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Pythagorean theorem

2007 Schools Wikipedia Selection. Related subjects: Mathematics

   In mathematics, the Pythagorean theorem or Pythagoras' theorem is a
   relation in Euclidean geometry among the three sides of a right
   triangle. The theorem is named after the Greek mathematician
   Pythagoras, who by tradition is credited with its discovery, although
   knowledge of the theorem almost certainly predates him. Pythagoras was
   noted for his obsession with ratios. He believed that ratios could be
   applied to all aspects of the world.
   The Pythagorean theorem: The sum of the areas of the two squares on the
   legs (a and b) equals the area of the square on the hypotenuse (c).
   Enlarge
   The Pythagorean theorem: The sum of the areas of the two squares on the
   legs (a and b) equals the area of the square on the hypotenuse (c).

   The theorem is as follows:

                In any right triangle, the area of the square whose side
                is the hypotenuse (the side of a right triangle opposite
                the right angle) is equal to the sum of areas of the
                squares whose sides are the two legs (i.e. the two sides
                other than the hypotenuse).

   If we let c be the length of the hypotenuse and a and b be the lengths
   of the other two sides, the theorem can be expressed as the equation

          a^2 + b^2 = c^2. \,

   This 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.

   This theorem may have more known proofs than any other. The Pythagorean
   Proposition, a book published in 1940, contains 370 proofs of
   Pythagoras' theorem, including one by American President James
   Garfield.

   The converse of the theorem is also true:-

                For any three positive numbers a, b, and c such that a² +
                b² = c², there exists a triangle with sides a, b and c,
                and every such triangle has a right angle between the
                sides of lengths a and b.

   Visual proof for the (3, 4, 5) triangle as in the Chou Pei Suan Ching
   500–200 BC
   Enlarge
   Visual proof for the (3, 4, 5) triangle as in the Chou Pei Suan Ching
   500–200 BC

History

   The history of the theorem can be divided into three parts: knowledge
   of Pythagorean triples, knowledge of the relationship between the sides
   of a right triangle, and proofs of the theorem.

   Megalithic monuments from 4000 BC in Egypt, and in the British Isles
   from circa 2500 BC, incorporate right triangles with integer sides but
   the builders did not necessarily understand the theorem . Bartel
   Leendert van der Waerden conjectures that these Pythagorean triples
   were discovered algebraically .

   Written between 2000– 1786 BC, the Middle Kingdom Egyptian papyrus
   Berlin 6619 includes a problem, whose solution is a Pythagorean triple.

   During the reign of Hammurabi, the Mesopotamian tablet Plimpton 322,
   written between 1790 and 1750 BC, contains many entries closely related
   to Pythagorean triples.

   The Baudhayana Sulba Sutra, written in the 8th century BC in India,
   contains a list of Pythagorean triples discovered algebraically, a
   statement of the Pythagorean theorem, and a geometrical proof of the
   Pythagorean theorem for an isosceles right triangle.

   The Apastamba Sulba Sutra (circa 600 BC) contains a numerical proof of
   the general Pythagorean theorem, using an area computation. Van der
   Waerden believes that "it was certainly based on earlier traditions".
   According to Albert Bŭrk, this is the original proof of the theorem,
   and Pythagoras copied it on his visit to India. Many scholars find van
   der Waerden and Bŭrk's claims unsubstantiated.

   Pythagoras, whose dates are commonly given as 569– 475 BC, used
   algebraic methods to construct Pythagorean triples, according to
   Proklos's commentary on Euclid. Proklos, however, wrote between 410 and
   485 AD. According to Sir Thomas L. Heath, there is no attribution of
   the theorem to Pythagoras for five centuries after Pythagoras lived.
   However, when authors such as Plutarch and Cicero attributed the
   theorem to Pythagoras, they did so in a way which suggests that the
   attribution was widely known and undoubted.

   Around 400 BC, according to Proklos, Plato gave a method for finding
   Pythagorean triples that combined algebra and geometry. Circa 300 BC,
   in Euclid's Elements, the oldest extant axiomatic proof of the theorem
   is presented.

   Written sometime between 500 BC and 200 AD, the Chinese text Chou Pei
   Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the
   Circular Paths of Heaven) gives a visual proof of the Pythagorean
   theorem — in China it is called the "Gougu Theorem" (勾股定理) — for the
   (3, 4, 5) triangle. During the Han Dynasty, from 200 BC to 200 AD,
   Pythagorean triples appear in The Nine Chapters on the Mathematical
   Art, together with a mention of right triangles.

