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Greece Chapter 5: Euclid of Alexandria
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Mar 18, 2018

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Page 1: Greece Part 2 - mathed.byu.edumathed.byu.edu/~williams/Classes/300W2012/PDFs/PPTs/Greece Part … · into one kingdom. After his death, his son, ... Euclid • The Library of Alexandria,

Greece

Chapter 5: Euclid of Alexandria

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The Library at Alexandria

• What do we know about it?

• Well, a little history…

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Alexander the Great

• In about 352 BC, the Macedonian King Philip II began to unify the numerous Greek city‐states into one kingdom. After his death, his son, Alexander (the Great) continued the conquest of everything between Macedonia and India.

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Alexander the Great’s Empire

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Alexandria

• Alexander planned and built the city of Alexandria in Egypt, on the west end of the Nile river delta.  Although Alexander died before the city was complete, it remained the capital of Egypt for nearly a thousand years. Ptolemy, one of Alexander’s generals, took over the Egyptian part of his empire.  A son, Ptolemy II, built a library and museum, and Ptolemy III populated it with books (scrolls, really). 

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Euclid

• The Library of Alexandria, founded about 300 BC, became the center of learning.  It had reading rooms, lecture rooms, meeting rooms, a dining hall, and gardens to walk in (sounds like a modern‐day college).

• It is to this library that Euclid came to study and teach. 

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Euclid’s Elements

• What do we know about the source of our current version of the Elements?

• Original (lost)• Commentary and copy by Theon of Alexandria (300‐400 CE)– Copy at Oxford (888 CE)

• 900 CE, copy from version earlier than Theon’s, in Vatican library 

• 1880’s, comparison by J. L. Heiberg• 1908, Thomas Heath, pretty much the standard today.  

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Euclid’s Elements

• What can we remember about the content of the Elements?

• Geometry, plane and solid• Ratios of numbers and of magnitudes• Geometric algebra• Number theory• Polyhedra

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Euclid’s Elements

• Book I:  Basic plane geometry• Book II:  “Geometric” algebra• Book III:  Circles • Book IV:  Inscribing and circumscribing figures• Book V:  Extending Eudoxus’ ideas of ratio• Book VI:  Similarity of figures• Books VII – IX:  Number theory• Book X:  Incommensurable magnitudes• Books XI – XIII:  Solid (3 dimensional) geometry

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Euclid’s Elements

• 13 books, 465 propositions in plane and solid geometry and number theory.  

• Few if any results are original to Euclid; it is likely a compendium of already‐known results.  In fact it has been suggested that the first books of Euclid’s Elementsmay have been taken from the lost Elements of Hippocrates of Chios.

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Euclid’s Elements

• What is important is the logical structure of the books.  

• He gave us an axiomatic development of geometry:– 23 definitions– 5 postulates– 5 common notions

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Euclid’s Elements

• Most of each book consisted of propositions which were proved using only the definitions, common notions, and postulates, as well as any propositions previously proved.  Thus Proposition I.3 may be proved using only the common notions, postulates, definitions, and Propositions I.1 and I.2.  

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Euclid’s Elements

• Some of Euclid’s definitions:– A point is that which has no part (1).– A line is breadthless length (2).– A straight line is a line which lies evenly with the points on itself (4).

– When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands (10).

– A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another (15).

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Euclid’s ElementsEuclid’s five postulates:

• To draw a straight line from any point to any point• To produce a finite straight line continuously in a straight 

line.• To describe a circle with any center and distance.• That all right angles are equal to one another.• That, if a straight line falling on two straight lines makes the 

interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

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Euclid’s Elements

Euclid’s Common Notions:• Things which are equal the same thing are also equal to one another

• If equals be added to equals, the wholes are equal 

• If equals be subtracted from equals, the remainders are equal

• Things which coincide with one another are equal to one another

• The whole is greater than the part. 

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Euclid’s ElementsProposition 1:  On a given finite straight line to construct an equilateral 

triangle.• Let AB be the given finite straight line. It is required to construct an 

equilateral triangle on the straight line AB. Describe the circle BCDwith center A and radius AB (Postulate 3). Again describe the circle ACE with center B and radius BA (Postulate 3). Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B (Postulate 1).

C

A BD

E

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Euclid’s Elements

• Now, since the point A is the center of the circle CDB, therefore ACequals AB (Definition 15). Again, since the point B is the center of the circle CAE, therefore BC equals BA (Definition 15). But AC was proved equal to AB, therefore each of the straight lines AC and BCequals AB. And things which equal the same thing also equal one another, therefore AC also equals BC (Common Notion 1). Therefore the three straight lines AC, AB, and BC equal one another. Therefore the triangle ABC is equilateral, and it has been constructed on the given finite straight line AB.  Being what it was required to do.

