Survival Guide Essays - Adolescence, Educational Psychology

Survival manual This book is actually what it says it is ? a Guidebook or Manual for the two guardians what's more, youngsters that offers knowledge and counsel on a wide scope of juvenile formative concerns. Sorted out into helpful topical segments for both parent and teenager perusers, the content can be handily counseled when looking for guidance on a specific issue, or essentially read spread to cover. In any case, the peruser will discover an abundance of functional guidance for both parent and youngster. The creators approach their subject in a thoughtful and touchy way with an end goal to improve average parent/adolescent encounters. Areas planned for high schooler perusers are featured in blue and written in a conversational style where the writer is conversing with the high schooler peruser. Various statements from different teenagers are sprinkled inside these segments to give the material a very easy to understand feel. Guardians will discover these segments valuable in that they give a steady and getting point of view. They allow teenagers to step into the shoes of their folks and consider their circumstance inside a wide scope of potential parental concerns. The essential substance of the book is expected for guardians and goes into more profundity than an ordinary adolescent segment. Represented with simple to follow realistic coordinators, the area planned for guardians is stuffed with various case chronicles, tables and outlines, surveys, discoursed, and issues basic to families with adolescents. Of specific significance are the situation accounts, discourse, and issue boxes. The case narratives are taken from genuine circumstances and present both issue and answer for a wide scope of circumstances. These accounts present the perspective of both parent and teenager, which the peruser can look into to their own circumstance. They moreover give knowledge into the reasonable justification of the contention just as clarifying how the issue was really settled. The issues segments present issues that guardians and youngsters will in general contend about. Issues, for example, style of dress and companionships are composed is such a route as to give an accommodating center ground position for guardians to take while battling with a headstrong high schooler. The discourse boxes present situations normal to families managing youngsters. Notwithstanding introducing an average parent/high schooler discussion, (for example, playing music excessively noisy or discusses drugs) these segments offer understanding on how to talk about specific issues with your teenagers. Having such models to follow can be exceptionally accommodating to a parent or educator scanning for the correct words or courses in which to move toward a disputable subject. Notwithstanding the discourse, the creators have likewise included guidance toward the finish of discourse segments that propose a few manners by which the expressed issue may be accommodated. The various tables found all through the content give advantageous, cutting-edge information on a wide assortment of issues. Guardians and educators the same can utilize these tables, specifically, the area in the primary part managing physical improvement. Entitled Achievements of Early (ages 11 to 14), Middle (ages 15 to 16), and Late Adolescence (ages 17 to 18), this area presents different formative achievements with regards to the ramification for the kid also as the possible impact on the family. Of the considerable number of issues and concerns communicated in this book, the main area which guardians may discover questionable concerns part eleven ? sexuality in puberty. The creators apparently approach this subject with a no limits strategy. They examine all conceivable sexual issues, from the more harmless dating and foreplay worries to disputable issues for example, homosexuality, masturbation and pre-marriage sex. Guardians with solid strict feelings may disagree with some of the creators' points of view and suggestions. For example, concerning the issue of ahead of schedule sexual experimentation the creators state: ...in itself early...sexual experience isn't really either irregular or degenerate... early sexual conduct ? in spite of the fact that it might stun or baffle you ? ought not be respected as a significant calamity. Pg. 127 Concerning homosexuality the creators state: ...you must choose the option to acknowledge that this (homosexuality) is the way things are. They can't resist being gay and they can't change ? regardless of whether they need to. (Pg. 131) On the issue of choosing to turn out to be explicitly dynamic the creators state: You (the youngster) are the main individual who can decide about your body, your sentiments, and your life. (Pg 132 out of an area proposed for high schooler perusers) On the issue of safe sex the creators state: Most likely the most significant interesting point before you begin engaging in sexual relations is the means by which to dodge undesirable pregnancy. (Pg. 135

A reflective account of the application of learning to clinical Essay

