Kitabı oku: «A course of plane geometry», sayfa 2

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1.4 Book plan

Chapter 2 is a preliminary chapter, necessary for the understanding of the rest of the book. It starts with a review of the methods for proving statements of the form “P implies Q”, and also of methods for proving other types of statements, with particular emphasis on the Induction Method, used for proving statements of the form “P(n) for n ≥ n₀”. Then there is a rapid introduction to the symbolism used in logic. After this the basics of the elementary theory of sets are reviewed, including a discussion of the notion of equivalence relation, and of the important fact that an equivalence relation defined on a set, determines a partition of the set into equivalence classes.

Chapter 3 introduces the notion of incidence geometry as a set together with a collection formed by some of its subsets, having three properties called axioms of incidence. Then examples of incidence geometries of various kinds are presented. Then the “main” examples of incidence geometries, namely the real cartesian plane, the hyperbolic plane and the elliptic geometry, are presented in complete detail. In particular, complete proofs, based only on the properties of addition and multiplication in the real number system, that the three axioms of incidence hold in these examples, are supplied. After this the subject of parallelism of lines is discussed, and an additional axiom, called Playfair’s axiom is studied. Playfair’s axiom is a refined version of Euclid’s fifth postulate. The chapter ends with a long discussion of how paralllelism behaves in all the examples of incidence geometries previously given. It is of particular importance the discussion of the behaviour of parallelism in the real cartesian plane, the hyperbolic plane and the elliptic geometry. It is shown that in the real cartesian plane, given any point A and any line l, with A not in l, there exists a unique line passing through A and being parallel to l; that in the hyperbolic plane, given any point A and any line l, with A not in l, there exists an infinite number of lines passing through A and being parallel to l; and that in the elliptic geometry, given any point A and any line l, with A not in l, there is no line passing through A and being parallel to l.

Chapter 4 treats the formalization of the notion that one point is between two other points, i.e., the concept of a betweenness structure for an incidence geometry. A betweenness structure is defined as a collection of ordered triples of points of the plane, having four properties called the betweenness axioms. We remark that the realization by Hilbert that one of the main deficiencies of Euclid’s axiomatics, making some of Euclid’s proofs in the Elements ultimate dependant on pictures, lied in the lack of axioms governing betweenness, constitutes perhaps his main contribution for saving Euclid’s work. The chapter begins with the definition of betweenness structure for an incidence geometry. This structure makes it possible to define segments, triangles and the convexity of a subset of the plane. Then it is shown that a line l divides the plane minus l into two parts, called sides of the plane divided by l; and also that a point A of a line l, divides l minus {A} into two parts, called the sides of l divided by A. As consequences the following interesting facts are proved to hold in any incidence geometry equipped with a betweenness structure, namely, that the endpoints of a segment are entirely determined by the segment, that each line is formed by an infinite number of points, and that there is a point between any two given points. Next, the important notion of ray is introduced, and a long theorem containing a bunch of facts about rays which prove very useful for the rest of the book, is stated and proved. Then the fundamental notion of angle is introduced, followed by a discussion of the important notion of interior of an angle. An important result called Crossbar Theorem is then presented at length. The chapter ends with a discussion of the usual betweenness structures carried by the real cartesian plane and the hyperbolic plane. Elliptic geometry is abandoned at this point for the rest of the book, due to the fact that it does not admit any betweenness structure. It does admit a modified betweenness structure though (see [2]).

Chapter 5 introduces the notion of structure of congruence of segments for an incidence geometry equipped with a betweenness structure. It is defined as a collection of ordered pairs of segments, having three properties called axioms of congruence of segments. Then a useful result, called subtraction of segments, is proven. Next, the notion that a segment is less than another segment, is introduced, and its main properties are stated and proven, using subtraction of segments as the main tool. The rest of the chapter is devoted to define the usual congruence of segment structures in the real cartesian plane and in the hyperbolic plane, and proving that these satisfy the three axioms.

