Math 102A Hw 2
P.92
5 (2 points)
Prop. 2.2 By IA3, there exist three non-collinear points, call them P , Q and
R. By IA1, there exist lines P Q, QR, and RP . These three lines are distinct, by
the definition of non-collinear. Further, these three lines are not parallel, since
there exists at least one point common to each pair chosen among the three
lines. Since QR and RP intersect at R, and since their point of intersection is
unique by Proposition 2.1, the three lines are concurrent only if line P Q passes
through R. It does not, again because P , Q, and R are not collinear.
6 (2 point)
a Prop 2.3 By IA3, there exist three non-collinear points, call them P , Q
and R. Let l be a line. Then at least one of P , Q or R is not incident with l, since
P , Q, and R all coincident with l contradicts the fact that they are non-collinear.
b Prop. 2.4 Let P be any point. By Proposition 2.2, there exist three
distinct lines that are not concurrent. P is incident with at most two of these
(in the case where P is their unique common point). Therefore, at least one of
these three lines does not pass through P .
c Prop. 2.5 Let P be a point, and consider once again the three noncollinear points, lets call them Q, R and S, given by IA3.
Case 1: P is one of the points Q, R or S. Without loss of generality, suppose
P = Q. Then lines P R and P S (given by IA1) are distinct, by the definition of
non-collinearity.
Case 2: P is distict from the points Q, R and S. Consider the lines P Q, P R,
and P S (given by IA1). At least two of these lines must be distinct, because if
they are all the same line, then Q, R, and S would all lie in it, again contradicting the fact that they are non-collinear.
7 (2 point)
a IA1 and IA2 Hold, not IA3 Let the set of points be {A, B} and the set
of lines be {{A, B}}, with incidence given by set membership. In other words,
there is one line with two points, A and B, lying on it.
A
B
b IA1 and IA3 Hold, not IA2 Let the set of points be {A,B,C} and the
set of lines be {{A, B}, {B, C}, {C, A}, {B}}. In other words, there are three
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non-collinear points A, B, and C, and in addition to the lines required by IA1,
thereis a line with one point (B) on it.
B
A
C
c IA2 and IA3 Hold, not IA1 Let the set of points be {A,B,C}. In other
words, there are only three non-collinear points A, B, and C with no lines at
all (here IA2 is vacuously satisfied).
B
A
C
8 (2 points)
First let us verify that the 4-point model of Example 3 satisfies the incidence
axioms. We can do this in general, without listing every case, if we note that
the set of lines is exactly the set of all subsets of the set of points having exactly
two elements.
For IA1, let P and Q be any two distinct points (chosen from the set {A,
B, C, D}). The set {P, Q} is a line incident with both these points, and there
is no other line incident with both of them because any other set with exactly
two elements would have to exclude either P or Q.
IA2 is satisfied because every line is a set of two elements.
IA3 is satisfied because there is no line with three elements, so any three
distinct points are non-collinear.
To prove the Euclidean Parallel Property, let l be any line and let P be
a point not on l. By definition (of line and incidence in this interpretation),
l = {R, S} for two distinct points R and S, chosen from the set {A, B, C, D},
each of which is also distinct from P . Thus there remains exactly one point, Q,
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distinct from P , R, and S, and {P, Q} is the unique line through P and parallel
to l.
For the 5-point model of Example 4, the proofs that the incidence axioms
hold are the same. To prove the Hyperbolic Parallel Property, let l be any line
and let P be a point not on l. As in the previous model, l = {S, T } for two distinct points S and T , chosen from the set {A, B, C, D, E}, each of which is also
distinct from P . In this model, in which the set of points has five elements,there
remain two distinct points, Q and R, each of which is distinct from P , S, and
T . Thus {P Q} and {P R} are distinct lines through P that are parallel to l.
[Note that simply showing that a given line has more than one line parallel to
it is not sufficient to prove the Hyperbolic Parallel Property; the lines must be
shown to pass through a common point.
9 (2 points)
a Only IA2 and IA3 hold. None of the 3 parallel properties hold.
b All three axioms hold. Elliptic.
c All three axioms hold. Hyperbolic.
d All three axioms hold. Elliptic.
10 (2 points)
b
No. Consider the following two models:
Model 1: Let the set of points be {A, B, C, D} and the set of lines be
{{A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}} and ’lies on’ is set membership. Incidence Axioms are readily verified.
Model 2: Let the set of points be {A, B, C, D} and the set of lines be
{{A, B, C}, {A, D}, {B, D}, {C, D}} and ’lies on’ is set membership. Incidence
Axioms are also readily verified.
So these are both Models of Incidence Geometry involving exactly four
’points’. They are not, however, isomorphic to each other, because model 1
has six ’lines’, while model 2 has only four ’lines’. Therefore, it is impossible to
have a one-to-one equivalence of ’lines’.
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c
With out loss of generality let O be the origin. Place the center of the sphere
at the origin. Given a line through the origin, intersecting it with the sphere
gives a pair of anti-podal points. This gives a one-to-one correspondence between lines through the origin and pairs of anti-podal points. Similarly, given
a plane through the origin, intersecting it with the sphere gives a great circle
(equator). Again, this is a one-to-one correspondence between planes through
the origin and great circles. It is clear that these correspondences preserve the
incidence relation. So model 9(b) and 9(d) are isomorphic.
11 (2 points)
Consider the geometry in which points are the elements of the set
{A, B, C, D, E} and lines are the elements of the set {{A, B, C}, {B, D}, {D, E},
{A, D}, {A, E}, {C, D}, {C, E}, {E, B}} and ’lies on’ is set membership. There
are parallel lines so the elliptic property doesnt hold; through D there is exactly
one line ({D, E}) parallel to {A, B, C} so the hyperbolic property doesnt hold;
and {A, E} and {E, B} are incident with E and parallel to {C, D} so the Euclidean property doesnt hold.
17 (1 points)
a Incorrect Argument: no frogs are poetic, gives us that all poetic things
cannot be frogs. Some ducks are unpoetical does not give us that there are
poetic ducks, it could be that all ducks are unpoetical.
b Correct Argument: because gold is heavy, anything that is light cannot be
gold. And since nothing but gold will silence him, nothing light will silence him.
c Incorrect Argument: all lions are fierce, yet one cannot presume that other
creatures that drink coffee are necessarily not fierce, unless the ferocity of these
other creatures has been qualified.
d Incorrect Argument: some pillows are soft does not give us that all pillows
are soft, it might be that all pokers are hard pillows. Or if there is no pokers, the
conclusion does not follow, since it requires a comparison between the softness
of pillows and pokers.
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