i know that ramanujam asked this in his primary class...but why isnt zero divided by zero equal to ONE?
600176
This is a Frequently Asked Question. Following is my previous answer:\n\nThis hypothesis is false:\nSuppose 0/0 = 1. Since 1 is the multiplicative identity, this implies (0/0)*a = a for all real numbers a. But (0/0)*a = (0*a)/0 = 0/0 = 1. This contradicts our hypothesis that 0/0 is the multiplicative identity, so the hypothesis must be false (proof by contradiction). QED\n0/0 is indeterminate as a limiting form. See http://mathworld.wolfram.com/Indeterminate.html .\nAs an algebraic expression over the real field, it is undefined: http://mathworld.wolfram.com/Undefined.html .\nThis is a Frequently Asked Question; the following is my previous answer:\n\nThis question may be asked so frequently (along with the infamous 0.999...) because no concrete axioms or definitions are ever given for the real number system until one studies higher algebra or real analysis; it is assumed that everyone knows what a real number is until then. The proliferation of these questions show that this assumption is no good.\nFor the sake of conciseness, I will not define what a real number is here, as only the algebraic structure of the real numbers is necessary to answer this question. The algebraic structure of the real numbers is what we call a field. A field is an algebraic structure that follows a finite set of axioms. These axioms define what statements are true and false in this system based on the laws of logic. You can find the field axioms here: http://mathworld.wolfram.com/Field.html .\nEach line is preceded by "For all real numbers a, b and c, the following is true...". Note that the proper form of the additive identity and multiplicative identity is that it prescribes the existence of some number 1 and some number 0 such that the equations hold. These same numbers are then used to prescribe the existence of additive and multiplicative inverses.\nDivision a/b is just shorthand for a*(b^(-1)). So in your question, you are looking for 0*0^(-1). It is easy to show that there is no such number as 0^(-1) in any field. Suppose 0^(-1) exists.\nThen 0*0^(-1) = 1. But multiplying both sides by 0^(-1) gives 0 = 0^(-1), which means 0*0 = 0 = 1. This is obviously false.\nSuppose 0 = 1. 0*a = a for all numbers a. But 0*a = (1 + (-1))*a = a + (-a) = 0. Thus 0 is not equivalent to 1 and our sole assumption that 0^(-1) exists must be false. Thus the expression 0*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs for trying to define 0/0 using limiting behavior, note that you are evaluating a limit over a plane of the function f(x,y) = x/y. As we approach (0,0) along the x-axis, this limit is 0, but as we approach (0,0) along the y-axis, the limit diverges. Thus, the limit of f(x,y) at (0,0) does not exist, so f(0,0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nThe expression 0/0 is said to be indeterminate as a limiting form because one can have many functions that approach the form 0/0 that have convergent limits of different values. For example, the limit as x approaches 0 of [sin(x)]/x yields the indeterminate form 0/0, but L'Hopital's rule reveals the limit to be 1. Similarly, multiplying the above limiting form by any real number will yield that number as the limit, even though the form is still 0/0. The expressions listed on this page: http://mathworld.wolfram.com/Indeterminate.html are called indeterminate because they do not have a fixed behavior when studied as limits of expressions.
665836
This question already discussed here so many times. For your information the answer is: "Indeterminate". \n\nTo say that "0/0" is an indeterminate form does not just mean that "0/0" by itself can represent any number, or may represent no number. Those points are true, in a certain sense, but of limited practical significance when stated in those terms. Unlike "0/0", "1/0" is not an indeterminate form.\n\nIt means also that the ratio of two functions f and g that approach zero can approach any member of a range of well-defined values, depending on which functions f and g are. Whether such a value exists, and what it might be, depends on how the functions approach zero.\n\nIn more formal language, the fact that the functions f(x) and g(x) both approach 0 as x approaches some limit c, is not enough information to evaluate the limit\n\nlim f(x)/g(x)\nx->c\n \nThat limit could be any number or plus or minus infinity, or might not exist, depending on what the functions f and g are.\n\nBy contrast "1/0" is not an indeterminate form because there is no range of different values that f/g could approach if f approaches 1 and g approaches 0. (The absolute value of the ratio will always approach positive infinite and so is not considered indeterminate.)
230430
The square root of -1 has no relation to zero.  It requires an imaginary number.  Division by zero is, unfortunately for our number system, impossible.  You cannot cut something into zero pieces, and in fact the very statement makes no sense.
