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Mathematicians don't argue about what an expression "really is" (or at least, real mathematics doesn't involve this). They define functions and use axioms to prove theories about them.

"No really". Mathematics just isn't concerned with this stuff. Sometimes infinity it defined as single point making the real number compact, sometimes a "positive infinity" and a "negative infinity" are defined. Sometimes you add points to a given function to make it more tractable and sometimes you don't. But none of this "means" anything. The real number line can be embedded in a number of topological spaces. At least two division rings and various things (the complex numbers are most common). The way you extend a given function (say e^x) is going to vary depending on what space you're looking at as well as what topic you're interested in.

Math works with definition systems and get theorems out of them. If you want to know what something "really is", consult philosophy or something.

Are you sure you're replying to right post? I ask, because nothing I see in yours can be seen as reply to mine -- it actually repeats my point.

joe's reply clarified things somewhat - at least for me. Are replies always supposed to be rebuttals?

Are replies always supposed to be rebuttals?

It turns out to be so. If a reply is not a rebuttal, it is usually preceded with something like "To clarify, ..." or "I wanted to add, that ...". I just got confused without it.

I felt the need to state what I thought was the case. Maybe originally it was a rebuttal but you don't disagree, feel free to take it as a clarification.

Anthropomorphizing math as having a "concern" is also fuzzy thinking.

>If you want to know what something "really is", consult philosophy or something.

You mean like the "foundations of mathematics"? http://en.wikipedia.org/wiki/Foundations_of_mathematics

This.

Math is a tool (and sometimes abused for pure pleasure, 200 years later applied to make hard crypto work). If your definition doesn't make sense for the application, fix your definition and get over it.

Another example I've recently often bitched about in discussions is modern measure theory and its application to probability calculations. People just don't get the concept of theorytically possible event, but probability 0, i.e. ignore this. But without Lebesgue integration L_p function spaces are not complete and an awful lot of stuff stops to work properly. Among them essentially all of modern physics.

The sane approach is to get over the "this doesn't make intuitive sense" bitchering and just use defintions to derive useful results. And after a few years of playing around with stuff and applying the un-intuitive definition, it's becoming intuitive ;-)

Interesting that you say that. I've skimmed but have been meaning to properly read Nelson's: Radically Elementary probability theory http://www.math.princeton.edu/~nelson/books/rept.pdf and http://www.stat.umn.edu/geyer/nsa/. They do away with that problem all together as well as infinite constructions (replaced by hyperfinite) by replacing measure theory with non standard analysis. The gain is at least increased intuitiveness.

What is "indeterminate form"?

http://en.wikipedia.org/wiki/Indeterminate_form

You see, 0^0 = 1, and it's obvious to a mathematician . . . we define 0^0 = 1, to be consistent with exponentiation rules

Well, you're going to be inconsistent with them no matter how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.

0^0 is simply an expression that doesn't make sense. There isn't an answer, and there certainly isn't something we could agree to define it as. It is gibberish, nothing more, nothing less. One cannot assume just because there are mathematical symbols on paper that they make sense.

By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one.

0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this. If you think that -e^(i pi) makes more sense than 0^0, please, explain me why.

Also, mathematicians agree in this, seriously. Go and ask one.

I haven't seen any reason why -e^(i pi)=1 could be considered incorrect. It's consistent with the Taylor Series expansion of e^x. It's consistent with the view of complex exponentiation as rotation. I don't know of any particular problems that arise from taking -e^(i pi)=1.

This is not the case with 0^0=1, which is inconsistent with many limits. That's why 0^0=1 is an agreed-upon convention sometimes. http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero... has a fairly nice summary of the issues involved in defining it.

This is not the case with 0^0=1, which is inconsistent with many limits.

So what? It's only a problem if you want the exponentation function to be continuous, so you escape the problem by leaving it undefined. You could place similar unbased requirements on complex exponentiation to make it seem incorrect. For instance, real exponentiation always gives a positive value for positive base, while complex does not, so e^(i pi) = -1 is wrong. I agree that this is ridiculous requirement, but leaving 0^0 undefined because the math is not as we want it to be (e.g. exponentation is not continuous) looks just as ridiculous and silly to me.