An intelligent record of the utilization of figuring out how to clinical practice - Essay Example Comprehensively, Ischaemic stroke influences a bigger number of individuals than different types of strokes. The stroke can occur or happen in two different ways; blood vessel apoplexy or cerebral embolism. Blood vessel apoplexy is the point at which a blockage shapes in the vein that provisions blood to the mind. As an outcome, blood flexibly to the cerebrum is blocked (Baker, 2008). Then again, cerebral embolism happens when a blood coagulation shapes in an alternate piece of the body (typically an enormous corridor or the heart) and in this way goes to the cerebrum and hinders the blood flexibly. The illness is generally common in more established grown-ups (individuals more than 65 years old). Considering this, the paper will utilize Gibbs’ intelligent cycle model. The model is best since it is genuinely straight forward. Furthermore, the model supporters for a brief depiction of the situation, assessment of the sentiments and experience, decision, and a reflection depende nt on experience of what I would do if a comparative circumstance came up once more. The condition that I am managing is known as the ischaemic stroke. It is the coagulating of the blood in the blood supply routes. In his book, a Schmer (1973) expressed that the mind is denied off food and the oxygen when coagulating happens in the veins. Accordingly, the synapses bite the dust. Albeit some platelets may pass on promptly at the beginning of the stroke, some may endure longer. On the off chance that the blood gracefully is recovered, the enduring cells may live beyond words. The condition may happen quick, in practically no time, or the patient may encounter side effects for certain hours. The patient should have been checked for the side effects of an ischaemic stroke. On occasion, the basic issues probably won't be the stroke yet something else (Wilbur 1997). A patient experiencing ischaemic stroke shows deadness or a few shortcomings of one side of the body. This is called authoritative hemiplegia. Then again, some may show imperfections of the face, for example, dropping eyelids, and aptosis. All things considered, others get visual

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).

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Booker T Washington Essay Free Essays

Booker T. Washington founder of Tuskegee Institute, he was a well know black educator. Tuskegee provider industrial training to African American. We will write a custom essay sample on Booker T Washington Essay or any similar topic only for you Order Now He believed that African Americans would gain respect from the white community if they had trade skills. Washington also believed that trade skills were key to economic security. He thought that economic progress was sufficient and that blacks can tolerate inequality. Booker urged blacks to accept discrimination and focus more on elevating themselves , educating themselves in crafting and industrial farming skills . W. E. B Du Bois was more of a political thinker a intellectual scholar . He help found NAACP, he also published a book called The Souls of Black Folks . Bois believed that academic education was more than trade education. He felt like Booker T. Washington was keeping the African Americans trapped in a lower social and economic class with is emphasis on industrial education. Du Bois wanted the black race to expand their mind in arts and science, he wanted blacks to work hard no matter of their career . They both was for African American to get their education and wanted us to have the same equal rights as whites. They both had different political views Du Bois encourage blacks to demand their equal rights ,Washington believed that it was important for blacks to develop good relationships with whites ignoring discrimination . Booker T. Washington had a more rational strategy it was more planed out. Washington plan was to change how other race perceived African American. His overall goal was to demonstrate to other race that African American were capable of being civilized, educated, and productive man and women in the work force. Although, DuBois and Washington had the same goal their methods of achieving their goals were completely opposite. Dubois felt that African Americans should demand respect because they deserved it. However, Washington stance was that African American should earn their respect by showing the world that they weren’t ignorant, thieves, or anything else negative that was commonly used to describe blacks. I agree with Washington approach because his strategy demands that black people become active in their own advancements. I think African American have to get back connect to their root before we can advance as a race in today society . Booker strategy can be modify because not everybody was book smart , but their some people who could work magic with their hands . If blacks could take advantage of the thoughts and ideas of our ancestors we can grow as a race . Ida B. Wells a civil right activist, she expressed