勾股定理

本條目仍然有文字尚未被翻譯成中文，條目是根據其他語言的維基百科內容進行翻譯。 歡迎您積極參與翻譯與修訂，目前翻譯的進度為5%。

直角邊的平方和等於斜邊的平方

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

勾股定理，西方称畢達哥拉斯定理或畢氏定理是一個基本的幾何定理，相传由古希腊的畢達哥拉斯首先證明。據說畢達哥拉斯證明了這個定理後，即斬了百頭牛作慶祝，因此又稱「百牛定理」。在中國，《周髀算經》記載了勾股定理的一個特例，相传是在商代由商高發現，故又有稱之為商高定理；三国时代的赵爽对《周髀算經》内的勾股定理作出了详细注释，作為一個證明。法国和比利时称为驴桥定理，埃及称为埃及三角形。

[编辑] 定理

一種證明方法的圖示：左右兩正方形面積相等，各扣除四塊藍色三角形後面積仍相等

勾股定理指出：

直角三角形兩直角邊（即“勾”，“股”）邊長平方和等於斜邊（即“弦”）邊長的平方。

也就是說，

設直角三角形兩直角邊為a和b，斜邊為c，那麼

a² + b² = c²

勾股定理現發現約有400種證明方法，是數學定理中證明方法最多的定理之一。

在公元前500-200年，周髀算經的圖解

[编辑] 勾股數组

勾股数组是滿足勾股定理a² + b² = c²的正整數組(a,b,c)，其中的a,b,c称为勾股数。例如(3,4,5)就是一組勾股數組。

任意一组勾股数(a,b,c)可以表示为如下形式：a = k(m² − n²),b = 2kmn,c = k(m² + n²)，其中。

[编辑] 歷史

這個定理的歷史可以被分成三個部份 knowledge of Pythagorean triples, knowledge of the relationship between the sides of a right triangle, and proofs of the theorem.

Megalithic monuments from circa 2500 BC in Egypt, and in the British Isles, incorporate right triangles with integer sides.^[1] Bartel Leendert van der Waerden conjectures that these Pythagorean triples were discovered algebraically.^[2]

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, the dates of which are given variously as between the 8th century BCE and the 2nd century BCE, 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; he further theorizes that Pythagoras visited Arakonam, India, and copied it.

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.

大約公元前400年Around 400 BC, 根據柏拉圖的方法according to Proklos, Plato gave a method for finding Pythagorean triples that 結合了代數與幾何combined algebra and geometry. Circa 300 BC, in Euclid's Elements, 是距今最早的公理化證明the oldest extant axiomatic proof of the theorem is presented.

Written sometime between 500 BC and 200 AD, the 中國的古書Chinese text Chou Pei Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) 已出現具體的畢氏定理的證明方式gives a visual proof of the Pythagorean theorem — 在中國則稱之為in China it is called the "Gougu Theorem" (勾股定理) — for the (3, 4, 5) triangle. During the 在漢朝Han Dynasty, 約前200年至公元200年之間，算數九章記載了這個定理from 200 BC to 200 AD, Pythagorean triples appear in The Nine Chapters on the Mathematical Art, together with a mention of right triangles.^[3]

有許多辯論是否畢氏定理早已不只一次被發現There is much debate on whether the Pythagorean theorem was discovered once or many times. B.L. van der Waerden asserts a single discovery, 有人說是西元前2000年從英國發現，然後傳播到達米亞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. 然而許多學者並不同意這 種說法Most scholars disagree however, and favor independent discovery.

最近，Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja在吠陀數學一書中聲稱古代印度教吠陀證明了畢達哥拉斯定理。More recently, Jagadguru Swami Sri Bharati Krishna Tirthaji Maharaja in his book Vedic Mathematics claimed ancient Indian Hindu Vedic proofs for the Pythagoras Theorem.

[编辑] Proofs

這個定理有許多證明的方法This theorem may have more known proofs than any other (the law of quadratic reciprocity being also a contender for that distinction); 路明思（Elisha Scott Loomis）的 Pythagorean Proposition, 一書中總共提到 367 證明方式。

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.)

[编辑] Proof using similar triangles

Proof using similar triangles

有許多畢氏定理的證明方式Like many of the proofs of the Pythagorean theorem, 都是基於三角形的兩邊長比例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:

因為As

所以so

可以寫成These can be written as

綜合這兩個方程式，我們得到Summing these two equalities, we obtain

換句話說In other words, the Pythagorean theorem:

[编辑] Euclid's proof

Proof in Euclid's Elements

In Euclid's Elements, the Pythagorean theorem is proved by an argument along the following lines. 設Let A, B, C 為一直角三角形be the vertices of a right triangle, 直角為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.

在正式的證明中，我們需要四個輔助定理如下：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 congruent.
2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
3. 任意一個正方形的面積等於其二邊長的乘積The area of any square is equal to the product of two of its sides.
4. 任意一個四方形的面積等於其二邊長的乘積（據輔助定理3）The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).

The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.

