Zero divided by Zero ??? I am wondering why 0/0 = indeterminate my reasoning is that anything divided by itself is 1 right? 2/2=1 1/1=1 so wouln't lim x->0 x/x = 1 if so the why wouldn't 0/0=1 I am not trying to argue that 0/0 is not indeterminate. I accept that it is, but I am still have to wonder, why?
Try x^{2}/x or x/x^{2}, as x goes to 0. In both cases, you have 0/0, but the first limit is 0 and the second is infinite. That's why 0/0 is indeterminate.
More generally, a = b/c is defined to mean that b = a*c. Given b and c with c not zero, there is a unique a such that b = a*c. For any b other than zero, there is no a for which b = a*0. So b/0 does not exist. Finally that leaves b = c = 0: But 0 = a*0 for every a. This is why 0/0 is indeterminant: you can get anything at all from a 0/0 situation. In addition to the two situations "Mathman" has supplied, lim_{x->0} ax/x = a for any number a. Indeed, if 0/0 = 1, then all derivatives would be constant, and all functions linear!
The difficulty with you logic is that you can't use "anything divided by itself is 1" to PROVE that something (in particular 0) divided by itself is 1! You COULD use "any non-zero number divided by itself is 1" to conclude that the limit of x/x as x-> 0 is 1 but unfortunately, as mathman pointed out, other sequences that wind up as "0/0" give other limits. (Actually, it is not "unfortunate". If 0/0 were equal to 1 calculus would be trivial and useless!
here's how i explain it sometimes. consider balls in boxes. 8/4=2 because it would take 2 boxes, 4 balls each, to total 8 balls. in general a/b=c if it takes c boxes, b balls each, to total a balls. now consider 0/0. the question is this: how many boxes, 0 balls each, would it take to total 0 balls? well, it could be 1, yes, but it might as well be a million: a million boxes, 0 balls each, would total 0 balls. incidentally, consider 1/0. how many boxes, 0 balls each, would it take to total 1 ball?
Let's say that 0/0=X. Then, 0=0*X. Well, there's more than one solution to that equation. In fact, there's infinitely many, and thus, indeterminant. Same thing with Infinity/Infinity; Infinity=Infinity*X has infinitely many solutions, thus indeterminant. ~Rashad
Sorry but your proof is flawed, you have assumed that 0/0 = 1 to show that 0/0 = any number. This is would suggest all numbers = 1.
Now, I would like to comment your reasoning here, that any number divided by itself is 1. (It is a very common reasoning, but flawed nonetheless) Let's review a few axioms concerning the real numbers: 1. There is a real number 0, so that for any real number a, we have a+0=a (This can be regarded as the definition of the real number commonly known as 0, we may prove, with a few other axioms, that only a single 0 exists.) 2. Given any real numbers a,b,c we have a distributive law connecting addition and multiplication: a*(b+c)=a*b+a*c 3. For any real number a, there exists a real number (-a), which has the property: a+(-a)=0 ((-a) is called the additive inverse of a; we may show that for any a, only one additive inverse exists; and we also have: (-(-a))=a, that is the additive inverse of the additive inverse of a is a itself) 4. Associative law for addition: given real numbers a,b,c, we have: a+b+c=a+(b+c) That is, by axiom, the order in which we sum together numbers are irrelevant. (The notation a+b+c means: Add b to a, then add c to to the result, whereas a+(b+c) means add the sum of b and c to a) Now, 1,2,3,4 , along with "trivial" axioms about how equalities can be manipulated, are enough to prove that for any a, we have a*0=0. Let z=a*0 Since, 0=0+0, we may manipulate our equation into: z=a*0=a*(0+0)=a*0+a*0=z+z Or we have: z=z+z We then have: z+(-z)=z+z+(-z) Or: 0=z+0, which means a*0=0 for any choice of real number a! But that means, if you want to define a real number 1/a such that a*(1/a) equals 1, you cannot have a real number 1/0 (so that 0*(1/0)=1), because we have proven that no such real number can exist! Hence, the 0/0 (=0*(1/0)) expression should be seen a danger signal which tells you to proceed with extreme care in order to avoid meaningless results..