herself about lynching through writing and her speeches . She help reduce the amount of lynching in the south. Ida B. Wells demanded that the whites murders of the innocent people be held accountable of their actions. She also was involved in many creation of several organizations encouraging the advancement of women and other minorities . Wells wasn’t afraid to speak her mind she was determine to make away for the black race . Mary Church Terrell was born into wealth, she was the first black women appointed to the District of Columbia Board of education . Mary was a charter member and first president of the National Association of Colored Women. She was nationally known for her support of women’s suffrage and opposition to racial segregation . Mary had many contributes, she taught at a black secondary school , worked with Fredrick Douglass and spent two years studying in Europe . She was a leader of a large protest against segregated eating places, and she was the only black women to speak at the International Congress of Women in 1904 that was held in Berlin, Germany. She was a powerful woman back in her days, took the steps that was necessary to succeed . Anna Julia Cooper was part of the feminist movement, she published a book call â€Å" A Voice from the South by a Black Women of the South â€Å". She is an African American scholar, she was the fourth black women to earn a doctoral degree. Anna help found the Colored Women’s League in Washington D. C, she was one of the few black women invited to speak on the Pan-African Conference in London. She was committed to the race and gender equality Cooper lived her life as an active vocal participant in the Women Era. Anna Julia Cooper believed that intelligent women’s voices brought balance to the struggle for the human race . Black women today should look as these women as role models, they lead the way for many generation. It’s up to us as women to take a chance to be heard and fight for what we believe in. No matter the outcome Ida, Mary and Anna took a stand as women and make sure their voice were heard that alone say a lot about their character . Many women of our generation is scared to speak out on the thing they believe afraid of being judge. Women are more powerful than they know , I think women can still uplift the black race if we came together to fix the things in the black community and work together instead of tearing each other down. If we change our mindset we can change the world. How to cite Booker T Washington Essay, Essay examples

Taxation Law Implications on Investment

Question: Discuss about the Taxation Law Implications on Investment. Answer: Introduction The following assignment presents tax implications on investment acquisition with respect to the provisions and regulations of Income Tax Assessment Act (ITAA), 1997 and code of conduct under TASA 2009. The report covers the discussion on permanent establishment in Australia through various brokers for Alex Tan and Ryan Tan, Singapore resident brothers. The report states the permanent establishment from Melbourne and Cambrai property operations along with the relevant tax implications on the business operation income. Further, the report presents the residential status of the brothers and the relevant tax liability in Australia with respect to the exposure in tax. Considering the acquisition of properties and investment in securities the report highlights the application of capital gain taxation rules as per ITAA 1997 for the sale of securities including the construction and demolition of townhouses and beef business. Another part of the report presents the tax implications on the disposal of the townhouses with respect to the provisions on income from other sources. The assessees planned to dispose the townhouses to raise the required funds for the improvement of property in Cambrai including the purchase of animals for re-establishment of commercial herd. In view of the principles of ITAA 1997, the report highlights the discussion on identification of operation from Cambrai property as business to avail the deduction of business losses for future years. For the purpose of contingency funding options, the report presents the analysis and tax implications for the interest deduction during the taxation year. Considering the acquisition of property and business operations of Alex and Ryan in Australia the assignment highlights the analysis of structure options for tax liability including the points of strengths and weaknesses of the possible options. Permanent establishment in Australia As per the regulations of ITAA 97, permanent establishment refers to the place at which the assessee carries the business operations or the person carries the business through an agent subjected to the permanence of the place. According to the section 6(1) ITAA, 1997 permanent establishment can be fixed business place, construction or an agency that occurs under the tax treaties of Australia. Fixed place of business consists of three components i.e. the business should be operated through a fixed or particular geographic area along with the level of