Chapter 6 is dedicated to the notion of a structure of congruence of angles for an incidence geometry equipped with a betweennes structure and a congruence of segments structure. It is defined as a collection of ordered pairs of angles, having three properties called axioms of congruence of angles. This structure allows for the defintion of congruence of triangles. Then the concepts of adjacent angles, supplementary angles and vertical angles are defined. The important results summarized as “angles which are supplementary of congruent angles, are congruent”, “a pair of adjacent angles which are congruent to a pair of supplementary angles, are supplementary” and “vertical angles are congruent”, are precisely stated and proved. Then the angle addition theorem and the angle subtraction theorem are discussed. The notion of an angle being less than another angle is introduced, and its main properties are proven. Right angles are then defined as angles which are congruent to any of its (two) supplementary angles, and the congruence of any two right angles is established. The study of the usual structures of congruence of angles for the real cartesian plane and the hyperbolic plane occupy the rest of the chapter.

Finally, Chapter 7 is dedicated to Hilbert Planes. Hilbert Planes are incidence geometries equipped with a betweenness structure, a congruence of segments structure and a congruence of angles structure. This requires that a total of thirteen axioms are satisfied, three incidence axioms, four betweenness axioms, three congruence of segments axioms and three congruence of angles axioms. The main examples of Hilbert Planes are the real cartesian plane and the hyperbolic plane. The rest of the chapter is dedicated to the study of Book I of Euclid’s Elements, a la Hilbert.

1.5 How to study this book

Your attitude, in order to really grasp the material, should be that of a hyperactive student. Leisurely studying the material will not do it! Read every sentence carefully. Read the examples and do as many exercises as possible. You may even try to create some exercises. In a first reading of the book, you may skip certain particularly long examples, like the one on parallelism in the hyperbolic plane.

2
Preliminaries

The goal of this chapter is to present the main proof methods used in mathematics, and the basic elements of the elementary theory of sets, in order to establish the necessary theoretical basis for the axiomatic development of geometry, both euclidean and noneuclidean. We will explain the basic methods for proving conditional claims, such as the direct method, the cases method, the contraposition method, and the contradiction method. Other important methods, such as mathematical induction, will also be studied. In the case of the elementary theory of sets, we will establish the most basic results concerning equivalence relations and the equivalence classes determined by them. For a deeper presentation of these matters, we advice the reader to consult texts such as [3],[5],[6]. As a consequence, in the following sections we will freely state and use some results without providing a rigorous derivation. Nevertheless, the definitions will be presented in a coherent way, and with the standard notation used in the corresponding areas of knowledge.

2.1 Proof methods

The following is an informal but very practical presentation of the main strategies used for proving mathematical theorems.

2.1.1 Methods for proving conditional statements

As most theorems in mathematics are of conditional type or implications, i.e., have the form “claim 1 implies claim 2” or what amounts to the same “If claim 1, then claim 2”, we will start by making clear the precise meaning of an implication, and then we will present the main strategies used in proving statements of this type.

Meaning of the logical implication

The saying “Dog that barks does not bite” has conditional form, because we can put it in the form “If p is a dog that barks, then p is a dog which does not bite”. In our daily use and understanding of natural language, we tend to infer from this saying that if a dog does not bark, then the dog bites. In the strict logical sense, which is the sense adopted in mathematics, the saying does not say anything in case claim 1 does not hold. In that case, claim 2 may or may not hold. Hence, the saying does not tell us anything about dogs which do not bark. There might be nonbarking dogs which bite, and nonbarking dogs which do not bite. The only thing claimed by the saying is that no dog barks and bites.

Here is another example. If a father tells his son, “If you pass the grade, I will give you a bicycle”, and the son does not pass the grade, but the father gives the bicycle to him anyway, many of us would consider that the father did not keep his promise. In strict logical sense, the father kept his promise, because the implication “If you pass the grade, I will give you a bicycle” does not say what action should the father take in case the son does not pass the grade. It only says the action the father must take in case the son passes the grade.

