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{{short description|Equation relating the side lengths of a right triangle}}{{about|classical geometry|the baseball term|Pythagorean expectation}}{{pp-move-indef|small=yes}}{{pp-semi-indef}}(File:Pythagorean.svg|thumb|260px|Pythagorean theoremThe sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).){{General geometry |concepts}}In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that 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 on the other two sides. This theorem can be written as an equation relating the lengths of the sides a, b and c, often called the "Pythagorean equation":BOOK, Roots to research: a vertical development of mathematical problems, Judith D. Sally, Paul Sally, 63, Chapter 3: Pythagorean triples,weblink 0-8218-4403-2, 2007, American Mathematical Society Bookstore,
a^2 + b^2 = c^2 ,
where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides. The theorem, whose history is the subject of much debate, is named for the ancient Greek thinker Pythagoras.The theorem has been given numerous proofs{{spndash}}possibly the most for any mathematical theorem. They are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. The theorem can be generalized in various ways, including higher-dimensional 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 has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound.

## Rearrangement proof

(File:Pythagoras-proof-anim.svg|thumb|The rearrangement proof (click to view animation))The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. Therefore, the white space within each of the two large squares must have equal area. Equating the area of the white space yields the Pythagorean theorem, Q.E.D.Benson, Donald. The Moment of Proof : Mathematical Epiphanies, pp. 172â€“173 (Oxford University Press, 1999).Heath gives this proof in his commentary on Proposition I.47 in Euclid's Elements, and mentions the proposals of Bretschneider and Hankel that Pythagoras may have known this proof. Heath himself favors a different proposal for a Pythagorean proof, but acknowledges from the outset of his discussion "that the Greek literature which we possess belonging to the first five centuries after Pythagoras contains no statement specifying this or any other particular great geometric discovery to him."{{harvtxt|Euclid|1956}}, pp. 351â€“352 Recent scholarship has cast increasing doubt on any sort of role for Pythagoras as a creator of mathematics, although debate about this continues.ENCYCLOPEDIA, Huffman, Carl, Pythagoras, The Stanford Encyclopedia of Philosophy (Winter 2018 Edition), Zalta, Edward N., Edward N. Zalta,weblink , "It should now be clear that decisions about sources are crucial in addressing the question of whether Pythagoras was a mathematician and scientist. The view of Pythagorasâ€™ cosmos sketched in the first five paragraphs of this section, according to which he was neither a mathematician nor a scientist, remains the consensus."

## Other forms of the theorem

If c denotes the length of the hypotenuse and a and b denote the lengths of the other two sides, the Pythagorean theorem can be expressed as the Pythagorean equation:
a^2 + b^2 = c^2 .
If the lengths of both a and b are known, then c can be calculated as
c = sqrt{a^2 + b^2}.
If the length of the hypotenuse c and of one side (a or b) are known, then the length of the other side can be calculated as
a = sqrt{c^2 - b^2}
or
b = sqrt{c^2 - a^2}.
The Pythagorean equation relates the sides of a right triangle in a simple way, so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides, but less than their sum.A generalization of this theorem is the law of cosines, which allows the computation of the length of any side of any triangle, given the lengths of the other two sides and the angle between them. If the angle between the other sides is a right angle, the law of cosines reduces to the Pythagorean equation.

## Other proofs of the theorem

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.{{Harv|Loomis|1968}}

### Proof using similar triangles

(File:Pythagoras similar triangles simplified.svg|thumb|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. 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 Î¸ 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:
frac{BC}{AB}=frac{BH}{BC} text{ and } frac{AC}{AB}=frac{AH}{AC}.
The first result equates the cosines of the angles Î¸, whereas the second result equates their sines.These ratios can be written as
BC^2 = AB times BH text{ and } AC^2=AB times AH.
Summing these two equalities results in
BC^2+AC^2=ABtimes BH+ABtimes AH=ABtimes(AH+BH)=AB^2 ,
which, after simplification, expresses the Pythagorean theorem:
BC^2+AC^2=AB^2 .
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.{{Harv|Maor|2007| p= 39}}BOOK, God created the integers: the mathematical breakthroughs that changed history, Stephen W. Hawking, 12,weblink 0-7624-1922-9, 2005, Running Press Book Publishers, Philadelphia, This proof first appeared after a computer program was set to check Euclidean proofs.

### Euclid's proof

(File:Illustration to Euclid's proof of the Pythagorean theorem.svg|thumb|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 follow.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:
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).
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.See for example Pythagorean theorem by shear mapping, Saint Louis University website Java applet{{Clear}}(File:Illustration to Euclid's proof of the Pythagorean theorem2.svg|thumb|Illustration including the new lines)(File:Illustration to Euclid's proof of the Pythagorean theorem3.svg|thumb|Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF)The proof is as follows:
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.
BOOK, Mathematics: from the birth of numbers, Jan Gullberg, 435,weblink 0-393-04002-X, 1997, W. W. Norton & Company,
1. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
2. Join CF and AD, to form the triangles BCF and BDA.
3. Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
4. 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.
5. Since AB is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.
6. Since A-K-L is a straight line, parallel to BD, then rectangle BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2)
7. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
8. Therefore, rectangle BDLK must have the same area as square BAGF = AB2.
9. Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
10. Adding these two results, AB2 + AC2 = BD Ã— BK + KL Ã— KC
11. Since BD = KL, BD Ã— BK + KL Ã— KC = BD(BK + KC) = BD Ã— BC
12. Therefore, AB2 + AC2 = BC2, since CBDE is a square.
This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1,Elements 1.47 by Euclid. Retrieved 19 December 2006. demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.Euclid's Elements, Book I, Proposition 47: web page version using Java applets from Euclid's Elements by Prof. David E. Joyce, Clark University
This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.The proof by Pythagoras probably was not a general one, as the theory of proportions was developed only two centuries after Pythagoras; see {{Harv |Maor|2007 |p= 25}}

