Chapter 15

1. Let ABCD be a Saccheri quadrilateral with right angles at A and B. Show, using only Euclidean propositions not requiring the parallel postulate, that ÐC=ÐD.

Triangles ABC and ABD are congruent, since (SAS) AC = BD, ÐBAC=ÐABD, and AB=AB. Thus BC=AD. Then for triangles ACD and BCD, (SSS) AC = BD, CD=CD, and AD=BC, so the corresponding angles ACD and BDC must also be equal.

2. Given the hypothesis of the acute angle, both Saccheri and Lambert showed that the sum of the angles of any triangle is less than two right angles. Let the difference between 180° and the angle sum of a triangle be the defect of the triangle. Suppose triangle ABC is split into two triangles by line BD. Show that the defect of triangle ABC is equal to the sum of the defects of triangles ABD and BDC.

The defect of triangle ABD equals 180 – (mÐBAD+mÐADB+mÐABD). The defect of triangle BDC equals 180-(mÐBCD+mÐBDC+mÐCBD). Since mÐADB+mÐBDC=180, defectABD+defectBDC =

180-(mÐBAD+mÐADB+mÐABD)+180-(mÐBCD+180-mÐADB+mÐCBD)

=180-mÐBAD-mÐADB-mÐABD+180-mÐBCD-180+mÐADB-mÐCBD

=180- mÐBAD-mÐABD-mÐBCD-mÐCBD. Rewriting BAD as BAC, BCD as BCA, and ÐABD+ÐCBD as ÐABC, this is 180-(mÐBAC+mÐABC+mÐBCA), the defect of triangle ABC.

4. Given that the angle sum of a triangle made of great circle arcs on a sphere (a spherical triangle) is greater than two right angles, define the excess of a triangle as the difference between its angle sum and 180°. Show that if a spherical triangle ABC is split into two triangles by an arc BD from vertex B to the opposite side, then the excess of triangle ABC is equal to the sum of the excesses of triangles ABD and BDC.

The excess of triangle ABD is mÐBAD+mÐADB+mÐABD-180; the excess of triangle BDC, mÐBCD+mÐBDC+mÐCBD-180. Rewriting mÐBDC as 180-mÐADB and adding the defects gives

mÐBAD+mÐADB+mÐABD-180+mÐBCD+180-mÐADB+mÐCBD-180

= mÐBAD+mÐABD+mÐBCD+mÐCBD-180. Rewriting BAD as BAC, BCD as BCA, and ABD+CBD as ABC gives mÐBAC+mÐABC+mÐBCA -180, which is the excess of triangle ABC.

Chapter 17

11. Let A be the set of numbers in (3/5,2/3) that have decimal expansions containing only finitely many zeros and sixes after the decimal point and no other integer. Find the least upper bound of A. (prove your answer)

The least upper bound is 2/3. All numbers with the above property are less than 2/3 (since 2/3 has a decimal expansion containing an infinite number of sixes), and for whatever ε>0 one chooses, one can find an x with the given property which differs from 2/3 by less than ε.

17. Define a natural ordering < on Dedekind's set of real numbers R defined by the notion of cuts. That is, given two cuts a=(A₁,A₂) and b=(B₁,B₂), define a<b. Show that this ordering < satisfies the same basic properties on R as on the set of rational numbers.

α<β if for any number a₁ in A₁, there exists a b₁ in B₁ such that b₁>a₁; and for any b₂ in B₂, there exists an a₂ in A₂ such that b₂>a₂.

Transitivity: if α<β and β<γ (where γ=(C₁,C₂)), then for any a₁ in A₁, there exists a b₁ in B₁ greater than a₁, and because β<γ, there exists a c₁ in C₁ greater than b₁. Thus for any a₁ in A₁, there exists a c₁ in C₁ such that c₁>a₁, so α<γ.

Density: If α<β, then for any a₁ in A₁, there's a b₁ in B₁ such that a₁<b₁ and there is also a real number d₁=a₁ + (b₁-a₁)/2, so that a₁<d₁<b₁ and this d₁ then determines another cut δ such that α<δ<β. Another real number can be found between α and δ in the same way, and another between those two, and so on, so that there exist infinitely many real numbers between any α and β such that α<β.

Cut: Any real number α divides the set of real numbers into two sets, A₁ ={xÎR|x£α} and A₂ ={xÎR|x>α}; if x₁ is in A₁ and x₂ is in A₂, then x₁ is no greater than α and x₂ is strictly greater than α, so x₁<x₂.

18. Define an addition on Dedekind's cuts. Show that a+b=b+a for any two cuts a and b.

If (A₁,A₂) is the pair of sets corresponding to the cut α and (B₁,B₂) the pair corresponding to β, then the α+β cuts the real numbers into the sets {a₁+b₁|a₁ÎA₁ and b₁ÎB₁} and {a₂+b₂|a₂ÎA₂ and b₂ÎB₂}. Commutativity follows from the commutativity of addition on the rationals.

Chapter 20

5. Arrange all two-letter combinations in alphabetic order, then all three-letter combinations and so on, and then eliminate all combinations that do not define a real number ("six" defines a real number, "sx" does not). Then the set of real numbers definable in a finite number of letters forms a denumerable, well-ordered set E={p₁,p₂,...}. Now define the real number s=.a₁a₂... between 0 and 1 by requiring a_n to be one more than the nth decimal of p_n if this decimal is not 8 or 9 and equal to 1 otherwise. Although s is defined by a finite number of letters, it is not in E, a contradiction. How can one resolve this paradox? How is this paradox related to the barber paradox or to Zermelo's original paradox?

A resolution of the paradox would require that a definition of a real number in a finite number of letters not include as a part of itself the totality of all such definitions.

Other paradoxes, including the barber paradox and Zermelo’s,, hinge on self-reference as well.

7. Show that Zermelo's axiom of separation resolves Russell's barber paradox as well as the Richard paradox in the sense that it excludes certain "sets" from discussion.

The axiom of separation requires that a set can be defined by a property of elements of an already defined set. This excludes such sets as “the set of all sets that are not members of themselves” and, in fact, the set of all sets. In the Richard paradox, the total set E would have to be formed before defining its element s.

Also the property defining the subset must be “definite”; there must be a way to determine whether or not it holds for a given element. In the barber paradox, the property of not cutting one’s hair turns out not be definite in this sense, because it is impossible to determine whether or not it holds for the barber.