The axioms of incidence geometry are: I1 There exist three distinct noncollinear

Question

The axioms of incidence geometry are: I1 There exist three distinct noncollinear points. I2 For any two distinct points, there is at least one line that contains them both. I3 For any two distinct points, there is at most one line that contains them both. I4 Every line contains at least two distinct points. The axioms of projective geometry are: P1 For any two distinct points, there is a unique line that contains them both. P2 For any two distinct lines, there is a unique point that lies on them both. P3 There exist four distinct points, no three of which are collinear. Problems 1. (a) Assume the axioms of projective geometry. Prove the axioms of incidence geometry. The only one that is not immediate is I4. (b) Find a model of incidence geometry that is not a model of projective geometry, and explain which projective axiom(s) it fails.

Explanation / Answer

The axioms of incidence geometry are:
I1 There exist three distinct noncollinear points.
I2 For any two distinct points, there is at least one line that contains them both.
I3 For any two distinct points, there is at most one line that contains them both.
I4 Every line contains at least two distinct points.
The axioms of projective geometry
are:
P1 For any two distinct points, there is a unique line that contains them both.
P2 For any two distinct lines, there is a unique point that lies on them both.
P3 There exist four distinct points, no three of which are collinear.

Problem 1:

a) We assume axioms of projective geometry.
From P1,we can assure that between 2 points there is a unique line containing both ,ie there is at most one line that contain both.Also this imply there is at least one line(that is at most here,as its unique) that contain both
Hence I3 and I2 respectively.
Now,assume we have two distinct lines(say l1 and l2).From P2,we have there will be unique point(say O) which line on it.Now take 2 different points(A,B say) one on l1 and other on l2.Now,from P1 ,we can there is unique line passing through A and O,say l3,hence we must have l3=l1 otherwise its not unique.
Similar way,we can prove line passing through B and O is also unique hence, there points A,B,O are unique.(l1 proved)
Now, we can prove I4 as :If we prove through a single point no unique line exist,and if we take two it becomes unique.
Now we know from P2 that any 2 distinct lines(l1,l2 say) has unique point(say O) lies on both,,so this says 2 lines pass through one point ,ie though single point no unique line der.Now Take a point A on l1,Hence using P1 we have a unique line passing through A and O.So line(l1) has 2 distinct points(A,O).We showed "through a single point no unique line exist,and if we take two it becomes unique."hence l4.

b)

It is impossible in incidence geometry to prove that parallel lines exist.Also we cannot prove any two lines meet(even parallel lines!).We can prove using projective planes.Two parallel lines meet at point at infinity in projective geometry.

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