### Proofs by dissection and rearrangement

We have already discussed the Pythagorean proof, which was a proof by rearrangement. The same idea is conveyed by the leftmost animation below, which consists of a large square, side {{nowrap|a + b}}, containing four identical right triangles. The triangles are shown in two arrangements, the first of which leaves two squares a2 and b2 uncovered, the second of which leaves square c2 uncovered. The area encompassed by the outer square never changes, and the area of the four triangles is the same at the beginning and the end, so the black square areas must be equal, therefore {{nowrap|a2 + b2 {{=}} c2.}}A second proof by rearrangement is given by the middle animation. A large square is formed with area c2, from four identical right triangles with sides a, b and c, fitted around a small central square. Then two rectangles are formed with sides a and b by moving the triangles. Combining the smaller square with these rectangles produces two squares of areas a2 and b2, which must have the same area as the initial large square.WEB,weblink Pythagorean theorem, proof number 10, Alexander Bogomolny, Cut the Knot, 27 February 2010, 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 â€“ or conversely the large square can be divided as shown into pieces that fill the other two. This way of cutting one figure into pieces and rearranging them to get another figure is called dissection. This shows the area of the large square equals that of the two smaller ones.{{Harv|Loomis|1968|loc= Geometric proof 22 and Figure 123| page= 113}}{|
 thumb|Animation showing proof by rearrangement of four identical right triangles) thumb|Animation showing another proof by rearrangement) thumb|Proof using an elaborate rearrangement)

### Einstein's proof by dissection without rearrangement

(File:Altitude of a right triangle.svg|thumb|Right triangle on the hypotenuse dissected into two similar right triangles on the legs, according to Einstein's proof)Albert Einstein gave a proof by dissection in which the pieces need not get moved.BOOK, Schroeder, Manfred Robert, Fractals, Chaos, Power Laws: Minutes from an Infinite Paradise, Courier Corporation, 2012, 3â€“4, 0486134784, harv, Instead of using a square on the hypotenuse and two squares on the legs, one can use any other shape that includes the hypotenuse, and two similar shapes that each include one of two legs instead of the hypotenuse (see Similar figures on the three sides). In Einstein's proof, the shape that includes the hypotenuse is the right triangle itself. The dissection consists of dropping a perpendicular from the vertex of the right angle of the triangle to the hypotenuse, thus splitting the whole triangle into two parts. Those two parts have the same shape as the original right triangle, and have the legs of the original triangle as their hypotenuses, and the sum of their areas is that of the original triangle. Because the ratio of the area of a right triangle to the square of its hypotenuse is the same for similar triangles, the relationship between the areas of the three triangles holds for the squares of the sides of the large triangle as well.

### Algebraic proofs

(File:Pythagoras algebraic2.svg|thumb|upright|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.WEB,weblink Cut-the-knot.org: Pythagorean theorem and its many proofs, Proof #3, Alexander Bogomolny, Cut the Knot, 4 November 2010,
The triangles are similar with area tfrac12ab, while the small square has side {{nowrap|b âˆ’ a}} and area {{nowrap|(b âˆ’ a)2}}. The area of the large square is therefore

(b-a)^2+4frac{ab}{2} = (b-a)^2+2ab = b^2-2ab+a^2+2ab = a^2+b^2.
But this is a square with side c and area c2, so
c^2 = a^2 + b^2.
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.WEB,weblink Cut-the-knot.org: Pythagorean theorem and its many proofs, Proof #4, Alexander Bogomolny, Cut the Knot, 4 November 2010, This results in a larger square, with side {{nowrap|a + b}} and area {{nowrap|(a + b)2}}. The four triangles and the square side c must have the same area as the larger square,
(b+a)^2 = c^2 + 4frac{ab}{2} = c^2+2ab,
giving
c^2 = (b+a)^2 - 2ab = b^2+2ab+a^2-2ab = a^2 + b^2.
(File:Garfield Pythagoras.svg|thumb|upright|Diagram of Garfield's proof)A related proof was published by future U.S. President James A. Garfield (then a U.S. Representative) (see diagram).Published in a weekly mathematics column: JOURNAL, The New England Journal of Education, Pons Asinorum, 3, 14, 161, 1876, James A Garfield, harv,weblink as noted in BOOK, The mathematical universe: An alphabetical journey through the great proofs, problems, and personalities,weblink William Dunham, 1997, Wiley, 0-471-17661-3, 96, and in A calendar of mathematical dates: April 1, 1876 {{Webarchive|url=https://web.archive.org/web/20100714153516weblink |date=July 14, 2010 }} by V. Frederick RickeyWEB, Garfield's proof of the Pythagorean Theorem, David, Lantz,weblink Math.Colgate.edu, 2018-01-14,weblink" title="web.archive.org/web/20130828104818weblink">weblink 2013-08-28, yes, Maor, Eli, The Pythagorean Theorem, Princeton University Press, 2007: pp. 106-107. 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
frac{1}{2}(b+a)^2.
The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of frac{1}{2}, 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 calculus.JOURNAL, 10.2307/2691395, The Pythagorean proposition: A proof by means of calculus, Mike Staring, Mathematics Magazine, Mathematical Association of America, 69, 1, 45â€“46, 1996, 2691395, harv, WEB,weblink Pythagorean Theorem, 2010-05-09, Bogomolny, Alexander, Interactive Mathematics Miscellany and Puzzles, Alexander Bogomolny, yes,weblink" title="web.archive.org/web/20100706200930weblink">weblink 2010-07-06, JOURNAL, Bruce C. Berndt, Ramanujan â€“ 100 years old (fashioned) or 100 years new (fangled)?, The Mathematical Intelligencer, 10, 24, 1988, 10.1007/BF03026638, harv, 3, The triangle ABC is a right triangle, as shown in the upper part of the diagram, with BC the hypotenuse. At the same time the triangle lengths are measured as shown, with the hypotenuse of length y, the side AC of length x and the side AB of length a, as seen in the lower diagram part.(File:Pythag differential proof.svg|thumb|Diagram for differential proof)If x is increased by a small amount dx by extending the side AC slightly to D, then y also increases by dy. These form two sides of a triangle, CDE, which (with E chosen so CE is perpendicular to the hypotenuse) is a right triangle approximately similar to ABC. Therefore, the ratios of their sides must be the same, that is:
frac{dy}{dx}=frac xy.
This can be rewritten as y , dy=x , dx , which is a differential equation that can be solved by direct integration:
int y , dy=int x , dx,,
giving
y^2=x^2+C.
The constant can be deduced from x = 0, y = a to give the equation
y^2 = x^2 + a^2.
This is more of an intuitive proof than a formal one: it can be made more rigorous if proper limits are used in place of dx and dy.

