Download MAT 3271: Selected solutions to problem sets 2 and 3 Chapter 1

MAT 3271: Selected solutions to problem sets 2 and 3
Chapter 1, Exercises:
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7. To show that AB ∪ BA ⊆ AB is easy; the definition of ray says in general that the points of ray
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P Q lie on line P Q . If it didn’t say that, we’d need an axiom saying that if a point is between
two other points, then all three points are collinear. We will indeed state such an axiom.
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Many of you missed the more subtle part of this problem, which is to show that AB ⊆ AB ∪ BA .
I will describe in class a model using spherical geometry in which this is not true! The point is,
how do you prove that every point on the line is in one ray or the other? To do this from the
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definition of ray, we need to know that any point on line AB not equal to A or B must be in some
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“betweenness relationship” with A and B: that is, if P lies on AB , then P ∗ A ∗ B or A ∗ P ∗ B or
A ∗ B ∗ P . Each case then fits the definition of one ray or the other (or both). (It is also necessary
to know that P ∗ Q ∗ R means the same as R ∗ Q ∗ P .)
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To prove that AB ⊆ AB ∩ BA is, of course, easy since the segment is a subset of each ray by
definition. (Here we again need to know that P ∗ Q ∗ R means the same as R ∗ Q ∗ P in order to
show that AB = BA.) To show the converse inclusion requires that P ∗ A ∗ B and A ∗ B ∗ P be
mutually exclusive, so that only A ∗ P ∗ B is possible if P 6= A and P 6= B. Thus, we need to
strengthen our axiom for three (distinct) collinear points P , Q, and R to say that exactly one of
P ∗ Q ∗ R, Q ∗ R ∗ P , or R ∗ P ∗ Q holds.
Carefully thinking about these questions allows us to anticipate the betweenness axioms, which
will be introduced in Chapter 3!
8. (i) ⇒ (ii) follows from Euclid’s fourth postulate (which we will actually prove as a theorem):
all right angles are congruent. (ii) ⇒ (i) requires the assumption that the angle sum of a convex
quadrilateral is 360◦ . (This assumption is a theorem of Euclidean geometry, but not non-euclidean
geometry.) Of course, we also need to establish that the measure of a right angle is 90◦ . (i) ⇒ (iii)
follows from the theorem that, if the sum of the interior angles on one side of a transversal is 180◦ ,
then the lines are parallel. (This theorem is valid in both Euclidean and hyperbolic geometry.)
(iii) ⇒ (i) requires the converse of that theorem: if two parallel lines are cut by a transversal, then
the the sum of the interior angles on one side of a transversal is 180◦ . (The converse is a theorem
of Euclidean geometry but not hyperbolic geometry.) Note that we do not have to separately
prove that (ii) ⇔ (iii), since implication is transitive. (Why? Prove it!)
9. (a) No.
(b) No.
13. There are many circles to choose from for calculating the ratio of circumference to diameter, so in
order to legitimately define π as this ratio, we would need to prove it is the same for all circles.
[Note: In order to do that, we would need to define the circumference as the limit of perimeters
of approximating polygons, so calculus concepts are involved!]
14. In my view, the constitution cannot be considered a set of axioms because its language is imprecise
and open to substantial interpretation. Indeed, the courts have changed their interpretations of
constitutional provisions over the years (as many of you pointed out). This flexibility is necessary
to accommodate the complexity and evolving nature of human society.
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Chapter 2, Exercises:
3. There exist two right angles that are not congruent.
4. There exist a line l and a point P not incident with l such that there is no line through P that is
parallel to l or there exist two distinct lines through P that are parallel to l.
5. (a) If a transversal t to lines l and m cuts out congruent alternate interior angles, then l and m
are parallel. [Note: ”cuts out” and ”alternate interior angles” will have to be defined.]
(b) If lines l and m meet on one side of a transversal t, then the sum of the interior angles on
that side of the transversal is less than 180◦ .
6. Your proofs may vary from the ones below and still be correct.
Proposition 2.1. If l and m are distinct lines that are not parallel, the l and m have a unique point
in common.
Proof. Since l and m are not parallel, by definition they have a point of intersection, call it P . Suppose
l and m also intersect at a point Q distinct from P . Then by Incidence Axiom 1 (uniqueness part),
l = m, contradicting the hypothesis that they are distinct. Thus l and m have a unique point of
intersection.
Proposition 2.2. There exist three distinct lines that are not concurrent.
Proof. By Incidence Axiom 3, there exist three non-collinear points, call them P , Q and R. By Incidence
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Axiom 1, there exist lines P Q , QR , and RP . These three lines are distinct, by the definition of
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non-collinear. Since QR and RP intersect at R, and since their point of intersection is unique by
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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.
Proposition 2.3. For every line, there is at least one point not lying on it.
Proof. Let l be a line. Consider the three non-collinear points given by Incidence Axiom 3. By definition,
they cannot all lie on l. Thus there is a point not lying on l.
Proposition 2.4. For every point, there is at least one line not passing through it.
Proof. Let P be a point. By Proposition 2.2, there are three lines that are not concurrent. By definition,
they cannot all pass through P . So there is a line not passing through P .
Proposition 2.5. For every point P there exist at least two lines through P .
Proof. Let P be a point, and consider once again the three non-collinear points, let’s call them Q, R
and S, given by Incidence Axiom 3. (Note that it is possible that P is one of these points, and possible
that it isn’t.)
Case 1: P is one of the points Q, R or S. Without loss of generality, suppose P = Q. Then lines
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P R and P S (given by Incidence Axiom 1) are distinct, by the definition of non-collinearity.
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Case 2: Consider the lines P Q , P R , and P S . 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.
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