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Incidence geometry in planes and space

by mathstudent88
Tags: geometry, incidence, planes, space
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mathstudent88
#1
Sep7-08, 04:17 PM
P: 27
Consider the system [S, L, P], where S contains exactly four points A, B, C, and D, the lines are the sets with exactly two points, and the planes are sets with exactly three points. This "space" is illustrates by the following figure:



Here it should be remembered that A, B, C, and D are the only points that count. Verify that all the incidence postulates hold in this system.

Incidence Postulates:
I0)All lines and planes are sets of points.
I1) Given any two different points, there is exactly one line containing them.
I2) Given any three different noncollinear points, there is exactly one plane containing them.
I3) If two points lie in a plane, then the line containing them lies in the plane.
I4) If two planes interesect, then their intersection is a line.
I5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.



This is what I have thought about doing so far to prove each incidence postulate but I'm not sure if it is right or not:

I0) This holds becuase the hypothesis states the lines are sets with exactly two points.

I1) This holds because the hypothesis states that the planes have exactly three points in them.

I3) This holds because if A lies in a plane and B lies in the same plane, then the line joining them must lie in the same plane.

I4) This holds because there is an exact point, D in S, which is both a line and a plane.

I5) The first statement hold because it is given in the hypothesis. The second statement holds because we are given A, B, C, D and no more than three are lying on the same line. The last one is satisfied because no four of the points are lying on the same plane.



This is my first class having to prove things, so please bare with me.

Thanks for your help!! :)
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HallsofIvy
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Sep7-08, 08:10 PM
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Quote Quote by mathstudent88 View Post
Consider the system [S, L, P], where S contains exactly four points A, B, C, and D, the lines are the sets with exactly two points, and the planes are sets with exactly three points. This "space" is illustrates by the following figure:



Here it should be remembered that A, B, C, and D are the only points that count. Verify that all the incidence postulates hold in this system.

Incidence Postulates:
I0)All lines and planes are sets of points.
I1) Given any two different points, there is exactly one line containing them.
I2) Given any three different noncollinear points, there is exactly one plane containing them.
I3) If two points lie in a plane, then the line containing them lies in the plane.
I4) If two planes interesect, then their intersection is a line.
I5) Every line contains at least two points. S contains at least three noncollinear points. Every plane contains at least three noncollinear points. And S contains at least four noncoplanar points.



This is what I have thought about doing so far to prove each incidence postulate but I'm not sure if it is right or not:

I0) This holds becuase the hypothesis states the lines are sets with exactly two points.
And because planes are defined as set of 3 points. Why are you using "I" instead of "1"?

I1) This holds because the hypothesis states that the planes have exactly three points in them.
?? 11 says nothing about planes! 11 is true because lines are defined as all possible sets of 2 points.

I3) This holds because if A lies in a plane and B lies in the same plane, then the line joining them must lie in the same plane.
Because "lying in the plane" is defined as "contains only points in that plane".

[/quote]I4) This holds because there is an exact point, D in S, which is both a line and a plane.[/quote]
That makes no sense! There is a point that is both a line and a plane? A point is defined as a single letter. A line is defined as a set of two points while a plane is defined as being a set of three points. No set satisfies both of those and certainly a point is not a set anyway. 14 is true because (1) If two planes had all three points in common they would be exactly the same plane, and (2) there are no planes in the model that have exactly one point in common.

I5) The first statement hold because it is given in the hypothesis. The second statement holds because we are given A, B, C, D and no more than three are lying on the same line. The last one is satisfied because no four of the points are lying on the same plane.
Indeed, a line is defined as a set of 2 points so 3 points can't "lie on a line" which is, in turn, defined as being in that set. Similarly, a plane is defined as a set of 3 points and so cannot have 4 points in it.


This is my first class having to prove things, so please bare with me.

Thanks for your help!! :)
Once again, someone wants us to "bare" with him. I think that's the third time this month. This forum is getting too kinky for me!
mathstudent88
#3
Sep7-08, 09:15 PM
P: 27
Thanks for your help!!

I accidently skipped I2) Given any three different noncollinear points, there is exactly one plane containing them.

Would this be right?
This holds because the hypothesis states that the planes have exactly three points in them.

Thank you!

Hurkyl
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Sep7-08, 09:43 PM
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Incidence geometry in planes and space

Quote Quote by HallsofIvy View Post
And because planes are defined as set of 3 points. Why are you using "I" instead of "1"?
Because he means "I". "I0" is short for "Incidence axiom 0". Hilbert's axioms for Euclidean geometry are usually split up into 5 categories. Wikipedia's names for the categories are incidence, order, parallels, congruence, and continuity, although these aren't universal. (e.g. I seem to recall 'betweenness' instead of 'order') Rather than just number all the axioms from 1 to 20, they are usually labelled by category and index within the category.


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