29 ms·

0^0

506 points by hansy 13y ago

mcweaksauce 13y ago
This is ridiculous. Doesn't this seem counter-intuitive for it to be anything else besides 0 or undefined?
- deleted 13y ago
  [deleted]
- chasing 13y ago
  Math and science are built upon many ideas that are both counter-intuitive and true.
  - cgs1019 13y ago
    "A deep truth is one whose opposite is also a deep truth" -Niels Bohr
- JelteF 13y ago
  Please read the article before posting a question that is thoroughly explained.
- baddox 13y ago
  If we don't have a specific mathematical context, then saying it's undefined is intuitive to me. Without context, 0 is no more intuitive to me than 1. These two statements are equally intuitive to me, but they give different results for 0^0: "Zero raised to any power is still just zero." "Any number raised to the zeroth power is one."
  - dllthomas 13y ago
    For me, intuition-wise, I'd order it "undefined, 1, 0". There are 3 cases for 1 and one case for 0 that immediately spring to my mind when considering the problem: 0) Limit of 0^x, as x approaches 0 (from above). 1a) Limit of x^0 as x approaches 0 (from either direction). 1b) Limit of x^x as x approaches 0 (from above). 1c) "What did you multiply by 3 once, to get 3^1? So, multiplying 1 by zero, zero times..." Limits here, simplified to intuition level, being "what would you need to fill that hole in the graph?" The fact that these disagree would be why I'd assume undefined, but the case for 1 seems stronger (to me).
    - baddox 13y ago
      The limit in 1b is 1 from below as well, right? I'm not sure how limits work with complex numbers, but the imaginary part of x^x approaches zero as x approaches zero from below, so can we say that the limit of x^x as x approaches zero from below is also zero?
      - dllthomas 13y ago
        Yes, but getting there steps out of the realm of "intuition" for me.
        baddox 13y ago
        Yes, perhaps "intuition" isn't the best word. Formal limits certainly aren't "intuitive" to me, at least by one definition of the word. I suppose I used "intuitive" to mean "according to my mathematical understanding, ignoring the mathematics explicitly dealing with 0^0."
        dllthomas 13y ago
        Sure. I was limiting it to things I could do in my head in tens of seconds.
  - lmm 13y ago
    0^2 and 0^1 may be zero, but 0^-1 and 0^-2 are infinite. So I don't think that intuition leads to 0^0=0. If anything 0^0=1 preserves the symmetry.
- nostrademons 13y ago
  The article is making a very important point: Even for things as "objective" as mathematics, definitions ultimately come down to what makes it convenient to manipulate symbols like an expert would. This comes up all the time in fields as diverse as software engineering, law, finance, business, even hard sciences like physics - things are "right" because they're convenient, and because the consequences of them being that way make it possible to build on those results with new constructs, while doing it a different, more intuitive way would result in those constructs being impossible. What the article is saying is that your intuition for it being 0 or undefined is because you've been exposed only to exponentiation as repeated multiplication or as the limit of some series; if you consider other theorems like the binomial theorem, and figure out what is necessary for them to hold without special casing, you'll decide otherwise.
- Bahamut 13y ago
  Not really - the explanation from the "mathematician" perspective gives some of the rationale, and it makes perfect sense & in line with intuition. Defining 0^0 as anything but 1 is weird since x^0 is 1 given the other definitions we have adopted, i.e. x^n = x * x * ... * x (n times) for positive integers, (x^n)^(-1) being defined as the unique number such that x^n * (x^n)^(-1) = 1, (x^n)^(-1) * x^n = 1 for non-zero numbers x, and the simple result that (x^n)^(-1) = (x^(-1))^n from proof by induction & the uniqueness condition of the multiplicative inverse, we then get the natural formula that 1 = x^n * x^(-n) = x^(n - n) = x^0 for non-zero numbers. Defining 0^0 = 0 or undefined doesn't agree with the formula given for all non-zero real numbers, and goes against the limit of x^x as x approaches 0 (as detailed in the article), making x^x a discontinuous function at x = 0 if it is defined as another value.
myhf 13y ago
For the continuous use cases, it helps to look at a graph: http://www.wolframalpha.com/input/?i=y%3Dx%5Ex http://www.wolframalpha.com/input/?i=y%3Dx%5Ex
quchen 13y ago
That's a rather long text to say "it's an arbitrary -- and conveniently chosen -- definition of a special case of the power function similar to how 1 is not prime". I also think the presentation was chosen poorly: lots of wrong information before the correct approach is presented. Wikipedia is probably a better source here: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer...
- ars 13y ago
  In no way was wikipedia better, it was confusing and basically incomprehensible unless you already understood the material. His presentation was much better because it leads you to the answer instead of dropping it on you from the sky. (The difference between learning by rote vs learning by understanding.)
  - Bahamut 13y ago
    Amusingly, Wikipedia cites this very webpage from the OP.
- forrestthewoods 13y ago
  I love Wikipedia. It's amazing. It makes the world a better place. I'm a pretty decent programmer. I do video games so I do lots of 3d math. I'd say I'm decent at that as well. I hate Wikipedia for math. Absolutely hate it. Unless you are a mathematician by trade Wikipedia is damn near useless for learning new math concepts. I don't even bother checking it anymore.
  - vlasev 13y ago
    I had an idea for a couple of years now of creating a wikipedia-style mathematics textbook that will be crowd-sourced, standardized and cover all of math in a way that's accessible to learn from on your own. It would have a kind of a zoom function where you can expand details on explanations and calculations to a depth that you prefer. Ideally this kind of thing would start off with basic math and get progressively further into mathematics like the roots of a tree.
gweinberg 13y ago
Missing Q and A: But if mathematicians insist it is 1, why do high school teachers act like they know more than the mathematicians do? A: They don't. The statement that mathematicians uniformly say it is 1 is simply false. My high school teacher had a PhD in math, I think it's fair to say she was a mathematician. And yes, she said it was undefined.
- joliv 13y ago
  I think it might be better to say it's indeterminant.
  - Sniffnoy 13y ago
    No, because there's no such thing. Either we can define 0^0 to have a value, or we can not do so. Either way, there's no separate option "indeterminate" that is different from "undefined". Now, it's common to teach in calculus classes that 0^0 is one of the "indeterminate forms", along with 0/0 and so forth, which, so the story goes, is a different thing from being undefined, like 1/0. Since, after all, if f(x) approaches 1 and g(x) approaches 0, then the limit of f(x)/g(x) is undefined, whereas if f(x) and g(x) both approach 0, then the limit of f(x)/g(x) cannot be predicted in advance. So 1/0 is undefined, but 0/0 is indeterminate. Now this certainly is getting at a real distinction! But it's not a distinction between the value of 1/0 and that of 0/0; both are undefined. "Indeterminate" is not some separate actual value. Rather, they are getting at the distinction of the behavior of the division function near the point (1,0) vs. how it behaves near the point (0,0). Not at the points! At both those points, the function is not defined. And similarly with 0^0. Of course, it's pretty common to define that 0^0=1, and it's a definition I'd agree with -- but this is not inconsistent with the calculus teacher's statement that 0^0 is "indeterminate", because the latter (once made sense of) is not really a statement about the value of 0^0 at all; it's a statement about how the exponentiation function behaves near the point (0,0) (not at it; at it, it's equal to 1, or at least by my definition it is, at any rate). In short, there's no such value as "indeterminate"; the calculus teacher's "indeterminate forms" (as opposed to "undefined"), while getting at a real destinction, is not actually about the value of the function at the point at all.
  - 13y ago
didgeoridoo 13y ago
Students: Let's come up with some crazy proofs based on our individual levels of understanding. Teachers: Let's do it by the book and come up (somehow) with conflicting answers. Mathematicians: Yeah, sorry guys. We made it all up. Pretty much captures most mathematicians I know.
- Bahamut 13y ago
  Definitions in mathematics are used to make the language describing abstract concepts elegant, as explained in the entry. However, definitions aren't chosen all willy-nilly - there are good arguments why definitions are adopted, as should have been seen in the article.
