Martin Bormann Essays (2187 words) - SS-Obersturmbannfhrer

Martin Bormann The evening of October 15, 1946, ten of the twelve significant war hoodlums, sentenced to death at the Nuremberg preliminaries, were executed. Of the two who evaded the executioner, one was ReichMarshal Hermann Goring, who ended it all by gulping a deadly vial of cyanide two hours before his execution. The other man was Reichsleiter Martin Bormann, who had figured out how to increase a gigantic measure of intensity inside the Nazi Party. He was for all intents and purposes obscure outside of the Party world class as he had worked in the shadows of Hitler. As the finish of the war moved close, huge numbers of the top Nazis were escaping. Hermann Goring had fled west, and had been caught by American officers, after the demise of Hitler had been declared. In Hitler's political will, Goring hosted been ousted from the gathering while Martin Bormann hosted been named Get-together Minister . As per Jochen Von Lang, Gobbels and Bormann had ?held a military instructions the evening of May 2, 1945. ? Gobbels had just chosen to end it all yet Bormann frantically needed to endure. The last section into his journal was ?get away from endeavor! ? Martin Bormann's whereabouts after this night is obscure. There are numerous theories concerning his destiny extending from the likely to the awesome. Reichsleiter Bormann who, as indicated by A. Zoller, ?practiced total authority over the entire structure of the Reich? but, for all intents and purposes obscure to people in general, was conceived June seventeenth, 1900. He was conceived in Saxon to a Postal Clerk. Bormann joined an enemy of Semitic association in 1920 and by 1923 he was an individual from the Freikorps. During this period, he was detained for a year for homicide and one year after his discharge Bormann joined the Nazi Party as a budgetary chairman. By 1933 he had worked his approach to being made a Reichsleiter, a General of the SS and the Chief of Staff to Rudolf Hess. At the point when Hess took off to England, Bormann readily acquired his position and turned into Hitler's representative. He hosted numerous foes in the Gathering and Goring clarified that even Goebbels dreaded him and his capacity . Bormann saw himself to be a serious respectable character and in a letter to his better half dated April second, 194 5 he composed that, ?in the event that we are ordained, similar to the Nebeliung, to die in King Attila's corridor, at that point we go to death gladly and with our heads held high.? For all his grandiosity, as an opportunity to battle showed up, Bormann made a wild eyed endeavor to endure. Toward the finish of the war, the unified pioneers chose to arraign top Nazis as War Criminals in Nuremberg. As Martin Bormann was missing, it was concluded that he would be attempted in absentia. Despite the fact that the partners had declaration expressing that Bormann was dead, they disregarded it in such a case that ?Bormann now was to be proclaimed dead by the court, and afterward to surface later, beyond words Nazis would speculate that maybe the Furher was alive as well.? All together for united believability to stay unblemished, Bormann was to be gone after for Crimes against Peace, War Crimes and Crimes against Humanity. Dr. Friedrich Bergold was named to this troublesome assignment of shielding a missing man. He thought about it ?an unnatural birth cycle of equity for the Tribunal to attempt his customer in absentia.? The International Tribunal condemned Reichsleiter Martin Bormann to death. The evening of May 1-2, 1945 is the last known whereabouts of Martin Bormann. The Reichsleiter was frantically attempting to leave Berlin alive. He had attempted to haggle with the Russians for a concise truce with the end goal for him to get a sheltered section through the adversary's lines. It had been dismissed. The survivors in the Fuhrerbunker were endeavoring to get away from the city and like clockwork a gathering left. Bormann rose wearing a SS uniform without rank and a cowhide jacket. His pocket contained a duplicate of Hitler's will, tying down him to control. His gathering, that included Axmann, Kempka and Stumpfegger, showed up at the Friedrichstrasse Subway station yet were held up at the Weidendammer Bridge. The Russians held the opposite side of the extension and in this manner made it difficult to cross without the front of tanks. Phenomenally, some German tiger tanks and a couple of reinforced

Friedrich Nietzsche Research Paper Example | Topics and Well Written Essays - 2000 words

Friedrich Nietzsche - Research Paper Example While German Expressionism was communicated in a wide assortment of aesthetic arrangements, today it is generally perceived for its appearance inside film. This article thinks about the authentic foundation of German Expressionism and furthermore breaks down it for its more profound significance inside German Expressionist movies, contending that German Expression exhibited an agnostic worry with industrialization and modernization. Foundation While Expressionism was a craftsmanship development that had gotten well known in Europe during the early piece of the twentieth century, in Germany it adopted on an especially remarkable strategy. While this is a direct result of an assortment of reasons, one of the essential reasons is on the grounds that the German movie producers didn't have the monetary sponsorship Hollywood movies had so they started building up their own one of a kind style. In thinking about German Expressionism, it’s clear there are various topical components th at exist. In such matters, it’s been noticed that visit German Expressionist subjects include craziness, disloyalty, and scholarly themes. It is maybe for this last explanation that the workmanship development remains so pervasive a worry inside University conditions. While German Expressionism was kept to a set period inside German workmanship history, its impact has been expanded well past this mid twentieth century time span. With Nazi’s taking force in Germany, making German Expressionist movie producers moved to the United States and impacted filmmaking. Gradually German Expressionist methods grabbed hold in American film and formed into film noir. Today noir stays a noticeable component in Hollywood film. What's more, to film, it’s likewise been contended that German Expressionism developed in engineering. In such matters, it’s contended that the sharp points and differences of film additionally explained into design. To be sure, German Expressionis m appears to similarly speak to these worries. From the initial credits it’s simple to perceive how American investigates could have confounded The Cabinet of Dr. Caligari (1919) as cubist. The inclined and sideways examples with the entertainers and executives names put on them appear to speak to a large number of points of view. Investigation While it’s been noticed that German Expressionism to a great extent created as a reaction to Hollywood movies, one should likewise think about it corresponding to the importance behind the aesthetic style. In such matters, numerous specialists contend that matches between the German Expressionism and profound German skeptical tension with industrialization are apparent (Elsasesser, 2004, p. 72). In one of the most well known German Expressionist movies, the Cabinet of Dr. Caligari this is exceptionally clear. The film starts with an iris-in of Francis in the haven. One notes that the worry with craziness is a significant topic o f German Expressionism. It is additionally strangely proper similarly as with the murkiness shutting in the iris appears to speak to an adaptive perspective on his disturbed soul. Indeed, the film utilizes altering to slice to the profound center set-up that highlights Jane strolling toward the camera and afterward back to the dim and upsetting iris encompassing Francis. We at first discover Caligari enlisting his somnambulist for the jamboree. He appears to be to some degree strange in the sufficiently bright setting and the mise-en-scene. indeed, even positions him well-beneath the enlistment center, as though the recorder speaks to the set up administrative power that Caligari is endeavoring to supplant. The

Hilberts Third Problem (A Story of Threes)

Hilbert’s Third Problem (A Story of Threes) 1 Introduction and History 18.304 (discrete math) is one of two communication intensive (CI-M, where the M means it’s within my major) seminars I need to take as part of my math major. (The second one, which I’m taking this semester, is 18.424, information theory. The other seminar topics we get to choose from are real analysis, analysis, discrete applied mathematics, physical mathematics, theoretical computer science, logic, algebra, number theory, topology, and geometry.) The math CI-Ms are small classes, taught almost entirely by the students: in my experience so far, after a few starter lectures by the professor every student presents three lectures as a combination of slide presentations and chalktalks. Sometimes there are also p-sets; sometimes there are not. At the end of the semester we write in-depth explorations of the topics of our final presentations. In 18.304, I got to transiently experience several dozen proofs, getting closely acquainted with two of them and reading books and books and books about a third. My first topic was Dijkstra’s algorithm for finding shortest paths, my second topic was (the inifinite-ish quantity of) proofs for the infinity of primes, and my third topic was Hilbert’s third problem, which we’ll get to know much more closely in this blog post.  