Euclid s Elements Part II

Transcription

1 Euclid s Elements Part II

2 The discovery of incommensurable magnitudes steered the ancient Greeks away from the study of number and towards the development of geometry. s a result, geometry was pushed in directions more easily accommodated by algebra.

3 Propositions (about alternating, separating, combining, converting, and compounding ratios) are easy to deal with algebraically, but pose a challenge geometrically. Euclid needed these results to prove claims about similar figures.

4 Proposition 5.36: Triangles and parallelograms with the same height are to each other as the bases. Proposition 5.37: line drawn parallel to the side of a triangle cuts the sides proportionally, and conversely, if a line cuts two side of a triangle proportionally, it is parallel to the third side. What do these propositions mean? How can the top one be used to prove the bottom one?

5 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. D E B

6 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. D E B

7 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. Then triangle BDE equals triangle DE since they are between the same parallels on the same base. D E B

8 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. Then triangle BDE equals triangle DE since they are between the same parallels on the same base. D E This means BDE/DE equals DE/DE. B

9 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. Then triangle BDE equals triangle DE since they are between the same parallels on the same base. B D E This means BDE/DE equals DE/DE. We know BDE/DE equals BD/D since triangles of the same height are to each other as the bases.

10 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. Then triangle BDE equals triangle DE since they are between the same parallels on the same base. B D E This means BDE/DE equals DE/DE. We know BDE/DE equals BD/D since triangles of the same height are to each other as the bases. Similarly DE/DE equals E/D.

11 laim: line drawn parallel to the side of a triangle cuts the sides proportionally. Given triangle B, draw DE parallel to B. We want to show that BD is to D as E is to E. Proof. Draw BE and D. Then triangle BDE equals triangle DE since they are between the same parallels on the same base. B D E This means BDE/DE equals DE/DE. We know BDE/DE equals BD/D since triangles of the same height are to each other as the bases. Similarly DE/DE equals E/D. So BD/D equals BDE/DE, which equals DE/DE, which equals E/D, as desired.

12 laim. If a line cuts two sides of a triangle proportionally, it is parallel to the third side. Given triangle B, draw DE so that BD is to D as E is to E. We want to show that DE is parallel to B. How can we show that one line is parallel to another? D E B

13 laim. If a line cuts two sides of a triangle proportionally, it is parallel to the third side. Given triangle B, draw DE so that BD is to D as E is to E. We want to show that DE is parallel to B. How can we show that one line is parallel to another? B D E Either show that the alternate interior angels are equal (Proposition 5.8), or show that the interior angles equal two right angles (definition of parallel lines).

14 Proposition 5.38: In triangle B, if angle B is bisected by D, then BD is to D as B is to. How can we use Proposition 5.37 to prove this? B D

15 Proposition 5.38: In triangle B, if angle B is bisected by D, then BD is to D as B is to. How can we use Proposition 5.37 to prove this? B D E Draw a line through parallel to D, and extend B to meet this line at E. What does 5.37 allow you to conclude? What do you have to prove next?

16 Proposition 5.39: In triangles whose corresponding angles are equal, the sides opposite the equal angles are proportional. How do we normally refer to this theorem? We can prove it using Proposition 5.37.

17 Proposition 5.37 is also used to develop algebra geometrically, beginning with arithmetic. How do we add, subtract, multiply and divide in a geometric setting? It is easy to add or subtract two real numbers (lengths) geometrically (using a compass and straightedge). It is also natural to multiply a given real number (length) by a whole number. With Proposition 5.40 Euclid shows how to do divide a real number by a whole number

18 What algebraic operation does Proposition 5.41 show us how to perform geometrically? B Proposition 5.41: Given lines B,, find E so B is to as is to E. E

19 What algebraic operation does Proposition 5.41 show us how to perform geometrically? B Proposition 5.41: Given lines B,, find E so B is to as is to E. E Take B to be a unit length, and let be the length you re interested in. What length does E have in relation to?

20 What algebraic operation does Proposition 5.41 show us how to perform geometrically? B Proposition 5.41: Given lines B,, find E so B is to as is to E. E Take B to be a unit length, and let be the length you re interested in. What length does E have in relation to? E = 2

21 How can we find the square of a number geometrically? Set up B and as sides of a triangle. B E B

22 How can we find the square of a number geometrically? Set up B and as sides of a triangle. B E D B Extend B to D so that BD =.

23 How can we find the square of a number geometrically? Set up B and as sides of a triangle. B B E D E Extend B to D so that BD =. Draw a line through D parallel to B, and extend to meet this line at E.