## Converse

The converse of the theorem is also true:BOOK, Roots to Research, Judith D. Sally, Paul J. Sally Jr., 54â€“55, Theorem 2.4 (Converse of the Pythagorean theorem).,weblink American Mathematical Societydate=2007-12-21, For any three positive numbers a, b, and c such that {{nowrap|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.An alternative statement is:For any triangle with sides a, b, c, if {{nowrap|a2 + b2 {{=}} c2,}} then the angle between a and b measures 90Â°.This converse also appears in Euclid's Elements (Book I, Proposition 48):Euclid's Elements, Book I, Proposition 48 From D.E. Joyce's web page at Clark University
{{blockquote|text="If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right."}}
It can be proven using the law of cosines or as follows:Let ABC be a triangle with side lengths a, b, and c, with {{nowrap|a2 + b2 {{=}} c2.}} Construct a second triangle with sides of length a and b containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length c = {{radic|a2 + b2}}, the same as the hypotenuse of the first triangle. Since both triangles' sides are the same lengths a, b and c, the triangles are congruent and must have the same angles. Therefore, the angle between the side of lengths a and b in the original triangle is a right angle.The above proof of the converse makes use of the Pythagorean theorem itself. The converse can also be proven without assuming the Pythagorean theorem.Casey, Stephen, "The converse of the theorem of Pythagoras", Mathematical Gazette 92, July 2008, 309â€“313.Mitchell, Douglas W., "Feedback on 92.47", Mathematical Gazette 93, March 2009, 156.A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Let c be chosen to be the longest of the three sides and {{nowrap|a + b > c}} (otherwise there is no triangle according to the triangle inequality). The following statements apply:BOOK, Plane trigonometry and applications, Ernest Julius Wilczynski, Herbert Ellsworth Slaught, 85,weblink Theorem 1 and Theorem 2, 1914, Allyn and Bacon,
< c2,{edih} then the triangle is obtuse.Edsger W. Dijkstra has stated this proposition about acute, right, and obtuse triangles in this language:
{{nowrap|sgn(Î± + Î² âˆ’ Î³) {{=}} sgn(a2 + b2 âˆ’ c2),}}
where Î± is the angle opposite to side a, Î² is the angle opposite to side b, Î³ is the angle opposite to side c, and sgn is the sign function.WEB,weblink On the theorem of Pythagoras, Edsger W., Dijkstra, Edsger W. Dijkstra, September 7, 1986, EWD975, E. W. Dijkstra Archive,

## Consequences and uses of the theorem

### Pythagorean triples

A Pythagorean triple has three positive integers a, b, and c, such that {{nowrap|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. Such a triple is commonly written {{nowrap|(a, b, c).}} Some well-known examples are {{nowrap|(3, 4, 5)}} and {{nowrap|(5, 12, 13).}}A primitive Pythagorean triple is one in which a, b and c are coprime (the greatest common divisor of a, b and c is 1).The following is a list of primitive Pythagorean triples with values less than 100:
(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)

### Reciprocal Pythagorean theorem

Given a right triangle with sides a,b,c and altitude d (a line from the right angle and perpendicular to the hypotenuse c). The Pythagorean theorem has,
a^2+b^2 = c^2
while the reciprocal Pythagorean theoremR. B. Nelsen, Proof Without Words: A Reciprocal Pythagorean Theorem, Mathematics Magazine, 82, December 2009, p. 370 or the upside down Pythagorean theoremThe upside-down Pythagorean theorem, Jennifer Richinick, The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 313-316 relates the two legs a,b to the altitude d,Alexander Bogomolny, Pythagorean Theorem for the Reciprocalsweblink
frac1{a^2}+frac1{b^2} = frac1{d^2}
The equation can be transformed to,
frac1{(xz)^2}+frac1{(yz)^2} = frac1{(xy)^2}
where x^2+y^2=z^2 for any non-zero real x,y,z. If the a,b,d are to be integers, the smallest solution a>b>d is then
frac1{20^2}+frac1{15^2} = frac1{12^2}
using the smallest Pythagorean triple 3,4,5. The reciprocal Pythagorean theorem is a special case of the optic equation
frac1{p}+frac1{q} = frac1{r}
where the denominators are squares and also for a heptagonal triangle whose sides p,q,r are square numbers.