permanence to the assessee (Chaudhri v FCT 2001 ATC 4214). Place is another essential component that involves facilities or sources to conduct the business activities while third component business refers to the activities to be carried on partly or wholly at the permanent place (Higgins, Omer Phillips, 2015). Similarly, construction site is regarded as permanent establishment only if the tenure of construction is more or equal to the am ount of time mentioned in the ITAA regulations. Additionally, agency refers to the permanent establishment that is controlled by the principal while activities are undertaken by the agents at a fixed place of business. Accordingly, mere use of brokers would not result in permanent establishment in Australia unless the criteria under section 6(1) ITAA 97 is duly complied with since brokers are individuals assist in acquiring the specific requirements at a given point of time against the brokerage charges (Chirico et al., 2015). Establishment and tax implication of Melbourne and Cambrai operations In the present case study, Alex and Ryan planned to acquire two properties in Australia consisting of rundown abandoned factory in a suburb of Melbourne while the other was 5000 hectare farm property near Cambrai. Considering the rules of ITAA 1997 section 6, permanent establishment refers to the place through which or the place at which a person carries the business operations. In present situation it has been observed that the two brothers decided to expand business operations in Australia and therefore planned to acquire two properties. It was decided that the property in Melbourne would be re- constructed to build townhouses while the other property in Cambrai would be repaired for improvements to carry the operations of commercial herd. It has been observed that both the properties to be acquired with the intention to create fixed place to operate business activities and not temporary. Further, the brothers themselves would conduct operations on both the properties and not throu gh any agent; therefore, ITAA 97 rule on controlling the business does not arise but regulation of TASA 2009 is applicable (Mason Harrison, 2015). Since the criteria of ITAA 1997 section 6(1) has been complied with in terms of business operations at a fixed place, the operations in Melbourne and Cambrai will be regarded as permanent establishment. Income from the business operations from permanent establishment is taxable in Australia as per ITAA 1997 if the taxpayer belongs to the treaty country if the business activity is conducted through fixed place in Australia. Further, gain or loss on sale of assets as a permanent establishment part would be taxable Under Capital Gain Taxation (CGT), ITAA 1997 including the filing of Australian income tax return (Bimonte Stabile, 2015). Since, Alex and Ryan belongs to Singapore which is a treaty country and the operations on property held as permanent establishment in Australia, the brothers are liable to pay taxes on income from such operations as per ITAA 1997. Residency of the brothers and tax exposure in Australia The brothers would be considered as Australian resident for taxation purpose if they are actually present in the country for 183 days or more either continuously or with splits whereas the residential status of permanent establishment would be separate entity (Johnson Poterba, 2016). In the given case, the length of the two brothers stay in Australia is not mentioned clearly hence it can be said that if their stay is more than 183 days then Alex and Ryan would be considered as Australian resident. Accordingly, the brothers would be liable to pay income taxes arise from operations in Australia including any other income arising in the country. Moreover, permanent establishment is regarded as a separate entity hence, the income from such operations would be taxable as business income under Australian taxation system. The asset income i.e. loss or gain from sale of assets employed in permanent establishment operations would be taxable under Capital Gain Taxation. Application of CGT rules Profit on sale of shares is taxable under CGT rules, ITAA 1997 if the shares held by the taxpayer as an investment either for short- term or for long- term. In case the shares are held for long term then the cost element is reduced by applying indexation method at a specific taxation rate as per ITAA 97. On the contrary, if the shares held for short- term period then the indexation method would not be applicable for reducing the cost of acquiring shares. Moreover, the income on sale of shares acquired under business trading then profit or loss from the sale of such shares would be considered as ordinary business income. Besides, the net assets of the brothers throughout South East Asia valued to $7.5 million and acquired from their father valued to $1 million that is located in Singapore would be taxable as per the Singapore taxation system (Pomeranz, 2015). For the purpose of demolition and construction of townhouses the brothers are eligible to claim deduction if they have main residence in Australia as per their choice