# Proving axiom of incidence

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• Aug 30th 2006, 08:44 AM
OReilly
Proving axiom of incidence
These are the axioms of incidence from the book that I have:

I.1: Every line contains at least two distinct points.
I.2: There exists at least one line that contains two points.
I.3: There exists at most one line that containts two distinct points.
I.4: Every plane contains at least three non-colinear points.
I.5: There exists at least one plane that contains three points.
I.6: There exists at most one plane that contains three non-colinear points.
I.7: If two distinct points of some line belongs to one plane, then every point of that line belongs to same plane.
I.8: If two distinct planes have one common point, then they have at least one more common point.
I.9: There are four non-coplanar points.

I have to prove this:
"If we substitute axiom I.4 by axiom I.4': 'Every plane contains at least one point', then prove that consequence from given group of axioms (which I have typed) is axiom I.4"

My proof goes like this:
From I.9 we have three non-colinear points A,B,C which forms one plane $\alpha$ by axiom I.6 and one point D not contained by plane $\alpha$.
Any two points of plane $\alpha$ can form with point D another plane $\beta$.
Any plane different then $\alpha$ and $\beta$ must intersect one of them. By axiom I.4' that plane contains at least one point and by intersecting one of two planes $\alpha$ or $\beta$ it has two more distinct points by axiom I.8 which means that every plane has three non-colinear points.

Is my proof correct?
• Aug 30th 2006, 10:16 AM
ThePerfectHacker
Quote:

Originally Posted by OReilly
My proof goes like this:
From I.9 we have three non-colinear points A,B,C which forms one plane $\alpha$ by axiom I.6 and one point D not contained by plane $\alpha$.

This is what I do not understand.
You can find 4 points A B C D that are non-coplanar. But then you state you can you can draw a plane from among three of them such that the fourth is not contained. The second part (4th not being contained is understood). But the first part is not (how can you draw a plane through these 3?)
• Aug 30th 2006, 11:09 AM
OReilly
Quote:

Originally Posted by ThePerfectHacker
This is what I do not understand.
You can find 4 points A B C D that are non-coplanar. But then you state you can you can draw a plane from among three of them such that the fourth is not contained. The second part (4th not being contained is understood). But the first part is not (how can you draw a plane through these 3?)

I have missed out word "three" in axiom I.6: There exists at most one plane that contains THREE non-colinear points.
• Aug 30th 2006, 12:11 PM
Plato
Even with that correction, this is still a poor collection of axioms.
There is not way to conclude that given three non-collinear points there is a plane that contains them. There statement that “There exists at most one plane that contains THREE non-collinear points.” does not insure the plane exists. It simply says that given three non-collinear points there is at most one plane that contains them. But it does not grantee the existence of such a plane.
• Aug 30th 2006, 12:23 PM
OReilly
Quote:

Originally Posted by Plato
Even with that correction, this is still a poor collection of axioms.
There is not way to conclude that given three non-collinear points there is a plane that contains them. There statement that “There exists at most one plane that contains THREE non-collinear points.” does not insure the plane exists. It simply says that given three non-collinear points there is at most one plane that contains them. But it does not grantee the existence of such a plane.

What do you mean by "poor collection of axioms"?

These are all Hilbert's axioms.

Existence of a plane is guarantied by axiom I.5: There exists at least one plane that contains three points.
Axiom I.6 guaranties uniqenes of plane that contains three non-colinear points.
• Aug 30th 2006, 01:04 PM
OReilly
Quote:

Originally Posted by OReilly
These are the axioms of incidence from the book that I have:

I.1: Every line contains at least two distinct points.
I.2: There exists at least one line that contains two points.
I.3: There exists at most one line that containts two distinct points.
I.4: Every plane contains at least three non-colinear points.
I.5: There exists at least one plane that contains three points.
I.6: There exists at most one plane that contains three non-colinear points.
I.7: If two distinct points of some line belongs to one plane, then every point of that line belongs to same plane.
I.8: If two distinct planes have one common point, then they have at least one more common point.
I.9: There are four non-coplanar points.

