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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.

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mat 360
MAT 360
  • Lecture 10 Hyperbolic Geometry
what is the negation of hilbert s axiom
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
Hyperbolic axiom
  • There exist a line l and a point P not in l such that at least two parallels to l pass through P.
lemma
Lemma:
  • If the hyperbolic axiom holds (and all the axioms of neutral geometry hold too) then rectangles do not exist.
proof of lemma
Proof of lemma
  • 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.
recall
Recall:
  • 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)
universal hyperbolic theorem
Universal Hyperbolic 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.
theorem
Theorem
  • If hyperbolic axiom holds then all triangles have angle sum strictly smaller than 180.
  • Can you prove this theorem?
recall1
Recall
  • Definition: Two triangles are similar if their vertices can be put in one-to-one correspondence so that the corresponding angles are congruent.
similar triangles
Similar triangles
  • 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?
theorem1
Theorem
  • 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 have angles 22 5 degrees 60 degrees and 90 degrees

Therefore,

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

corollary
Corollary
  • In hyperbolic geometry, there exists an absolute unit of length.
definition
Definition
  • Quadrilateral □ABCD is a Saccheri quadrilateral if
    • Angles <A and <B are right angles
    • Sides DA and BC are congruent
    • The side CD is called the summit.
lemma1
Lemma
  • In a Saccheri quadrilateral □ABCD, angles <C and <D are congruent

18

definition1
Definition
  • 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
Question
  • Question: If l and m are parallel lines, and A and B are points in l, are A and B equidistant from m?
theorem2
Theorem
  • 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.
lemma2
Lemma
  • 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
theorem3
Theorem
  • 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’.
theorem4
Theorem
  • 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’.
hyperbolic geometry exercises
Hyperbolic Geometry Exercises
  • 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?
theorem5
Theorem

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.
crossbar theorem recall
Crossbar theorem (Recall)
  • If the ray AD is between rays AC and AB then AD intersects segment BC
dedekind s axiom
Dedekind’s Axiom
  • 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.
definition2
Definition
  • 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.
question1
Question
  • Given a line l, are the “angles of parallelism” associated to this line, congruent to each other?
theorem6
Theorem
  • 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.
definition3
Definition
  • 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.
janos bolyai
Janos Bolyai
  • I can’t say nothing except this: that out of nothing I have created a strange new universe.
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