On the other hand, putting 0^0 = 1 makes it consistent many combinatoric formulas, and is also consistent with cardinal exponentation, where nobody objects to 0^0 = 1, when you look at 0 as the cardinal number.

It has nothing to do with wanting exponentiation to be continuous. It's simply a recognition that limits of the form 0^0 are indeterminate, which means a convention that 0^0=1 is not appropriate in the context of evaluating limits. This isn't an argument that it "seems" incorrect, like your bizarre argument about real vs complex exponentiation; it's an argument that it IS incorrect in that context. If you're evaluating limits, you have to treat 0^0 as indeterminate, not as 1.

Even the original article noted that we don't choose the 0^0 convention because it's "correct", but because it's "nice" -- which is why we define it that way in the contexts where it makes sense to define it that way. If you're working with combinatorics, 0^0=1. If you're working with cardinal exponentiation, 0^0=1. If you're taking limits, or working in the hyperreals, or in certain other contexts, the convention doesn't apply. In some circumstances, 0^0 isn't even a valid statement -- like if you're working directly with the field axioms of R.

Recognize what context you're working in, and what assumptions or conventions apply in that context. That's just good mathematics.

It's simply a recognition that limits of the form 0^0 are indeterminate, which means a convention that 0^0=1 is not appropriate in the context of evaluating limits.

But the whole point of distinguishing some "forms" as "indeterminate" is to work around the discontinuity of elementary functions! My absolutely first sentence in this thread is asking, what exactly the "indeterminate form" is. I'm asking this question, because this not a formal notion and you will not find any formal definition of it. Its existence is rooted in the fact that for no reason other than the tradition (and convenience) we use special notation for some functions. Instead of +: R x R -> R, +(2, 3) = 5, we write 2 + 3 = 5. The same goes for ^: [0, \infty] x R -> R. The only reason we have all those fancy limit evaluating laws is because these function are continuous most of the time. For instance, + is continuous everywhere, so lim +(a_n, b_n) = +(lim a_n, lim b_n), if both lim a_n and lim b_n make sense. Similarly, /: R x R - {0} -> R is also continuous everywhere, so lim /(a_n, b_n) = /(lim a_n, lim b_n), if the right hand expression makes sense. If it does not make sense, for instance when both lim a_n and lim b_n are equal to zero, we need cannot approach this problem in such a simple way. Now, some people would call /(0, 0) an "indeterminate form", which makes for me no more sense than calling f(0, 0) an indeterminate form, where f(x, y) = log_(1/x) (y) -- while f is continuous everywhere where defined, you cannot extend its domain to contain (0, 0) for it to stay continuous, just like you cannot do it with / function.

As I repeated many times, the whole affair is because ^ seem to be more familiar than beta function (we have a special notation for it, for instance), people want it to behave nicely, so that for instance it conforms to some arbitrary limit evaluating laws, missing the whole underlying concept of continuity.

The whole point of distinguishing some forms as "indeterminate" is to work around the fact that you're trying to conduct operations on the real numbers that are not defined under the field axioms of the real numbers (or the axioms of the extended reals [-inf,inf]). That's where its essence is rooted; that's why this whole affair exists -- the fact that 0/0, 0^0, 0xinf, inf-inf, etc. are not well defined by our axioms.

This actually relates to all three examples I've presented where the 0^0=1 convention fails. It should be treated as an indeterminate form in limits because it's not well-defined by the axioms of the real numbers; it's also not well-defined by the axioms of the hyperreals, but division of infinitesimals is well-defined in the hyperreals, which gives us an alternate method of computing limits that avoids the "indeterminate form" entirely.

Let me reiterate: 0^0 is not defined under the field axioms of the real numbers. The choice to define it as 1 is a convention which makes certain math easier, in certain areas of mathematics. It is by no means a universal convention; it is by no means the one and only correct definition of 0^0. You continue to argue for the convention, but miss the larger point that it is a convention which is chosen for convenience, and which is not always appropriate.