The first two presentations went objectively horribly. The third one went much better: I showed up early and stood in front of the bathroom mirror similing to myself for 20 seconds in a Wonder Woman power pose. On the way from the bathroom to the classroom I passed a girl from among my soon-to-be audience. We made eye contact and exchanged shy smiles and somehow that made everything better: I felt less alone, and I felt that I had an ally among the people I had been scared were going to judge me. (If you’re reading this, thank you so much, kind, mysterious friend-when-I-needed-one.) Over the course of the semester, I went from having barely any exposure to proofs beyond what I’d gotten from my computer science classes and general institute requirements to being able to follow, read, write, and (thanks to 6.046, which I took at the same time) formulate a proof. This was a huge amount of growth for me, and a big part of my goal in becoming a math major. My grandfather, who was very, very dear to me, emphasized to me that there is a lot of power at the intersection of fields. It is valuable to be able to be a translator between fields, to be an avenue for the ideas, tools, and strengths of one field to enter and innovate in another field. This has been one of my hopes for my technical education at MIT: as a 6-7/18 major, I’m split officially between three technical departmentsâ€"computer science, biology, and mathematicsâ€"and I’ve also had to take courses in physics and chemistry. My personal theory (and experience) is that computational biology, the thing I am ultimately studying, is young enough that depending on their backgrounds, people primarily approach it from the perspective of biology, computer science, or mathâ€"occasionally two of them, but very rarely all three. I’m hoping that my broader education at MIT will help me find a niche and be, as my grandfather said, a useful translator across those junctions. The rest of this blog post is my final paper, which from the very start  I wrote with the dual purpose of turning in and posting here, with an intended audience of high school students. (Feel free to skim or skip around if it helps you get a picture of the proof.) I’m hoping to write a post about Dijkstra’s algorithm someday as well, since it is one of my favorite algorithms (tied maybe with Edmonds-Karp, because flow networks are absurdly useful). I want to communicate the following things I have learned about math: Math is an incredibly diverse field, not a single path, and you may or may not have gotten a chance to actually see all the many branches in high school. Math is a human and even political field, in which a single problem can connect people across centuries. Math is not memorizationâ€"in fact, the human subject of this blog post chose math specifically for that reason. 1 Introduction and History Hilbert’s third problem, the problem of defining volume for polyhedra, is a story of both threes and infinities. We will start with some of the threes. Already in early elementary school we learn about two- and three-dimensional shapes and some of their interesting properties. We learn that a triangle is a two dimensional polygon with three edges. More generally, we learn that a polygon can be defined as a two-dimensional shape constrained to a plane, bounded by any number of straight, uncurved lines, the edges. The area contained between the edges is the single face of the polygon and the points where the edges meet are the vertices. All polygons can be cut up into a discrete number of triangles, and the area of these triangles can always be further distributed in discrete chunks to form a square of the same area as the original polygon. One method of doing this is to cut the polygon up into triangles, cut each triangle up into smaller triangles that can be reassembled into a rectangle, and cut this rectangle up and rearrange the pieces into another rectangle, this one with either its width or its height equal to the side length of the square we want to assemble. Finally, we can stack these rectangles to form a square with the same area as our original polygon, a square shared by all polygons of that area. As an example, here we chop up a two-dimensional house and rearrange it into a square of the same area. 1.1 Definition I: Equidecomposable The ability to chop up one figure and build a new one out of its pieces is the first property we are interested in: equidecomposability. Two shapes are equidecomposable if they can be divided up into congruent building blocks. For example, the house and the square that we looked at above are equidecomposable. Below is another of the many ways that they can be decomposed into congruent pieces. This relationship can be expressed mathematically. In the picture above, both the square S and the house H can be decomposed into the orange (O), purple (P), and blue (B) pieces. S = O ? P ? B = H Equidecomposability has also been called congruence by dissection and scissors congruence. 1.2 Definition II: Equicomplementable Another property that we are interested in is equicomplementability. Two shapes are equicomplementable if congruent building blocks can be added to both of them to create two equidecomposable supershapes. In addition to being equidecomposable, our house and square are also equicomplementable. The supershapes we have created are equidecomposable. We can express this relationship mathematically as well. We can add four piecesâ€"here a yellow piece Y, a pink piece P, an orange piece O, and a maroon piece Aâ€"to the square S in order to create a supershape M. We can add the same pieces to the house H to create a second supershape N. M = S ? Y ? P ? O ? A N = H ? Y ? P ? O ? A These supershapes can then be decomposed into the same pieces, pictured here as a red piece (R), a turquoise piece (T), a brown piece (B), a blue piece (L), and a green piece (G). M = R ? T ? B ? L ? G = N Equidecomposability, equicomplementability, and equality in area are intertwined for polygons in the second dimension. Throughout the 1800s there were many developments in these properties as they apply to two-dimensional polygons. Following preliminary work by William Wallace in 1807, independent proofs from Farkas Bolyai in 1832 and P. Gerwein in 1833 demonstrated that any polygon can be decomposed in such a way that its pieces can be reassembled into a square, as we illustrated earlier. This means that any pair of polygons of equal area is equidecomposable, since they can be decomposed and reassembled into the same square. In 1844, Gerling furthermore showed that it does not matter if reflections are allowed in the reassembly of the decomposed shapes. The following interesting theorems and lemmas were proved in the nineteenth century. They are outlined in the second chapter of  Boltíànskii’s 1978 Hilbert’s Third Problem: If two figures A and C are each equidecomposable with a third figure B, then A and C are also equidecomposable with each other. Every triangle is equidecomposable with some rectangle. Any two equal-area rectangles are equidecomposable. Any two equal-area polygons are equidecomposable. (This is the Bolyai-Gerwien theorem.) Any two figures that are equidecomposable are also equicomplementable. 1.3 Into the Third Dimension At the end of the nineteenth century, the question was settled in the second dimension and was expanded into the third. This is the topic that we will focus on: can the Bolyai-Gerwien theoremâ€"that any two equal-area polygons are equidecomposableâ€"be extended into the third dimension? Instead of looking at polygons, we will look at their counterparts in the third dimension, polyhedra. Polyhedra have been defined in many ways, and not all of the definitions are compatible. We will use the definition described in Cromwell’s Polyhedra. A polyhedron is the union of a finite number of polygons that has the following properties: The polygons can only meet at their edges or vertices. Each edge of each polygon is incident to exactly one other polygon. It is possible to travel from the interior of any of the polygons to the interior of any of the others without leaving the interior of the polyhedron. It is possible to travel over the polygons incident to a vertex without passing through that vertex. Within the constraints of this definition, polyhedra are diverse and varied. Below are some examples. (If the squished part of the second-to-last polyhedron were further squished into a single point, then by the third bullet point above it could no longer be a single polyhedron.) Just as any pair of polygons of equal area can be decomposed and reassembled into the same square, can any pair of polyhedra of equal volume be decomposed and reassembled into the same cube? Are polyhedra of equal volume equidecomposable? Are they equicomplementable? By the end of the nineteenth century there were several examples of equalvolume polyhedra that were both equidecomposable and equicomplementable, but there was no general solution. One simple example is prisms with the same height and equal area bases, stemming from the two-dimensional polygon result. In 1844 Gerling showed (and then Bricard proved again in 1896) that two mirror-image polyhedra are equidecomposable by cutting them up into congruent mirror-image pieces that can then be rotated into each other. There were also some specific tetrahedra equidecomposable with a cube, shown in 1896 by M.J.M. Hill. We can reduce the problem about polyhedra in general to a problem about tetrahedra. 1.4 Definition III: Tetrahedron A tetrahedron is a polyhedron with four triangular faces, six edges, and four vertices. In many current math textbooks the faces are required to be congruent. We are not going to require that any of the faces be congruent; our definition is closer to what many current math textbooks call a triangular pyramid. Just as any polygon can be cut up into triangles, any polyhedron can be cut up into tetrahedra. First, we can cut any polyhedron into a finite number of convex polyhedra. These can each then be cut up into a finite number of pyramids with polygonal bases. Because any polygon can be cut up into a finite number of triangles, each of these polygonal pyramids can be cut up into a finite number of triangular pyramids. This means that if we can prove or disprove the Bolyai-Gerwien theorem in the third dimension for tetrahedra, then we have also proved or disproved it more generally for polyhedra. Below is an example of a division of a polyhedron into triangular pyramids, based off of figures 13.12 and 13.13 in Rajwade’s 2001 Convex Polyhedra. The volume of a tetrahedron is ? the area of the base multiplied by the height; as described by Euclid, any two tetrahedra with bases of equal areas and with equal heights will also have the same volume. This definition of volume is reminiscent of the parallel result in two dimensions: the volume of a triangle is ½ the length of the base multiplied by the height. Unlike the area of a triangle, however, the volume of a tetrahedron and therefore the volume of a polyhedron is found through calculus, by dividing the three-dimensional polyhedron into infinitesimally thin two-dimensional cross sections and adding up their areas. If the Bolyai-Gerwien theorem can be expanded into the third dimension, we can define the volume of any three dimensional polyhedron the same way we define the area of a two dimensional polygon, by breaking it up into discrete building blocksâ€"tetrahedra in three dimensions and triangles in twoâ€"and reassembling the pieces into a cube or square. This would be an elementary solution, with no infinities (or calculus) required. This problem was posed by C.F. Gauss. In a letter in 1844, Gauss expressed that he wanted to see a proof that used finitely rather than infinitely many pieces. By Hilbert’s time it was not yet solved. 