   There is much debate on whether the Pythagorean theorem was discovered
   once or many times. B.L. van der Waerden asserts a single discovery, by
   someone in Neolithic Britain, knowledge of which then spread to
   Mesopotamia circa 2000 BC, and from there to India, China, and Greece
   by 600 BC. Most scholars disagree however, and favour independent
   discovery.

   More recently, Shri Bharati Krishna Tirthaji in his book Vedic
   Mathematics claimed ancient Indian Hindu Vedic proofs for the
   Pythagoras Theorem.

Proofs

   This theorem may have more known proofs than any other (the law of
   quadratic reciprocity being also a contender for that distinction); the
   book Pythagorean Proposition, by Elisha Scott Loomis, contains over 350
   proofs. James Garfield, who later became President of the United
   States, devised an original proof of the Pythagorean theorem in 1876.
   The external links below provide a sampling of the many proofs of the
   Pythagorean theorem.

   Some arguments based on trigonometric identities (such as Taylor series
   for sine and cosine) have been proposed as proofs for the theorem.
   However, since all the fundamental trigonometric identities are proved
   using the Pythagorean theorem, there cannot be any trigonometric proof.
   (See also begging the question.)

Geometrical proof

   Proof using similar triangles
   Enlarge
   Proof using similar triangles

   Like many of the proofs of the Pythagorean theorem, this one is based
   on the proportionality of the sides of two similar 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. The new triangle ACH is similar to
   our 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. By a similar
   reasoning, the triangle CBH is also similar to ABC. The similarities
   lead to the two ratios:

          \frac{AC}{AB}=\frac{AH}{AC}\ \mbox{and}\
          \frac{CB}{AB}=\frac{HB}{CB}.\,\!

   These can be written as:

          AC^2=AB\times AH\,\! and CB^2=AB\times HB.\,\!

   Summing these two equalities, we obtain:

          AC^2+CB^2=AB\times AH+AB\times HB=AB\times(AH+HB)=AB^2.\,\!

   In other words, the Pythagorean theorem:

          AC^2+BC^2=AB^2.\,\!

Euclid's proof

   Proof in Euclid's Elements
   Enlarge
   Proof in Euclid's Elements

   In Euclid's Elements, the Pythagorean theorem is proved by an argument
   along the following lines. Let P, Q, R be the vertices of a right
   triangle, with a right angle at Q. Drop a perpendicular from Q 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.

   We see here three similar figures, each being "a square with a triangle
   on top". Since the large triangle is made of the two smaller trangles,
   its area is the sum of areas of the two smaller ones. By similarity, it
   is clear that the same must apply also to the areas of the squares.

   For the formal proof, we require four elementary lemmata:

   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 equal in all respects.

   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 any square is equal to the product of two of its' sides.

   4. The area of any rectangle is equal to the product of two adjacent
   sides (follows from Lemma 3).

   The proof is as follows:

   Let QPR\,\! be a right-angled triangle; with the right angle at Q\,\!

   On each of the sides

   AC,\,\!

   CB\,\! and

   BA\,\!

   squares

   AFGC,\,\!

   CHKB,\,\! and

   BEDA\,\!

   are drawn,in that order.

   Q\,\! and D\,\! are joined.

   R\,\! and F\,\! are joined.

   Let the perpendicular from Q\,\! on DE\,\! be QZ\,\! .

   Let QZ\,\! intersect PR\,\! at Y\,\! .

   Then QZ\,\! is parallel to AD\,\! , since both of the angles QZE\,\!
   and ADL\,\! are right angles.

   Now, because both of the angles RQP\,\! and RQG\,\! are right angles,

   the points R, Q, G\,\! are collinear.

   Similarly, \angle RPD = \angle FRQ = \mbox{one right angle.}

   Adding \angle QPR\, to each side of the equality

          \angle RPD = \angle FRQ,\,

   we get

          \angle QPD = \angle FPR.\,

   Then the triangles

          QPD\,\! and FPR\,\!

   are congruent;

   because QP = FP,\,

   PD = PR,\,

   and

          \angle QPD = \angle FPR\,

   (from Lemma 1).

   Now the area of the rectangle

   PDZY\,\!

   is double the triangle

   QPR\,\! ,

   being on the same base

   PD\,\!

   and between the same parallels

   PD\,\!

   and QZ\,\! .

   Again, the area of the square

   PFGQ\,\!

   is double the triangle

   FPR\,\! ,

   as they both stand on the same base

   FP\,\!

   and between the same parallels

   FP\,\!

   and GR\,\! .

   Thus, the area of the rectangle

   PDZY\,\!

   is equal to the square

   PFQG\,\! ,

   ie,

   PD\cdot DZ = QP^2\,\!