  - kazagistar 13y ago
    Sort of like how we use terse symbols for everything because ink and papyrus is expensive.
    - na85 13y ago
      And because if you're Isaac Newton and you're writing pages and pages of proofs, you naturally want to use a shorthand in order to save yourself time, energy, and from repetitive strain injury.
    - marcosdumay 13y ago
      Not really. We have some very good reasons for using the definitions we use every day. Those reasons are more related to us than our technology, although our tech is getting so good that we already need to think about it. And by the way, people changed almost all the usual definitions at the XX century. Now, mathematicians do throw the usual definitions away all the time. It's important to know when to reuse other people's coding, or roll your own.
  - masterzora 13y ago
    My favourite algebra professor always said something to the effect of "we always have to be careful about abusing notation but if we make good choices of notation we can abuse it right." Definitions are sort of the same deal.
im3w1l 13y ago
It also plays nice with the convention of 0log 0=0, used in for example formulas for entropy.
- vlasev 13y ago
  It's perhaps more than a convention because this is also the limit that the function x log(x) naturally takes as x approaches 0 in the domain of the logarithm. Well, this is just a rehash of what the "cleverest" student does anyway.
- olalonde 13y ago
  Why? With the x^0=1 definition, wouldn't log 0 be undefined?
  - j2kun 13y ago
    0 log 0 = log (0^0) = log (1) = 0
ryannevius 13y ago
I understand the "math"...the numbers...the work on paper. But how does that translate to something useful in the real world? That, after all, is what useful math helps us do...solve problems for the real, tangible world. Saying that 0^0 = 1 is a cool math game; but translate 0 into something in the real world (i.e. nothing, none, etc.)...and trying to make something out of it other than 0 or "indeterminate" starts to make less sense.
- bhrgunatha 13y ago
  > But how does that translate to something useful in the real world? I think this is a by-product of the way we are taught maths at the very start; that it must somehow relate to real things. We start our understanding of maths by using real world objects like apples and we show how addition works and subtraction. I think perhaps it sticks in our head that everything must somehow relate to real objects and the real world. We somehow get past that when we are introduced to things like square roots and integrals and higher mathematical concepts, but even with those we often try and relate them back to the real world. I wonder of there are other ways to start teaching maths that doesn't start by using balls or apples? How would that work?
  - ryannevius 13y ago
    Square roots and integrals both relate back to the real world...
    - bhrgunatha 13y ago
      Perhaps not really great examples - I mean you can relate to the length of a diagonal line in a square room, but you can't really relate to the square root of three apples, can you? I'm not trying to say that those example don't relate to the physical world. My point is we abstract more and more away from the "real world" until you get something like this post. What about complex numbers - the square root of -1, incredibly, is useful in electronics and other real physical systems, but you cannot really relate it to physical objects - or at least not obviously. Or more incredibly something like Banach Tarski [1]. [1] https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox
Eleutheria 13y ago
Whenever you try to cross-breed an apple tree with a donkey, you get nonsense. But for convenience, whatever the offspring is, we may call it a donkapple. That's they beauty of math.
- noonespecial 13y ago
  The utility of math is that you can then do a bunch of calculations with donkapples and at the end, get a meaningful real-world answer.
  - LanceH 13y ago
    Well meaningful as soon as someone finds a donkapple.
    - dagw 13y ago
      Not at all. Complex numbers are used to solve real world engineering problems, despite the fact that you cannot "find" any complex numbers in nature.
      - JonnieCache 13y ago
        I thought quantum mechanics was riddled with complex numbers? What is it about QM which means we're "using" complex numbers without having "found" them in nature? (I realise that this is kindof a horrible question to try and answer on an internet forum, please try your best physicists ;)
azinman2 13y ago
Another good reminder on how math itself is arbitrary and made up by humans (often for what's simplest/easiest), and not handed down to us by God. Luckily it's an extremely useful and extendable made up system. I see this all the time with AI/machine learning. Most algorithms are based on assumptions that make the math work out better rather than being aligned with some "fundamental truth." The world is not linear, but it's much easier to approach it as if it were!
- dinkumthinkum 13y ago
  I don't think math is arbitrary at all.
  - lovamova 13y ago
    It is. We invented the arabic numerals because they were easy to draw and we could written any numbers with them. Just like we invented higher lever computer languages instead of using assembly. See what Fibonacci used to say in his first book Liber Abaci about using arabic numerals. http://en.wikipedia.org/wiki/Liber_Abaci http://en.wikipedia.org/wiki/Liber_Abaci
    - mason55 13y ago
      The symbols we use to represent math are arbitrary but that doesn't mean the rules behind them are. Many concepts in math are fundamental truths.
      - chunky1994 13y ago
        What exactly do you mean by fundamental truths here? Mathematics is an internally consistent (for the most part) logical framework that is extremely powerful in expressing our knowledge about the world. However, that doesn't mean that there is some intrinsic correctness about it or its concepts.
        mason55 13y ago
calcsam 13y ago
0 ^ any positive power = 0 0 ^ any negative power = 1/0 = undefined = +- inf So strictly only the right limit as n --> 0 of 0^n = 0. Not the limit.
- Bahamut 13y ago
  0 is not defined as 0 * x = 0 - that is a derived formula. 0 is defined as the additive identity, i.e. x + 0 = x = 0 + x. It is a unique number. 0 * x = 0 is proven by noting that 0 * x = (0 + 0) * x = 0 * x + 0 * x (distributive property) and then subtracting the additive inverse of 0 * x from both sides to get that 0 = 0 * x. However, this says nothing about 0^0, and one cannot talk about 0^(-n) for natural number n since 0 has no multiplicative inverse. Also your argument on limits is not correct - you chose a particular path of approach for the expression y^x fixed along y = 0. Looking at another angle, the limit of x^x as x approaches 0 is clearly 1 as reasoned in the article, so this causes a clear disagreement here since you can argue for different values to make sense by tweaking the path of approach of the two dimensional function y^x appropriately.
Tyr42 13y ago
I feel like in discrete mathematics (especially combinatorics), since we don't use continuous functions, it's useful to say 0^0 is 1, along with 0! = 1, and so on. Makes a lot of things around Binomial theorem and the like easier. I'm not so sure if it's safe to use that when doing and calculus proofs or anything along those lines, but there, you have more useful tools for dealing with limits that might approach 0^0.
- anonymoushn 13y ago
  0! really is 1 in a much more reasonable sense. The empty product is the multiplicative identity. The continuous notion of ! also agrees: http://en.wikipedia.org/wiki/Gamma_function http://en.wikipedia.org/wiki/Gamma_function
  - gordaco 13y ago
    And it's far easier to prove: 1!=1, (n+1)!=n!·(n+1), thus 0!=1!/1=1.
yaks_hairbrush 13y ago
A word from Knuth on the matter (warning: PDF): http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf See page 6.
- davyjones 13y ago
  Are PDF warnings relevant anymore? Of late Chromium and Firefox display PDF natively, sandboxed(?).
  - RexRollman 13y ago
    I expect it has more to do with filesize.
    - Dylan16807 13y ago
      Nobody says 'warning, images' and most pdfs are pretty lightweight. This one's 200KB.
  - waqf 13y ago
    On Android, both Chrome and Firefox still open an external tool and ruin your back button.
  - dragonwriter 13y ago
    There are browsers that are not the desktop versions of Chrom(e/ium) or Firefox, and which are widely used.
- Sniffnoy 13y ago
  Why link to the PDF? Better to just link to the abstract: http://arxiv.org/abs/math/9205211 http://arxiv.org/abs/math/9205211 Edit: To be clear, this is good practice when linking to arXiv in general. From the abstract, one can easily click through to the PDF; not so the reverse. And the abstract allows one to do things like see different versions of the paper, search for other things by the same authors, etc.