1.5 David Hilbert’s 23 Problems One of my favorite aspects of geometry is how seamlessly it can transition from elementary school math to the cutting edge. Modern geometry is built on fluid connections between the basic principles. This foundation was built largely by David Hilbert. The fundamentals of geometry were initially outlined by Euclid in Elements. In the nineteenth century, geometry was becoming increasingly abstract and less and less tied to the original shapes. As Hilbert said in a lecture: “One must be able to say at all timesâ€"instead of points, straight lines, and planesâ€"tables, chairs, and beer mugs.” In this context some of the unstated assumptions in Euclid’s Elements came uncovered. (As an example, one such unstated assumption was that if two lines cross, they must have a point in common.) Hlibert extended Elements, providing an axiomization of Euclidian geometry and proofs for the unchecked assumptions that stood in the way of geometry being as fully useful as algebra. David Hilbert was born in Wehla, Germany. His mother was interested in philosophy, and his father was a judge and wanted him to study law. He was homeschooled for two years and began school two years late, at age eight. The subject he went on to study was mathematics, because he did not like memorization. A turning point for Hilbert was at his first presentation, at the Technische Hochschule, where he impressed and befriended Klein, 13 years his senior. After his dissertation on invariants Hilbert went to Paris to meet the leading mathematicians upon Klein’s suggestion. In his letters back to Klein, he made comical judgements of some of the most important mathematicians of the time. In particular, he reflected that Poincar ´e, with whom Klein had a (mental breakdown inducing) rivalry, was quite shy, and that the reason he published a lot of papers was that he published even the smallest results. Hilbert continued his research in Germany, first at his hometown university and then at G ¨ottingen when he was invited over by Klein in 1894. In the beginning he continued to focus on invariant theory, a branch of abstract algebra examining how algebraic expressions change in response to change in their variables. Hilbert’s style of teaching was very different from Klein’s: unlike Klein’s prepared, perfectionist style of lecturing, Hilbert would prepare only an outline beforehand and work through the mathematics in front of his students, mistakes and all. Hilbert and Klein revolutionized the teaching of mathematics, incorporating visual aids and connecting mathematical concepts to their applications in the sciences, and elevated G ¨ottingen to a leading institution in mathematics. It was here, starting with lectures to his students, that Hilbert became interested in geometry. 1900 was the dawning of a new century and a new era of mathematics. At the second meeting of the International Congress of Mathematicians (to this day the largest math conference, meeting once every four years, and the conference at which the Fields Medal is awarded), that year in Paris, David Hilbert, then 38, presented a monograph of ten of the open problems that he considered the most important for the next century. He published all 23 later, in a report titled simply “Mathematical Problems.” Some of Hilbert’s 23 problems have been very influential in the subsequent century of mathematics, and almost all of them have been solved. (The two that have not been solved, the axiomization of physics and the foundations of geometry, are now considered less of a priority and too vague for a definitive solution, respectively.) 1.6 Hilbert’s Third Problem We are interested in the third of the 23 problems, which concerns the extension of the Bolyai-Gerwien theorem into the third dimension. Unlike Gauss, Hilbert did not believe that there was such a bridge: he asked simply for two tetrahedra that together formed a counterexample. Here we reproduce Hilbert’s third problem in its entirety, with the problem statement in bold. In two letters to Gerling, Gauss expresses his regret that certain theorems of solid geometry depend upon the method of exhaustion, i.e., in modern phraseology, upon the axiomization of continuity (or upon the axiom of Archimedes). Gauss mentions in particular the theorem of Euclid, that triangular pyramids of equal altitudes are to each other as their bases. Now the analogous problem in the plane has been solved. Gerling also succeeded in proving the equality of volume of symmetrical polyhedra by dividing them into congruent parts. Nevertheless, it seems to me probable that a general proof of this kind for the theorem of Euclid just mentioned is impossible, and it should be our task to give a rigorous proof of its impossibility. This would be obtained as soon as we succeed in exhibiting two tetrahedra of equal bases and equal altitudes which can in no way be split up into congruent tetrahedra, and which cannot be combined with congruent tetrahedra to form two polyhedra which themsel ves could be split up into congruent tetrahedra. Armed with our definitions, can find a more succinct and powerful statement of the challenge: Specify two tetrahedra of equal volume which are neither equidecomposable nor equicomplementable. Hilbert’s third problem was the first of the 23 to be solved. The first part of the problem, on equidecomposability, was solved by Hilbert’s student Max Dehn just a few months after the conference, before the full 23 problems were printed. The proof rests on a value describing a polyhedron, the Dehn invariant, which we will look at in more detail later. The Dehn invariant does not change when the polyhedron is cut apart and reassembled into a new shape: if two polyhedra are equidecomposable, then they must have same Dehn invariant (and they do). However, not all polyhedra with the same volume have the same Dehn invariant. Specifically, Dehn used the example of a regular tetrahedron and a cube of equal volume, and we will examine this case as well. Two years later Dehn showed in a second paper the second part of the problem, on equicomplementability. An incomplete and incorrect proof was published by R. Bricard four years previously in 1896. It was not cited by Hilbert, but Dehn based his proof largely on Bricard’s. Dehn’s paper was not easy to understand. It was refined by V.F. Kagan from Odessa in 1903. In the 1950s, Hadwiger, a Swiss geometer, together with his students found new properties of equidecomposability. This allowed for a more transparent presentation of Dehn’s proof. Further progress over the past century has made it even clearer and more concise. We will be presenting a very recent version of the proof, published in 2010 in Proofs from the Book. You may have noticed a theme; indeed, the story of the third of David Hilbert’s 23 problems, the quest to expand or blockade the Bolyai-Gerwien theorem from the second into the third dimension, is a story of threes. We will present the solution in three parts: three definitions, three proofs, and three examples. We have already gone through the definitions. A tetrahedron, as defined here, is a polyhedron with four (not necessarily congruent) triangular faces. Equidecomposability is a relationship between two shapes in which one can assemble one from all the pieces of another. Finally, equicomplementability is a relationship in which one can add congruent shapes to the two shapes to form two equidecomposable supershapes. Now, we will go through the three proofs, of the pearl lemma, the cone lemma, and Bricard’s condition. Afterward we will apply the result to three example tetrahedra. Two of these tetrahedra together form a counterexample, solving Hilbert’s third problem. 2 Three Proofs 2.1 Proof I: The Pearl Lemma Our first proof is by Benko. We start by defining the segments of an edge. Each edge in a decomposed shape consists of one or more segments that, placed end to end, make up the total length of that edge. In a decomposition of a polygon, the endpoints of segments are always vertices; in a decomposition of a polyhedron, the endpoints of segments can also be at the crossing of two edges. Otherwise, all non-endpoint points within any one segment belong to the same edge or edges. (In the decomposition of the square below, for example, the hypotenuse of the larger triangle is subdivided into two segments.) We examine two equidecomposable figures. They are broken up into the same pieces, which are rearranged and perhaps reflected in different ways. Imagine that we must distribute whole, indivisible tokensâ€"which we will call pearlsâ€" on all the segments in the two decompositions. The pearl lemma states that we can place the same positive number of pearls onâ€"or, in other words, assign the same positive integer toâ€"each segment in the two decompositions in such a way that each edge of a piece gets the same whole number of pearls no matter which of the two decompositions it is sitting in. Below, for example, is a correct distribution of pearls on the equidecomposable house and square we looked at earlier. For each edge of a piece, the sum of the pearls placed on the segments making up that edge in the decomposition of one figure must equal the sum of the pearls placed on the segments making up that edge in the corresponding decomposition of the other figure. In the second dimension, this simply means that the number of pearls we place on an edge in the first decomposition is equal to the number of pearls we place on that edge in the other figure’s decomposition. In the third dimension, an edge can consist of multiple segments that need not be consistent; an edge’s segments can be different, and even different in number, in the two decompositions. However, the number of pearls placed on a piece’s edge must still be the same between the two decompositions. We can express this idea as a system of linear equations. The variables to solve for are the numbers of pearls on each segment. All of the coefficients in our linear equation are positive integers; in fact, since each segment is represented once, all of the coefficients are 1. We can satisfy this system of linear equations by assigning a positive real number of pearls (which may or may not be an integer) to each segment. As an easy example, the number of pearls assigned to each segment can be equal to that segment’s length. Remember, though, that we cannot damage the pearls! We need to show that if our linear equation has positive real number solutions, then there are also positive integer solutions representing whole, unchopped pearls. This brings us to our second proof, a proof of the cone lemma. 