Soft Question - Definition of the Real Plane
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I have only seen the real plane to be defined as $mathbb R times mathbb R$.
In Euclidean geometry, the origin does not play any central role, but in this set theoretic definition, it does. Is there a way to define the real plane in a way that does not give any special importance to the origin, that is, is there a way to defined the real plane without using coordinates (or products)?
linear-algebra geometry euclidean-geometry
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
$endgroup$
add a comment |
$begingroup$
I have only seen the real plane to be defined as $mathbb R times mathbb R$.
In Euclidean geometry, the origin does not play any central role, but in this set theoretic definition, it does. Is there a way to define the real plane in a way that does not give any special importance to the origin, that is, is there a way to defined the real plane without using coordinates (or products)?
linear-algebra geometry euclidean-geometry
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
$endgroup$
2
$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
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What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57
add a comment |
$begingroup$
I have only seen the real plane to be defined as $mathbb R times mathbb R$.
In Euclidean geometry, the origin does not play any central role, but in this set theoretic definition, it does. Is there a way to define the real plane in a way that does not give any special importance to the origin, that is, is there a way to defined the real plane without using coordinates (or products)?
linear-algebra geometry euclidean-geometry
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
$endgroup$
I have only seen the real plane to be defined as $mathbb R times mathbb R$.
In Euclidean geometry, the origin does not play any central role, but in this set theoretic definition, it does. Is there a way to define the real plane in a way that does not give any special importance to the origin, that is, is there a way to defined the real plane without using coordinates (or products)?
linear-algebra geometry euclidean-geometry
linear-algebra geometry euclidean-geometry
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
edited Dec 6 '18 at 22:30
LinearGuy
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
asked Dec 6 '18 at 21:07
LinearGuyLinearGuy
13711
13711
2
$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
$begingroup$
What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57
add a comment |
2
$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
$begingroup$
What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57
2
2
$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
$begingroup$
What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57
$begingroup$
What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Euclid never defined the plane in his elements, and indeed never used numbers to measure length, angles, or area. In this sense, the Euclidean plane is not the same as the Cartesian plane. Euclid used a plane to mean a flat, two-dimensional surface that extends infinitely far, independent of numerical values. This definition gives no importance to the origin, as indeed there is no origin.
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
$endgroup$
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
add a comment |
$begingroup$
I think what you are interested in is exactly the difference between an affine space and a vector space. Read the first few paragraphs in that wikipedia article. Basically, you get rid of the "specialness of the origin" by thinking of two copies of the plane. One represents the group, $G$, of translations and the second is the geometric manifold, $X$, that represents your space. For any two points $x$ and $y$ in your space $X$, there is a unique translation $g$ in $G$ that takes $x$ to $y$. Thus, seen in this way, there are no "special" points in $X$. Notice that the origin is special in $G$ since it represents the identity translation, i.e. the one that keeps $X$ fixed. An even more general concept is that of a principal homogenous space.
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
$endgroup$
– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
add a comment |
$begingroup$
Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert
in 1899 in his book Grundlagen der Geometrie
Hilbert's axiom system is constructed with six primitive notions:
three primitive terms:[5]
point;
line;
plane;
and three primitive relations:[6]
Betweenness, a ternary relation linking points;
Lies on (Containment), three binary relations, one linking points and straight lines, one linking points and planes, and one linking
straight lines and planes;
Congruence, two binary relations, one linking line segments and one linking angles, each denoted by an infix ≅.
Line segments, angles, and triangles may each be defined in terms of
points and straight lines, using the relations of betweenness and
containment.[how?] All points, straight lines, and planes in the
following axioms are distinct unless otherwise stated.
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
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Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
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The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Euclid never defined the plane in his elements, and indeed never used numbers to measure length, angles, or area. In this sense, the Euclidean plane is not the same as the Cartesian plane. Euclid used a plane to mean a flat, two-dimensional surface that extends infinitely far, independent of numerical values. This definition gives no importance to the origin, as indeed there is no origin.
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
$endgroup$
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
add a comment |
$begingroup$
Euclid never defined the plane in his elements, and indeed never used numbers to measure length, angles, or area. In this sense, the Euclidean plane is not the same as the Cartesian plane. Euclid used a plane to mean a flat, two-dimensional surface that extends infinitely far, independent of numerical values. This definition gives no importance to the origin, as indeed there is no origin.
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
$endgroup$
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
add a comment |
$begingroup$
Euclid never defined the plane in his elements, and indeed never used numbers to measure length, angles, or area. In this sense, the Euclidean plane is not the same as the Cartesian plane. Euclid used a plane to mean a flat, two-dimensional surface that extends infinitely far, independent of numerical values. This definition gives no importance to the origin, as indeed there is no origin.
