Math 1230, Notes 2. Aug. 28, Math 1230, Notes 2 Aug. 28, / 17

Math 1230, Notes 2 Aug. 28, 2014 Math 1230, Notes 2 Aug. 28, 2014 1 / 17

This fills in some material between pages 10 and 11 of notes 1. We first discuss the relation between geometry and the quadratic formula. We can look for only real solutions. We start with Proposition 2.5. As stated above, it says that (a + b) (a b) = a 2 b 2. From the geometric figure in class, or pg. 122 of Boyer, figure 7.6, we see that this formula is only proved for 0 < b < a. Math 1230, Notes 2 Aug. 28, 2014 2 / 17

Now suppose we want to solve x 2 px + q 2 = 0. Since geometric solutions can only be real numbers, Euclid can only find the solution when q < 1 2p. We will also assume that q > 0. We draw AB of length p. Let a = 1 2p. Let C bisect AB. So AC = CB = a = 1 2p. Construct the perpendicular to AB from C, say down, of length q, to a point M. Draw the circle of radius a = 1 2 p around M. Since q < 1 2p, this circle will intersect CB at a point D between C and B. (Note that the diagonal from M to B must be of length longer than 1 2 p, while the vertical line from M to C is of length less than 1 2 p. I claim that x = DB solves x 2 px + q 2 = 0. To see this, we let b = CD. We already saw that a = 1 2p. From the Pythagorean theorem, b 2 + q 2 = a 2, so b = ( 1 2 p) 2 q 2. Then x = CD = a b = 1 2 p (1 2 p ) 2 q 2 which is one of the solutions we get using the quadratic formula. Math 1230, Notes 2 Aug. 28, 2014 3 / 17

We now repeat the statement of Book 2, proposition 6: If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole (with the added straight line) and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line. We will apply this in the next item. book 2, prop 11 To cut a straight line so that the rectangle contained by the whole and one of the pieces is equal to the square on the remaining piece. The construction is shown in the figure in the Elements, page 63 of the version on the class website, which will be explained in class. Math 1230, Notes 2 Aug. 28, 2014 4 / 17

prop 12 Math 1230, Notes 2 Aug. 28, 2014 5 / 17

prop 12 In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the sum of the squares on the other two sides by twice the rectangle contained by one of the sides around the obtuse angle to which a perpendicular falls and the straight line cut off outside the triangle by the perpendicular towards the obtuse angle. Math 1230, Notes 2 Aug. 28, 2014 5 / 17

book 9, prop 35 Math 1230, Notes 2 Aug. 28, 2014 6 / 17

book 9, prop 35 If there is any multitude whatsoever of continually proportional numbers, and numbers equal to the first are subtracted from both the second and the last, then as the excess of the second number is to the first, so the excess of the last will be to the sum of all the numbers before it. Math 1230, Notes 2 Aug. 28, 2014 6 / 17

Before continuing, make sure you have read the final parts of notes 1. Math 1230, Notes 2 Aug. 28, 2014 7 / 17

Hilbert s system Math 1230, Notes 2 Aug. 28, 2014 8 / 17

Hilbert s system Undefined: Point, Line, Incidence Lies on, contains, between, congruent Math 1230, Notes 2 Aug. 28, 2014 8 / 17

Hilbert s system Undefined: Point, Line, Incidence Lies on, contains, between, congruent Axioms: Math 1230, Notes 2 Aug. 28, 2014 8 / 17

Hilbert s system Undefined: Point, Line, Incidence Lies on, contains, between, congruent Axioms: Axioms of incidence Math 1230, Notes 2 Aug. 28, 2014 8 / 17

Hilbert s system Undefined: Point, Line, Incidence Lies on, contains, between, congruent Axioms: Axioms of incidence Postulate I.1. For every two points A, B there exists a line a that contains each of the points A, B. Postulate I.2. For every two points A, B there exists no more than one line that contains each of the points A, B. Math 1230, Notes 2 Aug. 28, 2014 8 / 17

Postulate I.4. For any three points A, B, C that do not lie on the same line there exists a plane α that contains each of the points A, B, C. For every plane there exists a point which it contains. Postulate I.5. For any three points A, B, C that do not lie on one and the same line there exists no more than one plane that contains each of the three points A, B, C. Postulate I.6. If two points A, B of a line l lie in a plane α then every point of l lies in the plane α. Math 1230, Notes 2 Aug. 28, 2014 9 / 17

Axioms of Order Math 1230, Notes 2 Aug. 28, 2014 10 / 17

Axioms of Order Postulate II.1. Math 1230, Notes 2 Aug. 28, 2014 10 / 17

Axioms of Order Postulate II.1. If a point B lies between a point A and a point C then the points A, B, C are three distinct points of a line, and B then also lies between C and A. Math 1230, Notes 2 Aug. 28, 2014 10 / 17

Axioms of Order Postulate II.1. If a point B lies between a point A and a point C then the points A, B, C are three distinct points of a line, and B then also lies between C and A. Postulate II.2. For two points A and C, there always exists at least one point B on the line AC such that C lies between A and B. Math 1230, Notes 2 Aug. 28, 2014 10 / 17