2.2 Proof II: The Cone Lemma The cone lemma is our name for the 1903 integrality argument by Kagan, which greatly simplified Dehn’s proof. We must show that if our system of linear equations has a positive real solution, then it also has a positive integer solution. We start with a set of homogenous linear equations Ax = 0, x 0 with integer coefficients (integer values in A) and positive real solutions (positive real values in x). This is a translation of our set of linear equations in the form Ax = b, x 0 required by the pearl lemma. We need to show that if the set of real solutions to Ax = 0, x 0 is not empty, then it also contains positive integer solutions. We required that our solutions be greater than zero. However, if there are solutions that are greater than zero, then there must also be solutions greater than or equal to one, since an all-positive solution vector x can always be multiplied by some positive value to produce an equivalent vector with all values at least 1. Therefore it will suffice to show that if the set of real solutions to Ax = 0, x = 1 is not empty, then it also contains integer solutions with all values greater than or equal to 1. We will prove this using a method devised by Fourier and Motzkin. We will use Fourier-Motzkin elimination to show that there exists a lexicographically smallest real solution x to Ax = 0, x = 1 and that if the coefficient matrix A is integral, then that lexicographically smallest solution is rational. Proving that if there is a real solution, then there must exist an integer solution to Ax = 0, x = 1 is equivalent to proving the same for Ax = b, x = 1. Therefore we will show that there exists a lexicographically smallest real solution x to Ax = b, x = 1 and that if A and b are integral, then that lexicographically smallest solution is rational. (Here, the lexicographically smallest vector results from a comparison of the elements in the vectors in order. If a vector has the smallest first element, then it is the lexicographically smallest vector. If there is a tie in the first element, then we compare the second element, followed by the third, and so on.) We will use a proof by induction on N, the size of any possible solution vector x = (x1, , xN). First we consider the base case, N = 1. There is only one variable x1 to solve for, and there are only inequalities that involve x1. We simply assign to x1 the smallest value that it can take on. Now we consider solutions in multiple variables, N 1. We examine those inequalities that involve xN. Because we have declared that all elements of the solution must be greater than or equal to 1, we know that there is at least a lower bound on xN: xN = 1. There might also be additional lower bounds or upper bounds. As before, we set xN to the smallest value it could take on. Now we are looking at a smaller system, in N - 1 variables, Ax = b, x = 1. This system contains all the inequalities of the previous system in N variables except for those involving xN , which we just resolved. In addition, we add a new constraint that all upper bounds on xN are at least as large as all lower bounds on xN. We continue by looking at the inequalities involving xN-1 and setting xN-1 to the smallest value it could take on, then examining xN-2, and so on. We established that the base case system in one variable has a smallest solution. The system that preceded it is a system in two variablesâ€"we created the N = 1 system out of the N = 2 system by setting the second variable x2 to its smallest possible value. This is definitely possible to do since each variable xi has at the very least an inequality xi = 1. We know then that the system in two variables has a smallest solution as well. We can follow the same logic to find that the system in three variables has a smallest solution, and a system in four variables, and so on. If the system in N variables has a smallest solution, then the preceding system in N + 1 variables has a smallest solution as well. All of these lexicographically minimal solutions are rational, since the inequalities that we use to set the minimal value of each variable have integer coefficients. Therefore, if a system of homogeneous linear equations has a positive real solutionâ€"as does the system generated by the pearl lemma (translated so that it is a system of homogeneous linear equations)â€"then it also has a positive integer solution. We have proved the cone lemma and with it the pearl lemma. (Why is this called the cone lemma? A set of homogenous linear equations Ax = 0, x 0 with integer coefficientsâ€"integer values in Aâ€"and positive real solutionsâ€"positive real values in xâ€"is called a rational cone.) 2.3 Proof III: Bricard’s Condition Our final proof allows us to connect the pearl lemma to concrete examples and finally solve Hilbert’s third problem. Bricard’s condition was claimed but incorrectly proved by Bricard in his 1896 paper. Dehn proved it successfully, and the proof has since been refined. From here on out we will focus on the dihedral angles of three-dimensional polyhedra, the angles between faces that share an edge. A square pyramid, for example, has eight dihedral angles, one for each of its eight edges. Two of them are illustrated below A tetrahedron has six dihedral angles (one for each of its six edges) and a cube has twelve. Bricard’s condition states that, looking at two three-dimensional polyhedra that are equidecomposable or, more generally, equicomplementable, their dihedral angles must be linear combinations of each other with a difference of some multiple of p. In other words, if we define the dihedral angles of one polyhedron a1, , ar and the dihedral angles of an equidecomposable or equicomplementable polyhedron ß1, , ßs, there must be some positive integers m1, , mr and n1, , ns and an integer k such that m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs + kp We will prove first that the property holds for any two equidecomposable polyhedra and then that it holds more generally for any two equicomplementable polyhedra. 2.3.1 Equidecomposability We start by assuming that two polyhedra are equidecomposable. As a (two-dimensional) example, we can look at the same house and square from earlier. We cut the two polyhedra up into two decompositions made up of the same pieces, and we follow the pearl lemma to assign a positive integer number of pearls to each segment in each of the two decompositions: each piece has the same total number of pearls on each of its edges in either of the two decompositions. At each pearl, we measure the dihedral angle made by the faces incident to the edge at the location of the perl. If the perl is in the interior of the figure, its dihedral angle will be p or 2p. If the pearl is on an edge of a piece but not on an edge of the figure being decomposed, then its dihedral angle will be p. We define a sum ?  1 as the sum of the dihedral angles at all of the pearls in the pieces of the first polyhedron, one angle per pearl. We also define a second sum ?  2, the sum of the dihedral angles at all of the pearls in the pieces of the second shape, again one angle per pearl. Multiple pearls can have the same dihedral angle, sometimes by being on the same edge and sometimes by being on different edges with the same dihedral angle, such as the two dihedral angles in the square pyramid illustrated earlier. In either case the dihedral angle will appear multiple times in our sum. The number of times a dihedral angle appears, and therefore its multiplier in either of the sums, must be a positive integer. Similarly, there will be an integer number of pearls with a dihedral angle of p or 2p, and therefore a nonnegative integer multiplier of p. If we define the dihedral angles of one polyhedron a1, , ar, we can represent the first sum as ?  1 = m1  a1 + + mr  ar + k1  p for some positive integers m1, , mr and a nonnegative integer k1. Similarly, if we define the dihedral angles of the second polyhedron ß1, , ßs, we can represent the second sum ?  2 = n1  ÃŸ1 + + ns  ÃŸs + k2  p for some positive integers n1, , ns and a nonnegative integer k2. We get ?  1 and ?  2 by adding the dihedral angles of each piece in the decompositions of the first polyhedron and the second polyhedron, respectively. These pieces are congruent, since the two polyhedra are equidecomposable and we are looking at decompositions that satisfy this relationship. Therefore by the perl lemma, which we used to distribute the pearls, each edge of each piece has the same dihedral angles in either of the two decompositions and the same number of pearls with each dihedral angle. In other words, ?  1 = ?  2 or, substituting in the definitions of ?  1 and ?  2, m1  a1 + + mr  ar + k1  p = n1  ÃŸ1 + + ns  ÃŸs + k2  p We define the integer k = k2 - k1 and find that our equation becomes Bricard’s condition. m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs + kp 2.3.2 Equicomplementability Now we assume that our two polyhedra are equicomplementable. As a two dimensional example, recall that we can add congruent pieces to our house and boxâ€" â€"such that the resulting polygons are equidecomposable. We can create two superfigures from our equicomplementable polyhedra by adding congruent pieces to the two polyhedra such that the superfigures are equidecomposable. In other words, we will be able to cut apart the superfigures so that their pieces are our two original polyhedra (one original polyhedra in each superfigure) and otherwise the same pieces. These are our first two decompositions of the two superfigures and our first decomposition of each individual superfigure. In addition, we can also cut apart the superfigures into an alternative decomposition such that they share all their pieces. These form the third and fourth decompositions of our superfigures, the second decomposition of each individual superfigure. We again apply the pearl lemma to distribute the pearls on the edges of all the pieces in all four decompositions. Because we used the cone lemma to prove the perl lemma, we can impose an additional constraint: each edge of each superfigure must have the same total number of pearls in each of the two decompositions it is involved in. As before, we compute the sums of the dihedral angles at all the pearls, ?1, ?2, ?1, and ?2. ?1 and ?2 are the first decomposition of each superfigure. ?1 includes our first original polyhedron, ?2 includes our second original polyhedron, and each other piece of ?1 is congruent to a piece in ?2 and vice versa. ?1 and ?2 are the alternative decompositions of the two superfigures. Each piece in ?1 has a congruent counterpart in ?2 and vice versa. First, we notice that ?1 and ?2 are decompositions into the same pieces. As in our proof for equidecomposable polyhedra, this means that ?1 and ?2 must be identical: ?1 = ?2. Next, we notice that ?1 and ?1 are decompositions of the same polyhedron, our first superfigure. Recall that we restricted our placement of pearls so that each edge of the superfigure needed to have the same number of pearls. Recall also that pearls that are placed inside of the superfigure or on the inside of one of its faces, rather than on one of its edges, yield dihedral angles of p or 2p. Though the two decompositions of our first superfigure might have different numbers of pearls, the edges of the superfigure itself must have the same numbers of pearls. This means that ?1 and ?1 can differ by an integer multiple of p, which we here call l1.                    ?1 = ?1 + l1  p By the same logic, with an integer l2,                    ?2 = ?2 + l2  p Since we deduced above that ?1 = ?2, we can restate these two equations as a relationship between ?1 and ?2, with l  =  l2 -  l1.                    ?2 = ?1 + lp We now have a statement describing the relationship between the original decompositions of the two superfigures, ?1 and ?2â€"we are almost done! Remember that the difference between ?1 and ?2 is that the first contains our first polyhedron and the second contains our second polyhedron. Each other piece in ?1 has a congruent counterpart in ?2 and vice versa. Since they are identical, we can subtract the contributions of these congruent pieces from each side of the equation. This leaves the first polyhedron’s contributions to its superfigure and the second polyhedron’s contributions to its superfigure. As before, they will differ by an integer multiple of pâ€"lp. m1  a1 + + mr  ar = n1  ÃŸ1 + + ns  ÃŸs +  lp This is Bricard’s condition as described in the opening of this proof, replacing the integer k with the integer l. We have now shown that Bricard’s condition holds for both equidecomposable and equicomplementable polyhedra. 2.3.3 The Dehn Invariant We do not use it in this version of the proof, but we will pause for a moment to formally define the Dehn invariant, which was used by Dehn in his original proof. You may notice a few parallels to Bricard’s conditionâ€"indeed, though it was published earlier, Bricard’s condition is a consequence of the relationship between the Dehn invariant and equidecomposability. We define in radians the dihedral angles of a polyhedron P: a1, a2, , ap. We also define the lengths l1,  l2, ,  lp of the corresponding edges in P. The Dehn invariant f(P) of the polyhedron P with respect to the function f is the sum: f(P) =  l1  f(a1) +  l2  f(a2) + +  lp  f(ap) f is an additive functionâ€"every linear dependence in the input values also exists  in the output values. We also add the additional constraint that f(p) = 0. In his original proof, Dehn showed that the Dehn invariant does not change when a polyhedron is cut apart and reassembled into a new shape: if two polyhedra P and Q are equidecomposable, then they have same Dehn invariant. In other words, for every additive function f that is defined for P’s and Q’s dihedral angles and for which f(p) = 0, f(P) = f(Q). Dehn found, however, that not all polyhedra with the same volume have the same Dehn invariant, demonstrating that not all polyhedra with the same volume are equidecomposable and thereby solving Hilbert’s third problem. This proof is more complicated than the three-part proof we presented here; it is described in detail in the corresponding chapter of the third edition of Proofs from the Book (otherwise we refer to the fourth edition). 3 Three Examples Finally, we can apply Bricard’s condition to the dihedral angles of three example tetrahedra and identify a counterexample, our solution to Hilbert’s third problem. 3.1 Example I: A Regular Tetrahedron Our first example is the same example used by Dehn in his first paper. Here we examine a regular tetrahedron, a tetrahedron in which each of its four faces is an equilateral triangle. All dihedral angles in a regular tetrahedron are arccos ?. In contrast, all dihedral angles in a cube are p/2. We will show that the regular tetrahedron cannot be equidecomposable nor equicomplementable with any cube. Because of the work we have already done, this is very easy to do! If a regular tetrahedron were equidecomposable or equicomplementable with a cube, then by Bricard’s condition, there must be some positive integers m1 and n1 and some integer k such that m1 arccos ? = n1p/2 + kp We can rearrange this equation to solve for k. k = 1/p (m1 arccos ? - n1 p/2) k = m1 1/p arccos ? - ½ n1 1/p arccos ? is irrational (we will not examine the proof here, but it is in chapter 7 of Proofs from the Book). k, then, must also be irrationalâ€"it cannot be an integer. Bricard’s condition does not hold for these two figures. This means that a regular tetrahedron cannot be equidecomposable nor equicomplementable with any cube, even a cube of the same volume. This example alone demonstrates that two polyhedra of equal volume are not necessarily equidecomposable or equicomplementable with each other. However, though we have gotten at the spirit of Hilbert’s third problem we have not solved it exactly. We will look at two more examples to produce two tetrahedra of equal volume that are neither equidecomposable nor equicomplementable. 3.2 Example II: A Tetrahedron with a Vertex Incident to Three Orthogonal Edges Our next example very closely parallels the logic we saw above. Now we look at a tetrahedron with three orthogonal edges of equal length u that share a vertex. Three of this tetrahedron’s six dihedral angles (the turquoise angles) are right angles (size p/2). The other three (the red angles) have size arccos v?. Like the regular tetrahedron in our previous example, this tetrahedron cannot be equidecomposable nor equicomplementable with a cube. If it were, then by Bricard’s condition there must be some positive integers m1, m2, and n1 and some integer k such that m1 arccos ? + m2 p/2 = n1 p/2 + kp As before, we can rearrange this equation to solve for k. k = 1/p (m1 arccos ? + (m2 - n1) p/2) k = m1 1/p arccos v? + ½ (m2 - n1) 1/p arccos v? is irrational (this proof is also covered in chapter 7 of Proofs from the Book). As in the first example, k must be irrational as well and therefore cannot be an integer. Bricard’s condition does not hold for this relationship. This means that a tetrahedron with a vertex incident to three orthogonal edges also cannot be equidecomposable or equicomplementable with a cube. 3.3 Example III: A Tetrahedron with an Orthoscheme Finally we look at a tetrahedron with a three-edge orthoscheme: a sequence of three edges that are mutually orthogonal and in this case of identical length u, the same length as each of the three orthogonal edges in the previous example. The dihedral angles in this tetrahedron are p/2 (the three turquoise angles), p/4 (the two orange angles), and p/6 (the blue angle). The dihedral angles of this tetrahedron are all rational multiples of p/2, the size of any dihedral angle of a cube. In fact, a cube can be decomposed into six congruent tetrahedrons with an orthoscheme. Notice that this tetrahedron has the same base and height as the tetrahedron in the second example, and therefore the same volume. However, it cannot be eqidecomposable nor equicomplementable with a tetrahedron with those dihedral angles (or with the regular tetrahedron in the first example). If it were, then by Bricard’s condition there must be some positive integers m1, m2, m3, n1, and n2, and some integer k such that m1 p/2 + m2 p/4 + m3 p/6 = n1 arccos v? + n2 p/2 + kp As in the previous two examples, we can rearrange this equation to solve for k. k = 1/p (m1 p/2 + m2 p/4 + m3 p/6 - n1 arccos v? - n2 p/2) k = ½ m1 + ¼ m2 + ? m3 - ½ n2 - n1 1/p arccos v? Recall from the second example that 1/p arccos v? is irrational. k, then, cannot be an integer and Bricard’s condition cannot hold for this relationship. A tetrahedron with an orthoscheme of three identical-length edges cannot be equidecomposable nor equicomplementable with a tetrahedron with a vertex incident to three orthogonal edges of that same length. We have found two tetrahedra with the same base and height (and therefore the same volume) that are neither equidecomposable nor equicomplementable. In them we have found a solution to Hilbert’s third problem. 4 Open Problems The theory of the equidecomposabililty and equicomplementability of polyhedra, revived by Hilbert’s third problem, is largely solved. When we look at polytopes in higher dimensions, however, some gaps do remain. Though we did not use it in this proof, we mentioned earlier the Dehn invariant and its relationship with the equidecomposability of polyhedra: if two polyhedra are equidecomposable, then they necessarily have the same Dehn invariant. In 1965, Sydler proved that this condition is not only necessary, but also sufficient for equidecomposability: if two polyhedra have the same Dehn invariant, then they are necessarily equidecomposable. This result has not yet been provedâ€"and its validity is unknownâ€"for spheres and hyperbolic spaces of at least three dimensions, and in general for dimensions of at least five. If we only allow translations while reconstructing the second polytope from the decomposition of the first, then this problem is open in four or more dimensions. In addition, the problem of finding the minimum number of allowed motions necessary for equidecomposability is also unsolved in four or more dimensions. The consequences of restricting the motions in equidecomposability (to translations, to translations and central inversions, or to all motions that preserve orientation) and the existing proofs in lower dimensions are explored in depth in Boltíànskii’s Hilbert’s Third Problem. 5 Conclusion Hilbert’s third problem is one example of the necessity and beauty of a rigorous mathematical proof. If the Bolyai-Gerwien theorem could have been expanded into the third dimension, then we could define the volume of any three dimensional polyhedron using the discrete methods from the second dimension. Instead, progressing from area in the second dimension to volume in the third dimension requires breaking the shape into continuous, infinitely small building blocks and the use of calculus, an entire other toolbox and an entire other field: we share the third dimension with unavoidable infinities. 6 References The chapter “Hilbert’s Third Problem: Decomposing Polyhedra” (pages 53-61) from the 4th edition of Proofs from the Book  by Martin Aigner and Günter M. Ziegler (published in 2010 by Springer in Berlin). That same chapter (pages 45-52) from the 3rd edition of Proofs from the Book (published in 2004)â€"this one is based on Dehn’s original proof. “A New Approach to Hilbert’s Third Problem” by David Benko, pages 665-76 of volume 114 of The American Mathematical Monthly  (2007). Hilbert’s Third Problem by V.G. Boltíànskii (published in 1978 by V.H. Winston Sons in Washington, D.C.). Polyhedra: One of the Most Charming Chapters of Geometry  by Peter R. Cromwell (published in 1997 by Cambridge University Press in New York). Scissors Congruences, Group Homology and Characteristic Classes by Johan L. Dupont  (published in 2001 by World Scientific in Singapore). Revolutions of Geometry by Michael O’Leary (published in 2010 by John Wiley Sons in Hoboken, NJ). Convex Polyhedra with Regularity Conditions and Hilbert’s Third Problem by A.R. Rajwade (published in 2001 by the Hindustan Book Agency in New Delhi). Hilbert’s Third Problem: Scissors Congruence by Chih-Han Sah (published in 1979 by Fearon Pitman in San Francisco).                             If you’re hungry for more college-level math, I recommend starting with Proofs from the Book, the source of this proof and the textbook we used in 18.304â€"they just came out with their fifth edition and you can read it for free online. Proofs is divided into five sections, each one containing a sample of a field of mathematics:  number theory,  geometry,  analysis,  combinatorics, and  graph theory. Together these cover a lot of what we learn in course 18  and, in the case of combinatorics and graph theory, course 6-3 (or the course 6 part of 6-7, in my case).