24 How can we find the square of a number geometrically? Set up B and as sides of a triangle. B B E D E Extend B to D so that BD =. Draw a line through D parallel to B, and extend to meet this line at E. Then B/BD = /E (Proposition 3.37), so B/ = BD/E (Proposition 5.31), and therefore B/ = /E, as desired.

25 What algebraic operations do Proposition 5.42 and 5.43 show us how to perform geometrically?

26 ertainly the Greeks could multiply, divide, square, and take the square root of numbers as usual. Why did they go through all the trouble of doing so with numbers interpreted as lengths of line segments?

27 Proposition 5.44: In two triangles with one equal angle, if the triangles are equal, then the side about the equal angle are reciprocally proportional, and so conversely. What does this tell mean? Suppose the equal angle is at. D B E

28 Proposition 5.44: In two triangles with one equal angle, if the triangles are equal, then the side about the equal angle are reciprocally proportional, and so conversely. What does this tell mean? Suppose the equal angle is at. /D D = B E E/B

29 What does Proposition 5.45 mean? D B E F Proposition 5.45: Similar triangles are to each other as the duplicate ratios on their corresponding sides.

30 What does Proposition 5.45 mean? D B E F If B and DEF are similar triangles, then B and EF are corresponding sides (as are B and DE, and and DF). The proof works with B and EF, though the argument works for any corresponding sides. The duplicate ratio on sides B and EF is a length S with the property that B/EF = EF/S. Proposition 5.45 says that if B and DEF are similar, then B/DEF = B/S.

31 Where does S appear in the proof? D B G E F Since B and DEF are similar, we have angles B = E. onstruct BG so that B/EF = EF/BG (5.41), and join G. Since B and DEF are similar, B/DE = B/EF. ombining the above, B/DE = EF/BG. BG and DEF have one angle equal and the sides about the equal angle reciprocally proportional, so they are equal (5.44). B and BG have the same height, so B/BG = B/BG (5.36). ombining the above B/DEF = B/BG = B/BG.

32 The ancient Greeks used geometric constructions to solve quadratic equations. What is a quadratic equation?

33 The ancient Greeks used geometric constructions to solve quadratic equations. What is a quadratic equation? n equation of the form ax 2 + bx + c = 0 has different types depending on the signs of a, b, and c. When solving equations geometrically, all numbers must be positive because they are represented as lengths of line segments. Given this constraint, what different types of quadratic equations are there?

34 The construction below is an example of a solution to the equation (a x)bx = c. area = c bx x a

35 The ancient Greeks could solve several types of quadratic equations geometrically. In order to understand how their solutions worked we need to know that Euclid could construct a figure with the same area as a given figure B and similar to a third figure (provided B had straight sides). This was the application of areas introduced by Pythagoras. B

36 We also need to know that if two parallelograms are similar, and we line them up, then their diameters will match (and vice versa).

37 We also need to know that if two parallelograms are similar, and we line them up, then their diameters will match (and vice versa).

38 What does Proposition 5.46 tell us about the parallelogram below? L D M E G N F H K B

39 What does Proposition 5.46 tell us about the parallelogram below? L D M E G N F H K B The meaning is easier to understand if we visualize F sliding along DB.

40 Which quadratic equation does Proposition 5.47 solve?

41 Given a line, a parallelogram, and a rectilinear figure. area = c b 1 a

42 To construct on the line a parallelogram equal to the rectilinear figure and deficient by a parallelogram similar to the given parallelogram. area = c b 1 a a x x bx a

43 To construct on the line a parallelogram equal to the rectilinear figure and deficient by a parallelogram similar to the given parallelogram. area = c b 1 a (a x)bx = c a x x bx a

44 This is the example we saw earlier. The construction determines the solution x. How? area = c bx x a

45 The book suggests we solve quadratic equations with rectangles rather than more general parallelograms. s you work through the construction given in the proof of Proposition 5.47, you will realize it boils down to the four steps given on page 119. How do we know step 2 is possible? Step 3 is possible for the same reason.

46 Let s see how the construction works by following Example 5.3 Example 5.3: Given a line 8 feet in length, a figure with an area of 14 square feet, and a rectangle whose sides have a ratio of 2 to 1, find the rectangle on the line. Which quadratic equation is being solved with this construction? What s the solution?