### Incommensurable lengths

File:Euclid Corollary 5.svg|thumb|The (spiral of Theodorus]]: A construction for line segments with lengths whose ratios are the square root of a positive integer)One of the consequences of the Pythagorean theorem is that line segments whose lengths are incommensurable (so the ratio of which is not a rational number) can be constructed using a straightedge and compass. Pythagoras's theorem enables construction of incommensurable lengths because the hypotenuse of a triangle is related to the sides by the square root operation.The figure on the right shows how to construct line segments whose lengths are in the ratio of the square root of any positive integer.BOOK, The Elements of Euclid: with many additional propositions, and explanatory notes, to which is prefixed an introductory essay on logic, Henry, Law, John Weale, 1853,weblink 49, Corollary 5 of Proposition XLVII (Pythagoras's Theorem), Each triangle has a side (labeled "1") that is the chosen unit for measurement. In each right triangle, Pythagoras's theorem establishes the length of the hypotenuse in terms of this unit. If a hypotenuse is related to the unit by the square root of a positive integer that is not a perfect square, it is a realization of a length incommensurable with the unit, such as {{radical|2}}, {{radical|3}}, {{radical|5}} . For more detail, see Quadratic irrational.Incommensurable lengths conflicted with the Pythagorean school's concept of numbers as only whole numbers. The Pythagorean school dealt with proportions by comparison of integer multiples of a common subunit.BOOK, Understanding the infinite, Shaughan Lavine, 13,weblink 0-674-92096-1, 1994, Harvard University Press,
According to one legend, Hippasus of Metapontum (ca. 470 B.C.) was drowned at sea for making known the existence of the irrational or incommensurable.
{{Harv|Heath|1921|loc= Vol I, pp. 65}}; Hippasus was on a voyage at the time, and his fellows cast him overboard. See JOURNAL, James R. Choike, The pentagram and the discovery of an irrational number, The College Mathematics Journal, 11, 312â€“316, 1980, harv, A careful discussion of Hippasus's contributions is found inJOURNAL, The Discovery of Incommensurability by Hippasus of Metapontum, Kurt Von Fritz, Annals of Mathematics, Second Series, 46, Apr 1945, 242â€“264, 1969021, 2, Annals of Mathematics, harv, 10.2307/1969021,

### Complex numbers

(File:Complex conjugate picture.svg|right|thumb|The absolute value of a complex number z is the distance r from z to the origin)For any complex number
z = x + iy,
the absolute value or modulus is given by
r = |z|=sqrt{x^2 + y^2}.
So the three quantities, r, x and y are related by the Pythagorean equation,
r^2 = x^2 + y^2.
Note that r is defined to be a positive number or zero but x and y can be negative as well as positive. Geometrically r is the distance of the z from zero or the origin O in the complex plane.This can be generalised to find the distance between two points, z1 and z2 say. The required distance is given by
|z_1 - z_2|=sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2},
so again they are related by a version of the Pythagorean equation,
|z_1 - z_2|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2.

### Euclidean distance

{{details|Euclidean distance}}The distance formula in Cartesian coordinates is derived from the Pythagorean theorem.BOOK, Mastering algorithms with Perl, Jon Orwant, Jarkko Hietaniemi, John Macdonald,weblink 426, Euclidean distance, 1-56592-398-7, 1999, O'Reilly Media, Inc, If {{nowrap|(x1, y1)}} and {{nowrap|(x2, y2)}} are points in the plane, then the distance between them, also called the Euclidean distance, is given by
sqrt{(x_1-x_2)^2 + (y_1-y_2)^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, by generalization of 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}.
If instead of Euclidean distance, the square of this value (the squared Euclidean distance, or SED) is used, the resulting equation avoids square roots and is simply a sum of the SED of the coordinates:
(a_1-b_1)^2 + (a_2-b_2)^2 + cdots + (a_n-b_n)^2 = sum_{i=1}^n (a_i-b_i)^2.
The squared form is a smooth, convex function of both points, and is widely used in optimization theory and statistics, forming the basis of least squares. In information geometry, more general notions of statistical distance, known as divergences, are used, and the Pythagorean identity can be generalized to Bregman divergences, allowing general forms of least squares to be used to solve non-linear problems.

### Euclidean distance in other coordinate systems

If Cartesian coordinates are not used, for example, if polar coordinates are used in two dimensions or, in more general terms, if curvilinear coordinates are used, the formulas expressing the Euclidean distance are more complicated than the Pythagorean theorem, but can be derived from it. A typical example where the straight-line distance between two points is converted to curvilinear coordinates can be found in the applications of Legendre polynomials in physics. The formulas can be discovered by using Pythagoras's theorem with the equations relating the curvilinear coordinates to Cartesian coordinates. For example, the polar coordinates {{nowrap|(r, Î¸)}} can be introduced as:
x = r cos theta, y = r sin theta.
Then two points with locations {{nowrap|(r1, Î¸1)}} and {{nowrap|(r2, Î¸2)}} are separated by a distance s:
s^2 = (x_1 - x_2)^2 + (y_1-y_2)^2 = (r_1 cos theta_1 -r_2 cos theta_2 )^2 + (r_1 sin theta_1 -r_2 sin theta_2)^2.
Performing the squares and combining terms, the Pythagorean formula for distance in Cartesian coordinates produces the separation in polar coordinates as:
begin{align}s^2 &= r_1^2 +r_2^2 -2 r_1 r_2 left( cos theta_1 cos theta_2 +sin theta_1 sin theta_2 right)
&= r_1^2 +r_2^2 -2 r_1 r_2 cos left( theta_1 - theta_2right)
&=r_1^2 +r_2^2 -2 r_1 r_2 cos Delta theta, end{align}
using the trigonometric product-to-sum formulas. This formula is the law of cosines, sometimes called the generalized Pythagorean theorem.BOOK, Plane Trigonometry and Tables, George, Wentworth, BiblioBazaar, LLC, 2009, 1-103-07998-0, 116,weblink , Exercises, page 116
From this result, for the case where the radii to the two locations are at right angles, the enclosed angle {{nowrap|Î”Î¸ {{=}} {{pi}}/2,}} and the form corresponding to Pythagoras's theorem is regained: s^2 = r_1^2 + r_2^2. The Pythagorean theorem, valid for right triangles, therefore is a special case of the more general law of cosines, valid for arbitrary triangles.