Analysis Of Robert Frost s Poem A Late Walk - 971 Words

The last days of each earthly season makes an impression on the conscious that has been pondered since the beginning of human evolution. Each season that passes represents change, new beginnings, and a better understanding of what was and will be. The poetry of Robert Frost particularly uses themes of nature, seasons, nostalgia and remembrance as means to reflect on the past and their implications on the future. â€Å"A Late Walk,† a poem inspired by his time on his farm during the turn of the 19th century, is one of many that show his melancholy affection for change. In this poem, Frost uses simple natural imaginary, written in quatrains and 3/4 beat line alterations, to take the reader on a stroll with a farmer as he returns home from the final days in the crop fields in the last days of fall and the associated nostalgia of every step. In the opening stanza, frost describes the narrator, a farmer, walking through the mowing fields back home. The phrase â€Å"The headless aftermath† produces imagery that connotes warfare, however, frost is portraying a field that has been recently harvested and the straw has been bailed (2). It is as if the farmer is a warrior that has battled the field all fall and now stands triumphant over his defeated enemies, now headless and still. Many have argued that Frost is supporting violence with this phrasing, however it is simply an observation the Narrator is making. Frost uses first person perspective to bring the reader along through the field thatShow MoreRelatedRobert Frost : A New England Poet3698 Words   |  15 PagesRobert Lee Frost Known for being a New England poet Robert Frost was born in San Francisco, California on March 26th, 1874. Born to a New England father William Prescott Frost Jr. and a Scottish mother Isabelle Moodie who moved to the west coast from Pennsylvania after marriage (Bailey). Both his parents were teachers and poets themselves, but his father later became a journalist with the San Francisco Evening Bulletin (Bailey). Frost spent 12 years of his life growing up in San Francisco, untilRead MoreLiterature and South Africa6682 Words   |  27 Pagesan autonomously functioning semiotic system. In this essay, the poem Mending Wall by Robert Frost is going to be used to describe how meaning are produced by codes, by recoding and overcoding according to Lotman’s semiotic theory. It is necessary to define codes and the process of interpretation before one delves in the semiotic analysis of the poem Mending Wall. As defined by Structuralist, literary codes that matter in our analysis per se are the literary signs, their overdetermination that amountRead MoreLiterature and South Africa6676 Words   |  27 Pagesan autonomously functioning semiotic system. In this essay, the poem Mending Wall by Robert Frost is going to be used to describe how meaning are produced by codes, by recoding and overcoding according to Lotman’s semiotic theory. It is necessary to define codes and the process of interpretation before one delves in the semiotic analysis of the poem Mending Wall. As defined by Structuralist, literary codes that matter in our analysis per se are the literary signs, their overdetermination that amountRead MoreLanguage of Advertising20371 Words   |  82 Pagesstylistic features. The eighth chapter introduces the role of parentheses insertion as additional information. Very important advertising text in sociolinguistic coverage and phonetic expressive means and stylistic devices in the ninth chapter (Galperin`s stylistic devices, the definitions of stylistic devices with examples). In the chapter of tenth we consider classification of language styles (Belles-Letters Style, Pablicist Style, Newspaper Style, Scientific Prose Style). And finally in the chapterRead MoreOrganisational Theory230255 Words   |  922 Pagesmanaging, organizing and reflecting on both formal and informal structures, and in this respect y ou will find this book timely, interesting and valuable. Peter Holdt Christensen, Associate Professor, Copenhagen Business School, Denmark McAuley et al.’s book is thought-provoking, witty and highly relevant for understanding contemporary organizational dilemmas. The book engages in an imaginative way with a wealth of organizational concepts and theories as well as provides insightful examples from theRead MoreStephen P. Robbins Timothy A. Judge (2011) Organizational Behaviour 15th Edition New Jersey: Prentice Hall393164 Words   |  1573 Pagesand permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, One Lake Street, Upper Saddle River, New Jersey 07458, or you may fa x your request to 201-236-3290. Many of the designations by manufacturersRead MoreLogical Reasoning189930 Words   |  760 PagesNorman Swartz, Simon Fraser University v Acknowledgments For the 1993 edition: The following friends and colleagues deserve thanks for their help and encouragement with this project: Clifford Anderson, Hellan Roth Dowden, Louise Dowden, Robert Foreman, Richard Gould, Kenneth King, Marjorie Lee, Elizabeth Perry, Heidi Wackerli, Perry Weddle, Tiffany Whetstone, and the following reviewers: David Adams, California State Polytechnic University; Stanley Baronett, Jr., University of Nevada-Las

There Were Several Reasons For Conducting This Needs...

There were several reasons for conducting this needs assessment. The first reason was to address the multi-sensory needs of preschool children. It is important to address this because during their early years, children learn by exploring their surroundings. Thus, preschoolers need a structured sensory environment that not only stimulate their senses but also teaches them new development skills such as cause and effect (Messauber, 2012). The second reason for this needs assessment was to learn more about the benefits of sensory rich environment for preschoolers. There is limited evidence based research about the effectiveness of MSE on preschoolers. Thus, the implementation of a sensory basket would provide us with information about how†¦show more content†¦However, there are clinics in farther developed areas such as Syracuse, NY that provide MSE rooms (â€Å"Pediatric Multisensory Environment,† n.d.). Nevertheless, it is important to take into consideration that pare nts do not have access to MSE environments in local towns such as Potsdam and Canton, NY. This need could be addressed by educating professionals in the North Country about the use and benefits of MSEs for children. The American Association of Multi-Sensory Environments also provides training and information for professionals who are interested in building a MSE (Messbauer, 2013). The overall purpose of the needs assessment was to learn about the multi-sensory needs of preschool children and whether a sensory rich environment would be beneficial for them. The current feel of the Smart Cookies daycare is that they have a limited amount of multi-sensory equipment (A. Cutler, personal communication, October 12, 2016). The problem identified with the needs assessment is that a MSE is not present for these preschoolers, therefore, they cannot explore and grow in their classroom environment (A. Cutler, personal communication, October 12, 2016). The preschool teacher indicated that during transition time, there is a decrease in appropriate behavior and participation skills in some of the students (A. Cutler, personal communication, October 12, 2016). The results of the needs assessment indicated that preschoolers would benefit from a MSE. The actionShow MoreRelatedMayor Schells Zero Homeless Family Pledge1240 Words   |  5 PagesContents Abstract: 2 Introduction: 2 Policy Choices: 2 Pre Implementation and Design Strategies 4 Steps Taken to Reengineer the Program 4 Importance of Conducting Assessments Prior to Implementation 5 References: 6 Abstract: The paper discusses Mayors Schells zero homeless family pledge. Mayor Schell was determined to eliminate the homelessness in Seattle when he became the mayor in 1998, to achieve his mission he and his team came up with some strategies and restructuring whichRead MorePad 500 Assignment 31608 Words   |  7 Pagesthe Program†¦..†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.†¦Ã¢â‚¬ ¦...†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.....†¦.†¦.6 Conducting Assessments.†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..8 References†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..†¦10 Abstract This assignment is about Mayor Paul Schell’s pledge for homeless families in Seattle, Washington. Mayor Schell on June 2, 1998 spoke to the press and pledge that there will be no homeless families with children and homeless single women on the streets of Seattle by Christmas 1998. This assignment will analyze the policies, strategiesRead MoreCase Study : Albany Law School1330 Words   |  6 PagesThe organization I have used throughout this project is Albany Law School (ALS). I’ve been employed by ALS for the last four years. Presently, I am the Assistant Dean for Admissions. I was recruited to this position to spearhead the admissions office and the overall enrollment initiatives for the law school. I am a direct report to the Dean of the law school and also serve of the Administrative Leadership Team. During my time at ALS, I have noticed that there is high turnover in leadership. OverRead MoreDisproportionate Minority Confinement ( Dmc ) Of Youth1422 Words   |  6 PagesAmerican and Hispanic decent. In the programs the application of the disproportionate minority confinement is to identification and assessment of the states involved. The program that determines whether DMC exists in secure facilities, it also identifies the causes and development of corrective strategies. The examination of compliance with the identification and assessment stages is the objective of the research in DMC. The program focuses on the overrepresentation of minorities in the juvenile justiceRead MoreRisk Assessment For A Financial Institution Essay1618 Words   |  7 PagesPart A For the purpose of this paper, I choose U.S Bank, a financial institute. A risk assessment for a financial institution measures and helps manage compliance, financial and operational risks associated with both internal and external, activities and events. In today’s operating environment, numerous regulations and standards make risk assessment not only logical but vital for financial institutions. These are examples of regulation and laws that governs financial institutions, Regulatory complianceRead MoreImportance of Educational Research1598 Words   |  7 PagesSenior Scientist  Why Research is ImportantSchool professionals have an increased awareness of the way the term research based is being used by publishers.   It has become a completely meaningless phrase in recent years.   Every type of intervention, assessment, or curriculum product now carries the label research based.   In particular, intervention publishers shamelessly proclaim that their products are bas ed upon research despite a complete absence of scientifically based research.   Reasonable care canRead MoreThen and Now: the Changing Paradigms of Special Education Assessments1637 Words   |  7 PagesRunning head: SPECIAL EDUCATION ASSESSMENT Then and Now: The Changing Paradigms of Special Education Assessments Michelle Walker Grand Canyon University: September 12, 2012 Then and Now: The Changing Paradigms of Special Education Assessments All school aged children who are currently enrolled among the many school districts and systems ranging from, ages 3 to 21, have been provided with an enormous opportunity to have rights, which ensure these children to receive a FreeRead MoreThe Problem Of Teen Pregnancy Rates951 Words   |  4 Pagesexperienced sexual and physical abuse by household member requiring the need for alternative housing options (Housing and Urban Development, 2016). Assessment of Effectiveness of Current Policy and Limitations of existing Research I would find it hard to say that the second chance home policy is effective simply based on the currently decreasing teen pregnancy rates. Since PRWORA was enacted in 1996, there have been several major policy changes that could also be affecting the decrease in teen pregnancyRead MoreEssay about Ligand Case1303 Words   |  6 Pagesprincipally changed the way auditing and financial reporting was being conducted. This act was prompted by high level frauds that public companies engaged in with regard to financial reporting and auditing practices. The act therefore recommended the setting up of a Public accounting Oversight board which was mainly to conduct regulatory and supervisory roles in auditing public audit firms and individual auditors. This was done through establishment of proper quality control measures on the work ofRead MoreEssay on Depression Inventory for the Elderly (Die)997 Words   |  4 Pagesmost of these researchers depicted that depression in the elderly people is very frequent and in spite of number of researches in this context, it is often undiagnosed or untreated. To add to this jeopardy, it has also been estimated that only 10% out of the total depressed elderly individuals receive proper diagnosis and treatment (Holroyd et al, 2000). And for that reason, an authentic evaluation of depression in the elderly people comes up with a challenge because as per their psychology, there are

