what is the history of the mathmatical sign division?
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The symbol for division\n\nThe Anglo-American symbol for division is of 17th century origin, and has long been used on the continent of Europe to indicate subtraction. Like most elementary combinations of lines and points, the symbol is old. It was used as early as the 10th century for the word est. When written after the letter "i", it symbolized "id est." When written after the word "it", it symbolized "interest." If written after the word "divisa", for "divisa est", this might possibly have suggested its use as a symbol of division. Towards the close of the 15th century the Lombard merchants used it to indicate a half, along with similar expressions such as this one on the right.\n\nThere is also a possibility that it was used by some Italian algebrists to indicate division. In a manuscript entitled Arithmetica and Practtica by Giacomo Filippo Biodi dal Aucisco, copied in 1684, this symbol stands for division, suggesting that various forms of this kind were probably used.\n\nThe Anglo-American symbol (above top) first appeared in print in the Teutsche Algebra by Johann Heinrich Rahn (1622-1676) which appeared in Zurich in 1659. This symbol was then made known in England by the translation of Rahn's work by Dr. John Pell in London in 1688. (Smith p406)\n\nAround the year 1200, both the Arabic writer al-Hassar, and Fibonacci (Leonardo of Pisa), symbolised division in fraction form with the use of a horizontal bar, but it is thought likely that Fibonacci adopted al-Hassar's introduction of this symbolisation.\n\nIn his Arithmetica integra (1544) Michael Stifel employed the arrangement 8)24 to mean 24 divided by 8. (NCTM p139)\n\nMichael Stifel (1486?-1567) was regarded as the greatest German algebrist of the 16th century. (Cajori p140)\n\nSMITH, D.E. "History of Mathematics" volume II, Dover Publications 1958\n\nTHE NATIONAL COUNCIL OF TEACHERS OF MATHEMATICS, "Historical Topics for the Mathematics Classroom", National Council of Teachers of Mathematics (USA) 1969\n\nCAJORI, FLORIAN "A History of Mathematics", The Macmillan Company 1926
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The order of mathmatical operations can be abbreviated as BEDMAS\n\nBrackets\nExponents\nDivision\nMultiplication\nAddition\nSubtraction\n\nYour question can be written SQRT (144 x 12 - 4)\n\nDo the brackets first, again following BEDMAS\n=1724\nThen the SQRT which is essentially an exponent.\n=41.52....\n\nIt's so nice to see so many eager to answer the question wrong.
99187
This was a form of math that gained a small following several years ago with the Hmong of Thiland. It's not used very much today due to its limited application.   Edited comment:  Pastor Dave is correct, it's still used. I think it was also used for a few months at Enron, Corp.
829392
just a piont of view, but normally to divide something symmetrically you have to do it at the widest piont along the centre line. where as assymetricaly is off the centre line at a narrower piont and therefore not a symmetry,this only applies to poloygons and symmetric shapes.  hope this helps
538012
modulus division is division that returns the remainder of a division\n\nSo for example, \n5/3 = 1 with remainder 2, so 5 mod 3 = 2\n\nit is basically returning the remainder for a division
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Oh, how dismal it was! They did not have a "decimal" or fraction system, but instead used words. Go here for more information:\n\nhttp://mathforum.org/dr.math/faq/faq.roman.html#calc
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At first glance, if the top left times the bottom right is A, and the other cross is B, then A > B, means the right fraction was larger.\n\nI couldn't come up with a quick counter.  Maybe you can.
495242
mathematic equations change that fast and before you know it, what you learned from school is already a thing of the past because so many formulas and equations have cropped up.  so the sign of the highest degree term in LCM is always positive which is correct.  next mathematical question please?
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Is it possible to have 2 equal signs in a problem?\nI am only in algebra but I haven't seen one like that before.\nSorry I couldn't help.
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You cant just add or substract whatever is left. You invalidate the function of "division" by partially dividing then filling in the rest of the numerator. The number left over after something has been divided is called the modulo, and is useful for programming functions. You cant do an incomplete division(only dividing 20/5 instead of the 2/5 also) and then speculate as to why it cant be negative when doing a full division. It invalidates the definition.
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The earliest know mention in literature of common denominators (least common multiple) was Thomas Blundeville in his 1594 publication "Exercises" where he says, "Multiply the Denominators the one into the other, and the Product thereof shall bee a common Denominator to both fractions." The term Least common multiple is found in 1823 in J. Mitchell, Dict. Math. & Phys. Sci.: "To find the least common Multiple of several Numbers"
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You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
197247
its impossible because there needs to be either one less sign or one more number look:\n\nin a the order:\n2 * 4 - 6 + 8 /\n              ^there is an extra sign!!
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The formula is credited to Heron of Alexandria in the 1st century, and a proof can be found in his book Metrica. It is now believed that Archimedes already knew the formula, and it is of course possible that it has been known long before.\n\nLet the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that\n S2 = p(p - a)(p - b)(p - c)\n\nW.Dunham analyzes the original Heron's proof in his Journey through Genius.\n\nFor the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that\n p - a = (- a + b + c)/2, p - b = (a - b + c)/2, p - c = (a + b - c)/2\n\nIt takes a little algebra to show that\n \n(a + b + c)(- a + b + c)(a - b + c)(a + b - c)\n= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nFor the right triangle, this expression is equal to 16S2 = 4a2b2. So we have\n 4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nTaking all terms to the left side and grouping them yields\n (a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0\n\nWith a little more effort\n (a2 + b2)2 - 2c2(a2 + b2) + c4 = 0\n\nAnd finally\n [(a2 + b2) - c2]2 = 0\nRemark\n\nFor a quadrilateral with sides a, b, c and d inscribed in a circle there exists a generalization of Heron's formula discovered by Brahmagupta. In this case, the semiperimeter is defined as p = (a + b + c + d)/2. Then the following formula holds\n S2 = (p - a)(p - b)(p - c)(p - d)\n\nSince any triangle is inscribable in a circle, we may let one side, say d, shrink to 0. This leads to Heron's formula.\n\nFor more information you can go to www.wikipedia.com  \nhttp://mathworld.wolfram.com/HeronsFormula.html\nhttp://www.cut-the-knot.org/pythagoras/herons.shtml\n\nI hope this helps! =-)
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The notation "i" or "a + bi" for complex numbers originated with Euler in 1777. Before that, sqrt(-1) was used.
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yes. when you multiply them on the divided sign, it gets rid of them. Example\n\na / b = xc\n\na =(xc)b
30751
The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.