No. of proofs for Pythogoras theorem?
350854
64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac
456169
There are many proofs but with fallacies. Please visit:\nhttp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html\n\nhttp://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html
1106346
You know there are 8 axioms in mathematics which we cannot prove\n\n1) if 'a' and 'b' are integers then 'a+b' is an integer and 'a.b' is an integer (Closure law)\n2) a + b = b + a and a..b = b.a (cumulative law)\n3) a + (b + c) = (a + b) + c (Associative law)\n4) a (b + c) = a.b + a.c (Distributive law)\n5) a + 0 = a = 0 + a (Additive Identity)\n6) a.1 = a = 1.a (Multiplicitive Identity)\n7) for all integer 'a' there is a unique integer (-a) such that a + (-a) = 0\n8) I can't remember...\n\nSo we can prove your requirement using above axioms\n\n0 + 0 = 0 (Obivious) \na ( 0 + 0 ) = a.0 (Multiplying by same factor)\na.0 + a.0 = a.0 (Distributive over addition)\na is an integer and 0 is an integer so a.0 is an integer, hence there exits a unique -(a.0) such that -(a.0) + a.0 = 0\n\nso adding -(a.0) both sides\n\n-(a.0) + (a.0 + a.0) = -(a.0) + (a.0) \n(-(a.0) + a.0 )+ a.0 = 0 (associative law)\n0 + a.0 = 0\na.0 = 0 (additive identity)
1121169
The  Pythagorean Theorem gives the length of the hypotenuse (the longest side) of a right triangle ( one of the angles is 90 degrees )\na^2 + b^2 = c^2\n\n\nApplying the Pythagorean Theorem twice you get a three dimensional version \n\na=x\nb=y\nx^2 + y^2 = c^2\nc^2 + z^2 = h^2 (using c as the side of a new triangle)\nx^2 + y^2 + z^2 = h^2\n\nwhich is the length of any line from (0,0,0) to (x,y,z)\n\nthe distance formula just finds the change in x, y, and z and puts it into this formula\nx = x1-x2\ny = y1-y2\nz = z1-z2\n\nThe midpoint formula is really unrelated to the other two.\nAll it does is average each of the coordinates of the other two points.  ( (x1+x2)/2 , (y1+y2)/2 , (z1+z2)/2 )\nThe only thing it has in common is the use of the same type of coordinate system
655583
The Pythagorean Theorem states that, in a right triangle, the sum of the squares of the lengths of the two sides which form the right angle is equal to the square of the length of the third side.\n\nThis is commonly stated as A^2 + B^2 = C^2, where A and B represent the lengths of the two sides which form a right angle, and where C represents the length of the other side.
806805
You can prove it by induction.\n\nIt is obviously true when n = 1.\n\nInduction pypothesis:  Suppose it is true for n\n\nWe need to show  is true for (n+1).\n\nDoes 24 divide (5^(2*(n+1))-1))?\n\nWe can rewrrite this as:\n\n25*(5^2n)-1  Subtract 24 from this to get 25*(5^2n)-25 \n\nThis is just equal to 25*(5^2n)-1).\n\nThis has to be divisible by 24 -- since one of the factors is (5^2n)-1) and our induction hypothesis says that it is.  \n\nAdd 24 to it and we get something that is also divisible by 24 -- but it is (5^(2*(n+1))-1)) -- so our induction is done.
1022973
"Pythagoras of Samos was a Greek philosopher responsible for important developments in mathematics, astronomy and the theory of music. He left Samos because of the tyrant who ruled there and went to southern Italy about 532 BC. He founded a philosophical and religious school in Croton that had many followers.\n\nAlthough the theorem now known as Pythagoras's theorem was known to the Babylonians 1000 years earlier he may have been the first to prove it.\n\nOf his actual work nothing is known. His school practised secrecy and communalism making it hard to distinguish between the work of Pythagoras and that of his followers. His school made outstanding contributions to mathematics.\n\nPythagoreans believed that all relations could be reduced to number relations. This generalisation stemmed from observations in music, mathematics and astronomy.\n\nThe Pythagoreans noticed that vibrating strings produce harmonious tones when the ratios of the lengths of the strings are whole numbers, and that these ratios could be extended to other instruments.\n\nThe most important discovery of this school was the fact that the diagonal of a square is not a rational multiple of its side. This result showed the existence of irrational numbers. Not only did this disturb Greek mathematics but the Pythagoreans' own belief that whole numbers and their ratios could account for geometrical properties was challenged by their own results.\n\nIn astronomy Pythagoras taught that the Earth was a sphere at the centre of the Universe. He also recognised that the orbit of the Moon was inclined to the equator of the Earth and he was one of the first to realise that Venus as an evening star was the same planet as Venus as a morning star. "\n\nThere is some more on Pythagoras at\nhttp://www-history.mcs.st-and.ac.uk/~history.\nGo to the search menu and search for "pythagoras". Hope this answers your question.
1156966
I can see two different interpretations of your question.\n\nInterpretation 1:  You are asking that if you have a right triangle with sides of length a and b, and hypotenuse of length c, then a^2 + b^2 = c^2.  No other powers work.\n\nInterpretation 2:  You are asking if there are solutions to a^n + b^n = c^n for n other than 2.  In this case, for n > 2 there are no nontrivial RATIONAL solutions.  This is the content of what is known as "Fermat's Last Theorem".