The Hunters Moonsong Chapter Forty Free Essays

Damon was moving fast, and Elena and the others had to almost race to keep up with him as they headed for the library. â€Å"Typical Stefan, sacrificing himself,† he muttered angrily. â€Å"He could have asked for help when he realized something was going on. We will write a custom essay sample on The Hunters: Moonsong Chapter Forty or any similar topic only for you Order Now † He stopped for a second to let the others catch up and glared at them al . â€Å"If Stefan can’t handle a few newly made vampires by himself, I’m ashamed of him,† he said. â€Å"Maybe we should just leave him after al . Survival of the fittest.† Elena touched his hand lightly, and, after a moment, Damon hurried on toward the library. She didn’t for an instant believe he would leave Stefan a captive. None of them did. The taut, strained lines of his face showed that Damon was entirely focused on the danger his brother was in, their rivalry temporarily forgotten. â€Å"It’s not just a few vampires,† Matt said. â€Å"There are about twenty-five of them. I’m sorry, you guys, I’ve been a moron.† He swung the stave Meredith had given him – Samantha’s stave – determinedly in one hand. â€Å"It’s not your fault,† Bonnie said. â€Å"You couldn’t have known your frat – or whatever – was evil, could you?† If anyone had spotted them as they crossed the campus, Elena was sure they would have been an alarming sight: she and Bonnie were clutching the large, sharp hunting knives Meredith had given them only half concealed under their jackets. Matt was holding the stave, and Meredith had her own stave in one hand. But it was past midnight, and the path they were fol owing was deserted. Only Damon wasn’t carrying a weapon, and he clearly was a weapon. His human fa?ade seemed to have lifted, and his angry expression could have been carved out of stone, except for the glimpse of sharp white teeth between his lips and the seemingly bottomless darkness of his eyes. When they reached the closed library, Damon didn’t pause, forcing its metal doors open with the grinding sound of splitting metal. Elena glanced around nervously. The last thing they needed was campus security showing up. But the paths near the library were dark and empty. They al fol owed Damon down to the basement and into the hal ways of administrative offices. Final y, he stopped outside the door marked Research Office where he and Elena had once met Matt. â€Å"This is the entrance?† he asked Matt and, at his nod, efficiently broke the lock on the door. â€Å"You’re al staying up here. Just Meredith and I are going down.† He looked at Meredith. â€Å"Want to kil some vampires, hunter? Let’s fulfil your destiny, shal we?† Meredith slashed her stave in the air, and a slow smile tugged at the corners of her mouth. â€Å"I’m ready,† she said at last. â€Å"I’m coming, too,† Elena said, keeping her voice steady. â€Å"I’m not waiting up here while Stefan’s in danger.† Damon drew a breath, and she thought he was going to argue with her, but instead he sighed. â€Å"Al right, princess,† he said, his voice gentler than it had been since Matt told them what had happened to Stefan. â€Å"But you do what I – or Meredith – tel you.† â€Å"I’m not waiting up here,† Matt said stubbornly. â€Å"This is my fault.† Damon turned on him, his mouth twisting into a sneer. â€Å"Yes, it is your fault. And you told us Ethan can control you. I don’t want to get your knife in my back while we’re fighting your enemies.† Matt dropped his head, defeated. â€Å"Okay,† he said. â€Å"Go down two flights of stairs, and you’l see the doors to the room they’re in.† Damon nodded sharply and pul ed up the trapdoor. Meredith fol owed him down the stairs, but Matt caught Elena’s arm as she headed after them. â€Å"Please,† he said quickly. â€Å"If any of the pledges stil seem rational, even if they’re vampires, try to get them out. Maybe we can help them. My friend Chloe†¦Ã¢â‚¬  In the grim lines of his face, his pale blue eyes were frightened. â€Å"I’l try,† Elena said, and squeezed his hand. She exchanged a glance with Bonnie, then fol owed Meredith through the trapdoor. When they reached the entrance to the Vitale Society’s chamber, Meredith and Damon pressed their backs against the elaborately carved wooden doors. Watching, Elena could see a similarity for the first time between them. Now that they were facing a battle, Meredith and Damon were both wearing eager smiles. One †¦ two †¦ came Damon’s silent count †¦ three. They pushed together. The double doors flew inward, and the chains that had held them closed went flying. Damon stalked in, stil smiling a vicious gleaming smile, Meredith erect and alert behind him, her stave poised. Dark figures rushed at them, but Elena was looking past them, searching for Stefan. Then her eyes found him, and al the breath rushed out of her. He was hurt. Tied firmly to a chair, he raised a pale face to greet her, his leaf-green eyes agonized. From his arm, dark red blood dripped steadily, pooling on the floor beneath his chair. Elena went a little mad. Charging across the room toward Stefan, she was only half aware of one of the hooded figures leaping at her, and of Damon catching it in midstride, casual y snapping its neck and letting the body fal to the floor. Absently, she registered the smack of wood against flesh as Meredith caught another attacker with her stave so that it fel in convulsions as the concentrated essence of vervain from the stave’s spikes hit its bloodstream. And then she was crouching next to Stefan, and, for a moment at least, nothing else mattered. He was shaking slightly, just the faintest tremors, and she stroked his hand, careful of the wound on his forearm. Raised red ridges ran around his wrists below the rope, spots of blood on their surface. â€Å"Vervain on the ropes,† he muttered. â€Å"I’m okay, just hurry.† And then, â€Å"Elena?† Below the pain in his voice, a dawning note of joy. She hoped he could read al the love she felt in her eyes as she met his gaze. â€Å"I’m here, Stefan. I’m so sorry.† She took out the knife Meredith had given her and began to saw at the ropes that held him, careful not to cut him, trying not to pul the ropes any tighter. He winced in pain, and then the ropes around his wrists snapped. â€Å"Your poor arm,† she said, and felt in her pockets for something to staunch the blood, final y just pul ing off her jacket and holding it against the cut. Stefan took the jacket from her. â€Å"You’l have to cut through the rest of the ropes, too,† he said, his voice strained. â€Å"I can’t touch them because of the vervain.† She nodded and went to work on the ropes holding his legs. â€Å"I love you,† she told him, concentrating on her work, not looking up. â€Å"I love you so much. I hurt you, and I never wanted to. Never, Stefan. Please believe me.† She finished cutting throu gh the ropes around his knees and ankles and chanced a glance up at Stefan’s face. Tears, she realized, were running down her own face, and she wiped them away. The thud of another body hitting the floor and a screech of rage came from behind them. But Stefan’s eyes held hers unwaveringly. â€Å"Elena, I†¦Ã¢â‚¬  he sighed. â€Å"I love you more than anything in the world,† he said simply. â€Å"You know that. No conditions.† She took a long, shuddering breath and wiped the tears away again. She had to be able to see, had to keep her hands from shaking. The ropes around his torso were looped and twisted together. She pul ed at them, finding where there was enough give to start cutting, and Stefan hissed in pain. â€Å"Sorry, sorry,† she said hurriedly, and began to slice through the rope as rapidly as she dared. â€Å"Stefan,† she began again, â€Å"the kiss with Damon – Well, I can’t lie and say I don’t feel anything for him – but the kiss wasn’t anything I’d planned on. I didn’t even mean to be with him that night, it just happened. And when you saw us, that kiss, he’d just saved my life†¦Ã¢â‚¬  She was stumbling over her words now, and she let them trail off. â€Å"I don’t have any real excuses, Stefan,† she said flatly. â€Å"I just want you to forgive me. I don’t think I can live without you.† The last of the ropes parted, and she eased them from around him before she looked up, frightened and hopeful. Stefan was gazing at her, his sculpted lips turning up in a half smile. â€Å"Elena,† he said and pul ed her to him in a brief, tender kiss. Then he pushed her to the wal . â€Å"Stay out of this, please,† he said, and limped toward the fight, stil weak from the vervain, but reaching to pul a vampire away from Meredith and sinking his own fangs into its neck. Not that she needed his help. Meredith was amazing. When had she gotten so good? Elena had seen her fight before, of course, and she’d been strong and quick, but now the tal girl was as graceful as a dancer and as deadly as an assassin. She was fighting three vampires, who circled her angrily. Spinning and kicking, moving almost as fast as the monsters she was fighting, despite the fact that their speed was supernatural, she knocked one off his feet, sending him flying, and, in a smooth fol ow-up blow, bashed another in the face, leaving the vampire staggering backward with his hands up, half blinded. There were bodies littered across the floor, evidence of Meredith’s skil and Damon’s vicious rage. As Elena watched, Stefan tossed down the drained body of the vampire he had been fighting and looked around. Only Ethan and the three vampires surrounding Meredith remained on their feet. Damon had Ethan on the run, backing nervously away as Damon stalked toward him, peppering him with sharp open-handed blows. â€Å"†¦ my brother,† she heard Damon muttering. â€Å"Insolent pup. You think you know anything, child, you think you want power?† With a sudden, violent movement, he grabbed Ethan’s arm and jerked. Elena could hear the bone snap. Stefan passed Elena, heading toward Meredith again, and paused for a moment. â€Å"Ethan was laying a trap for Damon,† he told her dryly. â€Å"I don’t know why I was worried. Clearly, he didn’t know what he was trying to catch.