HOMEWORK 1, MATH 431, SECTION 2, SOLUTIONS

HARM DERKSEN

Exercise 1 (Chapter 1, Exercises, 1).

(1) The midpoint of AB is the point M on←→AB such that AM and BM are congruent.

(2) The perpendicular bisector of a segment AB is the line ` through the midpoint M ofAB with the property that ∠CMB is a right angle for any point C on `. (Note that∠CMB is right if and only if ∠CMB is congruent to ∠CMA.)

(3) We say that−−→BD bisects angle ∠ABC if ∠ABD and ∠DBC are congruent, and

−−→BD

is in between−−→BA and

−−→BC. (Of course, “in between” here is again vague.)

(4) A,B, C are colinear if there exists a line ` such that A,B, C lie on `.(5) Lines `,m, n are colinear if there exists a point A which lies on `, m and n.

(Slightly different definitions might be correct as well.)

Exercise 2 ( Chapter 1, Exercises, 7). I’ll draw a picture in class.

For example, as additional axiom for “between” one could take: For any line ` and distinctpoints A,B, C on `, exactly one of the points lies in between the other 2. With this axiomone can prove the equalities. By definition (Definition, page 18) of

−−→AB, C lies on

−−→AB if and

only if (i) C is on←→AB, and (ii) C lies on AB or B lies in between A and C. Also, by definition

(Definition, page 16), C lies on AB if and only if (i) C lies on←→AB and (ii) C = A, C = B or

C lies in between A and B.

Suppose D is a point on←→AB. We have the following cases:

(1) D = A or D = B. Then D lies on−−→AB.

(2) D lies in between A and B. Then D lies on AB, hence on−−→AB.

(3) A lies in between D and B. Then D lies on−−→BA.

(4) B lies in between D and A. Then D lies on−−→AB.

In any case, D lies on−−→AB ∪

−−→BA.

This shows that−−→AB ∪

−−→BA =

←→AB.

If D lies on AB, then D lies on−−→AB and

−−→BA, so D lies on

−−→AB ∩

−−→BA.

Conversely, suppose that D lies on−−→AB ∩

−−→BA. We have the following cases.

(1) D = A or D = B. Then D lies on AB.(2) D lies in between A and B. Then D lies on AB,(3) A lies in between D and B. Then D is not equal to A or B, and D does not lie in

between A and B. Hence D does not lie on AB. Also, B does not lie in between D1

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and A, so D does not lie on−−→AB, and D does not lie on

−−→AB ∩

−−→BA. So this case is not

possible.(4) B lies in between D and A. As the previous case, this case is not possible.

We conclude that D lies on AB.We conclude that BA =

−−→AB ∩

−−→BA.

Exercise 3 (Chapter 1, Exercises, 12). I’ll draw a picture in class.

An accurate picture of a triangle which is not isosceles will reveal that either E lies on ABand F does not lie on AC, or F lies on AC and E does not lie on AB. (If you draw a picture,make AB twice as long as AC, say.) This case is not treated in the “proof”. Indeed, in thecases 2,3,4 on page 26, either E lies on AB and F lies on AC, or E does not lie on AB andF does not lie on AC. Figure 1.13 is not very accurate, because AB and AC have about thesame length. The proof is still correct as far as AE ∼= AF and BE ∼= CF . But we can nolonger conclude that AB ∼= AC.

Exercise 4 (Chapter 2, Exercises, 3).

Parallel postulate: For every two distinct points P and Q, there exists a unique linethrough P and Q.Negation: There exist two distinct points P and Q such that there is no line through P andQ or there exist at least 2 distinct lines through P and Q.

Exercise 5 (Chapter 2, Exercises, 6).

Proposition 2.3.

Proof. By Axiom I-3, there exist three distinct points A, B and C which are not colinear.Suppose that ` is a line. Suppose that A, B, C all lie on the line `. Then A,B, C are colinear.Contradiction. Therefore, at least one of the points A, B, C does not lie on `. So there existsa point which does not lie on `. We conclude that for every line there exists a point whichdoes not lie on that line. �

Proposition 2.4.