493260
Knowing what a proof is, is the first step in answering this question. From what I am familiar with...\n\nA proof is: a set of logical steps acquired through deductive (therefore, not making any giant leaps in logic, unless by definition), and hence, empirically (from the evidence provided) resulting in a direct equivalence (being, among other types of equivalence, but primarily, in permutation,  multiplicative/additively & negatively/positively & even/odd... meta-mathematically) of states, that's shortest distance is (in absolute terms), either infinity, zero, and/or, also, one.\n\nReally, the attempted 'proof' of 2 + 2 = 5 is based on a distorted type of Trigonometry, which was in essence the source of today's Calculus (just try to draw Tangent or Secant without running into the idea of Calculus' derivative & integral, respectively), and actually is the result of any additive equavalence of any two numbers' to being alike to any number, (because measuring hypotenuse of a given sides is essentially multiplicative, hence partially irrational). \n\n(Which makes me wonder... is there a 2 * 2 = 5 equivalent? and the answer is a resounding, yes! But first the 'proof' as written by Charles Seife.)\n\nLet a & b each be equal to 1. Since a ^ b are equal,\n\nb^2 = ab ...(eq.1)\n\nSince a equals itself, it is obvious that \n\na^2 = a^2  ...(eq.2)\n\nSubtract equation 1 from equation 2. This yeilds\n\n(a^2) - (b^2) = (a^2)-ab ...(eq. 3)\n\nWe can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get\n\n(a+b)(a-b) = a (a-b) ...(eq.5)\n\nSo far, so good. Now divide both sides of the equation by (a-b) and we get\n\na + b = a  ...(eq.5)\n\nb = 0 ...(eq.6)\n\nBut we set b to 1 at the very beginning of this proof, so this means that\n\n1 = 0 ...(eq.7)\n\n ...Anyways, getting that far gives us the jist of the proof, later in the proof, Charles Seife goes on to prove that Winston Churchill was a carrot! if you want to know how that is possible, I recommend you read the book.\n\nFrom equation 7, add a number to either side and get it equal to any other number, one greater than itself.\n\nMultiplying equation 7 after adding to it, and one can get: any number is equal to any other number. \n\nHence, conceptually, any number is equal to zero, and, theoretically, that includes infinity. But that's also the reason why when you divide by zero, it is 'Undefined.' Which, consequentially, is what is happening in this equation... just subsistute 1 into equation 3 and one will see that we are dividing by zero in equation 5. \n\nThis is what lead to the invention of calculus. Really, from here this segways into Hilbert Space... but that is best left for another entry, hopefully, on the actual subject of quantazation.\n\nThat's all I have time for... \n\nTHIS PROOF IS BY DEFINITION INCORRECT, but it provides a good tool as of why we define things in mathematics the way we do.\n\nA good question to ask from here would be (based on my previous tangent):\n\nDoes 1/3 plus 1/3 plus 1/3 = 1?\nOr, does it equal just zero point nine repeating?
1130099
(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
641478
a^2 + b^2 = c^2\n\nPythagorean theorem
324840
Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
301417
I'll try. I gave this proof to THREE people already:\n\nTake any two ordinary constants. (e.g. x and y)\nMake them alike, as in x = y = any non-zero number (in this case, 2)\nMultiply each by y, but x works equally well (y²=xy).\nSubtract x² from each equation (y²-x² = xy-x²).\nFactor y²-x² into (y+x)(y-x) to make ((y-x)(y+x) = xy-x²).\nFactor xy-x² into x(y-x) to make ((y-x)(y+x) = x(y-x)).\nDivide by (y-x) to get (y+x=x).\nSubtract x from both sides of this equation to get y=0.\nThis means that y=0.\n\nbut since, y equaled 2 in the first case, that means 2 = 0.\nIf you multiply this equation by 2.5, you get 5 = 0.\n\nSince 0 + 0 = 0, you can plug in 2 for the first two zeroes, and five for the last one, and then, TA DA! 2 + 2 = 5.\n\nP.S., this proof has a little falsity in it. you can't divide by (y-x) because that is zero!
1062456
proof positive\nyou can find it here\nhttp://www.scifi.com/proofpositive/
169166
well i think yoru referring to fermats last theorem. This was proven by Andrew Wiles of Britan in 1994-1995.  He squestered himself in a room for more or less 7 years and came up with a solution.  Originally his proof was rejected when he first presented it in 1993 because an error was found.  He fixed the error and we now know that Fermats last Theorem is true.
355573
Fermat claimed he had a proof 300 years ago but he\nnever demonstrated it. Andrew Wiles proved it in 1996\nbut had to use 20th century mathematics. There's an interesting book called Fermat's Enigma dedicated to\nDr. Wiles 7 year quest to prove it. There's also a \ndocumentary which has been aired on PBS which I've\nseen twice and wouldn't miss it if it came out again.
45454
Do you mean the Pythagorean theorem?\n\nIn any right triangle, the area of the square whose side is the hypotenuse (the side of the triangle opposite the right angle) is equal to the sum of the areas of the squares of the other two sides.