† Elena nodded again, suppressing a grin. The idea of any brand-new vampire getting the better of Damon, with al his experience and cunning, seemed ridiculous. Then the tide of the battle suddenly turned. One of the vampires Meredith was fighting dodged her blow and, half bent over, flung itself at her, knocking the slender girl into the air. There was an endless moment where Meredith looked like she was flying, arms akimbo, and then she slammed headfirst into the heavy altarlike table at the front of the room. The table wobbled and fel over with a heavy thud. Meredith lay stil , her eyes closed, unconscious. Elena ran to her and knelt down, cradling her head in her lap. The three vampires Meredith had been fighting were worse for the wear. One had blood steadily streaming down his face, another was limping, and the last was doubled over as if something had been injured inside her, but they could stil move fast. In an instant, they had surrounded Stefan. As Damon growled and turned, shifting his stance to help his brother, Ethan saw his chance and launched himself at Damon. Faster than Elena’s eye could fol ow, his teeth were gouging at Damon’s throat, bright spurts of blood flying up. He had a knife in one hand and was trying to cut at Damon at the same time as he bit. With a cry of pain and shock, Damon clawed at Ethan, trying to fling him away. Elena picked up her knife again and rushed toward them. But two of the remaining vampires were on Damon in a split second, pul ing his arms back. One caught Damon’s midnight dark hair in his hand, yanking the older vampire’s head back to expose his throat more ful y to Ethan’s teeth. Off balance, Damon staggered backward and for a moment caught Elena’s eye, his face soft with dismay. Terrified, Elena grabbed at the back of one of the vampires, and it threw her to the floor without even looking at her. Stefan, meanwhile, was caught in a struggle with another vampire, desperate to get to his brother. Damon was a better and a more experienced warrior than any of the vampires attacking him. But if they pushed their momentary advantage, used their superior numbers, they might bring him down before he could recover. She clutched her knife tighter and jumped to her feet again, knowing in her heart that she’d be too late to save him but that she needed to try. A snarling blur shot past her, and Stefan, free of his adversary, slammed into Ethan, throwing him across the room, sending his knife flying. Without pausing, he ripped one of the other vampires from Damon’s arm and snapped his neck. By the time the body hit the floor, Damon had neatly dispatched the other one. The brothers, both panting, exchanged a long look that seemed to carry a lot of unspoken communication. Damon wiped a smear of crimson blood from his mouth with the back of his hand. Suddenly an arm was around Elena’s throat, and the knife was wrenched out of her hand. She was being dragged upward. Something sharp was poking her in the tender hol ow at the bottom of her neck. â€Å"I can kil her before you could even get over here,† Ethan’s voice said, too loud by her ear. Elena flailed an arm backward, trying to grab at his hair or face, and he kicked viciously at her legs, knocking her off-balance, and pul ed her closer. â€Å"I could snap her neck with one arm. I could stab her with her own knife and let her bleed out. It would be fun.† He was holding her knife, Elena realized, pressed against her throat. His other arm hung loose, and curiously bent. Damon had broken it, Elena remembered. Stefan and Damon froze and then very slowly turned toward Elena and Ethan, both their faces shuttered and wary. Then Damon’s broke into a rictus of rage. â€Å"Let her go,† he snarled. â€Å"We’d kil you the second she hit the ground.† Ethan laughed, a remarkably genuine laugh for someone in a life-or-death standoff. â€Å"She’l stil be dead, though, so I think it might be worth it. You’re not planning to let me leave here anyway, are you?† He turned to Stefan, his voice mocking. â€Å"You know, I heard all about the Salvatore brothers from some of Klaus’s other descendants. They said you were aristocratic and beautiful and terribly hot tempered. That Stefan was moral, and that Damon was remorseless. But they also said that you were both fools for love, always for love. It’s your fatal flaw. So, yeah, I think my chances are a lot better when I’ve got your girlfriend in my power. Whose girlfriend is she, actual y? I can’t tel .† Elena flinched. â€Å"Wait a second, Ethan.† Stefan held out his hands placatingly. â€Å"Hold on. If you agree not to bring back Klaus and let Elena go safely, we’l give you whatever you want. Get out of town, and we won’t come after you. You’l be safe. If you know about us, you know we’l keep our word.† Behind him, Damon nodded reluctantly, his eyes on Elena’s face. Ethan laughed again. â€Å"I don’t think you have anything I want anymore, Stefan,† he said. â€Å"The rest of the Vitale Society, including our newest initiates, wil be coming back soon, and I think they’l tip the scales back in my favor.† He tightened his arm around Elena’s throat. â€Å"We’ve kil ed so many students on this campus. Surely one more won’t be missed.† Damon hissed in rage and started forward, but Ethan cal ed out, â€Å"Stop right there, or – â€Å" Suddenly, he jerked, and Elena felt a sharp, stinging pain in her throat. She squeaked in horror and grabbed at her own neck. But it was only a scratch from the knife. As Stefan and Damon stood helpless and furious, Ethan’s arm loosened from around her throat. He made a hideous gurgling noise. Elena yanked away as soon as his grip weakened. Blood was running in long thick rivulets from Ethan’s torso, and his mouth opened in shock as he clutched at himself and slowly fel forward, a round hole in his chest fil ing with blood. Behind him, Meredith stood, hair flying, her usual y cool gray eyes burning like dark coals in her face. Her stave was coated in Ethan’s blood. â€Å"I got him in the heart,† she said, her voice fierce. â€Å"Thank you,† Elena murmured politely. She was feeling†¦ real y †¦ very peculiar, and it wasn’t until she was actual y starting to fal that she thought, Oh no, I think I’m going to faint. Blurrily, she saw both Damon and Stefan rushing forward to catch her, and when she came to a moment later, she was held tightly in two pairs of arms. â€Å"I’m okay,† she said. â€Å"It was just †¦ for a second, I was†¦Ã¢â‚¬  She felt one pair of arms pul her closer for a moment, and then they released her, shifting her weight over to the other set. When she looked up, Stefan was clutching her tightly to him. Damon stood a few feet away, his face unreadable. â€Å"I knew you’d come to save me,† Stefan said, holding Elena but looking at Damon. Damon’s lips twitched into a tiny, reluctant smile. â€Å"Of course I did, you idiot,† he said gruffly. â€Å"I’m your brother.† They looked at each other for a long moment, and then Damon’s eyes flicked to Elena, stil in Stefan’s arms, and away. â€Å"Let’s put out the torches and go,† he said briskly. â€Å"We’ve stil got about fourteen vampires to find.† How to cite The Hunters: Moonsong Chapter Forty, Essay examples

Case Study of Smith and Jones Samples for Students Myassignment

Question: Discuss the Case Study of Smith and Jones. Answer: Introduction Tax system of the country plays important role in rising government fund. The fun is utilized for the development of the country as well as for the essential expenditure of the country. The essential expenditure such as the infrastructure development, defence budget, and expense for maintaining the law and order within the country, helping the deprived and old people the government fund is used. The money collecting from the tax help in this matter and for this reason the government use to bring amendment in the taxation policy according to the need of the country. Main study The assessable income includes income as per the ordinary concepts is called as ordinary income. Section of ITAA 1997 includes in the assessable income of the taxpayer where the taxpayer is the resident of Australia. All the ordinary income earned by the taxpayer in and out of the Australia would be included in the assessable income during that period of time[1]. The capital gain tax is identified in the case of the Pre CGT land where, the asset is depicted to be considered as the pre CGT tax and is also not assessable to the nature of the tax. The enhancement of the tax can be easily determined by showing that the Pre-CGT tax is not applicable by showing the enhancement of the nature. The consideration can be easily identified as the enhancement of the motion which is being used for showing the structure of the work and also the cost of the base income for the Smith and Jones are depicted to be increasing in nature. This the figure simply depicts the implementation of the paragraph 104-10(5)(a) of the ITAA 1997 which is applicable to the case as per the undertakings are made[2]. Therefore the enhancement of the structure can be depicted by implementing the capital gain taxes on the income made from the grazing of the sheep on the land and on the improvement of the properties that are made on the land. As per the Assessable Income Tax Law, th e relevant income tax will be made on the capital gains as it is being seen in the activities continued by the Smith and the Jones. Therefore the structure clearly describes the appropriate format of the study and also the enhancement of the study can be made by showing the appropriate explanations to the case study as it is undertaken in this case[3]. Apart from this, the threshold of the result can be made by showing the development costs and the involvement of the income through out six years made by the Smith and the Jones. The aforementioned case study it is observed that Smith and Jones are property developer. They have purchased a property eight years ago, which has huge potential. The property is used for grazing purpose for the last six years and thenSmith and Jones have subdivided the land into 5 hectors blocks and sold the land to one purchase[4]. In this scenario as Smith and Jones are involving in the property development profession and they use the property for grazing for the last six years and then they trade the property; thus, as per the taxation rule in Australian taxation Office (ATO) they have to pay tax on the income they earned from the grazing and the enhanced value of the property to the government[5]. According to the law it must be considered as the ordinary income and the income tax rule should be applied for this case. The capital gain tax is also relevant for this case as the property during this year developed a lot and thus, the rate of the property has been increased a lot. In this situation Smith and Jones have to pay capital gain tax for the profit they make from the capital gain of the property. Conclusion The taxation system supports the essential government expenditure. According to the tax consequences the various types of taxes use to be imposed on the tax payers [6]. According to the undertaken case study both ordinary income tax and capital gain tax will be applied. References Cooper, Graeme S,Executing Income Tax(Australian Tax Research Foundation, 2008) Hurley, Annette,Tax Laws Amendment (Personal Income Tax Reduction) Bill 2008(Standing Committee on Economics, 2008) Krishna, Vern,Income Tax Law(Irwin Law, 2015) Mason, Tom,Income Tax Law(Taxation Concepts, 2012) Whittenburg, Gerald E, Martha Altus-Buller and Steven L Gill,Income Tax Fundamentals(1st ed, 2013) Woellner, R. H,Australian Taxation Law 2012(CCH Australia, 2013) Graeme S Cooper,Executing Income Tax(Australian Tax Research Foundation, 2008). Annette Hurley,Tax Laws Amendment (Personal Income Tax Reduction) Bill 2008(Standing Committee on Economics, 2008). Vern Krishna, Income Tax Law (Irwin Law, 2015). Gerald E Whittenburg, Martha Altus-Buller and Steven L Gill,Income Tax Fundamentals(1st ed, 2013). Mason, Tom, Income Tax Law (Taxation Concepts, 2012) Woellner, R. H, Australian Taxation Law 2012 (CCH Australia, 2013)