- jacobolus 13y ago
elwell 13y ago
In any case, zero is a different animal than any other number. It can't actually exist in the physical world like other numbers; it's definition is non-existence. Zero is merely conceptual. Therefore, it can't easily fit into the picture of mathematics; at least not without requiring the definition of special cases.
SilasX 13y ago
- "0^0. Why? Because mathematicians said so. No really, it’s true." - [Detailed explanation of the tradeoffs involved in choosing different definitions of exponentiation.] So, it's not "because mathematicians said so", it's because of a deep review of the tradeoffs of defining how exponentiation generalizes, the kind of thing that mathematicians happen to study more than other identifiable groups.
- glhaynes 13y ago
  That's why we trust mathematicians' say-so on the topic. :) (Well, and because they can justify it, of course.)
- dhaivatpandya 13y ago
  It is still a "said so" because it is a definition. Even if thought went into the definition, it doesn't make any less of a definition.
vlasev 13y ago
This is exactly why you have things like 0! = 1, 0 choose 0 = 1, 0^0 = 1, the empty sum is 0 and the empty product is 1, and so on. These are DEFINITIONS and they make notation easier. They basically help the flow of mathematics. It's often difficult to watch someone try to explain "intuitively" why some of these things are the way they are and completely miss the point that they are like this because they help make other things easier.
Oculus 13y ago
A pretty interesting question(the value of 0^0), in a very understandable format for the layman (me). Very nice blog post!
analog31 13y ago
Perhaps a related question: How should it be defined in a math library for a programming language? Should it return 1, or throw an exception?
- biot 13y ago
  Why is that even a question? It should return the correct answer, of course. Edit: At the very least, that behaviour should be a configurable option for those who desire something other than what most mathematicians accept as being the correct answer.
  - my_username_is_ 13y ago
    The whole point of this is that there isn't necessarily a single right answer. As another user pointed out, mathematicians define 0^0 to be 1, but you don't necessarily have to accept that as truth in the way that 1+1=2.
    - xamuel 13y ago
      If the calculator is powerful enough (as in the case of Mathematica) to evaluate Taylor series, for example the Maclaurin series for e^x when x=0, then in doing so it implicitly admits 0^0=1. If it simultaneously says 0^0 is not 1, then the calculator is inconsistent. (Mathematica IS inconsistent in this example)
- kijin 13y ago
  In JavaScript, Math.pow(0, 0) == 1. In Python and Ruby, 0 asterisk asterisk 0 == 1. In Windows Calculator, 0 [x^y] 0 == 1. So I guess the consensus is that you should return 1.
- krilnon 13y ago
  Mathematica (or do we call it Wolfram Language now?) evaluates 0^0 as Indeterminate and warns: `Power::indet: "Indeterminate expression 0^0 encountered."`
- stephencanon 13y ago
haddr 13y ago
I think that the most intuitive way to get the idea of why 0^0 = 1 is to take the example from combinatorics. n^k is the number of distinct sequences for the sampling with replacement and ordering (for example ball picking from repository of n different balls and counting the number of distinct ways that k balls can be picked and ordered - with replacement). I think that there is only one way of ordering results of drawing zero balls from a set of 0 different balls :)
skrowyek 13y ago
While high schoolers try to prove their own intuitions about their understanding of exponents (intuition drilled into them through rote learning), mathematicians just say "we defined it that way". It speaks to the tragedy that is the high school math curriculum.
gameshot911 13y ago
I have a problem with this part of the explanation: >However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving y^{0} = 1, which holds for any y. Hence, when y is zero, we have 0^0 = 1. When y is zero, don't we have 0^0 = 1 x [y zero times]? Maybe I'm conceptualizing it incorrectly, but I'm envisioning an empty space where y would be, akin to an empty set. 1 times an 'empty set' is _not_ 1, it's one empty set, which rubs me as another way of saying nothing/zero, not 1. I know my language is imprecise and I'm probably describing empty sets and the definition of zero incorrectly. The point is that the last step of his explanation doesn't sit right with me. Just because Y exists zero times does not mean you can just throw it out of the multiplication.
- j2kun 13y ago
  Another convention is that the empty product is 1, and this is precisely what is being said by putting the 1 out front.
anonymoushn 13y ago
I did not encounter this convention while working on my math degree. I am surprised that none of the characters in the article said "0^0 is nothing, but limits of the form 0^0 can be any nonnegative real number or infinite".
- yaks_hairbrush 13y ago
  Yes, you did. It may not have been stated explicitly, but it was implied in a couple places: 1. Binomial theorem. Its statement does not conventionally include any caveats about the x^0 case. 2. D(x^n) = n x^(n-1). When n = 1 and x = 0, you get a 0^0 on the RHS. If 0^0 is taken to be 1, the derivative rule holds. Else, the statement gets messy: D(x^n) = n x^(n-1) if (x,n) != (0,1)
nemasu 13y ago
I don't get what the big deal is. I put 0^0 in Windows Calculator and it gives me 1 ... problem solved!
deleted 13y ago
[deleted]
stephencanon 13y ago
The real problem here is that x^y is a single shorthand which refers to a few fundamentally different mathematical concepts (which happen to have significant overlap with each other). First, it refers to a function f:C x N --> C, defined in terms of repeated multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention that f(0,0) is also 1. But it also refers to a function g:C x C --> C, defined as g(x,y) = exp(y log(x)), which has a branch cut on the negative real axis of the first argument and an essential singularity at (0,0), and so g(0,0) is necessarily undefined. The value of 0^0 depends entirely on what sort of mathematics one is doing at the time, and therefore which function one is referring to.
- Sniffnoy 13y ago
  But if you define 0^0=1 in general, it doesn't cause a problem here -- that definition never disagrees with x^y=exp(ylog(x)), it just defines it at the point 0^0, while the latter leaves it undefined. In other words, it's possible to make a common extension of the two; they don't actually give different values in any case. Of course, doing this makes exponentiation discontinuous at (0,0), but seeing as it already had an essential singularity there, this isn't really a loss.
  - stephencanon 13y ago
    Right, but when you're working with the function on C x C, it doesn't really add anything either to arbitrarily pick 0^0 = 1, except for consistency with that other function that happens to be written the same way. I'm not really opposed to saying 0^0 = 1; it's the only reasonable choice if we're going to insist on using the same notation for these two functions, and I don't expect that's going to change.
- brandonhsiao 13y ago
  I'm not a mathematician, but this seems the most reasonable/believable answer I've heard. The last sentence especially.
- rntz 13y ago
  In fact, exponentiation f : C x N -> C has a natural generalization to any monoid C, where f(x,n) is x "multiplied" by itself using the monoid operator n times. In this setting, the only sensible choice is that x^0 = 1 for any x, where 1 is the unit of the monoid in question. In particular, that is the only definition of exponentiation which is parametric in our choice of monoid. Naive Haskell example code (using Int for Nat): f :: Monoid m => (m,Int) -> m f (x,0) = mempty f (x,n) = x `mappend` f (x,n-1) -- or equivalently f (x,n) = msum (replicate n x)
textminer 13y ago
Had a set theory professor who taught us that for the non-negative integers, m^n was just the number of unique mappings from a set of cardinality n to one of cardinality m. Ergo, for all sets A such that |A| = k, k^0 is just all mappings from Ø, which is necessarily the one with empty image and pre-image. So 0^0 = 1.
EGreg 13y ago
Mathematics is about generalizing concepts and principles to ever larger domains. In this case, x^y is defined for all pairs (x, y) of real numbers except (0, 0). The question is what limit is "closer" to the set of outputs in the neighborhood of (0, 0) than any other. 0^x is defined for all x except 0, and same for x^0. We can define 0^0 as the limit of one or the other as x goes to 0. and one is constant and more "stable" than the other, so it is typically taken to be that, i.e. 1. Up next ... if P(X) = false, what is "P(X) for all X in Ø"?