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
$endgroup$
Euclid never defined the plane in his elements, and indeed never used numbers to measure length, angles, or area. In this sense, the Euclidean plane is not the same as the Cartesian plane. Euclid used a plane to mean a flat, two-dimensional surface that extends infinitely far, independent of numerical values. This definition gives no importance to the origin, as indeed there is no origin.
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
answered Dec 6 '18 at 21:54
shipwaymgshipwaymg
311
311
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
add a comment |
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
The type of this defintion that Euclid used is what I am looking for, except I am looking for a defintion that has actually mathematical meaning rather than just intuitive meaning.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:14
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
$begingroup$
Euclid's axioms and definitions do indeed have mathematical meaning. This was first shown, with full (modern) mathematical rigor, by Hilbert. If you want to see a recent treatment of Euclid to be convinced of this, one good place to start is the book by Hartshorne, "Geometry: Euclid and beyond".
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:54
add a comment |
$begingroup$
I think what you are interested in is exactly the difference between an affine space and a vector space. Read the first few paragraphs in that wikipedia article. Basically, you get rid of the "specialness of the origin" by thinking of two copies of the plane. One represents the group, $G$, of translations and the second is the geometric manifold, $X$, that represents your space. For any two points $x$ and $y$ in your space $X$, there is a unique translation $g$ in $G$ that takes $x$ to $y$. Thus, seen in this way, there are no "special" points in $X$. Notice that the origin is special in $G$ since it represents the identity translation, i.e. the one that keeps $X$ fixed. An even more general concept is that of a principal homogenous space.
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
$endgroup$
– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
add a comment |
$begingroup$
I think what you are interested in is exactly the difference between an affine space and a vector space. Read the first few paragraphs in that wikipedia article. Basically, you get rid of the "specialness of the origin" by thinking of two copies of the plane. One represents the group, $G$, of translations and the second is the geometric manifold, $X$, that represents your space. For any two points $x$ and $y$ in your space $X$, there is a unique translation $g$ in $G$ that takes $x$ to $y$. Thus, seen in this way, there are no "special" points in $X$. Notice that the origin is special in $G$ since it represents the identity translation, i.e. the one that keeps $X$ fixed. An even more general concept is that of a principal homogenous space.
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
$endgroup$
$begingroup$
Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
$endgroup$
– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
add a comment |
$begingroup$
I think what you are interested in is exactly the difference between an affine space and a vector space. Read the first few paragraphs in that wikipedia article. Basically, you get rid of the "specialness of the origin" by thinking of two copies of the plane. One represents the group, $G$, of translations and the second is the geometric manifold, $X$, that represents your space. For any two points $x$ and $y$ in your space $X$, there is a unique translation $g$ in $G$ that takes $x$ to $y$. Thus, seen in this way, there are no "special" points in $X$. Notice that the origin is special in $G$ since it represents the identity translation, i.e. the one that keeps $X$ fixed. An even more general concept is that of a principal homogenous space.
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
$endgroup$
I think what you are interested in is exactly the difference between an affine space and a vector space. Read the first few paragraphs in that wikipedia article. Basically, you get rid of the "specialness of the origin" by thinking of two copies of the plane. One represents the group, $G$, of translations and the second is the geometric manifold, $X$, that represents your space. For any two points $x$ and $y$ in your space $X$, there is a unique translation $g$ in $G$ that takes $x$ to $y$. Thus, seen in this way, there are no "special" points in $X$. Notice that the origin is special in $G$ since it represents the identity translation, i.e. the one that keeps $X$ fixed. An even more general concept is that of a principal homogenous space.
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
answered Dec 6 '18 at 22:13
zoidbergzoidberg
1,065113
1,065113
$begingroup$
Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
$endgroup$
– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
add a comment |
$begingroup$
Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
$endgroup$
– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
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– Daniel Schepler
Dec 6 '18 at 22:43
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Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
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– LinearGuy
Dec 6 '18 at 22:17
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Yes, I am more or less looking for a defintion that resembles and affine space, but, as far as I know, one still uses $mathbb R times mathbb R$ and then lets the group act on it.
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– LinearGuy
Dec 6 '18 at 22:17
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
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– zoidberg
Dec 6 '18 at 22:22
$begingroup$
OK, well if you already know what an affine space is, I don't know what else can satisfy you. In general, I think it's difficult to define specific "non-special things" since the very act of defining it privileges some choice. The way to get at the "generic" is by looking at all the possibilities at once. For instance, how do you get rid of the special "north pole" point on a sphere? By imagining the set of all rotations you can perform on it, i.e. all the symmetries.
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– zoidberg
Dec 6 '18 at 22:22
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
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– David K
Dec 6 '18 at 22:24
$begingroup$
@LinearGuy As far as I can see, the group $G$ is isomorphic to $mathbb Rtimesmathbb R$ but there is no need for $X$ to have intrinsic coordinates. Of course, once you declare that some point in $X$ has coordinates $(x_1,x_2)$ then all the points have coordinates and exactly one has coordinates $(0,0)$--so don't do that.
