Geometry and Proof: Course SummaryJohn T. BaldwinDepartment of Mathematics, Statistics and Computer ScienceUniversity of Illinois at ChicagoDecember 1, 2007This course is designed to explicate the following motto attributed to Hilbert.”It must always be possible to substitute ’table’, chair’ and ’beer mug’ for ’point’, line’ and ’plane’ in a systemof geometrical axioms.”We have considered this theme where ‘geometrical axioms’ are statements in Tarski’s world, various systems ofgeometry, and axioms for algebraic systems like the real numbers and integers.A key point is that for any given set of first order axioms for geometry, there are many many models of thoseaxioms.1 Basic LogicThe following key notions should be known: formal language, vocabulary, sentence, the compositional theorytruth, model/structure, truth in a model, validity, consistency, completeness theorem, compactness theorem.Know the difference between ` and |=. Note we first define M |= φ and then derive the usage Γ |= φ.The major skill is to be able to decide if a particular sentence in a formal language is true in a particular model.2 GeometryThere are really two different ways of categorizing geometry involved in this discussion.Synthetic geometry means that some set of axioms are fixed and proofs take place from those axioms. A metricgeometry has a notion of numerical length in its fundamental vocabulary. A synthetic geometry may or maynot be metric. And one can develop metric geometry axiomatically or not.2.1 Vertical AnglesThe ‘fact’ that vertical angle are equal requires some axiom. Euclid assumed essentially ‘all straight anglesare equal’. Some modern texts just assume ‘vertical angles’ are equal. But one cannot just read this from thepicture.1If one takes a metric approach and as sumes the ‘protractor postulate’ – each angle has a measure in degrees,then the property that all s traight angles are equal is immediate because, by definition, they measure 180oandall right angles are congruent because they each measure 180o. This is dual to Euclid’s approach, he explicitlyassumes any two right angles (recall Definition 10 of right angle) are congruent. Then in Propositions 13 and14 he essentially proves all straight angles are equal. (I say essentially because for Euclid all angles have lessthan 180o; he speaks obliquely of the sum of two angles being the same as the sum of two right angles.)2.2 Superposition and CongruencyEuclid’s proof of SAS is vitiated by his use of an ill-defined notion of ‘superposition’. This can be resolved inat least two ways. Hilbert assumes the axiom SAS. Other authors, e.g. Weinzweig, give a formal treatmentwith transformations of the plane as objects of the geometry and assume an explicit superposition axiom. TheMoise-Birkhoff version also assumes SAS.Note that even to discuss SAS, there must be a formal definition of angle. A major motivation for discussinghalf-planes is to provide a precise definition of angle [1].2.3 AreaProblem. What justifies the formula A =bh2for the area of a triangle?Definition 1 Two polygons are said to be of equal area when they can be decomposed into a finite number oftriangles which are respectively congruent to one another in pairs.Under Definition 1 one can easily se e that the area of triangle is one-half the area of a rectangle. If the sidesof the rectangle are commensurable then thinking of breaking the base into b units and the height into h unitsgives the area of the triangle as bh.Commensurability is essential for this result using Definition 1. In a non-Archimedean geometry (even satisfyingthe parallel postulate) there are triangles with same base and height that do not have the same area.But commensurablity is not essential if we slightly weaken the notion of area. The distinction between contentand area is found in [2]Definition 2 Two polygons are said to be of equal content when it is possible, by the ad dition of other polygonshaving equal area, to obtain two resulting polygons having equal area.We sketched in [10] the proof of the following theorem. It needs only axiom groups I-IV of Hilbert. The parallelpostulate is essential to prove that the opposite sides of parallelograms have the same length (which is animportant lemma). See the diagram labeled: Euclid’s diagram for area of a triangle or I.35 and I.38 in [3]Theorem 3 If tw o triangles have the same height and the same base then they have the same content.Remark 4 The distinction between content and area is very advanced. I would see no problem with a highschool course that defined area by Hilbert’s notion of content and then proved the theorem.This theorem needs only axiom groups I-IV of Hilbert. The parallel postulate is essential to prove that theopposite sides of parallelograms have the same length.22.4 Arithmetic and GeometryProofs of the following two theorems were sketched in [4] and homework assignments 8,9, and 10.Theorem 5 [Euclid/Hilbert)]: In a geometry satisfying Hilbert’s axioms it is possible to define using the vo-cabulary of geometry, operations ⊗, , ⊕ on the points of a line so that the line with these operations is afield.Fine point: Hilbert actually finds the structure of what I (idiosyncratic notation) will call a pseudofield: theelements are equivalence classes of segements under congruence. ⊕ is associative, commutative, and a⊕b = a⊕cimplies b = c. But there are no inverses. The multiplication is a commutative group. And the distributive lawholds. In [4], I choose a point 0 on a line and take the segment 0A as the representative of the equivalence classof segments congruent to 0A. Then I define an actual field.Theorem 6 (Descartes) Given an ordered field (e.g. the reals or rationals), it is possible to define using thea arithmetic operations a system of points and lines and relations of incidence, betweeness, and congruence tosatisfy Hilbert’s axioms.Thus one of the theory of geometry and the theory of fields is consistent if and only if the other is.2.5 ProportionalityWe want to understand the following fundamental theorem.Theorem 7 Corresponding sides of similar triangles are proportional.This is straightforward if the sides are commensurable. See Euclid VI.1 and VI.2 ([10, 3]). (In a nutshell, iftwo triangles ABD and ABE have the same height, and the bases are commensurable the area of ABD is to thearea of ABE as AD is to AE. (Just break them up into smaller triangle. If it