In the context of the natural numbers = {0, 1, 2, . . .}, the claim

“If n is odd, then n² is odd”

is an example of a claim of conditional type. In the same context, the claim

“The sum of any two consecutive numbers is odd”

is, even though it does not seem at first, a claim of conditional type, because it can be stated equivalently as,

“If a, b are consecutive numbers, then a + b is odd”.

There is an important difference between the example of the father that makes a promise to the son, and the other three examples of conditional claims we have given. The latter can be rephrased putting before them a universal quantification. In this way, the claim “If p is a dog that barks, then p is a dog which does not bite” means the same as the claim “For all dog p, if p barks, then p does not bite”. Likewise, the claim “If n is odd, then n² is odd” is the same as claiming “For all n, if n is odd, then n² is odd”. And the claim, “If a, b are consecutive, then a + b is odd” is the same as saying “For all a, b, if a, b are consecutive, then a +b is odd”. This kind of rephrasing is not possible with the father’s promise.

Let us consider now an issue which will be very important later: the negation of a claim of conditional type. The distinction made in the previous paragraph is crucial in order to correctly negate conditional type claims. The negation of the father’s promise would be “The son passes the grade and the father does not give the bike to him”. The negation of the claim “For all dog p, if p barks, then p does not bite” is “There is at least one dog p, such that p barks and p bites” or “There exists a dog p, such that p barks and p bites”. Notice the introduction of the phrase “There is at least one dog p, such that” or “There exists a dog p, such that”. Likewise, the negation of “For all natural n, if n is odd, then n² is odd” is “There exists a natural n, such that n is odd and n² is not odd”; and the negation of “For all natural numbers a, b, if a, b are consecutive, then a +b is odd” is “There exists natural numbers a, b, such that a, b are consecutive and a + b is not odd”.

In order to prove this type of claims, there exist four main strategies: the direct method, the method by cases, the contraposition method, and the contradiction method.

Direct method

Let us assume that we have a claim of the form “claim 1 implies claim 2”. The direct method consists in assuming that claim 1 is true, and, using claim 1, seeing that claim 2 has to be true, by the means of logical deduction. In the course of the argument claim 1 can be used as many times as necessary as well as any other fact which is an axiom or that has been proved previously.

Let us prove, using this method, the claim

“If a, b are consecutive natural numbers, then a + b is odd”.

In this case claim 1 is “a, b are consecutive natural numbers” and claim 2 is “a + b is odd”.

Before starting the proof, let us write in a more precise way claims 1 and 2. Claim 1 can be stated more precisely as “a, b are natural numbers such that a = b + 1 or b = a + 1”; and claim 2 can be stated as “there exists a natural number n such that a +b = 2n + 1”. Writing claims 1 and 2 in this way, helps a lot in creating a correct proof.

Let us now prove, by the direct method, that

“If a, b are natural numbers such that a =b + 1 or b = a + 1, then there exists a natural n such that a + b = 2n + 1”.

We begin by writing:

“Let a, b be natural numbers such that a = b + 1 or b = a + 1”.

Then we write:

“There are two possibilities: a = b + 1 or b = a + 1”.

And we go on writing:

“Let us assume that a = b+1. In this case a+b = (b+1)+b = 2b+1. Then there exists a natural number n (n = b), such that a + b = 2n + 1.

Next, we consider the other case. We write:

“Let us now assume that b = a + 1. In this case a + b = a + (a + 1) = 2a + 1. Then there exists a natural number n (n = a), such that a + b = 2n + 1”.

and we finish the proof by writing:

“Since in both cases we arrived to the conclusion that there exists a natural number n, such that a + b = 2n + 1, we have that in every case there exists a natural number n such that a + b = 2n + 1.”

