how 0/0 is indiscriminite ? it should be 1?
348468
This 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, 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. Tus the expression 0*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs to the limit argument, 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. 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.
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
766487
x(x^2+x+1)one of the critical numbers is 0 and the other two are imaginary.
997010
x^0 is equal to  x^(1-1) = x^1/x^1 = x/x = 1/1 = 1\n0 is actually undefined. (0^0) because you're dividing by 0.
238636
It is impossible to have 0 as a denominator.
932248
zero, since the ratio between any number and infinite is zero.
565951
the answer is zero b'coz at the end, X-X is equal to zero.
659301
Zero divided by any number (except zero) is zero.
606683
The factor (1/infinity) equals to zero.\n\nHence the term\n\n(infinity) - (1/infinity) = (infinity) - 0\n\nTherefore,\n\nX = infinity
415435
nope... 01 (in binary) = 1 (in decimal)\n\nand \n\n10 (on binary) = 2 (in decimal)
807068
(-1.6875) (81) (0.28125) = 38.443359375\n\nIf thats not right, sue me.
239035
Independence....\n\nAnd what =0\n\nBig I
647936
The answer is 1 - you're raising a quantity to a power of 0, which always results in 1. The quantity within the parentheses contains no division by zero, so I know it's a real number, too.
820085
We say that division by zero is undefined, for it is not consistent with division by other numbers.\n\nHere's a little experiment for you to try on your calculator. Observe the output when you try the following set of calculations:\n       1/1\n       1/.1\n       1/.01\n       1/.001\n       1/.0001\n       1/.00001\n       etc... \n\nuntil your calculator can't go any further or you get tired. You should notice that the answers continue getting larger and larger. \nAnother way of thinking of it is to imagine filling a box with apples. Say a box can hold 100 apples. Now try filling it with apples that are half the size of these apples. You can put 200 in the box. Now imagine a special, magic apple that takes up no room at all. How many can you put in the box?\n\nWell, the answer is... there is no answer! That is why mathematicians refer to numbers that are divided by 0 as "undefined." Some people tend to think of them as being infinite, but this isn't exactly true. There simply is no answer.
252229
Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
339198
2x0 = 0\nIt'll be the same as today... :P
68748
uhm, ZERO!  0x2=0  what, did you fail basic math?