C

A BD

E

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Euclid’s Elements

• Some problems:– Definitions:  The first few are vague and intuitive (that which has no part? breadthless length?).  Later definitions sometimes leave out parts (what does it mean for an angle to be “greater” than another?) Many  he never uses or even refers to later in the book.

– The fix:  some terms are defined, while some remain undefined.  Gaps are filled in definitions. 

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Euclid’s Elements

• More Problems– Proofs: Use unstated assumptions.  For example, in the proof of Proposition 1, how do we know that the point C exists at the intersection of two circles? What does it mean for one point to be “between” two other points on a line? Diagrams can make these ideas clear and convincing, but it does not meet modern standards of rigor. 

– The fix:  make all unstated assumptions explicit and prove them first if possible.  Add postulates if necessary.  

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Aside: Fixing Euclid’s Elements• In the late 1800's David Hilbert developed a set of 20 axioms that 

made explicit all the assumptions needed to complete Euclid’s program.  Like Euclid’s axioms, Hilbert’s were synthetic, that is, they did not depend on properties of the real numbers made explicit though coordinates (as in Descartes’ analytic geometry).  

• In 1932, Birkhoff developed a system of axioms that made explicit use of real numbers (e.g. made lines essentially equivalent to real‐number lines).  

• In the 1960's the School Mathematics Study Group (SMSG) developed set of axioms somewhat like Birkhoff’s, but especially suited to high school study.  They differed in that they were not a “minimal” list of axioms but included some axioms that could be proved from the others in order to avoid more tedious technical proofs.  

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Other Axiom Systems for GeometryDeveloper Axiom Set Comments

Hilbert 20 Axioms “Fixed” EuclidDoes not explicitly depend on Real Numbers, Set Theory, etc.

Many AxiomsDetailed Technical Proofs of the ObviousImplicitly depends of the real numbers and Set Theory

Birkhoff Only 4 AxiomsIncorporates Real Numbers and MeasurementMore accessible

Only Euclidean (fourth axiom is about similar triangles)

MacLane 14 Axioms including “continuity axiom”

SMSG 22 Axioms Designed for High School Geometry

Some unnecessary axioms

UCSMP Lots of Axioms RedundancyUses Reflection

Venema(a current text)

12 Axioms Explicitly uses real numbers and set theory.  Probably closest to Birkhoff’s.

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David Hilbert

• 1862 (Königsberg) – 1943 (Göttngen) 

• Grundlagen der Geometrie (1899) 

• 23 Unsolved Problems (1900)

• Modern Axiomatics and Proof Theory

• Wir müssen wissen. Wir werden wissen.

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Is it just me, or do these two look alike?

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OK, Back to the Elements

• As an overview of Book I, let’s look at how theorems build to the proof of the Pythagorean Theorem in Proposition 47.

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• Prop. I.4 – If two triangles have two sides of one triangle equal to two sides of 

the other triangle plus the angle between the sides that are equal in each triangle is the same, then the two triangles are congruent

Moving Toward I.47  ‐ I.4

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Moving Toward I.47 – I.14

• Prop. I.14– Two adjacent right angles make a straight line.

– Definition 10 asserted the converse, that a perpendicular erected on a straight line makes two right angles.

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• Prop. I.41– The area of a triangle is one half the area of a parallelogram with the same base and height.

Moving Toward I.47 – I.41

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Moving Toward I.47 – I.31

• Prop. I.31– Given a line and a point not on the line, a line through the point can be constructed parallel to the first line.

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• Prop. I.46– Given a straight line, a square can be constructed with the line as one side.

Moving Toward I.47 – I.46

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And Now: Proposition I.47

• In right‐angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

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Proof of Proposition I.47

• Draw a line parallel to the sides of the largest square, from the right angle vertex, A, to the far side of the square subtending it, L.

• Connect points F & C, and A & D, making ∆FBC and ΔABD

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• The two shaded triangles are congruent (by Prop. I.4) because the shorter sides are respectively sides of the constructed squares and the angle between them is an angle of the original right triangle, plus a right angle from a square.

Proof of Proposition I.47

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• The shaded triangle has the same base (BD) as the shaded rectangle, and the same height (DL), so it has exactly half the area of the rectangle, by Proposition I.41. 

Proof of Proposition I.47

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• Similarly, the other shaded triangle has half the area of the small square since it has the same base (FB) and height (GF). (Here is where he needed that G, A and C were all on one line. 

