at a different place in the plane, but also circles, or more exotic things) would require some alteration to the function description.
It is definitely possible, but it's not as pretty to describe exactly how the map works. You can't just interleave the digits and call it a day.
Besides, you have to be a bit careful with the whole $0.999ldots = 1$ issue. For instance,
$$
z = 0.5303030ldots
$$
corresponds to $x = frac12, y = frac13$, but so does
$$
z = 0.4393939ldots
$$
so it's not really a bijection as stated.
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
$endgroup$
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
add a comment |
$begingroup$
Yes, there exists a bijective function between any circle and any interval of $mathbb R$.
For example, for the basic circle $x^2+y^2=1$ and the interval $[0,1)$ you can construct bijective the mapping
$$t mapsto (cos(2pi t), sin(2pi t))$$
For any other circle and any other half-open interval, you can first map the circle to the unit circle, and the interval to the unit interval, and use the same function.
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
$endgroup$
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
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%2f3036710%2fbijective-function-from-a-circle-to-a-line%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
$begingroup$
No. That map is specifically constructed to work for points in the square $[0, 1]times[0, 1]$. Any other shape (including a square at a different place in the plane, but also circles, or more exotic things) would require some alteration to the function description.
It is definitely possible, but it's not as pretty to describe exactly how the map works. You can't just interleave the digits and call it a day.
Besides, you have to be a bit careful with the whole $0.999ldots = 1$ issue. For instance,
$$
z = 0.5303030ldots
$$
corresponds to $x = frac12, y = frac13$, but so does
$$
z = 0.4393939ldots
$$
so it's not really a bijection as stated.
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
$endgroup$
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
add a comment |
$begingroup$
No. That map is specifically constructed to work for points in the square $[0, 1]times[0, 1]$. Any other shape (including a square at a different place in the plane, but also circles, or more exotic things) would require some alteration to the function description.
It is definitely possible, but it's not as pretty to describe exactly how the map works. You can't just interleave the digits and call it a day.
Besides, you have to be a bit careful with the whole $0.999ldots = 1$ issue. For instance,
$$
z = 0.5303030ldots
$$
corresponds to $x = frac12, y = frac13$, but so does
$$
z = 0.4393939ldots
$$
so it's not really a bijection as stated.
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
$endgroup$
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
add a comment |
$begingroup$
No. That map is specifically constructed to work for points in the square $[0, 1]times[0, 1]$. Any other shape (including a square at a different place in the plane, but also circles, or more exotic things) would require some alteration to the function description.
It is definitely possible, but it's not as pretty to describe exactly how the map works. You can't just interleave the digits and call it a day.
Besides, you have to be a bit careful with the whole $0.999ldots = 1$ issue. For instance,
$$
z = 0.5303030ldots
$$
corresponds to $x = frac12, y = frac13$, but so does
$$
z = 0.4393939ldots
$$
so it's not really a bijection as stated.
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
$endgroup$
No. That map is specifically constructed to work for points in the square $[0, 1]times[0, 1]$. Any other shape (including a square at a different place in the plane, but also circles, or more exotic things) would require some alteration to the function description.
It is definitely possible, but it's not as pretty to describe exactly how the map works. You can't just interleave the digits and call it a day.
Besides, you have to be a bit careful with the whole $0.999ldots = 1$ issue. For instance,
$$
z = 0.5303030ldots
$$
corresponds to $x = frac12, y = frac13$, but so does
$$
z = 0.4393939ldots
$$
so it's not really a bijection as stated.
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
answered Dec 12 '18 at 13:54
ArthurArthur
116k7116199
116k7116199
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
add a comment |
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
$begingroup$
Is there any name for "That map"? Actually, the book does not mention, so it is limiting me to read about more. I guess "That map" somehow handle the 0.999 = 1 problem, but in my text there is no mention it. So actually the map is working or not?
$endgroup$
– arkhein
Dec 12 '18 at 14:31
add a comment |
$begingroup$
Yes, there exists a bijective function between any circle and any interval of $mathbb R$.
For example, for the basic circle $x^2+y^2=1$ and the interval $[0,1)$ you can construct bijective the mapping
$$t mapsto (cos(2pi t), sin(2pi t))$$
For any other circle and any other half-open interval, you can first map the circle to the unit circle, and the interval to the unit interval, and use the same function.
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
$endgroup$
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
add a comment |
$begingroup$
Yes, there exists a bijective function between any circle and any interval of $mathbb R$.
For example, for the basic circle $x^2+y^2=1$ and the interval $[0,1)$ you can construct bijective the mapping
$$t mapsto (cos(2pi t), sin(2pi t))$$
For any other circle and any other half-open interval, you can first map the circle to the unit circle, and the interval to the unit interval, and use the same function.
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
$endgroup$
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
add a comment |
$begingroup$
Yes, there exists a bijective function between any circle and any interval of $mathbb R$.
For example, for the basic circle $x^2+y^2=1$ and the interval $[0,1)$ you can construct bijective the mapping
$$t mapsto (cos(2pi t), sin(2pi t))$$
For any other circle and any other half-open interval, you can first map the circle to the unit circle, and the interval to the unit interval, and use the same function.
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
$endgroup$
Yes, there exists a bijective function between any circle and any interval of $mathbb R$.
For example, for the basic circle $x^2+y^2=1$ and the interval $[0,1)$ you can construct bijective the mapping
$$t mapsto (cos(2pi t), sin(2pi t))$$
For any other circle and any other half-open interval, you can first map the circle to the unit circle, and the interval to the unit interval, and use the same function.
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
answered Dec 12 '18 at 13:56
5xum5xum
91.1k394161
91.1k394161
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
add a comment |
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
I suspect that by "circle", the OP really means "disc". Not sure, though.
$endgroup$
– Arthur
Dec 12 '18 at 14:00
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
@Arthur Hmm. Possibly. In that case, of course, the answer is a little more complicated. Still a "yes", but more complicatedly so :)
$endgroup$
– 5xum
Dec 12 '18 at 14:03
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
I mean circe not disk.
$endgroup$
– arkhein
Dec 12 '18 at 14:27
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
$begingroup$
thanks the answer.
$endgroup$
– arkhein
Dec 12 '18 at 14:32
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%2f3036710%2fbijective-function-from-a-circle-to-a-line%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
1
$begingroup$
When you say "circle", do you mean just the edge of the circle, or the whole, filled-in disc? Not that it really matters for the final answer, but it may affect the details of any example people give you.
$endgroup$
– Arthur
Dec 12 '18 at 14:01
$begingroup$
I mean the just the edge.
$endgroup$
– arkhein
Dec 12 '18 at 14:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Jan 15 at 8:52