1 / 33

# MAT 360 - PowerPoint PPT Presentation

MAT 360 . Lecture 10 Hyperbolic Geometry. What is the negation of Hilbert’s Axiom?. There exists a line l and a point P not on l such that there are at least two parallels to l through p. Hyperbolic axiom.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'MAT 360' - schuyler

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Lecture 10 Hyperbolic Geometry

• There exists a line l and a point P not on l such that there are at least two parallels to l through p.

• There exist a line l and a point P not in l such that at least two parallels to l pass through P.

• If the hyperbolic axiom holds (and all the axioms of neutral geometry hold too) then rectangles do not exist.

• If hyp axiom holds then Hilbert’s parallel postulate does not hold because it is the negation of hyp. Axiom.

• Existence of rectangles implies Hilbert’s parallel postulate

• Therefore, rectangles do not exist.

• In neutral geometry, if two distinct lines l and m are perpendicular to a third line, then l and m are parallel. (Consequence of Alternate Interior Angles Theorem)

• In hyperbolic geometry, for every line l and every point P not in l there are at least two distinct parallels to l passing through p.

• Corollary: In hyperbolic geometry, for every line l and every point P not in l there are infinitely many parallels to l passing through p.

• If hyperbolic axiom holds then all triangles have angle sum strictly smaller than 180.

• Can you prove this theorem?

• Definition: Two triangles are similar if their vertices can be put in one-to-one correspondence so that the corresponding angles are congruent.

• Recall Wallis attempt to “fix” the “problem” of Euclid’s V:

• Add postulate: “Given any triangle ΔABC, and a segment DE there exists a triangle ΔDEF similar to ΔABC”

• Why the words fix and problem are surrounded by quotes?

• In hyperbolic geometry, if two triangles are similar then they are congruent.

• In other words, AAA is a valid criterion for congruence of triangles.

all triangles are congruent

All triangles have angles 22.5 degrees, 60 degrees, and 90 degrees.

The distance

we “see” in this picture is not the one given by Theorem 4.3

• In hyperbolic geometry, there exists an absolute unit of length.

• Angles <A and <B are right angles

• Sides DA and BC are congruent

• The side CD is called the summit.

• In a Saccheri quadrilateral □ABCD, angles <C and <D are congruent

18

• Let l’ be a line.

• Let A and B be points not in l’

• Let A’ and B’ be points on l’ such that the lines AA’ and BB’ are perpendicular to l’

• We say that A and B are equidistant from l’ if the segments AA’ and BB’ are congruent.

• Question: If l and m are parallel lines, and A and B are points in l, are A and B equidistant from m?

• In hyperbolic geometry if l and l’ are distinct parallel lines, then any set of points equidistant from l has at most two points on it.

• The segment joining the midpoints of the base and summit of a Saccheri quadrilateral is perpendicular to both the base and the summit and this segment is shorter than the sides

• In hyperbolic geometry, if l and l’ are parallel lines for which there exists a pair of points A and B on l equidistant from l’ then l and l’ have a common perpendicular that is also the shortest segment between l and l’.

• In hyperbolic geometry if lines l and l’ have a common perpendicular segment MM’ then they are parallel and MM’ is unique. Moreover, if A and B are points on l such that M is the midpoint of AB then A and B are equidistant from l’.

• Show that for each line l there exist a line l’ as in the hypothesis of the previous theorem. Is it the only one?

• Let m and l be two lines. Can they have two distinct common perpendicular lines?

• Let m and n be parallel lines. What can we say about them?

Where in the proof are we using the hyperbolic axiom?

• Let l be a line and let P be a point not on l. Let Q be the foot of the perpendicular from P to l.

• Then there are two unique rays PX and PX’ on opposite sides of PQ that do not meet l and such that a ray emanating from P intersects l if and only if it is between PX and PX’.

• Moreover, <XPQ is congruent to <X’PQ.

• If the ray AD is between rays AC and AB then AD intersects segment BC

• Suppose that the set of all points on a line is the disjoint union of S and T,

• S U T

• where S and T are of two non-empty subsets of l such that no point of either subsets is between two points of the other. Then there exists a unique point O on l such that one of the subsets is equal to a ray of l with vertex O and the other subset is equal to the complement.

• Let l be a line and let P be a point not in l.

• The rays PX and PX’ as in the statement of the previous theorem are called limiting parallel rays.

• The angles <XPQ and X’PQ are called angles of parallelism.

• Given a line l, are the “angles of parallelism” associated to this line, congruent to each other?

• Given lines l and m parallel, if m does not contain a limiting parallel ray to l then there exist a common perpendicular to l and m.

• Let l and m be parallel lines.

• If m contains a limiting parallel ray (to l) then we say that l and m are asymptotic parallel.

• Otherwise we say that l and m are divergently parallel.

• I can’t say nothing except this: that out of nothing I have created a strange new universe.