Mixing
Troublesome theorem in hyperbolic geometry
$begingroup$
I want to prove the following theorem without using Dedekind's axiom(i.e. only with axioms of hyperbolic plane)
Given arbitrary line $mathcal l$ and a point $mathcal P$ which does not lie on $mathcal l$, let $mathcal Q$ be the foot of the perpendicular to $mathcal l$ from $mathcal P$. Then for any segment $mathcal Amathcal B$ such that $mathcal Amathcal B$ $le$ $mathcal Pmathcal Q$, there exists a line $mathcal m$ which satisfies following three conditions. : (1) $mathcal m$ goes through $mathcal P$ (2) $mathcal l$ is divergently parallel to $mathcal m$ (3) For a common perpendicular segment $mathcal Mmathcal N$ between $mathcal l$ and $mathcal m$, $mathcal Mmathcal N$ $cong$ $mathcal Amathcal B$
(divergently parallel means that two parallel lines have a common perpendicular line)
hyperbolic-geometry noneuclidean-geometry
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
$endgroup$
|
show 2 more comments
$begingroup$
I want to prove the following theorem without using Dedekind's axiom(i.e. only with axioms of hyperbolic plane)
Given arbitrary line $mathcal l$ and a point $mathcal P$ which does not lie on $mathcal l$, let $mathcal Q$ be the foot of the perpendicular to $mathcal l$ from $mathcal P$. Then for any segment $mathcal Amathcal B$ such that $mathcal Amathcal B$ $le$ $mathcal Pmathcal Q$, there exists a line $mathcal m$ which satisfies following three conditions. : (1) $mathcal m$ goes through $mathcal P$ (2) $mathcal l$ is divergently parallel to $mathcal m$ (3) For a common perpendicular segment $mathcal Mmathcal N$ between $mathcal l$ and $mathcal m$, $mathcal Mmathcal N$ $cong$ $mathcal Amathcal B$
(divergently parallel means that two parallel lines have a common perpendicular line)
hyperbolic-geometry noneuclidean-geometry
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
$endgroup$
1
$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
3
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
1
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
1
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30
|
show 2 more comments
$begingroup$
I want to prove the following theorem without using Dedekind's axiom(i.e. only with axioms of hyperbolic plane)
Given arbitrary line $mathcal l$ and a point $mathcal P$ which does not lie on $mathcal l$, let $mathcal Q$ be the foot of the perpendicular to $mathcal l$ from $mathcal P$. Then for any segment $mathcal Amathcal B$ such that $mathcal Amathcal B$ $le$ $mathcal Pmathcal Q$, there exists a line $mathcal m$ which satisfies following three conditions. : (1) $mathcal m$ goes through $mathcal P$ (2) $mathcal l$ is divergently parallel to $mathcal m$ (3) For a common perpendicular segment $mathcal Mmathcal N$ between $mathcal l$ and $mathcal m$, $mathcal Mmathcal N$ $cong$ $mathcal Amathcal B$
(divergently parallel means that two parallel lines have a common perpendicular line)
hyperbolic-geometry noneuclidean-geometry
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
$endgroup$
I want to prove the following theorem without using Dedekind's axiom(i.e. only with axioms of hyperbolic plane)
Given arbitrary line $mathcal l$ and a point $mathcal P$ which does not lie on $mathcal l$, let $mathcal Q$ be the foot of the perpendicular to $mathcal l$ from $mathcal P$. Then for any segment $mathcal Amathcal B$ such that $mathcal Amathcal B$ $le$ $mathcal Pmathcal Q$, there exists a line $mathcal m$ which satisfies following three conditions. : (1) $mathcal m$ goes through $mathcal P$ (2) $mathcal l$ is divergently parallel to $mathcal m$ (3) For a common perpendicular segment $mathcal Mmathcal N$ between $mathcal l$ and $mathcal m$, $mathcal Mmathcal N$ $cong$ $mathcal Amathcal B$
(divergently parallel means that two parallel lines have a common perpendicular line)
hyperbolic-geometry noneuclidean-geometry
hyperbolic-geometry noneuclidean-geometry
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
asked Jan 28 at 12:20
Ki Yoon EumKi Yoon Eum
277
277
1
$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
3
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
1
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
1
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30
|
show 2 more comments
1
$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
3
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
1
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
1
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30
1
1
$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
3
3
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
1
1
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
1
1
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30
|
show 2 more comments
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$begingroup$
It appears that you can rephrase the question like this: For any (non-zero) lengths, there is a Lambert quadrilateral with opposite sides having those lengths, the longer side being adjacent to the acute angle. (The other sides determine the lines $ell$ and $m$.) In the case where the lengths are equal, those opposite sides coincide, and the quadrilateral reduces to a line segment, with $m$ being the line through $P$ perpendicular to $overline{PQ}$ (and thus divergently parallel to $ell$).
$endgroup$
– Blue
Jan 28 at 12:41
$begingroup$
@Blue how about the case where the lengths are not equal? Now I'm not even sure that it can be proved. SERIOUSLY have no idea at all
$endgroup$
– Ki Yoon Eum
Jan 28 at 13:38
3
$begingroup$
Since you have not included it in your question, I am unsure what axioms of the hyperbolic plane you DO allow yourself to use. I am also unsure what you mean by Dedekind's axiom, although I will guess that you are referring to some version of the completeness of the real numbers. If my guess is correct, and if you really wish to make no completeness assumption whatsoever, then you will run into the issue that even Euclidean geometry has unprovable theorems. It might interest you to know that modern treatments of Euclidean geometry do incorporate versions of completeness in their axioms.
$endgroup$
– Lee Mosher
Jan 28 at 18:11
1
$begingroup$
@KiYoonEum: You can solve the problem with a little hyperbolic right triangle trigonometry. Is that allowed?
$endgroup$
– Blue
Jan 28 at 18:12
1
$begingroup$
@LeeMosher Oh by the way a hyperbolic axiom I assumed is this : For every line $mathcal l$ and every point P not on $mathcal l$, a limmiting parallel ray $overrightarrow {PX}$ emanating from P exists and it does not make a right angle with$overrightarrow {PQ}$, where Q is the foot of the perpendicular from P to $mathcal l$. Is this hyperbolic axiom not assumed in general?
$endgroup$
– Ki Yoon Eum
Jan 28 at 22:30