Proof. By Proposition 2.2, there exist three lines, `,m, n which are not concurrent. Supposethat P is a point. Suppose that `,m, n all go through P . Then `,m, n are concurrent.Contradiction. Therefore, at least one of the lines `,m, n does not go through P . So thereexists a line which does not go through P . We conclude that for every point, there exists aline which does not go through that point. �

Proposition 2.5.

Proof. Suppose that P is a point. By Proposition 2.4, there exists a line ` which does notgo through P . By Axiom I-2, this line has at least 2 distinct points, say A1 and A2. ClearlyP is not equal to A1 or A2 because P does not lie on `. By axiom I-1, there exists a uniqueline m1 through P and A1 and a unique line m2 through P and A2. Suppose that m1 = m2.Then both m1 and ` go through A1 and A2. By the uniqueness in Axiom I-1, we get ` = m1.But then P lies on m1 = ` and we get a contradiction. Therefore, `1 6= `2. It follows thatthere exist two distinct lines through P . So for every point P , there exist at least 2 distinctlines through P . �

HOMEWORK 1, MATH 431, SECTION 2, SOLUTIONS 3

Exercise 6. Prove the formula (A⇒ B) ∨ (B ⇒ C) (Hint: start with B∨ ∼B).

Proof.B∨ ∼B

Assume BThen we have A⇒ BIt follows that (A⇒ B) ∨ (B ⇒ C)

Assume ∼BThen we have B ⇒ CFrom this follows that (A⇒ B) ∨ (B ⇒ C)

We conclude that (A⇒ B) ∨ (B ⇒ C). �

Remarks: Since A ⇒ B is equivalent to ∼A ∨ B, A ⇒ B follows from ∼A, and A ⇒ Bfollows from B.

One can prove A⇒ B from ∼A as follows:

Proof.∼AAssume AThen A& ∼A. Contradiction!We can deduct anything we want from a contradiction, in particular we can conclude B.We conclude A⇒ B. �

It easy to prove that A⇒ B follows from B:

Proof.BAssume AThen B holdsHence we have A⇒ B. �

Why is it that anything follows from a contradiction? Suppose we have proven A& ∼A andB is an arbitary statement. Then we prove:

Proof.A& ∼AAssume ∼BWe have a contradiction because A& ∼A.So we have ∼∼BAnd therefore, we get B. �

Exercise 7. Assume the incidence axioms. Show that, if every line contains at least 3 distinctpoints, then there must be at least 7 distinct points.

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Proof. By Axiom I-3, there exist at least 3 distinct points, not on a line. Label these 3 pointsby A1, A2 and C. There exists a unique line `A through A1 and A2. The point C does notlie on `A because A1, A2, C are not colinear. By the assumptions, this line contains a thirdpoint A3. Let `i be the line through Ai and C for i = 1, 2, 3. Besides Ai and C, the line`i contains a third point, call it Bi. We have to prove that A1, A2, A3, B1, B2, B3, C are alldistinct points. Clearly A1, A2, A3 are distinct, and C is not equal to any of the A’s becauseC does not lie on `A. If B1 lies on `A, then `1 and `A both go through A1 and B1, so `A = `1

(by uniqueness in Axiom I-1) and C lies on `1 = `A. Contradiction. So B1 does not lie on `A

and neither do B2 and B3 using a similar argument. It follows that Bi 6= Aj for all i, j. Also,we have Bi 6= C for all i by construction. Suppose that B1 = B2. Then `1 and `2 both gothrough C and B1. By uniqueness in Axiom I-1, `1 = `2. But then A1 and A2 lie on `1 = `2.so `1 and `A boh go through A1 and A2. By uniqueness in Axiom I-1, it follows that `1 = `A.This yields a contradiction because C lies on `1 but not on `A. We conclude that B1 6= B2.Similarly, we have B1 6= B3 and B2 6= B3. Therefore, all points A1, A2, A3, B1, B2, B3, C aredistinct. The the plane has at least 7 points. �