### Pythagorean trigonometric identity

(File:Trig functions.svg|thumb|Similar right triangles showing sine and cosine of angle Î¸)In a right triangle with sides a, b and hypotenuse c, trigonometry determines the sine and cosine of the angle Î¸ between side a and the hypotenuse as:
sin theta = frac{b}{c}, quad cos theta = frac{a}{c}.
From that it follows:
{cos}^2 theta + {sin}^2 theta = frac{a^2 + b^2}{c^2} = 1,
where the last step applies Pythagoras's theorem. This relation between sine and cosine is sometimes called the fundamental Pythagorean trigonometric identity.BOOK, PreCalculus the Easy Way, Lawrence S. Leff,weblink 296, 0-7641-2892-2, 7th, Barron's Educational Series, 2005,
In similar triangles, the ratios of the sides are the same regardless of the size of the triangles, and depend upon the angles. Consequently, in the figure, the triangle with hypotenuse of unit size has opposite side of size sinâ€‰Î¸ and adjacent side of size cosâ€‰Î¸ in units of the hypotenuse.

### Relation to the cross product

(File:Cross product parallelogram.svg|right|thumb|The area of a parallelogram as a cross product; vectors a and b identify a plane and {{nowrap|a Ã— b}} is normal to this plane.)The Pythagorean theorem relates the cross product and dot product in a similar way:JOURNAL, 10.2307/2323537, Cross products of vectors in higher-dimensional Euclidean spaces, WS Massey, The American Mathematical Monthly, 90, Dec 1983, 697â€“701, 2323537, 10, Mathematical Association of America, harv,
|mathbf{a} times mathbf{b}|^2 + (mathbf{a} cdot mathbf{b})^2 = |mathbf{a}|^2 |mathbf{b}|^2.
This can be seen from the definitions of the cross product and dot product, as
begin{align} mathbf{a} times mathbf{b} &= ab mathbf{n} sin{theta}
mathbf{a} cdot mathbf{b} &= ab cos{theta}, end{align}
with n a unit vector normal to both a and b. The relationship follows from these definitions and the Pythagorean trigonometric identity.This can also be used to define the cross product. By rearranging the following equation is obtained
|mathbf{a} times mathbf{b}|^2 = |mathbf{a}|^2 |mathbf{b}|^2 - (mathbf{a} cdot mathbf{b})^2.
This can be considered as a condition on the cross product and so part of its definition, for example in seven dimensions.BOOK, Clifford algebras and spinors, Pertti Lounesto,weblink 96, Â§7.4 Cross product of two vectors, 0-521-00551-5, 2001, Cambridge University Press, 2nd, BOOK, Methods of applied mathematics, Francis Begnaud Hildebrand, 24,weblink 0-486-67002-3, Reprint of Prentice-Hall 1965 2nd, Courier Dover Publications, 1992,

## Generalizations

### Similar figures on the three sides

A generalization of the Pythagorean theorem extending beyond the areas of squares on the three sides to similar figures was known by Hippocrates of Chios in the 5th century BC,Heath, T. L., A History of Greek Mathematics, Oxford University Press, 1921; reprinted by Dover, 1981. and was included by Euclid in his Elements:Euclid's Elements: Book VI, Proposition VI 31: "In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle."If one erects similar figures (see Euclidean geometry) with corresponding sides on the sides of a right triangle, then the sum of the areas of the ones on the two smaller sides equals the area of the one on the larger side.This extension assumes that the sides of the original triangle are the corresponding sides of the three congruent figures (so the common ratios of sides between the similar figures are a:b:c).Putz, John F. and Sipka, Timothy A. "On generalizing the Pythagorean theorem", The College Mathematics Journal 34 (4), September 2003, pp. 291â€“295. While Euclid's proof only applied to convex polygons, the theorem also applies to concave polygons and even to similar figures that have curved boundaries (but still with part of a figure's boundary being the side of the original triangle).The basic idea behind this generalization is that the area of a plane figure is proportional to the square of any linear dimension, and in particular is proportional to the square of the length of any side. Thus, if similar figures with areas A, B and C are erected on sides with corresponding lengths a, b and c then:
frac{A}{a^2} = frac{B}{b^2} = frac{C}{c^2}, , Rightarrow A + B = frac{a^2}{c^2}C + frac{b^2}{c^2}C, .
But, by the Pythagorean theorem, a2 + b2 = c2, so A + B = C.Conversely, if we can prove that A + B = C for three similar figures without using the Pythagorean theorem, then we can work backwards to construct a proof of the theorem. For example, the starting center triangle can be replicated and used as a triangle C on its hypotenuse, and two similar right triangles (A and B ) constructed on the other two sides, formed by dividing the central triangle by its altitude. The sum of the areas of the two smaller triangles therefore is that of the third, thus A + B = C and reversing the above logic leads to the Pythagorean theorem a2 + b2 = c2. (See also Einstein's proof by dissection without rearrangement){|thumbGeneralization for similar triangles,green area {{nowrap|A + B {{=}} blue}} area C)thumb|Pythagoras's theorem using similar right triangles)thumbGeneralization for regular pentagons)

### Law of cosines

File:Law of cosines2.svg|x200px|thumb|The separation s of two points {{nowrap|(r1, Î¸1)}} and {{nowrap|(r2, Î¸2)}} in polar coordinates is given by the law of cosineslaw of cosinesThe Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:BOOK, cited work, Lawrence S. Leff,weblink 326, 0-7641-2892-2, Barron's Educational Series, 2005-05-01,
a^2+b^2-2abcos{theta}=c^2,
where Î¸ is the angle between sides a and b.When Î¸ is 90 degrees ({{pi}}/2 radians), then cosÎ¸ = 0, and the formula reduces to the usual Pythagorean theorem.