- cbright 13y ago
  Correction: x^y is defined for all pairs of real numbers with x!=0. 0^x is not defined for negative x.
socrates1998 13y ago
I love math, but I have never really enjoyed these types of debates. I guess I have always been drawn to the application of the concepts in the real world rather than the abstract beauty of it.
rickhanlonii 13y ago
It's very important to note here that 0^0=1 is a shorthand and not a truth. Mathematicians are absolutely not stating that they have proven, or that it is true, that 0^0=1. It is a definition, not a claim of equality. They're not saying "0^0 is 1" in the sense that they say "1+1 is 2" or "0.999... is 1". They're saying "we define 0^0 to be 1". The difference is more than just pedantry, it strikes at the core of why mathematics is the most powerful tool for determining truth that humanity has either discovered or invented (hat tip to anyone familiar with that debate). One of the cold, austere, beauties of mathematics* is that if you do not accept a definition, you can reject it as false and reason with the result. To give the traditional example, if you accept as true that two parallel lines never intersect, then along with the other 4 of Euclid's Axioms you can prove all of Euclidian Geometry (what you learn in high school). If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry (one form of non-Euclidian Geometry). In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result; you will not break any mathematics. But you should know that mathematicians have never run into any issues with this convention--or can handle it trivially when they arise--so you're only adding a lot of work for yourself. When you really think about it, it fills you with awe. It's awesome. *“Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.” -Bertrand Russell
- philbarr 13y ago
  Can anyone recommend a really good book on mathematics that demonstrates the "beauty" of it? I'd love to really learn and understand about Fast Fourier Transforms and the like, but any book I have is from my old college days and is just so bloody tedious. I used to really love maths when I was younger but it got beaten out of me by endless repetition and "now do Questions 1 - 50" tedium. Now I get scared whenever I see the "summation" symbol.
  - coldpie 13y ago
    I found abstract algebra to be really interesting. It kind of removes the concept of "numbers" and transitions it to operations on objects like sets or functions. You redefine operations like "addition" and "multiplication" to apply to these new objects, and see what happens. A whole bunch of interesting things fall out, many of them important to information encoding and error correction in practice. For an example, see the concept of a Ring: http://en.wikipedia.org/wiki/Ring_%28mathematics%29 http://en.wikipedia.org/wiki/Ring_%28mathematics%29 Less directly related to math, I spent a joyous couple of hours back in college working through the Lambda Calculus wiki page, trying to every piece it. Once it "clicked" it became really fun to work through the examples. http://en.wikipedia.org/wiki/Lambda_calculus http://en.wikipedia.org/wiki/Lambda_calculus > Now I get scared whenever I see the "summation" symbol. I'm not sure why, but I guess for me, removing the idea of "numbers" makes math suddenly fascinating. It's like dropping down a level to redefine what you thought was a well-defined concept (addition of numbers) as a generic operation for which numbers are simply one special case. Weird. :)
rkowalick 13y ago
My favorite explanation: y^x is the number of functions from a set with x elements to a set with y elements. Since there is only one function from the empty set to the empty set, namely, the empty function, we get that 0^0 = 1.
blackrabbit 13y ago
Can anyone explain why this is a valid operation? http://wp.com/latex.php?latex=\lim_{x%20\to%200^{+}}%20x^{x}%20%20=%20\lim_{x%20\to%200^{+}}%20\exp(\log(x^{x}))&bg=ffffff&fg=000&s=0 http://wp.com/latex.php?latex=\lim_{x%20\to%200^{+}}%20x^{x}...
- my_username_is_ 13y ago
  it's using log to mean log base e, also known as the natural log or ln. you can do that because they're inverse operations (I think that's the right term for it, it's been a while since a formal math class) so they undo each other
  - clebio 13y ago
    Yes. It's an identity relationship. x = exp(log(x)) http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Canceling_exponentials http://en.wikipedia.org/wiki/List_of_logarithmic_identities#...
  - blackrabbit 13y ago
    Thanks!
deleted 13y ago
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olalonde 13y ago
Are there any examples where the x^0=1 definition turns out to make other definitions more complicated to write down? For example, you'd have a general definition and need a special definition for when you get an exponent that equals 0.
- xamuel 13y ago
  Suppose you want a compact, clever formula for the function: f(x)=1 if x!=0, f(x)=0 if x=0 If 0^0 were defined as 0, you could write the above function as f(x)=x^0. With 0^0 defined as 1, you're forced to write something like f(x)=1-0^|x| (absolute values to avoid division by zero), a bit more complicated. This is silly though, and of no importance anywhere.
Totient 13y ago
"...they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians." Along these lines, my preferred definition of the natural numbers" is "the set of all the positive integers and, when convenient, 0".
tel 13y ago
This question doesn't have an answer because, as this post suggests, the meanings of the first 0, the second 0, and the (^) are overloaded. The reason mathematicians don't mind the overloading is that they all have "kind of" similar meanings. So we can talk about a recursively defined function on naturals where it's convenient to "define" 0^0 = 1. We can talk about various kinds of limit techniques in the reals where 0^0 is not actually a value but instead a shorthand for a particular kind of limit. This probably forms the most interesting notion of what 0^0 means where we define it as the limit of a particular kind of path in the complex plane and then notice that 0^0 can take any value we choose---depending on exactly what path was taken to get there. [0] [0] http://en.wikipedia.org/wiki/Complex_logarithm http://en.wikipedia.org/wiki/Complex_logarithm
deepakjc 13y ago
I read upto "High School Teacher"... skimmed through "Calculus Teacher"... and scrolled through "Mathematician"...
bumbledraven 13y ago
Although standard mathematica notation glosses over the difference, it's important to distinguish between the following two versions of the power operation, which I'll call pow1 and pow2: pow1(x, y) = x^y where y is any integer pow2(x, y) = x^y where y is any real The value of 0^0 is 1 for (pow-a) and 0 for (pow-b).
- SamReidHughes 13y ago
  But lim x->0 (x^x) = 1. Why should 0^0 be 0 then?
emanuelsaringan 13y ago
Last time I checked, anything XOR'd with itself is 0. Kidding aside, 0 raised to 0 was said to be indeterminate according to my teachers.
t__r 13y ago
To me the matter looks similar to taking the conjunction of the empty set, which is defined to be TRUE. This, however, starts to make sense when you see that the taking the conjunction of a set X can be interpreted as "forall x in X, x is TRUE". If X is empty then this statement is trivially true.
bakhy 13y ago
best thing i read in quite a while!
vog 13y ago
I find it a bit disappointing that the article considers only limits that would justify "0^0 = 0" and "0^0 = 1". In fact, the termin "0^0" as a limit can reach any value - the same way as "0/0" can reach any value. Im stressing this because back in school, I really thought that although 0^0 is undetermined, it can only reach exactly 0 or 1, but nothing else. This is of course wrong, but no teacher was able to tell me why. For some time I even thought I found a new theorem and tried to prove it. Later, I told some math guru about this, he thought about a minute, and told me two functions f(x) and g(x) whose limit is each 0, but for which the limit of f(x)/g(x) is 2 (or any other value, if you adjust f(x) and g(x) accordingly). Having said that, in most cases "0^0 = 1" is a useful convention, especially in a purely algebraic context when polynomials are involved.
mxms 13y ago
Interesting article. I ran across a problem yesterday that was similar to P(x)^Q(x) = 1, which then asked me to find the sum of the solutions. I noticed that both P(x) and Q(x) share a root at some a. But I realized, 0^0 is most often defined as 1, and carried on. Checking the answer key later on showed that they chose to neglect that a, and call 0^0 undefined. I'm not sure how I really felt about it. Note, I also forgot to check when P(x) = -1, assuming that Q(x) is even there ;P