I have to prove this:
"If we substitute axiom I.4 by axiom I.4': 'Every plane contains at least one point', then prove that consequence from given group of axioms (which I have typed) is axiom I.4"

My proof goes like this:
From I.9 we have three non-colinear points A,B,C which forms one plane $\alpha$ by axiom I.6 and one point D not contained by plane $\alpha$.
Any two points of plane $\alpha$ can form with point D another plane $\beta$.
Any plane different then $\alpha$ and $\beta$ must intersect one of them. By axiom I.4' that plane contains at least one point and by intersecting one of two planes $\alpha$ or $\beta$ it has two more distinct points by axiom I.8 which means that every plane has three non-colinear points.

Is my proof correct?

I think I have a better proof.
From I.9 we have three non-colinear points A,B,C which forms one plane $\alpha$ by axiom I.6 and one point D not contained by plane $\alpha$.
Any two points of plane $\alpha$ can form with point D another plane so we can form two planes ABD and BCD so we have totally three planes with ABC.
Any plane must intersect at least two of those planes so they would have two common points with both of them which proves that every plane have at least three non-colinear points.
• Aug 30th 2006, 01:42 PM
Plato
Quote:

Originally Posted by OReilly
These are all Hilbert's axioms.
Existence of a plane is guarantied by axiom I.5: There exists at least one plane that contains three points.
Axiom I.6 guaranties uniqenes of plane that contains three non-colinear points.

Those are most definitely not Hilbert’s axioms.
The best translation of Hilbert’s axioms can be found in A Survey of Geometry by Howard Eves. A somewhat looser and more modern translation can be found in Modern Geometries by James R. Smart. Marvin Jay Greenberg has a thirty page discussion of Hilbert’s axioms in his book Euclidean and Non-Euclidean Geometries.

Hilbert’s Axioms:
Group I: Postulates of connection.
I-1 There is one and only one line passing through ant two given and distinct points.
I-2 Every line contains at least two distinct points, and for any given line there is at least one point not on the line.
I-3 Three non-collinear points a unique plane.
I-4 If two points A & B of a line a are in a plane alpha, then a is a subset of alpha.
I-5 If two planes alpha and beta have a point A in common then they have at least a second point B in common.
I-6 In every plane there exist at least three non collinear points, and in space there exist at least four non-coplanar points.

Actually R.L. Moore showed that Hilbert’s Axioms could be reduced in number. This was done as part of his 1905 PhD thesis written with Oswald Veblen and E. H. Moore.
• Aug 30th 2006, 02:02 PM
OReilly
Quote:

Originally Posted by Plato
Those are most definitely not Hilbert’s axioms.
The best translation of Hilbert’s axioms can be found in A Survey of Geometry by Howard Eves. A somewhat looser and more modern translation can be found in Modern Geometries by James R. Smart. Marvin Jay Greenberg has a thirty page discussion of Hilbert’s axioms in his book Euclidean and Non-Euclidean Geometries.

Hilbert’s Axioms:
Group I: Postulates of connection.
I-1 There is one and only one line passing through ant two given and distinct points.
I-2 Every line contains at least two distinct points, and for any given line there is at least one point not on the line.
I-3 Three non-collinear points a unique plane.
I-4 If two points A & B of a line a are in a plane alpha, then a is a subset of alpha.
I-5 If two planes alpha and beta have a point A in common then they have at least a second point B in common.
I-6 In every plane there exist at least three non collinear points, and in space there exist at least four non-coplanar points.

Actually R.L. Moore showed that Hilbert’s Axioms could be reduced in number. This was done as part of his 1905 PhD thesis written with Oswald Veblen and E. H. Moore.

I don't see inconsistency with group of axioms that I have typed with Hilbert's axioms that you have typed. They are the same, only difference is that my axioms are somewhat partitioned more.
• Aug 30th 2006, 02:51 PM
Plato
Quote:

Originally Posted by OReilly
I don't see inconsistency with group of axioms that I have typed with Hilbert's axioms that you have typed. They are the same, only difference is that my axioms are somewhat partitioned more.