under section 118-150 ITAA 97 (Ndikumana, 2015). Since the brothers do not own any main residence in Australia, they are entitled to claim exemption for the part of townhouse to be used for residential purpose. Further, disposal of townhouses and beef business would constitute the tax liability for the brothers according to ordinary business income under Australian Income Tax since the sale of townhouses and beef business would be done as a part of business operations. Share portfolio Amount $ (million) Sales proceeds 1.75 Less: Brokerage 0.02 Less: Initial value of shares 0.63 Gain/ (loss) on sale of shares 1.11 (Source: Created by author) Alex and Ryan are required to pay tax on the income from sale of shares under CGT ITAA 1997 at the rate of 33.33% as long-term assets. However, the brothers are eligible to claim 50% deduction on the taxable amount. Tax Implication on disposal of townhouses The purpose of acquiring the property, construction of townhouses including the disposal was to earn profit and to raise funds in regular intervals for the brothers in Australia. The intention of raising funds includes the generation of funds to carry out improvement and repair work on Cambrai property to re-establish commercial herd. Further, the intention to sell the townhouses involves profit earning and to raise funds at regular intervals hence, the income will be taxed as ordinary business income. Property valuation (profit/ loss) Cost of Fitzroy block 1.25 Brokerage and fees 0.02 Demolition cost 0.26 Cost of subdivision 0.04 Construction Cost 2.40 Interest cost (annual) 0.04 Total Cost 4.00 Less: Estimated sales proceeds 2 undeveloped blocks 0.65 6 town houses 4.80 2 residence town houses 0.68 Total Sales 6.13 Estimated Profit (Total Sales- Total Cost) 2.13 (Source: Created by author) Computation on expenses and expected income from Melbourne property reflected expected income, therefore the brothers are eligible to pay tax on income u/s 6-5 ITAA 97 in the year it is actually earned. Tax implication on operations from Cambrai property Alex and Ryan acquired the property at Cambrai in order to re-establish the operations of commercial herd therefore, it will be considered as ordinary business under permanent establishment. In view of the decided case of Liftronic Pty Ltd v FC of T (Liftronic), section 6-5 of ITAA 97, it was concluded that the income from the property as a business operation and with the intention to earn profit was subjected to ordinary income tax. Accordingly, all the expenses related to acquisition of property in Cambrai would be allowed as deduction under ordinary business taxation section 6-5 ITAA 97. Further, the losses from the commercial herd operation would be considered as deduction for business losses in subsequent years. Alex and Ryan planned and decided to consider contingency funding for the purpose of conducting business operations on the Australian properties that is created to meet the expenditure in urgent time period. Accordingly, amount of interest incurred to borrow contingency funding would be allowed as deduction from taxability provided the amount of funds are utilized only for considering business operation and not for any other personal purpose as per section 6-5 ITAA 97. In the case judgment of Dixon J in Sun Newspaper Ltd and Associated Newspapers Ltd v FC of T, CLR 337 it was decided that if expenses for acquiring funds are recurring in nature then it would be considered as revenue in nature. Purchase price of Cambrai property 3.25 Brokerage and fees 0.06 Repairs 0.85 Capital improvements 0.50 Depreciation @25% 25% on (3.25+0.85+0.50) 1.15 Interest expenses 0.06 Total cost 5.87 Total Revenue in the initial year Nil Total Loss 5.87 (Source: Created by author) Loss from the Cambrai Property during the initial years would be considered for adjusting with the profit from Melbourne property, since both constitutes the criteria of permanent establishment. Business structure options In the given case study, it has been stated that the brothers are considering different business structure to utilize the property for business operations activities in Australia along with the consideration of borrowing loan funds. One option of acquiring loan fund is acquisition from the Singapore head office while the other option is to borrow the loan funds from the Melbourne branch of Singapore bank. In case the brothers acquire funds from Singapore head office then they would be able to save interest charges whereas if the funds borrowed from Singapore bank, Melbourne branch they will be entitled to claim interest deduction. In case of John Fairfax Sons Pty Ltd v FC of T, CLR 30, advantages for borrowing the loan amount and deduction of interest charges was considered. Further, conducting the business operations under one company structure provides several benefits along with certain disadvantages with respect to the tax implications. In case the company is incorporated in Singapore, the brothers would still be entitled to pay tax in Australia due to the rules of permanent establishment (Du Zhang, 2015). However, if the company is incorporated