   Similarly, by joining QE\,\! and PK\,\! ,

   it can easily be shown that the rectangle YREZ\,\!

   is equal in area to the square QHKR\,\! ,

   which leads to

   RE \cdot EZ = QR^2\,

   Adding these two results, we see that

   PD \cdot DZ + RE\cdot EZ = QR^2 + QP^2\,\!

   or, PD\cdot DZ + ZE\cdot PD = QR^2 + QP^2\,\! (since PDER\,\! is a
   square)

   or, PD(DZ + ZE) = QR^2 + QP^2\,\!

   or, PD\cdot DE = QR^2\,\! + QP^2\,\!

   or, PD^2 + QR^2 + QP^2\,\!

   or, RP^2 = QR^2 + QP^2\,\! (since PDER\,\! is a square)

   And thus,

   RP^2 = QR^2 + QP^2\,\!

   which proves the statement that

   in any right-angled triangle, the square desribed on the hypotenuse is
   equal to the sum of the squares described on the other two sides.

   With the help of trigonometry, this theorem can be extended for all
   triangles.

   For any triangle QPR\,\! ,

          q^2 = p^2 + r^2 - 2pr\cos Q,\,

   where p, q\,\! and r\,\! denote the sides of the triangle opposite the
   angles P,Q\,\! and R\,\! ; and cyclic permutations.

Visual proof

   Proof using area subtraction
   Enlarge
   Proof using area subtraction

   A visual proof is given by this illustration. The area of each large
   square is (a + b)². In both, the area of four identical triangles is
   removed. The remaining areas, a² + b² and c², are equal. Q.E.D.

   This proof is indeed very simple, but it is not elementary, in the
   sense that it does not depend solely upon the most basic axioms and
   theorems of Euclidean geometry. In particular, while it is quite easy
   to give a formula for area of triangles and squares, it is not as easy
   to prove that the area of a square is the sum of areas of its pieces.
   In fact, proving the necessary properties is harder than proving the
   Pythagorean theorem itself. For this reason, axiomatic introductions to
   geometry usually employ another proof based on the similarity of
   triangles (see above).
   Proof using rearrangement
   Enlarge
   Proof using rearrangement

   A second graphic illustration of the Pythagorean theorem fits parts of
   the sides' squares into the hypotenuse's square. A related proof would
   show that the repositioned parts are identical with the originals and,
   since the sum of equals are equal, that the corresponding areas are
   equal. To show that a square is the result one must show that the
   length of the new sides equals c.

Algebraic Proof

   A square created by aligning four right angle triangles and a large
   square.
   Enlarge
   A square created by aligning four right angle triangles and a large
   square.

   A more algebraic variant of this proof is provided by the following
   reasoning. Looking at the illustration, the area of each of the four
   red, yellow, green and pink right-angled triangles is given by:

          \frac{1}{2} ab

   The blue square in the middle has side length c, so its area is c^2.
   Thus the area of everything together is given by:

          4\left(\frac{1}{2}ab\right)+c^2.

   However, as the large square has sides of length a+b, we can also
   calculate its area as (a+b)², which expands to a²+2ab+b². This can be
   shown by considering the angles.

   Thus, a^2 + 2ab + b^2= 4\left(\frac{1}{2}ab\right)+c^2.

   (Distribution of the 4) a^2 + 2ab + b^2=2ab + c^2

   (Subtraction of 2ab) a^2 + b^2=c^2

Proof by differential equations

   One can arrive at the Pythagorean theorem by studying how changes in a
   side produce a change in the hypotenuse in the following diagram and
   employing a little calculus.
   Proof using differential equations
   Enlarge
   Proof using differential equations

   As a result of a change in side a,

          \frac {da}{dc} = \frac {c}{a}

   by similar triangles and for differential changes. So

          c\, dc = a\, da

   upon separation of variables. A more general result is

          c\ dc = a\, da + b\, db

   which results from adding a second term for changes in side b.

   Integrating gives

          c^2 = a^2 +b^2 + \mathrm{constant}.\ \,\!

          a = b = c = 0 \Rightarrow \mathrm{constant} = 0\,\!

   So

          c^2 = a^2 +b^2.\

   As can be seen, the squares are due to the particular proportion
   between the changes and the sides while the sum is a result of the
   independent contributions of the changes in the sides which is not
   evident from the geometric proofs. From the proportion given it can be
   shown that the changes in the sides are inversely proportional to the
   sides. The differential equation suggests that the theorem is due to
   relative changes and its derivation is nearly equivalent to computing a
   line integral. A simpler derivation would leave b\ fixed and then
   observe that

          a=0 \Rightarrow c^2 = b^2 = \mathrm{constant}.\,\!