Method of proof by cases

This method can be considered as a special case of the direct method. It is used when we want to prove an implication “P implies Q”, in which the claim P is equivalent to a disjunction of several claims, i.e., when P is equivalent to a claim of the form “P₁ or P₂ or . . . or Pn”. The method is based on the equivalence between the claim “P implies Q” and the claim “P₁ implies Q, and P₂ implies Q, and . . ., and Pn implies Q”. In the example we gave when explaining the direct method, we used, without noticing it, the cases method, since claim 1 of that example was “a, b are consecutive natural numbers” which is equivalent to “a, b are natural numbers such that a = b+1 or b = a+1”, which in turn is equivalent to “a, b are natural numbers such that a = b+1, or, a, b are natural numbers such that b = a +1”. So what we did was to prove, separately, the two implications “If a, b are natural numbers such that a = b + 1, then a + b is odd” and “If a, b are natural numbers such that b = a + 1, then a + b is odd”.

Method of contraposition

This method is based on the observation that a claim of the form “claim 1 implies claim 2”, is logically equivalent to the claim “no(claim 2) implies no(claim 1)”. For example, to say that “if p is a dog that barks, then p is a dog which does not bite”, is the same as saying that “if p is a dog that bites, then p is a dog which does not bark”. Or to say “If you pass the grade, then I will give you a bicycle” is the same as saying “If I do not give you a bicycle, then you did not pass the grade”.

We say that “no(claim 2) implies no(claim 1)” is the contrapositive of “claim 1 implies claim 2”. An implication and its contrapositive say the same but in a different way.

Then, the method of contraposition consists in taking an implication we want to prove, and proving its contrapositive by, say, the direct method.

The constrapositive of

“If a, b are consecutive natural numbers, then a + b is odd”

“If a, b are natural numbers such that a + b not odd, then a, b are not consecutive”,

which put more precisely, is

“If a, b are natural numbers such that there is no natural n with a + b = 2n + 1, then a ≠ b + 1 and b ≠ a + 1”.

Let us prove the last claim.

We begin by assuming that a, b are natural numbers such that there is no natural number n for which a + b = 2n + 1. So, we write:

“Let a, b be natural numbers such that there is no natural number n for which a + b = 2n + 1.”

Then we change the form of this claim. We write:

“Then for every natural n, we have a + b ≠ 2n + 1”.

And then we write:

“But this means that for every natural number n, a + b ≠ n + (n + 1).”

And finish up by writing:

“This last claim tells us that a +b is different from all the possible sums of pairs of consecutive natural numbers, and therefore it cannot be that a, b are consecutive natural numbers.”

This concludes the proof.

The previous proof shows that “If a, b are natural numbers such that a + b is not odd, then a, b are not consecutive” and therefore “If a, b are consecutive natural numbers, then a + b is odd”.

Method of contradiction

This method is not only useful for proving implications, but for proving any type of claim. The idea is that if the negation of a claim A allows us to prove a contradiction, i.e., that some claim is true and that the same claim is false, then the claim A is necessarily true.

Let us prove by this method the claim “If a, b are consecutive natural numbers, then a + b is odd”.

We start by negating what we want to prove. In this case we obtain “There exists at least one pair of natural numbers a, b, such that a, b are consecutive and a + b is not odd”. We assume that this last claim is true and try to derive a contradiction. We write:

“Let a, b be natural numbers such that a, b are consecutive and a + b is not odd”.

Then we write:

“Since a, b are consecutive, then a = b + 1 or b = a + 1. In the first case a + b = (b+1)+b = (b+1)+b = 2b+1 and in the second case a+b = a+(a+1) = 2a + 1. In both cases it turns out that there exists a natural n such that a + b = 2n + 1.”

Now we try to also prove the negation of what we have just concluded. We write:

“On the other hand, since a + b is not odd, there is no natural number k such that a + b = 2k + 1.”

