Math!

Math question #1:

Write the answers on the blog.
 * **Are these statements true or false?** || **True** || **False**  ||   ||   ||
 * 1) In the expression 2n + 4, the coefficient of n is 4. ||  ||   ||   ||   ||
 * 2) "Five increased by twice a number" can be written as 10 + n. ||  ||   ||   ||   ||
 * 3) 2[[image:http://www.math.com/school/subject2/images/u1quizp3.gif width="7" height="6"]] is a real number. ||  ||   ||   ||   ||
 * 4) [[image:http://www.math.com/school/subject2/images/u1quizp4.gif width="125" height="10"]] ||  ||   ||   ||   ||
 * 5) "Seven less than a number equals nine" can be written as n - 7 = 9. ||   ||   ||   ||   ||
 * 6) "Twice a number is greater than 15" can be written as [[image:http://www.math.com/school/subject2/images/u1quizp6.gif width="40" height="10"]]. ||   ||   ||   ||   ||
 * 7) [[image:http://www.math.com/school/subject2/images/u1quizp7.gif width="140" height="18"]] ||   ||   ||   ||   ||
 * 8) "Five less than four times a number" can be written as 5 - 4n. ||   ||   ||

In [|mathematics], the **Pythagorean theorem** or **Pythagoras' theorem** is a relation in [|Euclidean geometry] among the three sides of a [|right triangle] (//right-angled triangle//). In terms of areas, it states: > In any right triangle, the area of the square whose side is the [|hypotenuse] (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a [|right angle]). The [|theorem] can be written as an [|equation] relating the lengths of the sides //a//, //b// and //c//, often called the //Pythagorean equation//:[|[1]] where //c// represents the length of the hypotenuse, and //a// and //b// represent the lengths of the other two sides. These two formulations show two fundamental aspects of this theorem: it is both a statement about //areas// and about //lengths//. [|Tobias Dantzig] refers to these as //areal// and //metric// interpretations. Some proofs of the theorem are based on one interpretation, some upon the other. Thus, Pythagoras' theorem stands with one foot in geometry and the other in algebra, a connection made clear originally by [|Descartes] in his work [|La Géométrie], and extending today into other branches of mathematics. The Pythagorean theorem has been modified to apply outside its original domain. A number of these generalizations are described below, including extension to many-dimensional Euclidean spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but //n//-dimensional solids. The Pythagorean theorem is named after the [|Greek] [|mathematician] [|Pythagoras], who by tradition is credited with its discovery and [|proof], although it is often argued that knowledge of the theorem predates him. (There is much evidence that [|Babylonian mathematicians] understood the formula, although there is little surviving evidence that they fitted it into a mathematical framework.) "[To the Egyptians and Babylonians] mathematics provided practical tools in the form of 'recipes' designed for specific calculations. Pythagoras, on the other hand, was one of the first to grasp numbers as abstract entities that exist in their own right." The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power. Popular references to Pythagoras' theorem in literature, plays, musicals, songs, stamps and cartoons abound.

Other forms As pointed out in the introduction, if //c// denotes the [|length] of the hypotenuse and //a// and //b// denote the lengths of the other two sides, Pythagoras' theorem can be expressed as the Pythagorean equation: or, solved for //c//: If //c// is known, and the length of one of the legs must be found, the following equations can be used: or The Pythagorean equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the [|law of cosines], which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle, the law of cosines reduces to the Pythagorean equation.

Proofs
This theorem may have more known proofs than any other (the law of [|quadratic reciprocity] being another contender for that distinction); the book //The Pythagorean Proposition// contains 370 proofs.[|[9]]

Proof using similar triangles
Proof using similar triangles This proof is based on the [|proportionality] of the sides of two [|similar] triangles, that is, upon the fact that the [|ratio] of any two corresponding sides of similar triangles is the same regardless of the size of the triangles. Let //ABC// represent a right triangle, with the right angle located at //C//, as shown on the figure. We draw the [|altitude] from point //C//, and call //H// its intersection with the side //AB//. Point //H// divides the length of the hypotenuse //c// into parts //d// and //e//. The new triangle //ACH// is [|similar] to triangle //ABC//, because they both have a right angle (by definition of the altitude), and they share the angle at //A//, meaning that the third angle will be the same in both triangles as well, marked as //θ// in the figure. By a similar reasoning, the triangle //CBH// is also similar to //ABC//. The proof of similarity of the triangles requires the [|Triangle postulate]: the sum of the angles in a triangle is two right angles, and is equivalent to the [|parallel postulate]. Similarity of the triangles leads to the equality of ratios of corresponding sides: The first result equates the [|cosine] of each angle //θ// and the second result equates the [|sines]. These ratios can be written as: Summing these two equalities, we obtain which, tidying up, is the Pythagorean theorem: This is a //metric// proof in the sense of Dantzig, one that depends on lengths, not areas. The role of this proof in history is the subject of much speculation. The underlying question is why Euclid did not use this proof, but invented another. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the //Elements//, and that the theory of proportions needed further development at that time.[|[10]][|[11]]