### Arbitrary triangle

File:TÃ¢bit ibn Qorra.svg|thumb|Generalization of Pythagoras's theorem by ThÄbit ibn Qurra|TÃ¢bit ibn Qorra]].BOOK, Great moments in mathematics (before 1650), Howard Whitley Eves,weblink Â§4.8:...generalization of Pythagorean theorem, 0-88385-310-8, 1983, 41, Mathematical Association of America,
( Lower panel: reflection of triangle ABD (top) to form triangle DBA, similar to triangle ABC (top).)
At any selected angle of a general triangle of sides a, b, c, inscribe an isosceles triangle such that the equal angles at its base Î¸ are the same as the selected angle. Suppose the selected angle Î¸ is opposite the side labeled c. Inscribing the isosceles triangle forms triangle ABD with angle Î¸ opposite side a and with side r along c. A second triangle is formed with angle Î¸ opposite side b and a side with length s along c, as shown in the figure. ThÄbit ibn Qurra stated that the sides of the three triangles were related as:JOURNAL, 10.1086/348837, 227603, Aydin Sayili, Mar 1960, ThÃ¢bit ibn Qurra's Generalization of the Pythagorean Theorem, Isis, 51, 1, 35â€“37, harv, BOOK, Roots to Research: A Vertical Development of Mathematical Problems, Judith D. Sally, Paul Sally, Exercise 2.10 (ii), 62,weblinkdate=2007-12-21,
a^2 +b^2 =c(r+s) .
As the angle Î¸ approaches {{pi}}/2, the base of the isosceles triangle narrows, and lengths r and s overlap less and less. When Î¸ = {{pi}}/2, ADB becomes a right triangle, r + s = c, and the original Pythagorean theorem is regained.One proof observes that triangle ABC has the same angles as triangle ABD, but in opposite order. (The two triangles share the angle at vertex B, both contain the angle Î¸, and so also have the same third angle by the triangle postulate.) Consequently, ABC is similar to the reflection of ABD, the triangle DBA in the lower panel. Taking the ratio of sides opposite and adjacent to Î¸,
frac{c}{a} = frac{a}{r} .
Likewise, for the reflection of the other triangle,
frac{c}{b} = frac{b}{s} .
Clearing fractions and adding these two relations:
cr +cs = a^2 +b^2 ,
the required result.The theorem remains valid if the angle theta is obtuse so the lengths r and s are non-overlapping.

### General triangles using parallelograms

(File:Pythagoras for scalene triangle.svg|thumb|Generalization for arbitrary triangles,green {{nowrap|area {{=}} blue}} area)(File:Pythagoras construction.svg|thumb|Construction for proof of parallelogram generalization)Pappus's area theorem is a further generalization, that applies to triangles that are not right triangles, using parallelograms on the three sides in place of squares (squares are a special case, of course). The upper figure shows that for a scalene triangle, the area of the parallelogram on the longest side is the sum of the areas of the parallelograms on the other two sides, provided the parallelogram on the long side is constructed as indicated (the dimensions labeled with arrows are the same, and determine the sides of the bottom parallelogram). This replacement of squares with parallelograms bears a clear resemblance to the original Pythagoras's theorem, and was considered a generalization by Pappus of Alexandria in 4 ADFor the details of such a construction, see BOOK, Modern geometry with applications: with 150 figures, George Jennings, Figure 1.32: The generalized Pythagorean theorem, 23,weblink 0-387-94222-X, 1997, 3rd, Springer, Claudi Alsina, Roger B. Nelsen: Charming Proofs: A Journey Into Elegant Mathematics. MAA, 2010, {{isbn|9780883853481}}, pp. 77â€“78 ({{Google books|mIT5-BN_L0oC|excerpt|page=77}})The lower figure shows the elements of the proof. Focus on the left side of the figure. The left green parallelogram has the same area as the left, blue portion of the bottom parallelogram because both have the same base b and height h. However, the left green parallelogram also has the same area as the left green parallelogram of the upper figure, because they have the same base (the upper left side of the triangle) and the same height normal to that side of the triangle. Repeating the argument for the right side of the figure, the bottom parallelogram has the same area as the sum of the two green parallelograms.