47 What restriction does this construction impose on a, b, and c? (a x)bx = c bx a x x a

48 What general type of quadratic equation does the construction for Proposition 5.48 solve?

49 Given a line, a parallelogram, and a rectilinear figure. area = c b a 1

50 To construct on the line a parallelogram equal to the rectilinear figure and exceeding the line by a figure similar to the given parallelogram. area = c b a 1 bx a x

51 To construct on the line a parallelogram equal to the rectilinear figure and exceeding the line by a figure similar to the given parallelogram. area = c b a 1 bx a x (a + x)bx = c

52 Do discussion questions 3, 5, and 6.

53 What does it mean to say that a number is perfect? What makes a perfect number Euclidean?

54 To obtain Euclidean perfect numbers, we have to compute sums of the form b + b 2 + b b n 1 (with b = 2). What did Euclid observe about such sums?

55 Proposition 5.50: ny composite number is measured by some prime. What does this mean? What is the difference between this proposition and the Fundamental Theorem of rithmetic?

56 Proposition 5.51: The number of primes exceeds any given number. What does this mean? Why is it true?

57 Book XII of the Elements contains results demonstrated using the method of exhaustion developed by Eudoxus. What was being exhausted with this method?

58 To apply the method of exhaustion we need Proposition 5.54, the axiom of rchimedes. What does it say? The axiom applies to magnitudes of any dimension (not just lengths). Why does it require that the quantity subtracted be more than half of the given, or remaining, magnitude at each step of the process?

59 To apply the method of exhaustion we also need the fact that similar polygons inscribed in a circle are to each other as the squares on their diameters.

60 Proposition 5.55: ircles are to each other as the squares on their diameters. In hapter 4 this proposition was stated as a conjecture. Prior to Eudoxus, geometers had observed the result empirically, but had been unable to prove it deductively. Euclid s proof appears in Book XII of the Elements, and uses the method of exhaustion.

61 Proposition 5.55: ircles are to each other as the squares on their diameters. Given circles with areas and B and diameters with lengths a and b, we will show /B = a 2 /b 2. The proof is by contradiction. a B b

62 Suppose that a 2 /b 2 /B. Then a 2 /b 2 = /S for some area S B. First suppose S < B. We will show that we can inscribe a polygon with area Q in the circle with area B, such that S < Q < B. To do so, we begin by inscribing a square. The area of the square is greater than half of B. Bisect the arcs on the circle with area B, and join them to the corners of the inscribed square. Each triangle formed inside a segment of the circle has area greater than half that of the segment. ontinuing this process yields a polygon Q with area greater than S. (The axiom of rchimedes makes this possible.) a B b S

63 Suppose that a 2 /b 2 /B. Then a 2 /b 2 = /S for some area S B. First suppose S < B. We will show that we can inscribe a polygon with area Q in the circle with area B, such that S < Q < B. To do so, we begin by inscribing a square. The area of the square is greater than half of B. Bisect the arcs on the circle with area B, and join them to the corners of the inscribed square. Each triangle formed inside a segment of the circle has area greater than half that of the segment. ontinuing this process yields a polygon Q with area greater than S. (The axiom of rchimedes makes this possible.) a B b S

64 Suppose that a 2 /b 2 /B. Then a 2 /b 2 = /S for some area S B. First suppose S < B. We will show that we can inscribe a polygon with area Q in the circle with area B, such that S < Q < B. To do so, we begin by inscribing a square. The area of the square is greater than half of B. Bisect the arcs on the circle with area B, and join them to the corners of the inscribed square. Each triangle formed inside a segment of the circle has area greater than half that of the segment. ontinuing this process yields a polygon Q with area greater than S. (The axiom of rchimedes makes this possible.) a B b S

65 Inscribe in the circle with area a polygon with area P similar to the polygon with area Q. We are assuming a 2 /b 2 = /S. We know similar inscribed polygons are to each other as the squares on their diameters, so P/Q = a 2 /b 2. This means /S = P/Q, which implies /P = S/Q (Proposition 5.31). Since is larger than P, we must also have S larger than Q. But we constructed Q so that it was larger than S. We have reached a contradiction. It must not be true that S < B. a B b S Q P

66 In a similar way we can show that S > B leads to a contradiction. This means the original assumption that /B a 2 /b 2 is false, and we can conclude that circles are to each other as the squares on their diameters. How does Proposition 5.55 suggest the existence of π? How can it be translated into an area formula for circles?

67 The method of exhaustion can also be used in three dimensions to compute the volumes of cylinders and cones. Instead of inscribing polygons, we inscribe prisms and pyramids.

68 How can our modern formula for the volume of a pyramid be derived from Proposition 5.59? Proposition 5.59: ny prism with a triangular base can be divided into three equal pyramids.

69 In the last book of the Elements, Euclid shows how to construct the five regular solids.

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