Proof of Proposition I.47

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• Since the triangles had equal areas, twice their areas must also be equal to each other (Common Notion 2), hence the shaded square and rectangle must also be equal to each other.

Proof of Proposition I.47

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• By the same reasoning, triangles constructed around the other non‐right vertex of the original triangle can also be shown to be congruent. 

Proof of Proposition I.47

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• And similarly, the other square and rectangle are also equal in area.

Proof of Proposition I.47

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• And finally, since the square across from the right angle consists of the two rectangles which have been shown equal to the squares on the sides of the right triangle, those squares together are equal in area to the square across from the right angle.

Proof of Proposition I.47

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Proof of Proposition I.47

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Just for fun, James Garfield’s proof:

• Start with a right triangle with legs of length b and a, and hypotenuse of length c.  Extend the side of length b to length a + b, and construct a perpendicular  of length b.  Form segments as shown.

a

cb

b

a c

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Garfield’s Proof

• Now the area of the trapezoid formed can be calculated in two different ways: as a trapezoid, the area is ½ the sum of the bases, times the height:

a

cb

b

a c

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Garfield’s Proof

• On the other hand, we can calculate the area of the half square of length c and add it to the sum of the two triangles: 

a

cb

b

a c

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Garfield’s Proof

So, 

So, ya think Bush or Obama or Mitt or Newt or… could do that? a

cb

b

a c

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Book II: Geometric Algebra

• An example of geometrical algebra is finding the square root of a magnitude by geometric construction:

• Let AB = a, and extend AB to C with BC = 1. 

• , so  .  

1a

x

D

MA

CB

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Proposition II‐5

• If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half.

• Huh?

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Proposition II‐5

• If a straight line 

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Proposition II‐5

• If a straight line is cut into equal and unequal segments,

a xy

MA BC

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Proposition II‐5

• If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole

x

y

x

a xy

MA BC

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Proposition II‐5

• If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section

a

a

x

y

x

a xy

MA BC

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Proposition II‐5

• If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half.

x

xa

a

x

y

x

a xy

MA BC

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Proposition II – 5 – Why?

• Well, since the length of segment AM is the same as x+a, this is an x by x+a rectangle.

x

xa

a

x

y

x

a xy

MA BC

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Proposition II – 5 – Why?

• This is also an x by x+a rectangle.

x

xa

a

x

y

x

a xy

MA BC

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Proposition II – 5 – Why?• So the rectangle on the two unequal segments, the left hand purple together with the brown rectangle, added to the small grey square, will add up to the square on the half (right‐hand purple together with brown and grey).

x

xa

a

x

y

x

a xy

MA BC

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OK, Where’s the Algebra?

• Suppose we have the two equations 

which are typical of Babylonian problems that gave rise to what we would call quadratic equations.  

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OK, Where’s the Algebra?

In the diagram, D is the area of the brown rectangle, and b is the length of the line segment  .  

a

a xy

MA BC

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OK, Where’s the Algebra?

Also,  and  .  Then II‐5 

says that  , or  .

a

a xy

MA BC

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OK, Where’s the Algebra?

So we have  and  written in terms of  (given) and  .  So, to find our  and  we just need to construct the value a somehow.  This can be done with ruler and compass.

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Finding  :

M

1D

x

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Finding a:

22

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Finally, 

• Since  and  it is now easy to construct segments of these two lengths.  

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• One notion, discussed extensively in Book V, was Eudoxus’ notion.  This was applied to magnitudes in Book VI where Euclid deals with similar figures. 

• For both magnitudes and numbers, there was a different notion, as discussed in Books VII and X.  This used Euclidean algorithm, and is our old friend anthyphairesis.

Ratios and Proportion in Euclid

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• Prime, composite.  • Every composite number is “measured” by some prime number.

• Primes are the basic “measures’ of other numbers.

• Many results of “elementary” number theory.• The infinitude of primes (IX – 20).

Number Theory, Books VII ‐ IX

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The Infinitude of Primes

• You should really know this proof for the good of your soul. 

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• Circles are to one another as the squares on their diameters.

• Spheres are to one another in the triplicate ratio of their respective diameters.

• Any cone is a third part of a cylinder which as the same base with it and equal height.

• Notice:  Not formulas.

“Areas” and “Volumes”

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Regular Polyhedra

TetrahedronCube

Octahedron

Dodecahedron Icosahedron

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The Elements

• Why are the Elements so important?• What legacy did they give us?

1. Axiomatic development of mathematics.2. Next to the Bible, probably the most 

published book ever.3. Numerous mathematicians first “met” 

mathematics by reading Euclid.