Mixing
Is an apeirogon contained in the euclidean plane?
$begingroup$
The question is self-explanatory. I think it is not, because when I create an apeirogon I am not using things that euclidean geometry allows. But then what exactly are the Hilbert Axioms that are being violated?
https://en.wikipedia.org/wiki/Apeirogon
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
geometry euclidean-geometry
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
$endgroup$
add a comment |
$begingroup$
The question is self-explanatory. I think it is not, because when I create an apeirogon I am not using things that euclidean geometry allows. But then what exactly are the Hilbert Axioms that are being violated?
https://en.wikipedia.org/wiki/Apeirogon
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
geometry euclidean-geometry
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
$endgroup$
$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37
add a comment |
$begingroup$
The question is self-explanatory. I think it is not, because when I create an apeirogon I am not using things that euclidean geometry allows. But then what exactly are the Hilbert Axioms that are being violated?
https://en.wikipedia.org/wiki/Apeirogon
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
geometry euclidean-geometry
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
$endgroup$
The question is self-explanatory. I think it is not, because when I create an apeirogon I am not using things that euclidean geometry allows. But then what exactly are the Hilbert Axioms that are being violated?
https://en.wikipedia.org/wiki/Apeirogon
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
geometry euclidean-geometry
geometry euclidean-geometry
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
asked Jan 22 at 0:42
creepyrodentcreepyrodent
50514
50514
$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37
add a comment |
$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37
$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37
add a comment |
1 Answer
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$begingroup$
If a straight line subdivided into segments of equal length counts as an apeirogon, or if you are fine with an irregular apeirogon, then these exist in the Euclidean plane.
(For an example of the latter, consider the apeirogon with vertices at coordinates $(k,k^2)$ for all integers $k$.)
However, no matter what, you will not be able to construct an apeirogon of any kind, because no finite construction can produce its infinitely many vertices. We might set ourselves more modest goals, such as being able to repeatedly extend a finite part of the apeirogon for any number of iterations.
If the straight line does not count, then a regular apeirogon (that is, one with equal sides and equal angles) does not exist in the Euclidean plane. For any regular polygon, being given three consecutive vertices lets you construct all the others in a unique way. We can show (using Hilbert's axioms, if you like, with the axiom of Archimedes playing a notable role) that if you're given three non-collinear points that are supposed to be vertices of an apeirogon and try to construct further vertices, eventually the perimeter will fold in on itself and you will fail.
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If a straight line subdivided into segments of equal length counts as an apeirogon, or if you are fine with an irregular apeirogon, then these exist in the Euclidean plane.
(For an example of the latter, consider the apeirogon with vertices at coordinates $(k,k^2)$ for all integers $k$.)
However, no matter what, you will not be able to construct an apeirogon of any kind, because no finite construction can produce its infinitely many vertices. We might set ourselves more modest goals, such as being able to repeatedly extend a finite part of the apeirogon for any number of iterations.
If the straight line does not count, then a regular apeirogon (that is, one with equal sides and equal angles) does not exist in the Euclidean plane. For any regular polygon, being given three consecutive vertices lets you construct all the others in a unique way. We can show (using Hilbert's axioms, if you like, with the axiom of Archimedes playing a notable role) that if you're given three non-collinear points that are supposed to be vertices of an apeirogon and try to construct further vertices, eventually the perimeter will fold in on itself and you will fail.
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
add a comment |
$begingroup$
If a straight line subdivided into segments of equal length counts as an apeirogon, or if you are fine with an irregular apeirogon, then these exist in the Euclidean plane.
(For an example of the latter, consider the apeirogon with vertices at coordinates $(k,k^2)$ for all integers $k$.)
However, no matter what, you will not be able to construct an apeirogon of any kind, because no finite construction can produce its infinitely many vertices. We might set ourselves more modest goals, such as being able to repeatedly extend a finite part of the apeirogon for any number of iterations.
If the straight line does not count, then a regular apeirogon (that is, one with equal sides and equal angles) does not exist in the Euclidean plane. For any regular polygon, being given three consecutive vertices lets you construct all the others in a unique way. We can show (using Hilbert's axioms, if you like, with the axiom of Archimedes playing a notable role) that if you're given three non-collinear points that are supposed to be vertices of an apeirogon and try to construct further vertices, eventually the perimeter will fold in on itself and you will fail.
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
add a comment |
$begingroup$
If a straight line subdivided into segments of equal length counts as an apeirogon, or if you are fine with an irregular apeirogon, then these exist in the Euclidean plane.
(For an example of the latter, consider the apeirogon with vertices at coordinates $(k,k^2)$ for all integers $k$.)
However, no matter what, you will not be able to construct an apeirogon of any kind, because no finite construction can produce its infinitely many vertices. We might set ourselves more modest goals, such as being able to repeatedly extend a finite part of the apeirogon for any number of iterations.
If the straight line does not count, then a regular apeirogon (that is, one with equal sides and equal angles) does not exist in the Euclidean plane. For any regular polygon, being given three consecutive vertices lets you construct all the others in a unique way. We can show (using Hilbert's axioms, if you like, with the axiom of Archimedes playing a notable role) that if you're given three non-collinear points that are supposed to be vertices of an apeirogon and try to construct further vertices, eventually the perimeter will fold in on itself and you will fail.
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
If a straight line subdivided into segments of equal length counts as an apeirogon, or if you are fine with an irregular apeirogon, then these exist in the Euclidean plane.
(For an example of the latter, consider the apeirogon with vertices at coordinates $(k,k^2)$ for all integers $k$.)
However, no matter what, you will not be able to construct an apeirogon of any kind, because no finite construction can produce its infinitely many vertices. We might set ourselves more modest goals, such as being able to repeatedly extend a finite part of the apeirogon for any number of iterations.
If the straight line does not count, then a regular apeirogon (that is, one with equal sides and equal angles) does not exist in the Euclidean plane. For any regular polygon, being given three consecutive vertices lets you construct all the others in a unique way. We can show (using Hilbert's axioms, if you like, with the axiom of Archimedes playing a notable role) that if you're given three non-collinear points that are supposed to be vertices of an apeirogon and try to construct further vertices, eventually the perimeter will fold in on itself and you will fail.
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 22 at 3:01
Misha LavrovMisha Lavrov
47.3k657107
47.3k657107
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$begingroup$
what do you mean by "I create an apeirogon"
$endgroup$
– Will Jagy
Jan 22 at 1:45
$begingroup$
create = construct. for instance, using the archimedean polygonal process to obtain $pi$, but not tending the measures of the sides to $0$, only the number of sides.
$endgroup$
– creepyrodent
Jan 22 at 2:37