$endgroup$
– David K
Dec 6 '18 at 22:24
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
$begingroup$
Furthermore, by viewing $mathbb{R} times mathbb{R}$ as a vector space (and often as an oriented inner product space) you get rid of the "specialness" of vectors pointing "along the $x$-axis or along the $y$-axis" since then no one (orthonormal) basis of $mathbb{R}^2$ is any better than another. This would reflect the isotropy (direction independence / rotation invariance) of Euclidean-style geometry, just as passing to the affine space reflects the homogeneity (location independence / translation invariance).
$endgroup$
– Daniel Schepler
Dec 6 '18 at 22:43
add a comment |
$begingroup$
Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert
in 1899 in his book Grundlagen der Geometrie
Hilbert's axiom system is constructed with six primitive notions:
three primitive terms:[5]
point;
line;
plane;
and three primitive relations:[6]
Betweenness, a ternary relation linking points;
Lies on (Containment), three binary relations, one linking points and straight lines, one linking points and planes, and one linking
straight lines and planes;
Congruence, two binary relations, one linking line segments and one linking angles, each denoted by an infix ≅.
Line segments, angles, and triangles may each be defined in terms of
points and straight lines, using the relations of betweenness and
containment.[how?] All points, straight lines, and planes in the
following axioms are distinct unless otherwise stated.
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
add a comment |
$begingroup$
Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert
in 1899 in his book Grundlagen der Geometrie
Hilbert's axiom system is constructed with six primitive notions:
three primitive terms:[5]
point;
line;
plane;
and three primitive relations:[6]
Betweenness, a ternary relation linking points;
Lies on (Containment), three binary relations, one linking points and straight lines, one linking points and planes, and one linking
straight lines and planes;
Congruence, two binary relations, one linking line segments and one linking angles, each denoted by an infix ≅.
Line segments, angles, and triangles may each be defined in terms of
points and straight lines, using the relations of betweenness and
containment.[how?] All points, straight lines, and planes in the
following axioms are distinct unless otherwise stated.
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
add a comment |
$begingroup$
Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert
in 1899 in his book Grundlagen der Geometrie
Hilbert's axiom system is constructed with six primitive notions:
three primitive terms:[5]
point;
line;
plane;
and three primitive relations:[6]
Betweenness, a ternary relation linking points;
Lies on (Containment), three binary relations, one linking points and straight lines, one linking points and planes, and one linking
straight lines and planes;
Congruence, two binary relations, one linking line segments and one linking angles, each denoted by an infix ≅.
Line segments, angles, and triangles may each be defined in terms of
points and straight lines, using the relations of betweenness and
containment.[how?] All points, straight lines, and planes in the
following axioms are distinct unless otherwise stated.
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
Hilbert's axioms are a set of 20 assumptions proposed by David Hilbert
in 1899 in his book Grundlagen der Geometrie
Hilbert's axiom system is constructed with six primitive notions:
three primitive terms:[5]
point;
line;
plane;
and three primitive relations:[6]
Betweenness, a ternary relation linking points;
Lies on (Containment), three binary relations, one linking points and straight lines, one linking points and planes, and one linking
straight lines and planes;
Congruence, two binary relations, one linking line segments and one linking angles, each denoted by an infix ≅.
Line segments, angles, and triangles may each be defined in terms of
points and straight lines, using the relations of betweenness and
containment.[how?] All points, straight lines, and planes in the
following axioms are distinct unless otherwise stated.
https://en.wikipedia.org/wiki/Hilbert%27s_axioms
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 6 '18 at 22:35
Ethan BolkerEthan Bolker
42.4k549112
42.4k549112
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
add a comment |
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
Yes, but he never defined the plane and only uses the intuition of it.
$endgroup$
– LinearGuy
Dec 6 '18 at 22:38
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
$begingroup$
The lines and points are the plane. With these axioms you can derive all the things Euclid proved, with no appeal to intuition, except to guess what you will be able to prove.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 0:43
add a comment |
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$begingroup$
$(0,0)$ is a couple as all others elements of $mathbb{R}times mathbb{R}$. Why you say that this couple play a special rule?
$endgroup$
– Emilio Novati
Dec 6 '18 at 21:17
$begingroup$
What axioms are you willing to start from? From your question, and from your comments to the various answers, it appears that you are not willing to start from the axioms of the real numbers, nor are you willing to start from the axioms of Euclidean geometry, both of which are fully rigorous axiomatic systems from the modern mathematical point of view. All of modern mathematics starts from some axiomatic system. So, what axiomatic system are you willing to start from?
$endgroup$
– Lee Mosher
Dec 10 '18 at 19:57