Postulate II.4. Math 1230, Notes 2 Aug. 28, 2014 11 / 17

Postulate II.4. Let A, B, C be three points that do not lie on a line and let a be a line in the plane ABC which does not meet any of the points A, B, C. If the line a passes through a point of the segment AB, it also passes through a point of the segment AC, or through a point of the segment BC. Math 1230, Notes 2 Aug. 28, 2014 11 / 17

Axioms of Congruence Math 1230, Notes 2 Aug. 28, 2014 12 / 17

Axioms of Congruence Postulate III.1. Math 1230, Notes 2 Aug. 28, 2014 12 / 17

Axioms of Congruence Postulate III.1. If A, B are two points on a line l, and A is a point on the same or on another line l then it is always possible to find a point B on a given side of the line l such that AB and A B are congruent. Postulate III.2. If a segment A B and a segment A B are congruent to the same segment AB, then segments A B and A B are congruent to each other. Postulate III.3. On a line a, let AB and BC be two segments which, except for B, have no points in common. Furthermore, on the same or another line l, let A B and B C be two segments which, except for B, have no points in common. In that case if AB A B and BCB C, then ACA C. Math 1230, Notes 2 Aug. 28, 2014 12 / 17

Postulate III.4. Math 1230, Notes 2 Aug. 28, 2014 13 / 17

Postulate III.4. If ABC is an angle and if B C is a ray, then there is exactly one ray B A on each side of line B C such that A B C = ABC. Furthermore, every angle is congruent to itself. Math 1230, Notes 2 Aug. 28, 2014 13 / 17

Postulate III.4. If ABC is an angle and if B C is a ray, then there is exactly one ray B A on each side of line B C such that A B C = ABC. Furthermore, every angle is congruent to itself. Postulate III.5. (SAS) If for two triangles ABC and A B C the congruences AB A B, AC A C and BAC B A C are valid, then the congruence ABC A B C is also satisfied. Math 1230, Notes 2 Aug. 28, 2014 13 / 17

Axiom of Parallels Math 1230, Notes 2 Aug. 28, 2014 14 / 17

Axiom of Parallels Postulate IV.1. Math 1230, Notes 2 Aug. 28, 2014 14 / 17

Axiom of Parallels Postulate IV.1. Let l be any line and A a point not on it. Then there is at most one line in the plane that contains l and A that passes through A and does not intersect l. Axioms of Continuity Postulate V.1. (Archimedes Axiom) If AB and CD are any segments, then there exists a number n such that n copies of CD constructed contiguously from A along the ray AB willl pass beyond the point B. Math 1230, Notes 2 Aug. 28, 2014 14 / 17

Postulate V.2. (Line Completeness) Math 1230, Notes 2 Aug. 28, 2014 15 / 17

Postulate V.2. (Line Completeness) An extension of a set of points on a line with its order and congruence relations that would preserve the relations existing among the original elements as well as the fundamental properties of line order and congruence (Axioms I-III and V-1) is impossible. Alternative to V2: Dedekind s axiom: Suppose that the set of all points on a line l is the union of two nonempty sets Σ 1 and Σ 2 such that no point of Σ 1 is between two points of Σ 2 and viceversa. Then there is a unique point O of l such that O lies between two points P and Q if and only if one of these points is in Σ 1 and the other is in Σ 2 and neither is equal to O. (Basis for a definition of real numbers Dedekind cut ) Math 1230, Notes 2 Aug. 28, 2014 15 / 17

Homework, due at beginning of class on 9/4. All problems are worth 5 points unless stated otherwise. 1. Suppose that l is a line and p a point not on l. Prove using Hilbert s axioms of incidence that there is a unique plane containing l and p. 2. Use the Euclidean algorithm to f ind the greatest common divisor of 567, 759, and 839. Comment on anything atypical that you run into. Math 1230, Notes 2 Aug. 28, 2014 16 / 17

3. On the next page you will find page 122 of Boyer s book. In the section beginning If a Greek scholar.. on line 10, he describes how to use Euclid s axioms and propositions to solve x 2 ax + b 2. Using the propositions, you could construct this solution using a straight edge and compass, if you were presented with line segments of lengths a and b. (Recall that you cannot measure anything numerically with straight edge and compass.) On page 3 of these notes I give another description of the same method. Note that I do not have to refer specifically to proposition 2.5. Now do problem 5 of Boyer, on page 132. It says: Given line segments a and b, construct with straight edge and compass alone a solution of the equation x 2 = ax + b 2. (You are not meant do actually do it. Explain how you would find x in the style of the book, or of my description on page 3. You do not have to use any result like propositions 2.5 or 2.6. You just need to use the theorem of Pythagoras. Stated geometrically, this says that if A,B,C is a right triangle, then the square with the hypotenuse as one side is equal (in area) to the (sum of the areas of) two squares on the other two sides.) Math 1230, Notes 2 Aug. 28, 2014 17 / 17