### Solid geometry

(File:Pythagoras 3D.svg|thumb|Pythagoras's theorem in three dimensions relates the diagonal AD to the three sides.)(File:Tetrahedron vertfig.svg|thumb|A tetrahedron with outward facing right-angle corner)In terms of solid geometry, Pythagoras's theorem can be applied to three dimensions as follows. Consider a rectangular solid as shown in the figure. The length of diagonal BD is found from Pythagoras's theorem as:
overline{BD}^{,2} = overline{BC}^{,2} + overline{CD}^{,2} ,
where these three sides form a right triangle. Using horizontal diagonal BD and the vertical edge AB, the length of diagonal AD then is found by a second application of Pythagoras's theorem as:
overline{AD}^{,2} = overline{AB}^{,2} + overline{BD}^{,2} ,
or, doing it all in one step:
overline{AD}^{,2} = overline{AB}^{,2} + overline{BC}^{,2} + overline{CD}^{,2} .
This result is the three-dimensional expression for the magnitude of a vector v (the diagonal AD) in terms of its orthogonal components {vk} (the three mutually perpendicular sides):
|mathbf{v}|^2 = sum_{k=1}^3 |mathbf{v}_k|^2.
This one-step formulation may be viewed as a generalization of Pythagoras's theorem to higher dimensions. However, this result is really just the repeated application of the original Pythagoras's theorem to a succession of right triangles in a sequence of orthogonal planes.A substantial 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 (like a corner of 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. This result can be generalized as in the "n-dimensional Pythagorean theorem":BOOK, Matrix analysis, Rajendra Bhatia,weblink 21, 0-387-94846-5, 1997, Springer, {{Blockquote|text=Let x_1, x_2, ldots,x_n be orthogonal vectors in â„n. Consider the n-dimensional simplex S with vertices 0,x_1,ldots, x_n. (Think of the (n âˆ’ 1)-dimensional simplex with vertices x_1,ldots,x_n not including the origin as the "hypotenuse" of S and the remaining (n âˆ’ 1)-dimensional faces of S as its "legs".) Then the square of the volume of the hypotenuse of S is the sum of the squares of the volumes of the n legs.}}This statement is illustrated in three dimensions by the tetrahedron in the figure. The "hypotenuse" is the base of the tetrahedron at the back of the figure, and the "legs" are the three sides emanating from the vertex in the foreground. As the depth of the base from the vertex increases, the area of the "legs" increases, while that of the base is fixed. The theorem suggests that when this depth is at the value creating a right vertex, the generalization of Pythagoras's theorem applies. In a different wording:For an extended discussion of this generalization, see, for example, Willie W. Wong {{Webarchive|url=https://web.archive.org/web/20091229111557weblink |date=2009-12-29 }} 2002, A generalized n-dimensional Pythagorean theorem.{{blockquote|text=Given an n-rectangular n-dimensional simplex, the square of the (n âˆ’ 1)-content of the facet opposing the right vertex will equal the sum of the squares of the (n âˆ’ 1)-contents of the remaining facets.}}

### Inner product spaces

{{See also|Hilbert space}}(File:Parallelogram equality.svg|thumb|Vectors involved in the parallelogram law)The Pythagorean theorem can be generalized to inner product spaces,BOOK, Classification, parameter estimation, and state estimation, Ferdinand van der Heijden, Dick de Ridder,weblink 357, 0-470-09013-8, 2004, Wiley,
which are generalizations of the familiar 2-dimensional and 3-dimensional Euclidean spaces. For example, a function may be considered as a vector with infinitely many components in an inner product space, as in functional analysis.
BOOK, Finite element methods: accuracy and improvement, Qun Lin, Jiafu Lin,weblink 23, 7-03-016656-6, 2006, Elsevier, In an inner product space, the concept of perpendicularity is replaced by the concept of orthogonality: two vectors v and w are orthogonal if their inner product langle mathbf{v} , mathbf{w}rangle is zero. The inner product is a generalization of the dot product of vectors. The dot product is called the standard inner product or the Euclidean inner product. However, other inner products are possible.BOOK, Elementary Linear Algebra: Applications Version, Howard Anton, Chris Rorres,weblink 336, 0-470-43205-5, 2010, 10th, Wiley, The concept of length is replaced by the concept of the norm ||v|| of a vector v, defined as:BOOK, Beginning functional analysis, Karen Saxe, Karen Saxe,weblink 7, Theorem 1.2, 0-387-95224-1, 2002, Springer,
lVert mathbf{v} rVert equiv sqrt{langle mathbf{v},mathbf{v}rangle} , .
In an inner-product space, the Pythagorean theorem states that for any two orthogonal vectors v and w we have
left| mathbf{v} + mathbf{w} right|^2 = left| mathbf{v} right|^2 + left| mathbf{w} right|^2 .
Here the vectors v and w are akin to the sides of a right triangle with hypotenuse given by the vector sum v + w. This form of the Pythagorean theorem is a consequence of the properties of the inner product:
left| mathbf{v} + mathbf{w} right|^2 =langle mathbf{ v+w}, mathbf{ v+w}rangle = langle mathbf{ v}, mathbf{ v}rangle +langle mathbf{ w}, mathbf{ w}rangle +langlemathbf{ v, w }rangle + langlemathbf{ w, v }rangle = left| mathbf{v}right|^2 + left| mathbf{w}right|^2,
where the inner products of the cross terms are zero, because of orthogonality.A further generalization of the Pythagorean theorem in an inner product space to non-orthogonal vectors is the parallelogram law :
2|mathbf v|^2 +2 |mathbf w|^2 = |mathbf {v + w} |^2 +| mathbf{v-w}|^2 ,
which says that twice the sum of the squares of the lengths of the sides of a parallelogram is the sum of the squares of the lengths of the diagonals. Any norm that satisfies this equality is ipso facto a norm corresponding to an inner product.{{anchor|Decomposition}}The Pythagorean identity can be extended to sums of more than two orthogonal vectors. If v1, v2, ..., vn are pairwise-orthogonal vectors in an inner-product space, then application of the Pythagorean theorem to successive pairs of these vectors (as described for 3-dimensions in the section on solid geometry) results in the equationBOOK, Douglas, Ronald G., Banach Algebra Techniques in Operator Theory, 2nd edition, Springer-Verlag New York, Inc, 1998, New York, New York, 60â€“61,weblink 978-0-387-98377-6,
left|sum_{k=1}^nmathbf{v}_kright|^2=sum_{k=1}^n|mathbf{v}_k|^2