Exponential function is not defined by the axioms of real numbers, as opposed to addition and multiplication, so this point is irrelevant -- you can define it in any way you want. There's no inconvenience in defining 0^0 = 1, apart from misunderstanding the concept of limits by some people. Defining 0^0 = 1 is universal convention -- people either do it like this, or do not define 0^0 at all, which I'm fighting against.

The whole point of distinguishing some forms as "indeterminate" is to work around the fact that you're trying to conduct operations on the real numbers that are not defined

I am not. Are you? Let me reiterate: the whole concept of "indeterminate forms" (which, I repeat, is not formal at all) stems from misunderstanding the process of taking limits.

> "people either do it like this, or do not define 0^0 at all, which I'm fighting against."

I think it's silly to fight against it. There are circumstances in which leaving it undefined is good, and in which trying to define it as 1 would lead to either misunderstandings (in the case of beginners doing limits, a case you are too quick to dismiss) or actually incorrect (an equivalent problem in the hyperreals could violate the transfer principle).

It's a broad convention, but it is not universal, and it shouldn't be.

> Also, mathematicians agree in this, seriously. Go and ask one.

Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-p....

More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-p..., if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is simply no good answer—unlike $e^{i\pi}$, which also lives in the continuous setting, but has a perfectly good, unambiguous answer. (lotharbot gives a nice explanation below of the ways that this is consistent with existing mathematics; but it can also be derived from the definition of the exponential function, with no further arbitrary conventions needed.)

But my point is, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not. 0^0 = 1 does not make calculus theorems more complicated to state or prove with modern language. It was a problem in XIX century, when mathematicians did not have a solid foundation with concepts like limit or continuity, but it's over now.

Mathematician here; we do not.

It seems I was a little too bold with my claim. All the mathematicians I know (and I'm a mathematician as well) agree with 0^0 = 1. It's a folklore specific thing, I guess.

I'm a mathematician as well

As am I, by training if not by profession. As is lotharbot. You're in a thread full of mathematicians. :)

Which is what I would expect on this site, actually. I'm always timid making technical claims here unless I'm sure I'm correct; it seems to be a place frequented by arbitrarily large fish.

To answer your question, if 0^0 = 1 is _the_ answer in discrete setting, and it's _an_ answer in continuous setting, why don't we just agree that 0^0 = 1 and stop creating confusing situation where sometimes it's defined and sometimes it's not.

I'm not persuaded it is always the answer. I think the fact that it is an indeterminate form in limits is a forceful enough demonstration of that. It all depends on context. If I came across a 0^0 in, say, an engineering context, my first instinct would be to check whether the formula was defined in that case, not to just assume that 1 would work.

I mean, it's like 1/0. If you're working in R, that's simply illegal. If you're working in R*, it's the infinite point. If you're taking a limit, it means "unbounded". If you're working in my favorite field, the hyperreals, it could be any number of flavors of infinity depending on the flavor of zero it was.

It would be foolhardy to try to define the symbol; without a context to supply some sort of sense, it is nonsense. And that is how I feel about 0^0 as well.

Please, treat exponentiation just like every other function out there. I don't get the whole limit argument at all. Given any function f: R x R -> R, if it happens that a_n -> a, b_n -> b, but lim f(a_n, b_n) != f(a, b), people just say that f is not continuous in (a, b), and the case is over. However, if f happens to be exponentiation function, people instead argue that f should not be defined in (a, b), forgetting about the fact that the theorem which lets you take a limit of an argument instead of a limit of a function values works only under assumption that f is continuous in a proper point. Instead of noting that there's no contradiction because the assumptions are not satisfied, people just run away from it, declaring 0^0 as undefined.

From this point of view, the whole notion of "indeterminate form" makes just as little sense as distinguishing some arbitrary class of functions and calling them "elementary". Why are some points of discontinuity of some functions more special than other points of discontinuity of other functions? Why sin is more elementary than gamma? Historical heritage of confusion, I guess.

