Minor detail, but the arcsin of 2 might be defined. It would be a complex number. I’ll check when I get home.
A few years ago, I copyedited a math book in which the author finally gave definitions that I could both remember and understand:
x/0 is undefined, because there is no y such that y*0 = x.
0/0 is indeterminate, because for every y, y*0 = 0.
I had learned about undefined vs. indeterminate previously, but could never keep them straigh because I had never heard such a clear explanation before.
The results of dividing by zero depend on which zero you’re using. Let’s say you wish to evaluate sin x / x where x=0. Since sin 0 = 0, this is 0/0. However, by l’Hopital’s rule, we can state that the limit as x approaches zero of f(x)/g(x) = f’(x) / g’(x). In this case, f’(x) = cos x, g’(x) = 1. Thus the limit of sin x / x as x approaches zero = cos 0 / 1 = 1/1 = 1.
Let me explain this a bit. We would like to have functions q(a, b) and r(a, b) defined on pairs natural numbers such that a = q(a, b)b + r(a, b), with 0 < r(a, b) < b. As of right now, these functions are partial functions, because q(a, 0) and r(a, 0) are undefined. Some people define q(a, 0) = 0 and r(a, 0) = a. These will satisfy the equation, but zero becomes a special case for the second clause. And most of us don’t like special cases.
Mr Filter, you ARE a special case 
Per my high school AP Physics teacher:
Dividing 0 by 0 is like saying, “I got no money and no one to give it to. What do I do?”
I never thought a single number could be completely and totally fascinating.
Then I read Charles Seife’s ZERO: The Biography of a Dangerous Idea.
Division by zero is only the start – the concept of “zero” has some fairly mindboggling implications. A highly recommended read.
[schoolhouse rock]
My hero zero…
[/schoolhouse rock/
Zev Steinhardt
I prefer to think of myself as a pathological case. 
When I was in High School, we had an old electronic calculator that divided by counting successive subtractions. It kept subtracting the denominator from the numerator until it was less than the denominator. If you tried to divide by zero, the gears would just spin until you unplugged it.
That’s all well and good. However, the fact remains that f(x) = sin x/x is not defined for x = 0. Lim (x->a) f(x) = L NOT imply that f(a) = L or even that it is defined. The limit is defined at x = 0, but the function is not. Look at the definition of limit. It says nothing about the value of the function at x = a.
In general, there are two sorts of problems you can get with algebraic operations. Some problems, such as arcsin(2) or sqrt(-4), make no sense in the real numbers, but do make sense in the complex numbers (note, however, that there are multiple complex solutions in each of these cases). All of the other “undefined” things in algebra are essentially division by zero problems. For instance, tan([symbol]p[/symbol]/2) is undefined, because tangent is sine over cosine, and cos([symbol]p[/symbol]/2) = 0.
Now here’s the thing. We can consistently define square roots of negative numbers in a way which makes sense, for all (or very nearly all) applications where we would want to use them. The same method works in all cases. But we can’t do that in the case of division by zero: There is no way to define “1/0” which would make sense in all applications.
We had at least one model that, when you plugged it back in, would pick up precisely where it left off, and continue chugging. The only way to cure the sucker involved considerable dismantling to get at the offending gear. It takes very little to amuse a 15-year-old.
How can you write 0[sup]0[/sup] in terms of division by zero?
0[sup]0[/sup] = 1, or the binomial theorem fails. Try expanding (a + 0)[sup]1[/sup].
Ah that’s right I remember. Well, I never liked the binomial theorem anyway, but that’s okay.
0[sup]0[/sup] is, as I thought, undefined. x[sup]0[/sup] is 1 for all x except when x is 0, and 0[sup]y[/sup] is zero for all y except when y is 0.
Possible proof:
Let
f = 0[sup]0[/sup]
Then:
log(f) = log(0[sup]0[/sup] )
log(f) = 0(log(0))
So whatever the value of f is, it’s 0 times the log of 0. The log of 0 is itself undefined as log(y) grows asymptotically towards minus infinity as y approaches 0.
Hm. if f is 1, then:
log(1) = 0(log(0))
0 = 0(log(0))
Interesting. I can almost buy this given that 0d = 0 for all d, but d being log(0) which is undefined really really bugs me.
Anybody else got anything on 0[sup]0[/sup] ?
See my last post in this thread.
I’m curious, though, whether that’s a convenience for the binomial theorem alone. I wonder if there’s a way to justify it outside of the binomial theorem or a way to show it leads to a contradiction like 0/0 can be made to show.
Notation: log denotes natural log, frequently written ln.
Let f(x) = x[sup]x[/sup]. Then log f = x log x = log x/x[sup]-1[/sup].
By L’Hôpital’s Rule, the limit of log f as x -> 0 is the same as the limit of (1/x)/(-1/x[sup]2[/sup]) = -x ( provided the latter limit exists) and this second limit is plainly 0.
Thus f = e[sup]log f[/sup] -> e[sup]0[/sup] = 1 as x -> 0, as log is comtinuous.
So certainly x[sup]x[/sup] tends to 1 as x tends to 0. I prefer to regard 0[sup]0[/sup] as undefined, though.
Truth be told, the only places you generally run into 0[sup]0[/sup] are in the multinomial theorem (a generalized version of the binomial theorem) and in the bivariate exponential function x[sup]y[/sup]. The multinomial theorem is very important, and elegant if you don’t need a special case. So that’s a compelling argument for 0[sup]0[/sup] = 1. Also, you can’t really define an arithmetic operation only in the context of a certain theory.
The most compelling argument that 0[sup]0[/sup] != 1 is that it makes the bivariate exponential function discontinuous at the origin. So what? Have you seen the damn thing to the left of the origin?
In fact, even for fixed values of x, the exponential function can’t be single-valued once you extend the domain to C. So trying to save its continuity at a given point is just a waste of time.