Talk:Euclidean plane

These properties 1,2,3 are the definition of a metric space. Metric spaces are very, very diverse in their geometric properties. Few examples are given in our "metric space" article; a lot more can be given. The Euclidean plane is just one of them (one up to isometry, I mean). Properties 1,2,3 are a tiny part of its characterization. It looks for me rather like "Homo sapience is a piece of live matter. It has the properties to be born, to eat, to breed an to die." :-) --Boris Tsirelson 07:37, 22 January 2012 (EST)


 * Please add the extra axioms that make a metric space into a Euclidean plane.--Paul Wormer 10:55, 22 January 2012 (EST)


 * I would be glad to do so, but it is a hard way. No wonder that people prefer "2-dimensional affine space with inner product". In principle one can define a straight line in terms of the metric only, and then require the numerous axioms of the 2-dim Euclidean geometry treating straight lines as above. --Boris Tsirelson 13:32, 22 January 2012 (EST)


 * Less elegant but more practical, if you like: Euclidean plane is a metric space whose points admit coordinates (x,y) such that the distance is $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ (and of course, the correspondence between the plane and all pairs of reals must be bijective). --Boris Tsirelson 14:39, 22 January 2012 (EST)


 * I extended the article along these lines.--Paul Wormer 08:02, 24 January 2012 (EST)


 * As far as I understand, a point R belongs to the straight line containing P and Q (these being different) if and only if R satisfies at least one of the following three equalities:
 * &rho;(R,P) + &rho;(P,Q) = &rho;(R,Q);
 * &rho;(P,R) + &rho;(R,Q) = &rho;(P,Q);
 * &rho;(P,Q) + &rho;(Q,R) = &rho;(P,R).
 * But I do not understand what does "$$\lambda \rho(P,R) +(1-\lambda)\rho(R,Q) = \rho(P,Q)$$" mean. (Maybe rather "$$R=\lambda P+(1-\lambda)Q$$"? but this is in terms of affine space rather than metric space.)
 * See also Line (geometry). Boris Tsirelson 11:45, 24 January 2012 (EST)


 * Yes I screwed up. What do you think if I say two equations:
 * &rho;(P,R) = &lambda;&rho;(P,Q)
 * &rho;(R,Q) = (1-&lambda;)&rho;(P,Q)
 * determine a straight line with 0 &le; &lambda; &le; 1? (BTW, I don't know what &lambda;P means.) --Paul Wormer 12:04, 24 January 2012 (EST)


 * Yes, the two equations can be used; but isn't it a longer way to say just "&rho;(P,R) + &rho;(R,Q) = &rho;(P,Q)"?
 * Also, "Note that a part of a line is not a line. In particular, a line segment is not a line." (quoted from Line (geometry))
 * BTW, yes, a linear combination in general is well-defined for points of a linear space, not of an affine space; however, when the sum of coefficients is equal to 1 (so-called affine combination), it is well-defined for points of an affine space. Thus, &lambda;P means nothing, but still, &lambda;P+(1-&lambda;)Q means a point. --Boris Tsirelson 12:31, 24 January 2012 (EST)


 * "It follows from the triangle inequality that this path of minimum length is unique" – really, follows? Here is a counterexample. Consider the four-point metric space {A,B,C,D} with &rho;(A,B)=1, &rho;(A,C)=1, &rho;(B,C)=1, &rho;(B,D)=1, &rho;(C,D)=1, &rho;(A,D)=2. It is a metric space, indeed. However, both B and C are "between" A and D. --Boris Tsirelson 15:13, 25 January 2012 (EST)
 * And here is a better counterexample: just the plane with the metric $$|x_1-x_2|+|y_1-y_2|$$ rather than $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$. There are a lot of shortest paths from (0,0) to (1,1). One of them goes through (1,0); another one through (0,1). --Boris Tsirelson 05:04, 26 January 2012 (EST)
 * I gave up trying to define straight lines. After all, points and angles were defined neither. --Paul Wormer 05:27, 27 January 2012 (EST)