EUCLID:  THE ELEMENTS
History
 

1. BACKGROUND TO THE PERIOD

The death of Alexander the Great led his generals to fight with each other and the Greek territory was divided up. Egypt came under the control of Ptolemy I, Seleucus and Lysimachus took Syria and the East while Antigonus and Cassander each ruled Macedon. Ptolemy set up a school in Alexandria known as the Museum, and here he managed to bring together the leading scholars of the day. The centre of culture moved from Athens to Egypt. Euclid, who was among the best teachers at the school, was also the author of the most renowned mathematics textbook ever written: The Elements.

Little is known of Euclid's life. He is known as Euclid of Alexandria as this is the place where he developed his work. In this work he took on the task of gathering together all the fundamentals of elementary mathematics and gave an explanation of each one. His intention was to include those results which could be deduced, through logical reasoning, using the least number of statements possible and which always had a didactic purpose.

The Elements is divided into 13 books: 6 on elementary plane geometry (4 focus on triangles, parallel lines, area, geometric algebra, circles and inscribed and circumscribed shapes in a circle. The other 2 focus on the theory of proportion  and similar shapes), 4 on the theory of numbers (which includes one book about incommensurability) and 3 on solid geometry (geometric solids, the measurement of figures and regular solids).

The book opens with 23 definitions (a point, a line, a surface area, a straight line, a right angle, an acute angle, an obtuse angle, a perpendicular etc.), a list of 5 postulates and another list of 5 common notions (we will call the 10 assumptions, which are so obvious that they hardly require any explanation, axioms). From this point on all the books include propositions which can be proved and are done so by referring back to these axioms or earlier propositions.

Let's see a few examples:

2.- THE CONSTRUCTION OF AN EQUILATERAL TRIANGLE ON A GIVEN SEGMENT: BOOK1 PROPOSITION 1

In this window you will see how an equilateral triangle can be constructed from a given segment.

Abbreviations are used in the demonstrations to refer to earlier questions, making them easier to follow. The Roman numerals refer to the book, Def n indicates a definition and its number, Post n indicates a postulate and its number, Prop n indicates a proposition and its number and CN n indicates a common notion and its number. You can change the values of A and BC.

1.- THE PROPOSITION:  If AB is the given segment construct an equilateral triangle ABD from the segment AB.

2.- THE PROOF:

Draw a circle of centre A and radius AB Post 3 

Draw a circle of centre B and radius AB  Post 3 

Draw segments BC and AC, C being the point where the circles intersect      Post 1

AC=AB and

BC=BA as they are both radii of the same circle  def 15

and AB=BA as the ends of the lines are the same points Post 1

And as things which are equal to the same thing are also equal to one another.                          CN 1

Therefore AC=BC =AB and the triangle is equilateral.


3. THE CONSTRUCTION OF A SEGMENT FROM ANY POINT A WHICH IS EQUAL TO ANOTHER SEGMENT BC: BOOK 1 PROPOSITION 2
In this window you can see the construction of a rectilinear segment AH from a given point A which is equal to another given segment BC.
1.- THE PROPOSITION:  If BC is the given segment and A is the given point construct a rectilinear segment AH which is equal to segment BC.

2.- THE PROOF:

Draw segment AB                         Post 1

Draw the equilateral triangle ABD          Prop1 

Extend the segments DA and DB            Post 2

Construct the circle of centre B and radius BC and the circle of centre D and radius DG              Post 3

BC=BG and

DH=DG as they are both radii of the same circle   def 15

BG=AH  as we are subtracting equal segments from equal segments        CN 3

But things which are equal to the same thing are also equal to one another.                      CN 1

Therefore AH=BC is the required segment


4. THE DISTRIBUTIVE LAW: BOOK 2 PROPOSITION 1

If there are two straight lines A and BC and BC is cut into any number of segments whatever, then the rectangle contained by the two lines is equal to the sum of the rectangles contained by the uncut straight line A and each of the segments of line BC.

In this window you can see an example given of this law with two segments. It is the most commonly used number-theory law found in text books today. Euclid explained it geometrically but its algebraic equivalent is: a(b+c)=ab+ac

To change the values of A and BC drag the blue points with the left-hand button on the mouse. If you click on the right-hand button on the mouse a menu will open which allows you to centre the diagram.

THE PROOF: Start with lines A and BC and divide BC into 2 parts

Draw a right angle from segment BC  I,11

Make BG equal to A and draw GH parallel to BC      I,3

From D and C draw DE and CH which are parallel to BG     I,31

Therefore the rectangle BH is equal to the sum of rectangles BE and DH since:

Its sides are BC and A so BG=A   II, Def 1; I, 34

The sides of BE are BD and A so GE=BD and BG=A

Similarly the sides of DH are DC and A 

Therefore the rectangle with sides A & BC is equal to the sum of the rectangles of sides A & BD and A & DC


       
           
  Rosa Jiménez Iraundegui
 
Spanish Ministry of Education. Year 2001