0^0

506 points by hansy 13y ago

mcweaksauce 13y ago

This is ridiculous. Doesn't this seem counter-intuitive for it to be anything else besides 0 or undefined?

deleted 13y ago
[deleted]
chasing 13y ago
Math and science are built upon many ideas that are both counter-intuitive and true.
- cgs1019 13y ago
  "A deep truth is one whose opposite is also a deep truth" -Niels Bohr
JelteF 13y ago
Please read the article before posting a question that is thoroughly explained.
baddox 13y ago
If we don't have a specific mathematical context, then saying it's undefined is intuitive to me. Without context, 0 is no more intuitive to me than 1. These two statements are equally intuitive to me, but they give different results for 0^0: "Zero raised to any power is still just zero." "Any number raised to the zeroth power is one."
- dllthomas 13y ago
  For me, intuition-wise, I'd order it "undefined, 1, 0". There are 3 cases for 1 and one case for 0 that immediately spring to my mind when considering the problem: 0) Limit of 0^x, as x approaches 0 (from above). 1a) Limit of x^0 as x approaches 0 (from either direction). 1b) Limit of x^x as x approaches 0 (from above). 1c) "What did you multiply by 3 once, to get 3^1? So, multiplying 1 by zero, zero times..." Limits here, simplified to intuition level, being "what would you need to fill that hole in the graph?" The fact that these disagree would be why I'd assume undefined, but the case for 1 seems stronger (to me).
  - baddox 13y ago
    The limit in 1b is 1 from below as well, right? I'm not sure how limits work with complex numbers, but the imaginary part of x^x approaches zero as x approaches zero from below, so can we say that the limit of x^x as x approaches zero from below is also zero?
    - dllthomas 13y ago
      Yes, but getting there steps out of the realm of "intuition" for me.
      - baddox 13y ago
        Yes, perhaps "intuition" isn't the best word. Formal limits certainly aren't "intuitive" to me, at least by one definition of the word. I suppose I used "intuitive" to mean "according to my mathematical understanding, ignoring the mathematics explicitly dealing with 0^0."
        dllthomas 13y ago
        Sure. I was limiting it to things I could do in my head in tens of seconds.
- lmm 13y ago
  0^2 and 0^1 may be zero, but 0^-1 and 0^-2 are infinite. So I don't think that intuition leads to 0^0=0. If anything 0^0=1 preserves the symmetry.
nostrademons 13y ago
The article is making a very important point: Even for things as "objective" as mathematics, definitions ultimately come down to what makes it convenient to manipulate symbols like an expert would. This comes up all the time in fields as diverse as software engineering, law, finance, business, even hard sciences like physics - things are "right" because they're convenient, and because the consequences of them being that way make it possible to build on those results with new constructs, while doing it a different, more intuitive way would result in those constructs being impossible. What the article is saying is that your intuition for it being 0 or undefined is because you've been exposed only to exponentiation as repeated multiplication or as the limit of some series; if you consider other theorems like the binomial theorem, and figure out what is necessary for them to hold without special casing, you'll decide otherwise.
Bahamut 13y ago
Not really - the explanation from the "mathematician" perspective gives some of the rationale, and it makes perfect sense & in line with intuition. Defining 0^0 as anything but 1 is weird since x^0 is 1 given the other definitions we have adopted, i.e. x^n = x * x * ... * x (n times) for positive integers, (x^n)^(-1) being defined as the unique number such that x^n * (x^n)^(-1) = 1, (x^n)^(-1) * x^n = 1 for non-zero numbers x, and the simple result that (x^n)^(-1) = (x^(-1))^n from proof by induction & the uniqueness condition of the multiplicative inverse, we then get the natural formula that 1 = x^n * x^(-n) = x^(n - n) = x^0 for non-zero numbers. Defining 0^0 = 0 or undefined doesn't agree with the formula given for all non-zero real numbers, and goes against the limit of x^x as x approaches 0 (as detailed in the article), making x^x a discontinuous function at x = 0 if it is defined as another value.

myhf 13y ago

For the continuous use cases, it helps to look at a graph: http://www.wolframalpha.com/input/?i=y%3Dx%5Ex http://www.wolframalpha.com/input/?i=y%3Dx%5Ex

quchen 13y ago

That's a rather long text to say "it's an arbitrary -- and conveniently chosen -- definition of a special case of the power function similar to how 1 is not prime". I also think the presentation was chosen poorly: lots of wrong information before the correct approach is presented. Wikipedia is probably a better source here: https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zero https://en.wikipedia.org/wiki/0%5E0#Zero_to_the_power_of_zer...

ars 13y ago
In no way was wikipedia better, it was confusing and basically incomprehensible unless you already understood the material. His presentation was much better because it leads you to the answer instead of dropping it on you from the sky. (The difference between learning by rote vs learning by understanding.)
- Bahamut 13y ago
  Amusingly, Wikipedia cites this very webpage from the OP.
forrestthewoods 13y ago
I love Wikipedia. It's amazing. It makes the world a better place. I'm a pretty decent programmer. I do video games so I do lots of 3d math. I'd say I'm decent at that as well. I hate Wikipedia for math. Absolutely hate it. Unless you are a mathematician by trade Wikipedia is damn near useless for learning new math concepts. I don't even bother checking it anymore.
- vlasev 13y ago
  I had an idea for a couple of years now of creating a wikipedia-style mathematics textbook that will be crowd-sourced, standardized and cover all of math in a way that's accessible to learn from on your own. It would have a kind of a zoom function where you can expand details on explanations and calculations to a depth that you prefer. Ideally this kind of thing would start off with basic math and get progressively further into mathematics like the roots of a tree.

gweinberg 13y ago

Missing Q and A: But if mathematicians insist it is 1, why do high school teachers act like they know more than the mathematicians do? A: They don't. The statement that mathematicians uniformly say it is 1 is simply false. My high school teacher had a PhD in math, I think it's fair to say she was a mathematician. And yes, she said it was undefined.

joliv 13y ago
I think it might be better to say it's indeterminant.
- Sniffnoy 13y ago
  No, because there's no such thing. Either we can define 0^0 to have a value, or we can not do so. Either way, there's no separate option "indeterminate" that is different from "undefined". Now, it's common to teach in calculus classes that 0^0 is one of the "indeterminate forms", along with 0/0 and so forth, which, so the story goes, is a different thing from being undefined, like 1/0. Since, after all, if f(x) approaches 1 and g(x) approaches 0, then the limit of f(x)/g(x) is undefined, whereas if f(x) and g(x) both approach 0, then the limit of f(x)/g(x) cannot be predicted in advance. So 1/0 is undefined, but 0/0 is indeterminate. Now this certainly is getting at a real distinction! But it's not a distinction between the value of 1/0 and that of 0/0; both are undefined. "Indeterminate" is not some separate actual value. Rather, they are getting at the distinction of the behavior of the division function near the point (1,0) vs. how it behaves near the point (0,0). Not at the points! At both those points, the function is not defined. And similarly with 0^0. Of course, it's pretty common to define that 0^0=1, and it's a definition I'd agree with -- but this is not inconsistent with the calculus teacher's statement that 0^0 is "indeterminate", because the latter (once made sense of) is not really a statement about the value of 0^0 at all; it's a statement about how the exponentiation function behaves near the point (0,0) (not at it; at it, it's equal to 1, or at least by my definition it is, at any rate). In short, there's no such value as "indeterminate"; the calculus teacher's "indeterminate forms" (as opposed to "undefined"), while getting at a real destinction, is not actually about the value of the function at the point at all.
- 13y ago

didgeoridoo 13y ago

Students: Let's come up with some crazy proofs based on our individual levels of understanding. Teachers: Let's do it by the book and come up (somehow) with conflicting answers. Mathematicians: Yeah, sorry guys. We made it all up. Pretty much captures most mathematicians I know.

