Is it still undefined? On a related note, is there a reason that numbers raised to the 0th power are 1 besides “because things don’t work otherwise”?
I was taught that 0/0 is “indeterminate”, as opposed to 1/0, which is “undefined”.
I think the best way to think of the definition of division is in terms of it actually being the inverse of multiplication.
In other words, we define a/b to be the unique number (if such a thing exists) x such that a = bx.
In this context, it’s clear that 0/0 is undefined: There is no unique number x such that 0=0x (it’s true for all real x).
As for raising things to the zeroth power, there are various justifications for it (loosely speaking). The one that’s really always satisfied me is this:
a[sup]n[/sup], in “loose” terms, is meant to be a multiplied by itself n times. In particular, a[sup]0[/sup] means a is to be multiplied by itself 0 times–In other words, we’re really not multiplying anything at all! What does it mean if we multiplying zero things together? The multiplicative identity is one; in other words, take a number, multiply it by one, and you’ve got the same number back. In this sense, the number one is the “nothingness” of multplication. So an “empty product” (a multiplicative product of zero things) should be defined as 1.
Similarly, an “empty sum” will give you the additive identity, which is zero. In other words na means you add a to itself n times. What does 0a mean? It means you add a to itself zero times–an “empty sum”. So we define it to be zero–the “nothingness” of addition.
I’m fully aware this is not a rigorous description, but the idea is sound, and I’m only trying to appeal to intuition.
Right. It’s pretty easy to see why, too. When you divide one number by another, you can check your math by reversing the calculation; you multiply the quotient by the divisor and if the result is the dividend, you know your math was right. But, if you divide zero by zero, you can set the quotient to any number at all, and the math will check, since anything multiplied by zero is equal to zero.
Calling 0/0 indeterminate is really only helpful in limits. Outside of a limit, it’s just as correct to say that it’s undefined.
to the zeroth power question:
x^n / x^m = x^(n-m) therefore, if n = m, x^n / x^n = x^0 = 1
You can actually make the limit (basically the value of f(x) as x goes to some number. In other words, its the value one would expect to find at that number) of 0/0 to be any number that you want it to. Say you have a function f(x)=ax/x and you take the limit as x->0 you will find that f(x)=a when you plug in 0.
What the ****, how does your post prove that any number raised to the zero power is one? All you’ve done is shown that the law of exponents relating to division.
And as an even easier demonstration of zeroth powers, consider our humble base of ten:
10[sup]5[/sup] = 100,000
10[sup]4[/sup] = 10,000
10[sup]3[/sup] = 1,000
10[sup]2[/sup] = 100
10[sup]1[/sup] = 10
10[sup]0[/sup] = 1
10[sup]-1[/sup] = 0.1
10[sup]-2[/sup] = 0.01
etc.
Makes perfect sense now. 
No, there are good reasons. You want the law of exponents that says x[sup]a[/sup]x[sup]b[/sup] = x[sup]a + b[/sup] to hold outside of positive integers. The only way that works is if x[sup]0[/sup] = 1.
You can’t plug 0 into that f, because that results in division by 0.
On another view to the problem of x^0=1, we have Roger Penrose’s definition on page 91 of The Road to Reality, in which he states that x^0=1 is required if we are to preserve the rule of exponentiation of B^(w+z) = B^w + b^z.
I also was more than a little hasty in my earlier condemnation of JustAnotherGeek after realizing that he was proving his point by using the identity of x/x=1.
I think there is a subtle but nonetheless distinguishable difference between 0/0 and any number divided by zero or n/0.
Speaking informally here, let’s take the number 5 as n. 5/1 = 5. When the denominator gets smaller (e.g. 5/.1 = 50), the result gets larger. So, intuitively, we can “conclude” that as the denominator gets smaller, the result gets much larger and approaches infinity.
With 0/0 we could say
8/8 = 1
23/23=1 therefore, one number divided by the same number equals one and so 0/0 must equal 1.
Then again we can say, zero divided by 10 is zero, zero divided by 2 is zero. Since every number divided by zero equals zero then 0/0 = 0
Add to that, the fact that any number times zero equals zero, we could (incorrectly) conclude that 0/0 = 97 because 97*0 = 0
Since 0/0 has these extra complications I can see why it is classified separately from a non-zero number divided by zero,
Any number (other than zero) divided by itself is 1. Incidently, this is why 0^0 is undefined.
There are several ways of getting around that problem, the easiest of which would be to use L’Hopital’s rule. (or here, you could just divide the x from numerator and denominator, but that doesn’t always work).
My response to the OP is that you can assume that it’s undefined, unless you’re far enough in math to bother with limits, and even then, you’ll only bother trying to get a non-undefined answer if you’re working in a situation where limits are relevant.
Well, yeah thats the point of the example. It shows that, depending on the circumstances, 0/0 literally could be any number. Clearly this is a problem in defining what 0/0 would be, hence it is undefined.
This bit is incorrect; every number divided by zero is not zero.
An earlier thread on 0/0.
WAG
“at the end of the day”
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OP - Zer0
Posters - Zer0!**
Think ya mean = B^w ***** B^z. 
As for 0/0… 0 is an indefinite number. I mean, you can’t actually show me 0 anything, right? (You can only show me the absence of something, which may sound like splitting hairs, but it’s important.) This is only useful in limits, as some folks have mentioned.
Some examples:
1.) What would be the limit of x/(y^2)? as x&y->+0 In this case, it looks like 0/0 will tend towards infinity.
2.) Well, how about sin(x)/x? This one took people a while to figure it out. Now the function is called sinc and the value when x=0 is 1.
3.) Lastly, contemplate the limit of something like y^2/x as both x & y -> +0. (Basically, this is the inverse of the first example.) This time, the limit goes to zero.
So, in short, it’s indeterminate.