其證明如下：

1. 設Let ABC 為一直角三角形，其直角為CBAbe a right-angled triangle with right angle CAB.
2. 其每一邊為On each of the sides BC, AB, and CA, 依序繪成四方形squares are drawn, CBDE, BAGF, and ACIH, in that order.
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角 CAB 和角 BAG 都是直角are both right angles; 因此 C, A, and G 都是線性對應的are colinear. Similarly同理可證 for B, A, 和 H.
6. 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.
7. Since AB 和 BD 各自相等於 FB 和 BC, respectively, 三角形 ABD 必須相等於三角形 FBC.
8. 因為 A is colinear with K 和 L, 四方形 BDLK 必須二倍面積於三角形 ABD.
9. 因為 C 與 A 和 G 有共同線性, 正方形 BAGF 必須二倍面積於三角形 FBC.
10. 因此四邊形 BDLK 必須有相同的面積 BAGF = AB².
11. 同理可證, 四邊形 CKLE 必須有相同的面積 ACIH = AC².
12. 加入這兩個結果值會到, AB² + AC² = BD BK + KL KC
13. 由於 BD = KL, BD BK + KL KC = BD(BK + KC) = BD BC
14. 因此Therefore AB² + AC² = BC², 由於since CBDE is a square.

This proof is mentioned in Euclid's Elements as proposition 1.47.^[4]

[编辑] Similarity proof

From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.

[编辑] Proof by rearrangement

以面積減算法證明Proof using area subtraction

A proof by rearrangement 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 (see Lebesgue measure). For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).

以重新排列法證明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. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.^[5]

[编辑] Algebraic proof

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

An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by

The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C². Thus the area of everything together is given by:

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.

(Distribution of the 4)

(Subtraction of 2AB)

[编辑] 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.^[6]

Proof using differential equations

As a result of a change in side a,

by similar triangles and for differential changes. So

upon separation of variables. A more general result is

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

Integrating gives

So

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 fixed and then observe that

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.

[编辑] Rational trigonometry

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

[编辑] Converse

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.

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 a² + b² = c². 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.

A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Where c is chosen to be the longest of the three sides:

• If , 則是直角三角形then the triangle is right.
• If , 則是銳角三角形then the triangle is acute.
• If , 則是頓角三角形then the triangle is obtuse.

[编辑] Consequences and uses of the theorem

[编辑] Pythagorean triples

主条目：Pythagorean triple

A Pythagorean triple consists of three positive integers a, b, and c, such that a² + b² = c². 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), some well-known examples are (3, 4, 5) and (5, 12, 13).

[编辑] 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.^[7]

[编辑] Distance in Cartesian coordinates

The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x₀, y₀) and (x₁, y₁) are points in the plane, then the distance between them, also called the Euclidean distance, is given by

More generally, in Euclidean n-space, the Euclidean distance between two points and , is defined, using the Pythagorean theorem, as:

[编辑] 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:

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:

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 v₁, v₂,…, v_n be vectors in an inner product space such that <v_i, v_j> = 0 for 1 ≤ i < j ≤ n. Then

The generalization 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, a² + b² > c² holds.

In a triangle with an obtuse angle, α + β < γ holds. Therefore, a² + b² < c² holds.

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

sgn(α + β − γ) = sgn(a² + b² − c²)

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.^[8]

[编辑] 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 .

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

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

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.

In hyperbolic geometry, for a right triangle one can also write,

where is the angle of parallelism of the line segment AB that μ(AB) = a where μ is the multiplicative distance function (see Hilbert's Arithmetic of Ends).

In hyperbolic trigonometry, the sine of the angle of parallelism satisfies

Thus, the equation takes the form

where a, b and c are multiplicative distance of the sides of the right triangle (Hartshorne, 2000).

[编辑] Cultural references to the Pythagorean theorem

• In The Wizard of Oz, when the Scarecrow receives his diploma from the Wizard, he immediately exhibits his "knowledge" by reciting a mangled and incorrect version of the theorem: "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh, joy, oh, rapture. I've got a brain!" The "knowledge" exhibited by the Scarecrow is incorrect. The accurate statement would have been "The sum of the squares of the legs of a right triangle is equal to the square of the remaining side."^[9]
• In an episode of The Simpsons, homage is paid to the Oz Scarecrow's quote, thus turning the theorem into a cultural reference to a cultural reference. After finding a pair of Henry Kissinger's glasses in a toilet at the Springfield Nuclear Power Plant, Homer puts them on and quotes the scarecrow's mangled formula. A man in a nearby toilet stall then yells out "That's a right triangle, you idiot!"
• In the English version of the seventeenth Asterix book "The Mansions of the Gods", Julius Caesar uses the services of an architect named "Squaronthehypotenus" to develop the estate near the village, through which he hopes to absorb the Gaulish village into the Roman culture.
• In 2000, Uganda released a coin with the shape of a right triangle. The tail has an image of Pythagoras and the Pythagorean theorem, accompanied with the mention "Pythagoras Millennium".^[10]
• In the Major-General's Song, "About binomial theorem I'm teeming with a lot o' news, With many cheerful facts about the square of the hypotenuse."

[编辑] 參見

• 勾股數
• 餘弦定理

[编辑] 相關網頁

• Pythagorean Theorem (MathWorld)

“勾股定理”是一个關於數學、还没写完的小作品，希望您能继续编辑或者是修订这个条目。