in Australia then the brothers will be entitled to receive benefits on DTAA being the owners from a treaty country, Singapore. On the contrary, if the brothers consider the business property as two separate organizations as wholly owned subsidiary then the compliance of accounting standards on consolidation financial statements would be mandatory. It has been mentioned that the brothers do not plan to consolidate two companies but if they are willing to operate the business as wholly owned subsidiary then it is mandatory to follow the consolidation accounting principles. Additionally, in case of partnership business structure the business income is taxable in the hands of the partners while the partnership tax return is required to be filed to declare the relevant incomes and deductible expenses. The partners are liable to pay taxes on income from partnership business as per the percentage of share that increases their individual tax liability as per Australian Taxation System. Moreover, in case of capital gain taxes each partner is liable to pay taxes according to the share of asset each partner owns as the verdict formed in Scott v FC of T (2002) 50 ATR 1235ATC. Accordingly, the tax liability of the individual partners becomes higher while the deduction for partners salary is also not allowable in the partnership business income. Considering the trust structure in Australia, it is not regarded as a separate taxable entity whereas it is important to file the tax return. According to ITAA 97, beneficiaries of the trust entitled to declare the income and required to pay the tax liability in their books in compliance with Division 30 TASA 2009. Recommendation In view of all the business structure provided above, it has been observed that organization as two separate entities under wholly owned subsidiary is not recommended since it incorporates several accounting and taxation principles. Additionally, incorporation of business as one company requires compliance of several regulations but the taxation rate is comparatively lower i.e. 30%. On the other side, the tax rate for individuals as applicable in case of partnership business is based on progressive rates between 0 to 45% including the tax liability under trust business structure. Considering the tax implications of all the business structures it can be said that the tax liability for a single company is the lowest however, it includes compliance of several regulations as per accounting principles and standards. Besides, partnership business has several advantages with respect to the acquisition of fund, business-controlling responsibility, management of business expenses and other sh are of profits and liability. Moreover, one of the major drawbacks of partnership business is the burden of tax liability on partners that can be up to 45%. Therefore, it is recommended that Alex and Ryan should consider one company business structure to operate the business operations from the properties in Australia. Conclusion In view of the tax implications on operations in Australia, it can be concluded that Alex and Ryan are eligible to claim the tax liability under ITAA as permanent establishment. Further, the brothers are eligible to claim deduction on loss as ordinary business from Cambrai property since it is permanent establishment while the profit on sale of shares is taxable under CGT ITAA 97. Considering several business structure on operations from property, incorporation of property as single company has been recommended since the tax liability is comparatively lower along with the accounting and expense deduction advantages. Reference List Bimonte, S., Stabile, A. (2015). Local taxation and urban development. Testing for the side-effects of the Italian property tax.Ecological Economics,120, 100-107. Chirico, M., Inman, R. P., Loeffler, C., MacDonald, J., Sieg, H. (2015). An Experimental Evaluation of Notification Strategies to Increase Property Tax Compliance: Free-Riding in the City of Brotherly Love. InTax Policy and the Economy, Volume 30. University of Chicago Press. Du, Z., Zhang, L. (2015). Home-purchase restriction, property tax and housing price in China: A counterfactual analysis.Journal of Econometrics,188(2), 558-568. Higgins, D., Omer, T. C., Phillips, J. D. (2015). The influence of a firm's business strategy on its tax aggressiveness.Contemporary Accounting Research,32(2), 674-702. Johnson, W. T., Poterba, J. M. (2016). The effect of taxes on shareholder inflows around mutual fund distribution dates.Research in Economics,70(1), 7-19. Mason, C. M., Harrison, R. T. (2015). Business angel investment activity in the financial crisis: UK evidence and policy implications.Environment and Planning C: Government and Policy,33(1), 43-60. Ndikumana, L. (2015). International Tax Cooperation and Implications of Globalization.Global Governance and Rules for the Post-2015 Era: Addressing Emerging Issues in the Global Environment.Bloomsbury Academic, 73-106. Pomeranz, D. (2015). No taxation without information: Deterrence and self-enforcement in the value added tax.The American Economic Review,105(8), 2539-2569.