Bahamut 13y ago
Definitions in mathematics are used to make the language describing abstract concepts elegant, as explained in the entry. However, definitions aren't chosen all willy-nilly - there are good arguments why definitions are adopted, as should have been seen in the article.
- kazagistar 13y ago
  Sort of like how we use terse symbols for everything because ink and papyrus is expensive.
  - na85 13y ago
    And because if you're Isaac Newton and you're writing pages and pages of proofs, you naturally want to use a shorthand in order to save yourself time, energy, and from repetitive strain injury.
  - marcosdumay 13y ago
    Not really. We have some very good reasons for using the definitions we use every day. Those reasons are more related to us than our technology, although our tech is getting so good that we already need to think about it. And by the way, people changed almost all the usual definitions at the XX century. Now, mathematicians do throw the usual definitions away all the time. It's important to know when to reuse other people's coding, or roll your own.
- masterzora 13y ago
  My favourite algebra professor always said something to the effect of "we always have to be careful about abusing notation but if we make good choices of notation we can abuse it right." Definitions are sort of the same deal.

im3w1l 13y ago

It also plays nice with the convention of 0log 0=0, used in for example formulas for entropy.

vlasev 13y ago
It's perhaps more than a convention because this is also the limit that the function x log(x) naturally takes as x approaches 0 in the domain of the logarithm. Well, this is just a rehash of what the "cleverest" student does anyway.
olalonde 13y ago
Why? With the x^0=1 definition, wouldn't log 0 be undefined?
- j2kun 13y ago
  0 log 0 = log (0^0) = log (1) = 0

ryannevius 13y ago

I understand the "math"...the numbers...the work on paper. But how does that translate to something useful in the real world? That, after all, is what useful math helps us do...solve problems for the real, tangible world. Saying that 0^0 = 1 is a cool math game; but translate 0 into something in the real world (i.e. nothing, none, etc.)...and trying to make something out of it other than 0 or "indeterminate" starts to make less sense.

bhrgunatha 13y ago
> But how does that translate to something useful in the real world? I think this is a by-product of the way we are taught maths at the very start; that it must somehow relate to real things. We start our understanding of maths by using real world objects like apples and we show how addition works and subtraction. I think perhaps it sticks in our head that everything must somehow relate to real objects and the real world. We somehow get past that when we are introduced to things like square roots and integrals and higher mathematical concepts, but even with those we often try and relate them back to the real world. I wonder of there are other ways to start teaching maths that doesn't start by using balls or apples? How would that work?
- ryannevius 13y ago
  Square roots and integrals both relate back to the real world...
  - bhrgunatha 13y ago
    Perhaps not really great examples - I mean you can relate to the length of a diagonal line in a square room, but you can't really relate to the square root of three apples, can you? I'm not trying to say that those example don't relate to the physical world. My point is we abstract more and more away from the "real world" until you get something like this post. What about complex numbers - the square root of -1, incredibly, is useful in electronics and other real physical systems, but you cannot really relate it to physical objects - or at least not obviously. Or more incredibly something like Banach Tarski [1]. [1] https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox

Eleutheria 13y ago

Whenever you try to cross-breed an apple tree with a donkey, you get nonsense. But for convenience, whatever the offspring is, we may call it a donkapple. That's they beauty of math.

noonespecial 13y ago
The utility of math is that you can then do a bunch of calculations with donkapples and at the end, get a meaningful real-world answer.
- LanceH 13y ago
  Well meaningful as soon as someone finds a donkapple.
  - dagw 13y ago
    Not at all. Complex numbers are used to solve real world engineering problems, despite the fact that you cannot "find" any complex numbers in nature.
    - JonnieCache 13y ago
      I thought quantum mechanics was riddled with complex numbers? What is it about QM which means we're "using" complex numbers without having "found" them in nature? (I realise that this is kindof a horrible question to try and answer on an internet forum, please try your best physicists ;)

azinman2 13y ago

Another good reminder on how math itself is arbitrary and made up by humans (often for what's simplest/easiest), and not handed down to us by God. Luckily it's an extremely useful and extendable made up system. I see this all the time with AI/machine learning. Most algorithms are based on assumptions that make the math work out better rather than being aligned with some "fundamental truth." The world is not linear, but it's much easier to approach it as if it were!

dinkumthinkum 13y ago
I don't think math is arbitrary at all.
- lovamova 13y ago
  It is. We invented the arabic numerals because they were easy to draw and we could written any numbers with them. Just like we invented higher lever computer languages instead of using assembly. See what Fibonacci used to say in his first book Liber Abaci about using arabic numerals. http://en.wikipedia.org/wiki/Liber_Abaci http://en.wikipedia.org/wiki/Liber_Abaci
  - mason55 13y ago
    The symbols we use to represent math are arbitrary but that doesn't mean the rules behind them are. Many concepts in math are fundamental truths.
    - chunky1994 13y ago
      What exactly do you mean by fundamental truths here? Mathematics is an internally consistent (for the most part) logical framework that is extremely powerful in expressing our knowledge about the world. However, that doesn't mean that there is some intrinsic correctness about it or its concepts.
      - mason55 13y ago

calcsam 13y ago

0 ^ any positive power = 0 0 ^ any negative power = 1/0 = undefined = +- inf So strictly only the right limit as n --> 0 of 0^n = 0. Not the limit.

Bahamut 13y ago
0 is not defined as 0 * x = 0 - that is a derived formula. 0 is defined as the additive identity, i.e. x + 0 = x = 0 + x. It is a unique number. 0 * x = 0 is proven by noting that 0 * x = (0 + 0) * x = 0 * x + 0 * x (distributive property) and then subtracting the additive inverse of 0 * x from both sides to get that 0 = 0 * x. However, this says nothing about 0^0, and one cannot talk about 0^(-n) for natural number n since 0 has no multiplicative inverse. Also your argument on limits is not correct - you chose a particular path of approach for the expression y^x fixed along y = 0. Looking at another angle, the limit of x^x as x approaches 0 is clearly 1 as reasoned in the article, so this causes a clear disagreement here since you can argue for different values to make sense by tweaking the path of approach of the two dimensional function y^x appropriately.

