Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Partial Solutions for 2nd Midterm I. Biographical Sketch – yours to do II. Proofs of Euclid’s Propositions: 4, 5, 10, 11, 16, 27 III. 1. Describe the Saccheri quadrilateral and prove, in neutral geometry, that the summit angles are congruent. A Saccheri quadrilateral has one distinguished side called the base from which the adjacent and equal edges are perpendicular to the base. These sides are called the arms of the quadrilateral. The side opposite the base is the summit. The angles formed by the arms and the base are the base angles – both are right angles. The other two angles are the summit angles. summit summit angle s arm arm base angles base Give n: Sacche ri quadrilate ral ABCD with base AB. To prove : C D Proof: Construct the diagonals AC and BD (Pos t. 1). Note that, by definition, A and B are both right angles and are congruent (Pos t. 4). Also, by definition, AD=BC. We have ABC BAD (SAS) AC = BD. So, ADC BCD D C. D C A B 2. What was Saccheri’s plan for “proving” the 5th postulate? Saccheri knew that Euclid’s 5th postulate and the existence of a rectangle were logically equivalent statements. Therefore, he reasoned, if one could prove in neutral geometry that a rectangle exists, then the 5th postulate would follow as a consequence and so would not be independent of the first four. Saccheri chose the quadrilateral that we have called a Saccheri quadrilateral to prove his case. He showed, as we did above, that the summit angles were congruent. If they were also right angles, then the SQ would be a rectangle. He posed the three possible hypotheses for the summit angles: 1) Both angles are obtuse, 2) both angles are acute, or 3) both angles are right angles. If he could show that cases 1 and 2 were impossible then, by elimination, the third case would have to hold. Saccheri easily disposed of the obtuse case, but worked much harder in coming up with a (as it turns out flawed) contradiction for the acute angle case. 3. State the Saccheri-Legendre Theorem and describe how Saccheri used it in his attempt to “prove” the 5th postulate. S-L Theorem: The angle sum of any triangle can not exceed two right angles. Saccheri noted that if the summit angles of a Saccheri quadrilateral ABCD were obtuse then at least one of the triangles ABC or ADC would have an angle sum exceeding 2 right angles, contrary to the S-L Thm. 4. Assuming Saccheri’s Acute Angle Hypothesis, show that the summit of a Saccheri quadrilateral is longer than the base. Give n: Sacche ri quadrilate ral ABCD with base AB. To prove : DC > AB. Proof: Construct the diagonals midline of the S.Q. MN which we know to be pe rpe ndicular to the summit and the base . Since M and N are midpoints of CD and AB, respective ly, then AB will be re late d (<, =, >) to CD as CM is related to BN. The re are thre e possibilitie s to conside r: 1) CM = BN (summit equals the base ) 2) CM < BN (summit shorter than the base ) 3) CM > BN (summit gre ater than the base ) We will e liminate the first two possibilitie s. Suppos e CM = BN. The n MNBC is a S.Q. with base MN. But this implie s that C B. But C is acute and B is right. This is a contradiction, cas e 1) does not hold. Suppos e CM < BN. The n the re is a point P on NB such that NP = MC. The n MNPC is a S.Q. with bas e MN. This implie s that MCP NPC. Since C is acute , then MCP is as well. On the other hand, mNPC > B (as it is an e xte rior angle of PBC, Prop. 16). So NPC is obtus e. Since an acute angle ( MCP) can't be congrue nt to an obtus e angle ( NPC), we have a contradiction. Thus, case 2) doe sn't hold eithe r. The refore, CM > BN and the s ummit is longer than the base . IV. D M A N C P B 1. Assuming you are given segments of length a, b, and 1, sketch out the basic constructions for each of the following: a ab etc. (See these notes ) 2. Define and give two examples for each of the following: Constructible number: A real number c is constructible if a segment of length |c| can be constructed from a unit segment using only a compass and straightedge. Examples: -3, 1, 5/2, √12 Field: A field is a triple (F, +,*) consisting of a set F and two operations that satisfies the closure, commutative, associative, identity and inverse properties for both operations and right and left distributivity for * over +. Alternatively, for our purposes, a field is a subset of real numbers that is closed under the four arithmetic operations (excluding division by 0). Examples: (Q, +,*) (R, +, *) and multiplication mod 5 Z5 with addition Quadratic field extension: Let F be a field and k an element of F for which √k is not in F. Then the quadratic extension of F, denoted F(√k ) is …. … the smallest field containing both F and √k . or … the intersection of all fields containing both F and √k . or … the field { a + b√k : a, b are in F }. Examples: Q(√2) is a quadratic extension of the rationals C (the complex numbers) is a quadratic extension of R 3. Describe pictorially the three classical impossible constructions. See these notes 4. Outline the proof that the cube root of 2 is not constructible. a. Suppose 3 2 is cons tructible , the n it lies in the terminal field of some quadratic exte ns ion se ries Q = F0 F1 F2 ... Fn. b. Show that if 3 2 lie s in the quadratic exte ns ion of some fie ld F, the n it lies in F as well. c. Conclude that s ince d. 3 e. The refore, 3 2 lie s in Fn the n by descent it is in Q. 2 is not rational, however, so this is a contradiction. 3 2 is not constructible.