   It is doubtful that the Pythagoreans would have been able to do the
   above proof but they knew how to compute the area of a triangle and
   were familiar with figurate numbers and the gnomon, a segment added
   onto a geometrical figure. All of these ideas predate calculus and are
   an alternative for the integral.

   The proportional relation between the changes and their sides is at
   best an approximation, so how can one justify its use? The answer is
   the approximation gets better for smaller changes since the arc of the
   circle which cuts off c more closely approaches the tangent to the
   circle. As for the sides and triangles, no matter how many segments
   they are divided into the sum of these segments is always the same. The
   Pythagoreans were trying to understand change and motion and this led
   them to realize that the number line was infinitely divisible. Could
   they have discovered the approximation for the changes in the sides?
   One only has to observe that the motion of the shadow of a sundial
   produces the hypotenuses of the triangles to derive the figure shown.

Rational trigonometry

   For a proof by the methods of rational trigonometry, see Pythagoras'
   theorem proof (rational trigonometry).

Proof of the converse

          For any three positive numbers a, b, and c such that
          a^2 + b^2 = c^2, there exists a triangle with sides a, b and c,
          and every such triangle has a right angle between the sides of
          lengths a and b.

   This converse also appears in Euclid's Elements. It can be proven using
   the law of cosines (see below under Generalizations), or by the
   following proof:

   Let ABC be a triangle with side lengths a, b, and c, with
   a^2 + b^2 = c^2. We need to prove that the angle between the a and b
   sides is a right angle. We construct another triangle with a right
   angle between sides of lengths a and b. By the Pythagorean theorem, it
   follows that the hypotenuse of this triangle also has length c. Since
   both triangles have the same side lengths a, b and c, they are
   congruent, and so they must have the same angles. Therefore, the angle
   between the side of lengths a and b in our original triangle is a right
   angle.

Consequences and uses of the theorem

Pythagorean triples

   A Pythagorean triple consists of three positive integers a, b, and c,
   such that a^2 + b^2 = c^2. In other words, a Pythagorean triple
   represents the lengths of the sides of a right triangle where all three
   sides have integer lengths. Evidence from megalithic monuments on the
   British Isles shows that such triples were known before the discovery
   of writing. Such a triple is commonly written (a, b, c), and a
   well-known example is (3, 4, 5).

   A Pythagorean triple is primitive if a, b, and c have no common divisor
   other than 1. There are infinitely many primitive triples, and all
   Pythagorean triples can be explicitly generated using the following
   formula: choose two integers m and n with m > n, and let a = m^2 − n^2,
   b = 2mn, c = m^2 + n^2. Then we have a^2 + b^2 = c^2. Also, any
   multiple of a Pythagorean triple is again a Pythagorean triple.

   Pythagorean triples allow the construction of right angles. The fact
   that the lengths of the sides are integers means that, for example,
   tying knots at equal intervals along a string allows the string to be
   stretched into a triangle with sides of length three, four, and five,
   in which case the largest angle will be a right angle. This methods was
   used to step masts at sea and by the Egyptians in construction work.

   A generalization of the concept of Pythagorean triples is a triple of
   positive integers a, b, and c, such that a^n + b^n = c^n, for some n
   strictly greater than 2. Pierre de Fermat in 1637 claimed that no such
   triple exists, a claim that came to be known as Fermat's last theorem.
   The first proof was given by Andrew Wiles in 1994.

The existence of irrational numbers

   One of the consequences of the Pythagorean theorem is that irrational
   numbers, such as the square root of two, can be constructed. A right
   triangle with legs both equal to one unit has hypotenuse length square
   root of two. The Pythagoreans proved that the square root of two is
   irrational, and this proof has come down to us even though it flew in
   the face of their cherished belief that everything was rational.
   According to the legend, Hippasus, who first proved the irrationality
   of the square root of two, was drowned at sea as a consequence.

Distance in Cartesian coordinates

   The distance formula in Cartesian coordinates is derived from the
   Pythagorean theorem. If (x[0], y[0]) and (x[1], y[1]) are points in the
   plane, then the distance between them, also called the Euclidean
   distance, is given by

                \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2}.

   More generally, in Euclidean n-space, the Euclidean distance between
   two points A=(a_1,a_2,\dots,a_n) and B=(b_1,b_2,\dots,b_n) , is
   defined, using the Pythagorean theorem, as:

          \sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} =
          \sqrt{\sum_{i=1}^n (a_i-b_i)^2}

Generalizations

   The Pythagorean theorem was generalised by Euclid in his Elements:

          If one erects similar figures (see Euclidean geometry) on the
          sides of a right triangle, then the sum of the areas of the two
          smaller ones equals the area of the larger one.