Are you serious? You can no difference?
In the axioms you stated the is absolutely now way to be assured that given two points there exist a line that contains them. To do that you would have to change the following: “I.3: Given any two distinct points, there exists one and only one line that contains them.”

As TPH pointed out to you the same can be said about three non-collinear points determining a plane. To do that you must change “I.6: There exists one and only one plane that contains three non-collinear points.”

The quantifier “There exists at most one” logically means ‘none or one’! It does not imply existence.
• Aug 30th 2006, 03:29 PM
OReilly
Quote:

Originally Posted by Plato
Are you serious? You can no difference?
In the axioms you stated the is absolutely now way to be assured that given two points there exist a line that contains them. To do that you would have to change the following: “I.3: Given any two distinct points, there exists one and only one line that contains them.”

As TPH pointed out to you the same can be said about three non-collinear points determining a plane. To do that you must change “I.6: There exists one and only one plane that contains three non-collinear points.”

The quantifier “There exists at most one” logically means ‘none or one’! It does not imply existence.

You are looking at isolated axioms.
For line first you must look at I.2: "There exists at least one line that contains two points" which insures existence of line for given two points.
I.3 is just insuring uniquenes of line.
Same goes for plane. First look at I.5 then at I.6.
• Aug 30th 2006, 03:54 PM
Plato
Have you done a course in mathematical logic?
The sentence "There exists at least one line that contains two points" means that there exist a line and it contains two points. It is a simple matter of quantification.

One the other hand, if you want to say that two points determine exactly one line then we can do it with one or two statements. In keeping with your list we will do with two.
1. Given two points there is a line that contains both.
2. Any two points are in at most one line.
• Aug 30th 2006, 04:01 PM
OReilly
Quote:

Originally Posted by Plato
Have you done a course in mathematical logic?
The sentence "There exists at least one line that contains two points" means that there exist a line and it contains two points. It is a simple matter of quantification.

One the other hand, if you want to say that two points determine exactly one line then we can do it with one or two statements. In keeping with your list we will do with two.
1. Given two points there is a line that contains both.
2. Any two points are in at most one line.

Your statement "1. Given two points there is a line that contains both." is exactly what axiom I.2 is all about, and your statement 2. is axiom I.3!

I.2 has assumption that there are given two points and there is at least one line that contains them.
• Aug 30th 2006, 04:10 PM
Plato
You really don't understand quantification do you?
• Aug 30th 2006, 04:36 PM
OReilly
Quote:

Originally Posted by Plato
You really don't understand quantification do you?

I don't understand you!

First you are saying that "There exists at least one line that contains two points" means that there exist a line and it contains two points and then you said that it must be 1. Given two points there is a line that contains both..

You said "there exist a line" and then "there is a line" ! It's the same!!!

Do you think that "There exists at least one line that contains two points" means that every line contains two points? That is not what I.2 is about. I.1 is about how many points line contains.
• Aug 31st 2006, 09:57 AM
topsquark
Quote:

Originally Posted by OReilly
I.1: Every line contains at least two distinct points.
I.2: There exists at least one line that contains two points.
I.3: There exists at most one line that contains two distinct points.

I think I know what this is about, but some statements are poorly made. I hope I can explain clearly what I am seeing here.

I.1 Every line contains at least two distinct points.
I take this as a given: in fact a line is made from a continuum of distinct points, so a line contains an infinite number of them.

Now look at I.3. I don't agree with the statement as it is written. I think what it is trying to say is "Given two distinct points there exists at most one line containing these points." This is the only statement I can write that doesn't either repeat or contradict I.1.

However, I don't understand why I.2 is there at all. I can't find any reason for it. If I am correct in my guess about I.3 we can simply rewrite I.3 (or I.2) to explicitly state "Given two distinct points there exists exactly one line containing these points." This seems to me to be a much more explicit axiom, it replaces two of the given axioms by one, and we would want the fewest axioms to work with in building such a logical system. Otherwise all I.2 is stating is that a line exists that contains two points; points which may not even be distinct (distinctness is not mentioned in I.2). This is already taken as a given in I.1. I am open to another interpretation but, as written, this one just looks bad to me.

-Dan
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