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Young's Finite Geometry

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by

Sasha Smith

on 20 November 2013

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Transcript of Young's Finite Geometry

Young's Finite Geometry
The Five Axioms...
Axioms:
1 There exists at least one line.
2 Every line of the geometry has exactly three points on it.
3 Not all points of the geometry are on the same line.
4 For two distinct points, there exists exactly one line on both of them.
5 If a point does not lie on a given line, then there exists exactly one line on that point that does not intersect the given line.

...and a few Theorems
T1 For every point, there is a line not on that point.
T2 For every point, there are exactly 4 lines on that point.
T3 Each line is parallel to exactly 2 lines.
T4 There are three lines, no two of which intersect.
T5 There are exactly 12 lines.
T6 There are exactly 9 points.

By: Sasha Smith
With Young's Geometry, it can be shown that there are exactly 9 points and thus exactly 12 lines.
Old Facts about Young
born in 1773 in Sommerset, England
very intelligent child, learning to read at age 2
1793, studied medicine in London and became a physician.
moved on to studying light -> introduced the original theory of color
In 1816, helped ascertain the precise length of the seconds or seconds pendulum
one of the first to try deciphering Egyptian hieroglyphs by studying the Rosetta Stone
first to define 'energy' in scientific terms
His description of elasticity came to be known as Young's Modulus
Established a finite geometry
Young's Finite Geometry
Finite Geometry consists of any geometric system consisting only of a finite number of points.
Hoo Says these are true??
Proof of T5:
- By A1 there is a line
l
, which is on three points p1, p2, and p3, by A3.
- On each of these points there are exactly three other distinct lines by T2 & A4
- So far we have 10 lines, and there can be no other line which intersects
l
.
- By T3, there are exactly two lines parallel to
l
.
- Any 13th line must either intersect or be parallel to
l
, but this is impossible as noted. Thus, there exist exactly 12 lines.
Proof of T6:
- Let
l
1 (on p1,p2,p3),
l
2 (on p4,p5,p6)), and
l
3 (on p7,p8,p9)), be three lines as in T4.
- Suppose there exists a 10th point, q,
- Then by A4 there is a line m on q and p1.
- Since we already have a line through p1 parallel to
l
2, and a line through p1 parallel to
l
3 (in both cases, it's
l
1), m must be on points of
l
2 and
l
3.
- This would force m to have at least four points, violating A2.
Thus there exists exactly 9 points.
The End!
Full transcript