Tyr42 13y ago

I feel like in discrete mathematics (especially combinatorics), since we don't use continuous functions, it's useful to say 0^0 is 1, along with 0! = 1, and so on. Makes a lot of things around Binomial theorem and the like easier. I'm not so sure if it's safe to use that when doing and calculus proofs or anything along those lines, but there, you have more useful tools for dealing with limits that might approach 0^0.

anonymoushn 13y ago
0! really is 1 in a much more reasonable sense. The empty product is the multiplicative identity. The continuous notion of ! also agrees: http://en.wikipedia.org/wiki/Gamma_function http://en.wikipedia.org/wiki/Gamma_function
- gordaco 13y ago
  And it's far easier to prove: 1!=1, (n+1)!=n!·(n+1), thus 0!=1!/1=1.

yaks_hairbrush 13y ago

A word from Knuth on the matter (warning: PDF): http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf See page 6.

davyjones 13y ago
Are PDF warnings relevant anymore? Of late Chromium and Firefox display PDF natively, sandboxed(?).
- RexRollman 13y ago
  I expect it has more to do with filesize.
  - Dylan16807 13y ago
    Nobody says 'warning, images' and most pdfs are pretty lightweight. This one's 200KB.
- waqf 13y ago
  On Android, both Chrome and Firefox still open an external tool and ruin your back button.
- dragonwriter 13y ago
  There are browsers that are not the desktop versions of Chrom(e/ium) or Firefox, and which are widely used.
Sniffnoy 13y ago
Why link to the PDF? Better to just link to the abstract: http://arxiv.org/abs/math/9205211 http://arxiv.org/abs/math/9205211 Edit: To be clear, this is good practice when linking to arXiv in general. From the abstract, one can easily click through to the PDF; not so the reverse. And the abstract allows one to do things like see different versions of the paper, search for other things by the same authors, etc.
jacobolus 13y ago

elwell 13y ago

In any case, zero is a different animal than any other number. It can't actually exist in the physical world like other numbers; it's definition is non-existence. Zero is merely conceptual. Therefore, it can't easily fit into the picture of mathematics; at least not without requiring the definition of special cases.

SilasX 13y ago

- "0^0. Why? Because mathematicians said so. No really, it’s true." - [Detailed explanation of the tradeoffs involved in choosing different definitions of exponentiation.] So, it's not "because mathematicians said so", it's because of a deep review of the tradeoffs of defining how exponentiation generalizes, the kind of thing that mathematicians happen to study more than other identifiable groups.

glhaynes 13y ago
That's why we trust mathematicians' say-so on the topic. :) (Well, and because they can justify it, of course.)
dhaivatpandya 13y ago
It is still a "said so" because it is a definition. Even if thought went into the definition, it doesn't make any less of a definition.

vlasev 13y ago

This is exactly why you have things like 0! = 1, 0 choose 0 = 1, 0^0 = 1, the empty sum is 0 and the empty product is 1, and so on. These are DEFINITIONS and they make notation easier. They basically help the flow of mathematics. It's often difficult to watch someone try to explain "intuitively" why some of these things are the way they are and completely miss the point that they are like this because they help make other things easier.

Oculus 13y ago

A pretty interesting question(the value of 0^0), in a very understandable format for the layman (me). Very nice blog post!

analog31 13y ago

Perhaps a related question: How should it be defined in a math library for a programming language? Should it return 1, or throw an exception?

biot 13y ago
Why is that even a question? It should return the correct answer, of course. Edit: At the very least, that behaviour should be a configurable option for those who desire something other than what most mathematicians accept as being the correct answer.
- my_username_is_ 13y ago
  The whole point of this is that there isn't necessarily a single right answer. As another user pointed out, mathematicians define 0^0 to be 1, but you don't necessarily have to accept that as truth in the way that 1+1=2.
  - xamuel 13y ago
    If the calculator is powerful enough (as in the case of Mathematica) to evaluate Taylor series, for example the Maclaurin series for e^x when x=0, then in doing so it implicitly admits 0^0=1. If it simultaneously says 0^0 is not 1, then the calculator is inconsistent. (Mathematica IS inconsistent in this example)
kijin 13y ago
In JavaScript, Math.pow(0, 0) == 1. In Python and Ruby, 0 asterisk asterisk 0 == 1. In Windows Calculator, 0 [x^y] 0 == 1. So I guess the consensus is that you should return 1.
krilnon 13y ago
Mathematica (or do we call it Wolfram Language now?) evaluates 0^0 as Indeterminate and warns: `Power::indet: "Indeterminate expression 0^0 encountered."`
stephencanon 13y ago

haddr 13y ago

I think that the most intuitive way to get the idea of why 0^0 = 1 is to take the example from combinatorics. n^k is the number of distinct sequences for the sampling with replacement and ordering (for example ball picking from repository of n different balls and counting the number of distinct ways that k balls can be picked and ordered - with replacement). I think that there is only one way of ordering results of drawing zero balls from a set of 0 different balls :)

skrowyek 13y ago

While high schoolers try to prove their own intuitions about their understanding of exponents (intuition drilled into them through rote learning), mathematicians just say "we defined it that way". It speaks to the tragedy that is the high school math curriculum.

gameshot911 13y ago

I have a problem with this part of the explanation: >However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving y^{0} = 1, which holds for any y. Hence, when y is zero, we have 0^0 = 1. When y is zero, don't we have 0^0 = 1 x [y zero times]? Maybe I'm conceptualizing it incorrectly, but I'm envisioning an empty space where y would be, akin to an empty set. 1 times an 'empty set' is _not_ 1, it's one empty set, which rubs me as another way of saying nothing/zero, not 1. I know my language is imprecise and I'm probably describing empty sets and the definition of zero incorrectly. The point is that the last step of his explanation doesn't sit right with me. Just because Y exists zero times does not mean you can just throw it out of the multiplication.

j2kun 13y ago
Another convention is that the empty product is 1, and this is precisely what is being said by putting the 1 out front.

anonymoushn 13y ago

I did not encounter this convention while working on my math degree. I am surprised that none of the characters in the article said "0^0 is nothing, but limits of the form 0^0 can be any nonnegative real number or infinite".

yaks_hairbrush 13y ago
Yes, you did. It may not have been stated explicitly, but it was implied in a couple places: 1. Binomial theorem. Its statement does not conventionally include any caveats about the x^0 case. 2. D(x^n) = n x^(n-1). When n = 1 and x = 0, you get a 0^0 on the RHS. If 0^0 is taken to be 1, the derivative rule holds. Else, the statement gets messy: D(x^n) = n x^(n-1) if (x,n) != (0,1)

nemasu 13y ago

I don't get what the big deal is. I put 0^0 in Windows Calculator and it gives me 1 ... problem solved!

deleted 13y ago

[deleted]

stephencanon 13y ago

The real problem here is that x^y is a single shorthand which refers to a few fundamentally different mathematical concepts (which happen to have significant overlap with each other). First, it refers to a function f:C x N --> C, defined in terms of repeated multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention that f(0,0) is also 1. But it also refers to a function g:C x C --> C, defined as g(x,y) = exp(y log(x)), which has a branch cut on the negative real axis of the first argument and an essential singularity at (0,0), and so g(0,0) is necessarily undefined. The value of 0^0 depends entirely on what sort of mathematics one is doing at the time, and therefore which function one is referring to.

