The Elements, Book I – Propositions 1 – 10 MONT 104Q – Mathematical Journeys: Known to Unknown September 25, 2015

Preliminaries, 1 Following the Common Notions (Axioms) and Postulates comes a sequence of Propositions and their proofs. Two traditional types of Propositions: The “ problems ” show how to construct something (and show the construction works) The “ theorems ” claim something (and prove those claims) Copies of the Elements also contained extensive scholia, or commentaries, added by later scholars. We won't read them, but they were effectively part of the Elements for later readers.

Preliminaries, 2 Want to follow Euclid's logical structure, without necessarily sticking that closely to the specific ways he said things. So in particular, we'll freely use modern symbolic forms of geometric statements. Example : Congruence of figures (line segments, angles, triangles, etc.) We'll write something like ΔAEB ≅ ΔBDA to say triangles are congruent. Euclid doesn't do this – just says line segments, angles, triangles are “equal.” So, this might be a bit different from the online translations of Euclid you're reading.

Proposition 1 To construct on a given line segment AB an equilateral triangle.

The Proof 1. AC = AB, since radii of a circle are all equal (this was stipulated in the definition of a circle). 2. Similarly BC = AB. 3. Therefore, AC = BC (Common Notion 1) 4. Hence ∆ABC is equilateral (again, this was given as a definition previously). QEF

A Question Do we actually know that such a point C exists? The answer: strictly speaking, we don't (!) It does not follow from any of the Postulates that Euclid set out at the start – we would need additional postulates to assure this. Euclid is pretty clearly appealing to our intuition about physical circles here and either does not appreciate that there is something missing, and/or does not want to address that point at this stage. This is definitely a flaw of a sort, but perhaps justified from educational/pedagogical orientation of the text. (He doesn't come back to it later either(!))

Propositions 2 and 3 These are somewhat technical constructions aimed at showing that the straightedge and “collapsing” compass are sufficient for routine tasks such as measuring off a given length on a given line. Proposition 2. Given a line segment AB and a point P, construct a point X such that PX = AB. The construction uses Proposition 1 and Postulate 3 Proposition 3. Given two unequal line segments, lay off on the greater a line segment equal to the smaller. This construction uses Proposition 2 and Postulate 3 again, but without using the compass to “transfer” the length

A natural Question It’s natural to ask: Why did Euclid go to the trouble of making these somewhat involved constructions for relatively simple tasks that would be easy if we had an implement like the modern compass that can be used to measure and transfer lengths in a construction? The answer seems to be that his goal was to show that a very small set of simple starting assumptions was sufficient to develop the basic facts of geometry. So some technical stuff would be acceptable at the start to establish “routines” for those tasks under the more restrictive working conditions or hypotheses.

The “SAS” Congruence Criterion Proposition 4. Two triangles are congruent if two sides of one triangle are congruent with two sides of the other triangle, and the included angles are also congruent. The proof given amounts to saying: move the first triangle until the sides bounding the two equal angles coincide, then the third sides must coincide too. This idea gives a valid proof, of course, but it again raises a question: What in the Postulates says we can move any figure? Again, there is physical intuition and/or an unstated assumption being used here(!)

The “ Pons Asinorum” Proposition 5. In any isosceles triangle, the base angles are equal; also the angles formed by the extensions of the sides and the base are equal.

Start of proof of Proposition 5 By the construction CD = CE, and CA = CB by hypothesis. Moreover ACE = BCD. Hence ΔDCB ≅ ΔECA (Proposition 4) It follows that BD = AE, CBD = CAE, and CDB = CEA (corresponding parts of congruent triangles) Then since CE = CD and CA = CB, we also have AD = BE (Common Notion 3) Therefore, ΔAEB ≅ ΔBDA (Proposition 4 – note the angle at D is the same as the angle at E from the third line above.)

Conclusion of proof of Proposition 5 This shows that DAB = EBA, which is the second part of the statement of the Proposition. To finish the proof we must show that the base angles in ΔCAB are equal. EAB = DBA from the congruence of the small triangles at the bottom of the figure. But also EAC = DBC by the first congruence established before. Hence BAC = EAC - EAB = DBC - DBA = ABC (Common Notion 3). QED

Propositions 6 and 7 First, a converse of Proposition 5: Proposition 6. If two angles of a triangle are equal, then the sides opposite those angles are equal. Euclid gives a proof by contradiction, using Propositions 3 and 4. Next: Proposition 7. If in the triangles ΔABC and ΔABD, with C and D on the same side of AB, we have AC = AD and BC = BD, then C = D. Another proof by contradiction, using Proposition 5; Euclid gives only one case out of several.

Proposition 8 Proposition 8. If the three sides of one triangle are equal to the three sides of another triangle, then the triangles are congruent. This is the “SSS” congruence criterion Proof is based on Proposition 7; in fact can almost see that Euclid wanted to present the reasoning “broken down” into easier steps by doing it this way. Modern mathematicians call a result used primarily to prove something else a “lemma.”

Next, a sequence of “bread-and-butter” constructions Proposition 9. To bisect a given angle.

Angle Bisection – The Proof Proof: AB = AC by construction. BD = CD since the triangle ΔBCD is equilateral AD is common to the two triangles ΔABD and ΔACD Therefore, ΔABD and ΔACD are congruent (Proposition 8). Hence <ADB = <ADC, and we have done what we set out to do – the angle at A is bisected. QEF

Line Segment Bisection Proposition 10. To bisect a given line segment. Construction: Let AB be the given line segment Using Proposition 1, construct the equilateral triangle ΔABC Using Proposition 9, bisect the angle at C Let D be the intersection of the angle bisector and AB. Then D bisects AB.

Line Segment Bisection – Proof Proof: AC = BC since ΔABC is equilateral <ACD = <BCD since CD bisects <ACB The side CD is in both triangles ΔACD and ΔBCD Therefore, ΔACD and ΔBCD are congruent by Proposition 4. Hence AD = BD since the corresponding parts of congruent triangles are equal. QEF