Download Euclidean Geometry exercise solutions 4 and more Exercises Analytical Geometry and Calculus in PDF only on Docsity!
MAT 141
Homework 4 โ Solutions
Formal Logic: Suppose that P, Q, R, S are true state- ments and decide whether each of the following is true or false. a: โผ (P V Q) => R true b: โผ (P โง Q) =>โผ R true c: (โผ (P โจ Q) => R) => S true d: (โผ (P V Q) => R) => (SV โผ Q) true e: (โผ (P V Q) => R) => (โผ SV โผ Q) false
Consider the two models of incidence geometry discussed in class and show that they fail the Betweenness Axioms.
Model a: Points are A, B, C; lines are l, m, n; l is incident to A, B; m is incident to A, C; n is incident to B, C.
This interpretation is not a model for the Betweenness Axioms. B-2 is violated. For instance, on m, there is no point P between A and C.
Model b: Points are A, B, C, D; lines are l, m, n, o, p, q; l is incident to A, B; m is incident to A, C; n is incident
to A, D; o is incident to B, C; p is incident to B, D; q is incident to C, D.
This interpretation is not a model for the Betweenness Axioms. Again, B-2 is violated. For instance, here too, on m, there is no point P between A and C.
- The Fano plane is the following interpretation of the terms point, line, incidence: Points are A, B, C, D, E, F, G, lines are {A, B, D}, {A, C, G}, {D, E, G}, {B, C, E}, {A, E, F }, {B, F, G}, {D, C, F }. A line is incident to a point and vice versa if the point is one of the triple of points that are the line.
Prove that the Fano plane does not satisfy the Betweenness Axioms.
The Fano plane is not a model for the Betweenness Ax- ioms. Again, B-2 is violated. For instance, here too, on the line {A, C, G}, there is no point P between A and C.
- Prove Propositions 3.5 through 3.9.
You will be quizzed on Prop. 3.
Prop 3.5: Given A โ B โ C. Then AC = AB โช BC and B is the only point common to segments AB and BC.
Proof: (โ) By Proposition 3.4, every point P on the line
Case 2: P is in BC. A similar argument shows that in this case too, P is in AC.
Thus AC = AB โช BC.
To see that B is the only point common to AB and BC, suppose that P is in both AB and BC. Since P is in AB, either P = A, or P = B, or A โ P โ B. (RAA) If P = A, then A is in BC, implying that either A = B, A = C, or B โ A โ C. Since A, B, C are distinct, it follows that BโAโC, which is equivalent to C โ A โ B. Hypothesis A โ B โ C together with Prop. 3.3 then gives C โ A โ C, which is a contradiction. Thus P โ A. A similar argument shows that P โ C. (RAA) If A โ P โ B, then hypothesis A โ B โ C together with Prop. 3.3 gives P โ B โ C. Since P is also in BC and P โ A, the definition of seg- ment gives B โ P โ C. This contradicts Betweenness Axiom B-3. Hence P = B. โ
Prop 3.6: Given A โ B โ C. Then B is the only point
common to rays
รโ
BA and
รโ
BC, and
รโ
AB =
รโ
AC.
Proof: To see that B is the only point common to
รโ
BA and รโ BC, suppose that P is in both
รโ
BA and
รโ
BC.
Since P is in
รโ
BA, by the definition of ray, either P is in
AB or B โ A โ P. Since P is in
รโ
BC, either P is in BC or B โ C โ P. Case 1: P is in AB and BC. In this case Prop. 3.5 gives P = B. Case 2: P is in AB and B โ C โ P. In this case, either P = A, P = B, or A โ P โ B. If P = A, then B โ C โ A while A โ B โ C by hypothesis. This contradicts Betweenness Axiom B-3. If A โ P โ B, then by Prop. 3.3, P โ B โ C. Since, in the case under consideration, B โ C โ P , this contradicts Betweenness Axiom B-3. Thus P = B. Case 3: (RAA) B โ A โ P and B โ C โ P. By Hypothesis, A โ B โ C, hence by the Corollary to Prop. 3.3, B โ C โ P implies that A โ B โ P. Thus B โAโP and AโB โP , contradicting Betweenness Axiom B-3. Thus in all possible cases, P = B.
To see that
รโ
AB =
รโ
AC, consider a point P. (โ) If P is in
รโ
AB, then either P is in AB or A โ B โ P. If P is in AB, then by Prop. 3.5, P is in AC and hence
in
รโ
AC.
If A โ B โ P , then we consider how P lies with respect to C. By Betweenness Axiom B-3, either P โAโC, or AโP โC, or A โ C โ P. (RAA) If P โ A โ C, then since A โ B โ P (which is equivalent to P โ B โ A), Prop. 3.3 gives B โ A โ C.
Hence B โ C โ D. (โ) Conversely, suppose that B โ D โ C. By the definition of angle and Proposition 3.6, A, B, C are not collinear. Then consider line
BC.
By Proposition 2.1,
BC meets
AB in the single point B. By Prop. 3.5, BC = BD โช DC and BD meets DC in the single point D.
Thus segment DC on line
BC does not contain B and
thus does not meet
AB.
Hence D lies on the same side of
AB as C. A similar argument shows that D lies on the same side of
AC as B. โ
Prop 3.8: If D is in the interior of โขCAB, then (a) so is
every other point on ray
รโ
AD except A; (b) no point on the
opposite ray to
รโ
AD is in the interior of โขCAB; and (c) if C โ A โ E, then B is in the interior of โขDAE.
Proof: (a) The line
AD meets the lines
AB and
AC in the point A. Since D is on a particular side of each of these lines, the
line
AD is not equal to either of these lines. By Prop. 2.1, A is the only point common to
AD and either one of these lines. In particular, segment of
AD meets lines
AB and
AC
only if A lies on this segment. Let P be a point on ray
รโ
AD other than A or D.
(RAA) If segment P D meets
AB, then segment P D includes A. Since P โ A and P โ D, we have P โ A โ D. Since P is a point on ray
รโ
AD other than A or D, we have either A โ P โ D (if P is in the segment AD) or A โ D โ P. This contradicts Betweenness Axiom B-3. Thus segment P D does not meet
AB, hence P lies on
the same side of
AB as D. A similar argument shows that P lies on the same side of
AC as D. Thus P is in the interior of โขCAB.
(b) The argument in Part (a) shows that the points on a ray all lie on the same side of a given line.
Points on the opposite ray to
รโ
AD lie on the opposite side of the lines as D. Thus they do not lie in the interior of โขCAB.
(c) By definition of interior of โขCAB, B lies on the same
side of
AC as D. Since C โ A โ E, C, A, E are collinear. Thus
AC =
AE.
So B lies on the same side of
AE as D. By Prop 3.7, since BโDโC, B does not lie in the interior of โขCAD. Since B lies on the same side of
AE =
AC as D, it follows that B lies on the opposite side
of
AD as C. The distinct lines
AD,
AC =
AE =
CE meet (only) in A.