### Sets of m-dimensional objects in n-dimensional space

Another generalization of the Pythagorean theorem applies to Lebesgue-measurable sets of objects in any number of dimensions. Specifically, the square of the measure of an m-dimensional set of objects in one or more parallel m-dimensional flats in n-dimensional Euclidean space is equal to the sum of the squares of the measures of the orthogonal projections of the object(s) onto all m-dimensional coordinate subspaces.JOURNAL, 10.2307/2319528, Generalized Pythagorean Theorem, Donald R Conant, William A Beyer, yes, The American Mathematical Monthly, 81, Mar 1974, 262â€“265, 2319528, 3, Mathematical Association of America, harv, In mathematical terms:
mu^2_{ms} = sum_{i=1}^{x}mathbf{mu^2}_{mp_i}
where:
• mu_m is a measure in m-dimensions (a length in one dimension, an area in two dimensions, a volume in three dimensions, etc.).
• s is a set of one or more non-overlapping m-dimensional objects in one or more parallel m-dimensional flats in n-dimensional Euclidean space.
• mu_{ms} is the total measure (sum) of the set of m-dimensional objects.
• p represents an m-dimensional projection of the original set onto an orthogonal coordinate subspace.
• mu_{mp_i} is the measure of the m-dimensional set projection onto m-dimensional coordinate subspace i. Because object projections can overlap on a coordinate subspace, the measure of each object projection in the set must be calculated individually, then measures of all projections added together to provide the total measure for the set of projections on the given coordinate subspace.
• x is the number of orthogonal, m-dimensional coordinate subspaces in n-dimensional space ({{math|Rn}}) onto which the m-dimensional objects are projected (m â‰¤ n):

x = binom{n}{m} = frac{n!}{m!(n-m)!}

### Non-Euclidean geometry

{{See also|Hilbert's axioms}}The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, were the Pythagorean theorem to fail for some right triangle, then the plane in which this triangle is contained cannot be Euclidean. More precisely, the Pythagorean theorem implies, and is implied by, Euclid's Parallel (Fifth) Postulate.BOOK, CRC concise encyclopedia of mathematics, Eric W. Weisstein,weblink 2147, The parallel postulate is equivalent to the Equidistance postulate, Playfair axiom, Proclus axiom, the Triangle postulate and the Pythagorean theorem., 2nd, 1-58488-347-2, 2003, BOOK, The principle of sufficient reason: a reassessment, Alexander R. Pruss, We could include...the parallel postulate and derive the Pythagorean theorem. Or we could instead make the Pythagorean theorem among the other axioms and derive the parallel postulate., 0-521-85959-X, 2006, Cambridge University Press, 11,weblink
Thus, right triangles in a non-Euclidean geometry
BOOK, cited work, Stephen W. Hawking, 4,weblink 0-7624-1922-9, 2005, do not satisfy the Pythagorean theorem. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to {{pi}}/2, and all its angles are right angles, which violates the Pythagorean theorem because
a^2 + b^2 = 2 c^2 > c^2 .
Here two cases of non-Euclidean geometry are consideredâ€”spherical geometry and hyperbolic plane geometry; in each case, as in the Euclidean case for non-right triangles, the result replacing the Pythagorean theorem follows from the appropriate law of cosines.However, the Pythagorean theorem remains true in hyperbolic geometry and elliptic geometry if the condition that the triangle be right is replaced with the condition that two of the angles sum to the third, say A+B = C. The sides are then related as follows: the sum of the areas of the circles with diameters a and b equals the area of the circle with diameter c.JOURNAL, Victor Pambuccian, Maria Teresa Calapso's Hyperbolic Pythagorean Theorem, The Mathematical Intelligencer, 32, December 2010, 2, 10.1007/s00283-010-9169-0, 4,

#### Spherical geometry

(File:Triangle sphÃ©rique.svg|thumb|Spherical triangle)For any right triangle on a sphere of radius R (for example, if Î³ in the figure is a right angle), with sides a, b, c, the relation between the sides takes the form:BOOK, Elementary differential geometry, Barrett O'Neill,weblink Exercise 4, 441, 0-12-088735-5, 2006, 2nd, Academic Press,
cos left(frac{c}{R}right)=cos left(frac{a}{R}right)cos left(frac{b}{R}right).
This equation can be derived as a special case of the spherical law of cosines that applies to all spherical triangles:
cos left(frac{c}{R}right)=cos left(frac{a}{R}right)cos left(frac{b}{R}right) +sinleft(frac{a}{R}right) sinleft(frac{b}{R}right) cos gamma .
By expressing the Maclaurin series for the cosine function as an asymptotic expansion with the remainder term in big O notation,
cos x = 1 - frac{x^2}{2} + O(x^4) text{ as } x to 0 ,
it can be shown that as the radius R approaches infinity and the arguments a/R, b/R, and c/R tend to zero, the spherical relation between the sides of a right triangle approaches the Euclidean form of the Pythagorean theorem. Substituting the asymptotic expansion for each of the cosines into the spherical relation for a right triangle yields
1-frac{1}{2}left(frac{c}{R}right)^2 + Oleft(frac{1}{R^4}right) = left[1-frac{1}{2}left(frac{a}{R}right)^2 + Oleft(frac{1}{R^4}right) right]left[1-frac{1}{2}left(frac{b}{R}right)^2 + Oleft(frac{1}{R^4}right) right] text{ as }Rtoinfty .
The constants a4, b4, and c4 have been absorbed into the big O remainder terms since they are independent of the radius R. This asymptotic relationship can be further simplified by multiplying out the bracketed quantities, cancelling the ones, multiplying through by âˆ’2, and collecting all the error terms together:
left(frac{c}{R}right)^2 = left(frac{a}{R}right)^2 + left(frac{b}{R}right)^2 + Oleft(frac{1}{R^4}right)text{ as }Rtoinfty .
After multiplying through by R2, the Euclidean Pythagorean relationship c2 = a2 + b2 is recovered in the limit as the radius R approaches infinity (since the remainder term tends to zero):
c^2= a^2 + b^2 + Oleft(frac{1}{R^2}right)text{ as }Rtoinfty .
For small right triangles (a, b

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