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0/0 is a discontinuity.\n\nIt is not specifically defined in the axioms of normal arithmetics. but, you can modify any set of axioms according to your own need, as long as they remain consistent and do not self contradict. U can define 0/0 to be 0, or u can define it to be 1 or 2 or 3... but before u proceed ahead, u have to make sure, by defining it as 1, your axioms do not self contradict. say,\n\n0/0=1\n\nwhat is (0/0) x0?\n\nif normal arithmetical rules apply... that (a/b)c=(ac)/b\n\n(0/0)x0=1x0=0\n(0x0)/0=0/0=1\n\nby setting 0/0=1, it contradict one of the rules, (a/b)c=(ac)/b, hence, in light of the original axioms, u cannot have 0/0=1.\n\nThe point is... as long as other rules r not broken, u can add definitions to mathematical objects to create a new object. U can still define 0/0=1, and change the (a/b)c=(ac)/b rule. u then have to make sure by removing this, other contradictions don't arise. Or u can set a rule saying, (a/b)c=(ac)/b applies except when (0/0)x0. again, u gotta check for contradictions. u can adjust everything until finally u have a new mathematical object that is wholly consistent with 0/0=1. Then, with this object, 0/0 is defined. Now, however, this new object would not behave like the ordinary number field we r familiar with.
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
357910
hello!!!!!!!!!! \n1+1 in binary is 10! it's not ten but one zero!!!
1115436
The number is not defined.\n\nLots of people are saying things like...\n"0.000...00001"\n"1/x as x approaches infinity"\n\nThese numbers, technically, are the same as 0.\n\n"n^0" --- ok that's just not thinking. If n <> 0 or n <> inf, then any n^0 = 1.\n\nProof:\nLet 0* be a number infinitely close to 0.\n0* = 1 - .999....\n.999... = 3 * (.333...) // Because 3*3 = 9 no matter how many time you do it.\n.999... = 3 * (1/3 )  // Exact fractional representation\n.999... = 1\nTherefore, 0* = 1 - 1, and 0* = 0.\n\nAs for the answer, the "smallest", non-zero real number does not exist. Well, we know that it "should" exist. It just doesn't.\n\nExcept on computers. That's another problem entirely. Because computers have a data limit, they do actually have a smallest real number >0
153279
Zero (0) is a paradoxical number that has high philosophical and scientific importance. Context of zero in absolute & applied Mathematics is different. In applied mathematics we use it according to our convenience, for example 0 degree celsius, 0!(factorial), initial Velocity, initial position etc. And in Absolute Mathematics we think of it transcendently. Absolute zero is beyond our experience.Therefore we are using and thinking of zero in diffrent context. \nIn the fraction a/o, at present the numerator `a' is considered as something that is considerable. And the denominator `0' is not used in the absolute sense insted it  is taken in the concept of `infinitesimally small'.That is the symbol `0' here stands for `infinatesimally small'.For instance a `point' is considered as magnitude and dimensionless which is relatively true and wrong in the absolute sense.\nContext a/0=a is also correct iff same meanig is strected all through. ie. a/0 implies `a' is `undivided'.Then a*0 means a is not at all multiplied, hence a*0=a only in the absolute sense. Here `a' which is the quotient and  the product is same to both expressions implies that `a' is individual and Independent.This concept is non-pragmatic hence not in use.\nMathematically,a=(1+1+1...`a' times).\nGeneraly, if a/m =n, a=m*n = (m+m+m+...`n' times) \nAnd, if m is infinatesimally small,a=(m+m+m+... infinite  times).At present the parodoxical symbol `0' is used to denote the axiom `infintesimaly small'.consequently. it is written as a/0= `Infinity'\nNote that divisibility exists iff something exists in the denominator.And in the same way multiplicability exits iff something exits as multiplier.
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I've seen one variant that goes like this:\nLet a and b be equal non-zero quantities\n\n          a = b\n\nMultiply through by a\n\n          a2 = ab\n\nSubtract b2\n\n          a2 − b2 = ab − b2\n\nFactorize both sides\n\n          (a − b)(a + b) = b(a − b)\n\nDivide out (a − b)\n\n          a + b = b\n\nObserving that a = b\n\n          b + b = b\n\nCombine like terms on the left\n\n          2b = b\n\nDivide by the non-zero b\n\n          2 = 1\n\nQ.E.D.\n\nThe problem here is that dividing by (a-b) is dividing by zero, which is an undefined operation.
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a/a = 1 (axiom: any thing except zero divided by the same yields 1 all the time)\n\nso, a/a = 1\nnow multiplying everyside by (a/1) we get,\n\na/a * (a/1) = 1* (a/1)\n\nor a^2/a = a/1\n\nor a = a/1  (axiom: anything except zero multiplied by 1 yields the same number)\n\nso it is proved that a = a/1
469511
To quote dilbert: "Early Civilisations never had a concept of zero" (hey, you wanted it thorough right?) And they started the number system at one. .... Hmm, pretty much the panel cuts out... \nRemember, the ones and zeros are really the positive and negative polaritys on a magnetic disk. That is the key to your computer informations. Do not take a magnet to your computer.
750843
you cannot divide by 0 = a-b\n&#92;
659301
Zero divided by any number (except zero) is zero.
574280
a number to the zero exponent is always 1 as a answer
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
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My old math teacher showed me this.\n\nSince A=B, then A-B=0, and you can't divide by 0
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!