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: |
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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.
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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.
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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 | ||