   The Pythagorean theorem is a special case of the more general theorem
   relating the lengths of sides in any triangle, the law of cosines:

                a^2+b^2-2ab\cos{\theta}=c^2, \,

          where θ is the angle between sides a and b.
          When θ is 90 degrees, then cos(θ) = 0, so the formula reduces to
          the usual Pythagorean theorem.

   Given two vectors v and w in a complex inner product space, the
   Pythagorean theorem takes the following form:

                \|\mathbf{v}+\mathbf{w}\|^2 = \|\mathbf{v}\|^2 +
                \|\mathbf{w}\|^2 +
                2\,\mbox{Re}\,\langle\mathbf{v},\mathbf{w}\rangle

   In particular, ||v + w||^2 = ||v||^2 + ||w||^2 if and only if v and w
   are orthogonal.

   Using mathematical induction, the previous result can be extended to
   any finite number of pairwise orthogonal vectors. Let v[1], v[2],...,
   v[n] be vectors in an inner product space such that <v[i], v[j]> = 0
   for 1 ≤ i < j ≤ n. Then

                \left\|\,\sum_{k=1}^{n}\mathbf{v}_k\,\right\|^2 =
                \sum_{k=1}^{n} \|\mathbf{v}_k\|^2

   The generalisation of this result to infinite-dimensional real inner
   product spaces is known as Parseval's identity.

   When the theorem above about vectors is rewritten in terms of solid
   geometry, it becomes the following theorem. If lines AB and BC form a
   right angle at B, and lines BC and CD form a right angle at C, and if
   CD is perpendicular to the plane containing lines AB and BC, then the
   sum of the squares of the lengths of AB, BC, and CD is equal to the
   square of AD. The proof is trivial.

   Another generalization of the Pythagorean theorem to three dimensions
   is de Gua's theorem, named for Jean Paul de Gua de Malves: If a
   tetrahedron has a right angle corner (a corner like a cube), then the
   square of the area of the face opposite the right angle corner is the
   sum of the squares of the areas of the other three faces.

   There are also analogs of these theorems in dimensions four and higher.

   In a triangle with three acute angles, α + β > γ holds. Therefore, a^2
   + b^2 > c^2 holds.

   In a triangle with an obtuse angle, α + β < γ holds. Therefore, a^2 +
   b^2 < c^2 holds.

   Edsger Dijkstra has stated this proposition about acute, right, and
   obtuse triangles in this language:

                sgn(α + β − γ) = sgn(a^2 + b^2 − c^2)

   where α is the angle opposite to side a, β is the angle opposite to
   side b and γ is the angle opposite to side c.

The Pythagorean theorem in non-Euclidean geometry

   The Pythagorean theorem is derived from the axioms of Euclidean
   geometry, and in fact, the Euclidean form of the Pythagorean theorem
   given above does not hold in non-Euclidean geometry. (It has been shown
   in fact to be equivalent to Euclid's Parallel (Fifth) Postulate.) For
   example, in spherical geometry, all three sides of the right triangle
   bounding an octant of the unit sphere have length equal to π / 2; this
   violates the Euclidean Pythagorean theorem because
   (\pi/2)^2+(\pi/2)^2\neq (\pi/2)^2 .

   This means that in non-Euclidean geometry, the Pythagorean theorem must
   necessarily take a different form from the Euclidean theorem. There are
   two cases to consider -- spherical geometry and hyperbolic plane
   geometry; in each case, as in the Euclidean case, the result follows
   from the appropriate law of cosines:

   For any right triangle on a sphere of radius R, the Pythagorean theorem
   takes the form

                \cos \left(\frac{c}{R}\right)=\cos
                \left(\frac{a}{R}\right)\,\cos \left(\frac{b}{R}\right).

   By using the Maclaurin series for the cosine function, it can be shown
   that as the radius R approaches infinity, the spherical form of the
   Pythagorean theorem approaches the Euclidean form.

   For any triangle in the hyperbolic plane (with Gaussian curvature −1),
   the Pythagorean theorem takes the form

                \cosh c=\cosh a\,\cosh b

          where cosh is the hyperbolic cosine.

   By using the Maclaurin series for this function, it can be shown that
   as a hyperbolic triangle becomes very small (i.e., as a, b, and c all
   approach zero), the hyperbolic form of the Pythagorean theorem
   approaches the Euclidean form.
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