Math 409/409GHistory of Mathematics

Book I of the Elements

Part I

In Book I of the Elements, Euclid presented and proved 48 propositions dealing with straight lines, triangles, and parallelograms.

As you will see, not all of Euclid’s proofs were flawless. However, the validity of the 465 propositions in the 13 Books of the Elements have withstood the test of time.

In the last lesson you saw that Euclid’s use of Common Notion #4 in his proof of Proposition #4 was so objectionable to some mathematicians, that today, many take Proposition #4 as Axiom #6.

CN 4 Things which coincide with one another are equal to one another.

Ax. 6 SAS: If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the triangles are congruent.

Here’s another example of a proof by Euclid that resulted in the need for another axiom.

Proposition #1

Given a segment AB, there exists (you can construct) an equilateral triangle having AB as one of its sides.

As I present Euclid’s proof of this proposition, try to see if you can find the flaw.

Euclid’s proof of Prop. 1

Let segment AB be given.

Construct a circle having center B and radius AB.

Construct a circle having center A and radius AB.

BA

BA

Let C be the point where the two circles intersect.

Construct segments AC and BC.

C

BA

C

BA

It then follows from the definition of a circle that

and

So ABC is equilateral.

Def. 15: A circle is the set of all points that are a given distance from a given point.

Where’s the flaw in this proof?

C

BA

AC AB .BC AB

Let’s revisit this proof using the “Statement/Reason” display for a proof. Bear in mind that since this was Euclid’s first proof, the only reasons he could use were his axioms, common notions, and definitions.

Statement Reason1. Let segment AB be given. Given2. Construct a circle having Ax. 3 center B and radius AB.3. Construct a circle having Ax. 3

center A and radius AB.4. Let C be the point where the two

circles intersect.5. Construct segments AC and BC. Ax. 16. and Def: circle7. Triangle ABC is equilateral. Def: equil.

AC AB .BC AB

C

BA

Where’s the problem?

It’s in the step where Euclid assumed that the two circles he constructed would intersect. He has no axioms or common notions to back this up. All he had was a figure which misled him into believing that the circles must intersect.

An axiom is needed to justify the existence of this point of intersection.

C

BA

Postulate of Continuity

Ax. 7 If a circle or line has one point inside and one point outside another circle, then the circle or line intersects the other circle in two points.

DD

C

C

B

B

AA

In our construction, the blue circle has one point inside and one point outside the green circle. So by axiom 7, the two circles intersect in two points, one of which is point C.

outside

inside

C

BA

Statement Reason1. Let segment AB be given. Given

2. Construct a circle having Ax. 3

center B and radius AB.

3. Construct a circle having Ax. 3 center A and radius AB.

4. Let C be the point where the two Ax. 7 circles intersect.

5. Construct segments AC and BC. Ax. 1

6. and Def: Circle

7. Triangle ABC is equilateral. Def: Equil.

AC AB .BC AB

C

BA

Euclid’s straightedge & compass

Ancient Greeks required that all constructions be done using only a straightedge and a compass.

The straightedge was a ruler without markings and Euclid’s first axiom guaranteed that you could use a straightedge to construct lines and segments.

Euclid’s third axiom (a circle can be described with any center and radius) guaranteed that you could construct any circle.

But this axiom didn’t guarantee that when you picked up the compass it would remain rigid, thus allowing you to transfer a length.

For example: today, if you want to transfer the length BC onto the segment AP, you’d place a compass at B, open it to the length BC, pick up the compass, place it on point A, and then construct a circle having center A and radius BC.

Euclid had no axiom guaranteeing that the compass would not collapse when lifted.

Q

C

BA

P

But Euclid didn’t need such an axiom since his second and third propositions cleverly showed how you could transfer a length without requiring that your compass remained rigid when lifted and moved to a new center.

P1.2 Given a point A and a segment BC, you can construct a segment AF such that

P1.3 Given two segments AP and BC with there is a point Q on AP such that

.AF = BC

,AP > BCAQ = BC.

Proof of Proposition #2

Given a point A and a segment BC, you can construct a segment AF such that

C

BA

.AF = BC

1. Construct segment AB. (Ax. 1)

2. Construct equilateral triangle ABD. (P1.1)

3. Then . (Def: equilateral .)DA = DB

3. DA = DBD

AB

C

4. Construct a circle centered at B and having radius BC. (Ax. 3)

5. Construct E, the point of intersection of ray DB and the circle. (Ax. 7)

6. Then . (Def: Circle)BC = BE

3. DA = DB

6. BC = BE

E

C

BA

D

7. Construct the circle centered at D and having radius DE. (Ax. 3)

8. Let F be the intersection of ray DA and the circle. (Ax. 7)

9. Then . (Def: Circle)DF = DE

3. DA = DB

6. BC = BE

9. DF = DE

F

D

AB

C

E

10.Then . (Steps 9 & 3 and CN 3)

11.So . (Steps 10 & 6 and CN 1)

11. AF = BC

9. DF = DE

E

C

BA

D

F

3. DA = DB

6. BC = BE

10. AF = BE

AF = BE

AF = BC

So we have shown that:

Given a point A and a segment BC, you can construct a segment AF such that .AF = BC

A

B

C

F

D

AB

C

AF = BC

E

Proof of Proposition #3

Given two segments AP and BC with there is a point Q on AP such that

,AP > BC

AQ = BC.

P

AB

C

1. Using point A and segment BC, construct a segment AF having the same measure as BC. (P1.2)

P

1. AF = BC

F

D

AB

C

E

P

AB

C

2. Construct a circle centered at A and having radius AF. (Ax. 3)

3. Let Q be the intersection of segment AP and the circle. (Ax. 7)

1. AF = BC

F

D

AB

C

Q

P

4. . (Def: Circle)

5. Then . (Steps 4 & 1 and CN 1)

AQ = AF

AQ = BC

5. AQ = BC

P

Q

C

BA

D

F

1. AF = BC

4. AQ = AF

This not only proves Proposition #3 (Given two segments AP and BC with

there is a point Q on AP such that ), it also justifies that you can assume that a compass is rigid. Q

C

BA

P

,AP > BCAQ = BC.

This ends the lesson on

Book I of the Elements

Part I