[6] Use Hilbert's axioms of incidence and betweenness to complete the proof of the following theorem. A picture has been provided for your convenience, but be careful not to rely on it. Theorem: If A and B are any two distinct points, then there exists a point X between A and B. Hypothesis: Points A and B. Conclusion: There exists a point X so that A-X-B. Statements 1) Points A and B 2) There exists a point E not on AB 3) There exists a point F so that A-E-F 4) There exists a point G so that F-B-G 5) EG intersects BF at G, hence cannot intersect at another point 6) EG contains a point X between A and B PROOF 1) Hypothesis 2) 3) 4) 5) 6) Reasons EX B

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question

Q20. 

[6] Use Hilbert's axioms of incidence and betweenness to complete the proof of the following
theorem. A picture has been provided for your convenience, but be
careful not to rely on it.
Theorem: If A and B are any two distinct points, then there
exists a point X between A and B.
Hypothesis: Points A and B.
Conclusion: There exists a point X so that A-X-B.
Statements
1) Points A and B
2) There exists a point E not on AB
3) There exists a point F so that A-E-F
4) There exists a point G so that F-B-G
5) EG intersects BF at G, hence cannot
intersect at another point
6) EG contains a point X between A and B.
PROOF
1) Hypothesis
2)
(3)
4)
5)
Reasons
B
G
Transcribed Image Text:[6] Use Hilbert's axioms of incidence and betweenness to complete the proof of the following theorem. A picture has been provided for your convenience, but be careful not to rely on it. Theorem: If A and B are any two distinct points, then there exists a point X between A and B. Hypothesis: Points A and B. Conclusion: There exists a point X so that A-X-B. Statements 1) Points A and B 2) There exists a point E not on AB 3) There exists a point F so that A-E-F 4) There exists a point G so that F-B-G 5) EG intersects BF at G, hence cannot intersect at another point 6) EG contains a point X between A and B. PROOF 1) Hypothesis 2) (3) 4) 5) Reasons B G
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