Consider this related case: if you evaluate a limit and you get 0/0, you recognize that you need to do more work to find the actual limit. It could be 1, -1, 0, infinite, etc. depending on how you reached it (say, sin(x)/x versus sin(x)/x^2). The issue is not the continuity of x/x; the issue is whether setting a convention for 0/0 would give you the right value for a limit. Since it doesn't always, we call it "indeterminate".

Similarly, if you're evaluating a limit and you get 0^0 you need to do more work. You can't just stop and say "oh, that's 1". It depends on what function you used to get there -- x^x will give you a different answer from ( e^(-1/x) )^x. Again, it has nothing to do with the continuity of exponentiation. The issue is whether the convention of 0^0=1 is correct in the specific part of mathematics you're working in.

The same argument can be made if you're working in the hyperreals, or if you're working with field axioms -- the convention 0^0 doesn't work in that context.

Please, by all means, use the convention 0^0=1 when it's appropriate. But understand that it's not always appropriate. Not every mathematician works in the particular subschool that you do; not every mathematician is going to find your convention appropriate.

Consider this related case: if you evaluate a limit and you get 0/0

What do you mean by "getting 0/0" in the process of evaluating limits?

The issue is not the continuity of x/x; the issue is whether setting a convention for 0/0 would give you the right value for a limit.

Please, tell me - what is the relation between lim f(a_n) and f(lim a_n) ?

> "why don't we just agree that 0^0 = 1"

Because sometimes it's better not to. Sometimes it's inconsistent with our definitions.

Just like sometimes we agree that you can't divide by zero, and sometimes we agree that you can. Sometimes infinity is an actual value (say, in the extended reals), and sometimes it's just a symbol for "unbounded". Sometimes we agree that you can't take the square root of a negative number, and sometimes you can. Sometimes we use the axiom of choice, and sometimes we don't (and you can have an awful lot of fun either way!)

Mathematics is contextual. How various operations behave depends on which axioms and conventions are being used.

> Because sometimes it's better not to. Sometimes it's inconsistent with our definitions.

I'd love to see even one example of 0^0=1 being inconsistent with a definition. The closest I've ever seen is that it bothers people that for reasons of their own had their hearts set on (x,y) -> x^y having no discontinuities...

Fair question.

Perhaps it's more precise to say "Because sometimes it's better not to. Sometimes there is no canonical choice that follows from our definitions, and it doesn't help to assign an arbitrary value that doesn't help solve any related problems."

What is "x" equal to? In general, I mean, not in the context of any equation like "x+1=2". You could say "x=7 in the study of free variables over integers when no other constraints are given", and that is completely consistent with the rest of mathematics, and yet would not be particularly useful and introduces an ugly (philosophical weasel word, yes) asymmetry in the theory (I'd say it introduces a gauge invariance (https://secure.wikimedia.org/wikipedia/en/wiki/Gauge_theory), but I'm really not qualified to discuss that in a rigorous way.)

Someone like Scott Aaronson could put this claim on more solid footing, but I would state that, intuitively, "assigning a value to an indeterminate form leads to a more complex definition of a mathematical system" in some formal complexity-theory sense.

I agree with you and lotharbot that if you view things from a continuous perspective, 0^0 is indeterminate.

Thinking back on my math education, part of the difference in viewpoint may be the first time I rigorously met the continuous-domain exponential.

This was in real analysis. Exponentiation is defined first for positive integer exponents, and then for rational exponents. All elementary. Then it's extended to real-valued exponents by taking the limits of rational numbers, and appealing to continuity.

I just looked, this is exercise 6 in chapter 1 of baby Rudin.

So, because the notion of limit and continuity is embedded in this definition of the exponential function, it's natural to "approach" (groan) 0^0 as a special case, because the conditions of this definition (continuity) don't hold.

> x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.

Technically, 0^y is 0 only if you approach it from the right: y>0. To the left of 0 it is indeterminate or infinity depending on how you look at it. x^0, however, makes sense for all x and is always 1

Yes. Tying to define f(0) as 0 or 1 won't make it continuous, as approaching from the lines x = 0 and y = 0 will make the limits differ (the definition of lack of a limit).