What is the full form of "BODMAS", related to the hierchy of arithmatic operators?
959125
Brackets, Orders, Division, Multiplication, Addition, Subtraction.\nIn some countries its BEDMAS, where E is exponents.
1060963
Aryabhatta - and other ancient Indian mathematicians.\n\nThe concept of zero, or Shunya was then taken by the Arabs to Europe.
584069
http://www.mathcymru.org/english/thatsoddans.html
877407
In the following <op> means either +, -, * (multiply) or / (divide).\n\nclass your-class {\n\nvoid your-method()\nnumber a, b, c;\nc = a <op> b;\n}\n}
747794
Okay, it's "beginning algebra", not "begging algebra".  For a good start, check out "Elmentary Algebra" in Wikipedia, and check out its external links.
703992
120(14-6)\n\nStep 1: 120(8) <---- because in order of operations, the "p" which is parentheses in "pemdas" is first\nSteo 2: 120 x 8= 960 <------------- because when a number is outside the parentheses yet doesn't have any operation sign between them, u do multiplication
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
573115
The Order of Operations rules as we know them could not have existed \nbefore algebraic notation existed; but I strongly suspect that they \nexisted in some form from the beginning - in the grammar of how people \ntalked about arithmetic when they had only words, and not symbols, to \ndescribe operations. It would be interesting to study that grammar in \nGreek and Latin writings and see how clearly it can be detected.\n\nAt the other end, I think that computers have influenced the subject, \nso that it is taught more rigidly now than it used to be, since \nprogramming languages have had to define how every expression is to be \ninterpreted. Before then, it was more acceptable to simply recognize \nsome forms, like x/yz, as ambiguous and ignore them - something I \nthink we should do more often today, considering some of the questions \nwe get on such issues.\n\nI spent some time researching this question, because it is asked \nfrequently, but I have not found a definitive answer yet. We can't say \nany one person invented the rules, and in some respects they have \ngrown gradually over several centuries and are still evolving.\n\nHere are my conclusions, perhaps in more detail than you want:\n\n1. The basic rule (that multiplication has precedence over addition) \nappears to have arisen naturally and without much disagreement as \nalgebraic notation was being developed in the 1600s and the need for \nsuch conventions arose. Even though there were numerous competing \nsystems of symbols, forcing each author to state his conventions at \nthe start of a book, they seem not to have had to say much in this \narea. This is probably because the distributive property implies a \nnatural hierarchy in which multiplication is more powerful than \naddition, and makes it desirable to be able to write polynomials with \nas few parentheses as possible; without our order of operations, we \nwould have to write\n\n     ax^2 + bx + c\nas\n     (a(x^2)) + (bx) + c\n\nIt may also be that the concept existed before the symbolism, perhaps \njust reflecting the natural structure of problems such as the \nquadratic.\n\nYou can see an example of early notation in "Earliest Uses of Grouping \nSymbols" at:\n\n   http://members.aol.com/jeff570/grouping.html   \n\nwhere the use of a vinculum (an early version of parentheses) shows, \nboth in its presence (around an additive expression) and its absence \n(around the multiplicative term "B in D") that the rules were \nimplicitly followed:\n                                                 ________________\n   In Van Schooten's 1646 edition of Vieta, B in D quad. + B in D\n   is used to represent B(D^2 + BD). \n\n2. There were some exceptions early in this development; in \nparticular, math historian Florian Cajori quotes many writers for \nwhom, in the special case of a factorial-like expression such as\n\n     n(n-1)(n-2)\n\nthe multiplication sign seems to have had some of the effect of an \naggregation symbol; they would write\n\n     n * n - 1 * n - 2\n\n(using a dot or cross where I have the asterisks) to express this. Yet \nCajori points out that this was an exception to a rule already \nestablished, by which "nn-1n-2" would be taken as the quadratic \n"n^2 - n - 2." \n\nThere was also an early notation in which a multiplication would be \nreplaced by a comma to indicate aggregation: \n\n     n, n - 1 \n\nwould mean\n     \n     n (n - 1) \n\nwhereas \n     \n     nn-1 \n\nmeant\n\n     n^2 - 1.