At this point one remarks that we have reached to a conclusion and also to its negation:

“We have thus arrived on one hand to the existence of a natural n such that a + b = 2n + 1, and on the other to the nonexistence of a natural k such that a + b = 2k + 1. This proves that our initial assumption that there exist natural numbers a, b which are consecutive and whose sum is not odd, cannot be true, and therefore its negation, “for all natural numbers a, b, if a, b are consecutive, then a +b is odd” must be true.

And this finishes the proof.

2.1.2 Methods for proving other types of statements

Since not all theorems of mathematics are claims of conditional type, we must consider other important types of claims and have strategies to prove them.

Proving “claim 1 implies claim 2”, when claim 1 is false

Sometimes when trying to prove an implication “claim 1 implies claim 2”, one notices that claim 1 is false. In this case there is nothing to do, because the implication is automatically true!

Proving biconditional claims

A claim of the type “claim 1 if and only if claim 2” is said to be an equivalence or of biconditional type. What this type of claim means is that either claim 1 and claim 2 hold, or neither claim 1 nor claim 2 hold. Sometimes the phrase “if and only if” is abbreviated as “iff”. This type of claim is equivalent to the claim “If claim 1, then claim 2, and if claim 2, then claim 1”. According to this, in order to prove “claim 1 if and only if claim 2”, one proves separately that “claim 1 implies claim 2” and that “claim 2 implies claim 1”.

Proving disjunctions

A disjunction is a claim of the type “P₁ or P₂ or . . . or Pn”. Let us first consider the case n = 2, i.e., claims of the type “P₁ or P₂”. This claim is equivalent to the claim “If no P₁, then P₂”. It is of course also equivalent to “If no P₂, then P₁”. Hence, one would prove any of these two reformulations, which being of conditional type, can be proved by using any of the methods we have already studied. Now, if n ≥ 3, the disjunction is equivalent to the claim “[no(P₁) and no(P₂) and . . . and no(P_n−1)] implies Pn”.

Proving existence claims

An existence claim is a claim of the type “There exists x, such that x has property 1”. There are essentially two ways to prove a claim of this type: by constructing or exhibiting an x having property 1, or by contradiction, i.e., assuming the negation of the existence, which is “For all x, x does not have property 1”, and deriving a contradiction.

Proving universal claims

A universal claim is one of the type “For all x, x has property 1”. These claims can be proved in at least two ways. One is by saying “Let x be a fixed but arbitrary (of the x one is talking about)” and then proving that x has property 1. The other way is by contradiction, i.e., assuming the negation of “For all x, x has property 1”, which is “There exists x, such that x does not have property 1”, and use this assumption to derive a contradiction.

Proving that “For all natural n ≥ 1, P(n) holds”

Let us suppose that for each natural n ≥ 1, we have a claim P(n), i.e., that we have an infinite list of claims P(1), P(2), P(3), . . ., and that we want to prove that each one of them is true. There is a very ingenious way to do this, known as Induction method. It is enough to prove the following two claims:

1. P(1)

2. “For all k ≥ 1, P(k) implies P(k + 1)”.

The reason why just proving these two claims proves that P(n) es true for each n ≥ 1 is as follows. What the second claim says is that each one of the infinite number of implications, “P(1) implies(2) implies P(3)”, “P(3) implies P(4)”, “P(4) implies)”, etc. is true. Let us see why, for example, P(4) must be true. We know that P(1) is true. Since “P(1) implies P(2)” is true, then P(2) must be true. Now, since “P(2) implies P(3)” is true, we have that P(3) has to be true. Again, since “P(3) implies P(4)” is true, P(4) has to be true.

Let us consider an example. Let us prove that for any natural n ≥ 1,

In this case P(n) is “1 + 2 + . . . + n = ”.

Let us verify that P(1) holds. P(1) is “”, which is true. Now, let k ≥ 1 be fixed but arbitrary. We wish to see that P(k) implies P(k+1), i.e., that implies 1+2+. . .+k+(k+1) = . Let us prove this implication by the direct method.