Euclid's proof
Proof in Euclid's //Elements// In outline, here is how the proof in [|Euclid's] //[|Elements]// proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details are next. Let //A//, //B//, //C// be the [|vertices] of a right triangle, with a right angle at //A//. Drop a perpendicular from //A// to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs. For the formal proof, we require four elementary [|lemmata]: Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.[|[12]] Illustration including the new lines The proof is as follows: Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF This proof, which appears in Euclid's //Elements// as that of Proposition 47 in Book 1,[|[14]] demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.[|[15]] It is therefore an //areal// proof in the sense of Dantzig, one that depends on areas, not lengths. This makes it quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.[|[11]][|[16]]
 * 1) If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent ([|side-angle-side]).
 * 2) The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
 * 3) The area of a rectangle is equal to the product of two adjacent sides.
 * 4) The area of a square is equal to the product of two of its sides (follows from 3).
 * 1) Let ACB be a right-angled triangle with right angle CAB.
 * 2) On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.[|[13]]
 * 3) From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
 * 4) Join CF and AD, to form the triangles BCF and BDA.
 * 5) Angles CAB and BAG are both right angles; therefore C, A, and G are [|collinear]. Similarly for B, A, and H.
 * 1) Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
 * 2) Since AB and BD are equal to FB and BC, respectively, triangle ABD must be congruent to triangle FBC.
 * 3) Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD, since it shares a height with BK and a base with BD and a triangle's area is half the product of its base and height.
 * 4) Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
 * 5) Therefore rectangle BDLK must have the same area as square BAGF = AB2.
 * 6) Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
 * 7) Adding these two results, AB2 + AC2 = BD × BK + KL × KC
 * 8) Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
 * 9) Therefore AB2 + AC2 = BC2, since CBDE is a square.

Proof by rearrangement
In the animation at the left, the total area and the areas of the triangles are all constant. Therefore, the total black area is constant. But the original black area of side //c// can be divided into two squares with sides //a//, //b//, demonstrating that //a//2 + //b//2 = //c//2. A second proof is given by the middle animation. An initial large square is formed of area //c//2 by adjoining four identical right triangles, leaving a small square in the center of the big square to accommodate the difference in lengths of the sides of the triangles. Two rectangles are formed of sides //a// and //b// by moving the triangles. By incorporating the center small square with one of these rectangles, the two rectangles are made into two squares of areas //a//2 and //b//2, showing that //c//2 = //a//2 + //b//2. The third, rightmost image also gives a proof. The upper two squares are divided as shown by the blue and green shading, into pieces that when rearranged can be made to fit in the lower square on the hypotenuse – or conversely the large square can be divided as shown into pieces that fill the other two. This shows the area of the large square equals that of the two smaller ones.[|[17]]

Proof by rearrangement of four identical right triangles || Animation showing another proof by rearrangement[|[18]] || Proof using an elaborate rearrangement ||
 * [[image:http://upload.wikimedia.org/wikipedia/commons/6/65/Pythag_anim.gif width="200" height="200" link="http://en.wikipedia.org/wiki/File:Pythag_anim.gif"]][[image:http://bits.wikimedia.org/skins-1.5/common/images/magnify-clip.png width="15" height="11" link="http://en.wikipedia.org/wiki/File:Pythag_anim.gif"]]

[[|edit]] Algebraic proofs
Diagram of the two algebraic proofs. The theorem can be proved algebraically using four copies of a right triangle with sides //a//, //b// and //c//, arranged inside a square with side //c// as in the top half of the diagram.[|[19]] The triangles are similar with area, while the small square has side //b// − //a// and area (//b// − //a//)2. The area of the large square is therefore But this is a square with side //c// and area //c//2, so A similar proof uses four copies of the same triangle arranged symmetrically around a square with side //c//, as shown in the lower part of the diagram.[|[20]] This results in a larger square, with side //a// + //b// and area (//a// + //b//)2. The four triangles and the square side //c// must have the same area as the larger square, giving Diagram of Garfield's proof. A related proof was published by [|James A. Garfield].[|[21]][|[22]] Instead of a square it uses a [|trapezoid], which can be constructed from the square in the second of the above proofs by bisecting along a diagonal of the inner square, to give the trapezoid as shown in the diagram. The [|area of the trapezoid] can be calculated to be half the area of the square, that is The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of, which is removed by multiplying by two to give the result.

Proof using differentials
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse and employing a little [|calculus].[|[23]][|[24]] This proof is a //metric// proof in the sense of Dantzig, as it uses lengths, not areas. In the figure, triangle PBC is the original right triangle and triangle ABC is the modification of PBC when side PB is extended by increasing //a// to //a// + Δ//a//. The circular arcs have radii //c// and //c// + Δ//c// where Δ//c// is the change in hypotenuse //c// that occurs as a result of the change Δ//a// in side //a//. Constructions to determine upper and lower bounds upon [|[23]] The figure shows two constructions, right triangles ADP and AQP, in the upper and lower panels which will be used to find respectively [|upper and lower bounds] of the [|ratio] Δ//c///Δ//a//. Then the limit will be taken as Δ//a//, Δ//c// → 0, and the resulting expression for the [|derivative] //dc// ///da// will be used to establish Pythagoras' theorem. From triangle ABC (upper panel), Construct right triangle ADP (upper panel). Then, The last inequality results from //AD// > Δ//c//, as shown in the upper panel of the figure. Combining the above expressions for cos //θ//, Next construct right triangle AQP (lower panel). Since both triangles AQP and PBC have an angle , The last inequality results from //PQ// < Δ//c//, as shown in the lower panel of the figure. Combining the two inequalities that were obtained using triangles ADP and AQP, We now have upper and lower bounds for the ratio Δ//c// /Δ//a//. As Δ//a//, Δ//c// → 0, the ratio Δ//c// /Δ//a// becomes the derivative //dc// ///da// and the upper bound becomes the same as the lower bound //a// ///c//. Consequently, or: which has the [|integral]: When //a// = 0 then //c// = //b//, so the "constant" is //b//2. Hence, Pythagoras' theorem is established: Using this expression, the [|total differential] is: This result shows that the increase in the square of the hypotenuse is the sum of the independent contributions from the squares of the sides. I got this from wikipedia.