\n\n3. Some of the specific rules were not yet established in Cajori's own \ntime (the 1920s). He points out that there was disagreement as to \nwhether multiplication should have precedence over division, or \nwhether they should be treated equally. The general rule was that \nparentheses should be used to clarify one's meaning - which is still \na very good rule. I have not yet found any twentieth-century \ndeclarations that resolved these issues, so I d
349526
:}\nShort for KR^ishna
1099580
Ah they are Common operator notation, we use it to to express mathematical sequences using tokens.\n\nPostfix and prefix are mainly used for computer programming, that is how the computer thinks when doing operational sequences of numbers. We use postfix and prefix to do lexical analysis as well. Take a look at the webpage and you will know what I mean:)\n\nTake a look at Wikipedia they have a great article
651564
hemidemisemiquaver\n\nhttp://en.wikipedia.org/wiki/Musical_notation
457175
2 + (a + b)\ngiven\n\n2 + (b + a)\ncommutative prop. of addition\n\n(2 + b) + a\nassociative prop. of addition\n\n(b + 2) + a\ncommutative prop. of addition\n\nb + (2 + a)\nassociative prop. of addition\n\n^_^
241809
Quote: "The binary arithmetic operators are +, -, *, /, and the modulus operator %.", more abstractly an arithmetic operator is an operator which is applied to an arithmetic type.  Hope this helps.
538579
One starts with the basic concept of unity: 1\n\nOne then defines the successor of 1: it is called 2.\n\nOne then defines addition-of-one:  Anything plus 1 is the successor of that anything; declare that "successor" is commutative in this definition\n\nNEXT...\n\nDefine successor of two as being named 3.\n\nDefine successor of three as being 4.\n\nBUT: 4 is 3 + 1\n\nAND: since 3 is 2 + 1,\n\nTHEN: 4 = (2+1)+1 \n\nBUT then, remembering that 2 is the name for successor-of-1\nand using commutation,\n\n4 = (2+1) +1 = 2+(1+1) = 2 + 2\n\nSO, 2+2 = 4\n\n....  Actually had to do this as a homework problem in an advanced Math course, along about 1972 or so.  I actually had to prove it commuted, but I don't recall how I did that.\n\nObnoxiously simple, really, once you realize that 4 is just the name of successor-of-successor-of-successor-of-1, and all that jazz...
991775
Cantor used a diagonalization argument. We show that the open interval (0,1) is uncountable.  Suppose that f:N->(0,1) is any function. We how it cannot be onto by constructing an element of (0,1) that is not in the image of f. Write the decimal expansions\n\nf(1)=.a_11 a_12 a_13 a_14 .....\nf(2)=.a_21 a_22 a_23 a_24 ....\nf(3)=.a_31 a_32 a_33 a_34 ...\nf(4)=.a_41 a_42 a_43 a_44 ...\n.........\n.........\n\nSo each a_ij is an integer between 0 and 9. There is a slight ambiguity because of the non-uniqueness of decimal expansions. For such cases, we use the expansion with infinitely many 0's, so\n.25000000...\nrather than\n.2499999.....\n\nNow construct a new real number between 0 and 1 as follows:\nLet b_1 be different from a_11 and either 3 or 7.\nLet b_2 be different from a_22 and either 3 or 7.\nLet b_3 be different from a_33 and either 3 or 7.\n.....\nand let\nx=.b_1 b_2 b_3 b_4 .....\nThen x is in (0,1). But by construction, b is not equal to f(1) (since the first decimal digits are different). It is different from f(2) since the second decimal digit is different. Continue and see that x is not f(n) for *any* natural number n. In other words, x is not in the image of the function f!\n\nSince countability requires a function which is both one-to-one and onto, and we have shown that no function from N to (0,1) can be onto, we see that (0,1) is uncountable. Since R is even larger, R is uncountable.\n\nNote: I used the choice of 3 or 7 to make sure to avoid the ambiguity of decmal expansions.
846821
the way you communicate to one another using algebraic symbols?
1125943
The order of operations are :\nParenthesis\nExponents\nMultiply\nDivide\nAdd\nSubtract