Sniffnoy 13y ago
But if you define 0^0=1 in general, it doesn't cause a problem here -- that definition never disagrees with x^y=exp(ylog(x)), it just defines it at the point 0^0, while the latter leaves it undefined. In other words, it's possible to make a common extension of the two; they don't actually give different values in any case. Of course, doing this makes exponentiation discontinuous at (0,0), but seeing as it already had an essential singularity there, this isn't really a loss.
- stephencanon 13y ago
  Right, but when you're working with the function on C x C, it doesn't really add anything either to arbitrarily pick 0^0 = 1, except for consistency with that other function that happens to be written the same way. I'm not really opposed to saying 0^0 = 1; it's the only reasonable choice if we're going to insist on using the same notation for these two functions, and I don't expect that's going to change.
brandonhsiao 13y ago
I'm not a mathematician, but this seems the most reasonable/believable answer I've heard. The last sentence especially.
rntz 13y ago
In fact, exponentiation f : C x N -> C has a natural generalization to any monoid C, where f(x,n) is x "multiplied" by itself using the monoid operator n times. In this setting, the only sensible choice is that x^0 = 1 for any x, where 1 is the unit of the monoid in question. In particular, that is the only definition of exponentiation which is parametric in our choice of monoid. Naive Haskell example code (using Int for Nat): f :: Monoid m => (m,Int) -> m f (x,0) = mempty f (x,n) = x `mappend` f (x,n-1) -- or equivalently f (x,n) = msum (replicate n x)

textminer 13y ago

Had a set theory professor who taught us that for the non-negative integers, m^n was just the number of unique mappings from a set of cardinality n to one of cardinality m. Ergo, for all sets A such that |A| = k, k^0 is just all mappings from Ø, which is necessarily the one with empty image and pre-image. So 0^0 = 1.

EGreg 13y ago

Mathematics is about generalizing concepts and principles to ever larger domains. In this case, x^y is defined for all pairs (x, y) of real numbers except (0, 0). The question is what limit is "closer" to the set of outputs in the neighborhood of (0, 0) than any other. 0^x is defined for all x except 0, and same for x^0. We can define 0^0 as the limit of one or the other as x goes to 0. and one is constant and more "stable" than the other, so it is typically taken to be that, i.e. 1. Up next ... if P(X) = false, what is "P(X) for all X in Ø"?

cbright 13y ago
Correction: x^y is defined for all pairs of real numbers with x!=0. 0^x is not defined for negative x.

socrates1998 13y ago

I love math, but I have never really enjoyed these types of debates. I guess I have always been drawn to the application of the concepts in the real world rather than the abstract beauty of it.

rickhanlonii 13y ago

It's very important to note here that 0^0=1 is a shorthand and not a truth. Mathematicians are absolutely not stating that they have proven, or that it is true, that 0^0=1. It is a definition, not a claim of equality. They're not saying "0^0 is 1" in the sense that they say "1+1 is 2" or "0.999... is 1". They're saying "we define 0^0 to be 1". The difference is more than just pedantry, it strikes at the core of why mathematics is the most powerful tool for determining truth that humanity has either discovered or invented (hat tip to anyone familiar with that debate). One of the cold, austere, beauties of mathematics* is that if you do not accept a definition, you can reject it as false and reason with the result. To give the traditional example, if you accept as true that two parallel lines never intersect, then along with the other 4 of Euclid's Axioms you can prove all of Euclidian Geometry (what you learn in high school). If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry (one form of non-Euclidian Geometry). In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result; you will not break any mathematics. But you should know that mathematicians have never run into any issues with this convention--or can handle it trivially when they arise--so you're only adding a lot of work for yourself. When you really think about it, it fills you with awe. It's awesome. *“Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.” -Bertrand Russell

philbarr 13y ago
Can anyone recommend a really good book on mathematics that demonstrates the "beauty" of it? I'd love to really learn and understand about Fast Fourier Transforms and the like, but any book I have is from my old college days and is just so bloody tedious. I used to really love maths when I was younger but it got beaten out of me by endless repetition and "now do Questions 1 - 50" tedium. Now I get scared whenever I see the "summation" symbol.
- coldpie 13y ago
  I found abstract algebra to be really interesting. It kind of removes the concept of "numbers" and transitions it to operations on objects like sets or functions. You redefine operations like "addition" and "multiplication" to apply to these new objects, and see what happens. A whole bunch of interesting things fall out, many of them important to information encoding and error correction in practice. For an example, see the concept of a Ring: http://en.wikipedia.org/wiki/Ring_%28mathematics%29 http://en.wikipedia.org/wiki/Ring_%28mathematics%29 Less directly related to math, I spent a joyous couple of hours back in college working through the Lambda Calculus wiki page, trying to every piece it. Once it "clicked" it became really fun to work through the examples. http://en.wikipedia.org/wiki/Lambda_calculus http://en.wikipedia.org/wiki/Lambda_calculus > Now I get scared whenever I see the "summation" symbol. I'm not sure why, but I guess for me, removing the idea of "numbers" makes math suddenly fascinating. It's like dropping down a level to redefine what you thought was a well-defined concept (addition of numbers) as a generic operation for which numbers are simply one special case. Weird. :)

rkowalick 13y ago

My favorite explanation: y^x is the number of functions from a set with x elements to a set with y elements. Since there is only one function from the empty set to the empty set, namely, the empty function, we get that 0^0 = 1.

blackrabbit 13y ago

Can anyone explain why this is a valid operation? http://wp.com/latex.php?latex=\lim_{x%20\to%200^{+}}%20x^{x}%20%20=%20\lim_{x%20\to%200^{+}}%20\exp(\log(x^{x}))&bg=ffffff&fg=000&s=0 http://wp.com/latex.php?latex=\lim_{x%20\to%200^{+}}%20x^{x}...

my_username_is_ 13y ago
it's using log to mean log base e, also known as the natural log or ln. you can do that because they're inverse operations (I think that's the right term for it, it's been a while since a formal math class) so they undo each other
- clebio 13y ago
  Yes. It's an identity relationship. x = exp(log(x)) http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Canceling_exponentials http://en.wikipedia.org/wiki/List_of_logarithmic_identities#...
- blackrabbit 13y ago
  Thanks!

deleted 13y ago

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olalonde 13y ago

Are there any examples where the x^0=1 definition turns out to make other definitions more complicated to write down? For example, you'd have a general definition and need a special definition for when you get an exponent that equals 0.

xamuel 13y ago
Suppose you want a compact, clever formula for the function: f(x)=1 if x!=0, f(x)=0 if x=0 If 0^0 were defined as 0, you could write the above function as f(x)=x^0. With 0^0 defined as 1, you're forced to write something like f(x)=1-0^|x| (absolute values to avoid division by zero), a bit more complicated. This is silly though, and of no importance anywhere.

Totient 13y ago

"...they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians." Along these lines, my preferred definition of the natural numbers" is "the set of all the positive integers and, when convenient, 0".

tel 13y ago

This question doesn't have an answer because, as this post suggests, the meanings of the first 0, the second 0, and the (^) are overloaded. The reason mathematicians don't mind the overloading is that they all have "kind of" similar meanings. So we can talk about a recursively defined function on naturals where it's convenient to "define" 0^0 = 1. We can talk about various kinds of limit techniques in the reals where 0^0 is not actually a value but instead a shorthand for a particular kind of limit. This probably forms the most interesting notion of what 0^0 means where we define it as the limit of a particular kind of path in the complex plane and then notice that 0^0 can take any value we choose---depending on exactly what path was taken to get there. [0] [0] http://en.wikipedia.org/wiki/Complex_logarithm http://en.wikipedia.org/wiki/Complex_logarithm

deepakjc 13y ago

I read upto "High School Teacher"... skimmed through "Calculus Teacher"... and scrolled through "Mathematician"...

bumbledraven 13y ago

Although standard mathematica notation glosses over the difference, it's important to distinguish between the following two versions of the power operation, which I'll call pow1 and pow2: pow1(x, y) = x^y where y is any integer pow2(x, y) = x^y where y is any real The value of 0^0 is 1 for (pow-a) and 0 for (pow-b).

SamReidHughes 13y ago
But lim x->0 (x^x) = 1. Why should 0^0 be 0 then?

emanuelsaringan 13y ago

Last time I checked, anything XOR'd with itself is 0. Kidding aside, 0 raised to 0 was said to be indeterminate according to my teachers.

t__r 13y ago

To me the matter looks similar to taking the conjunction of the empty set, which is defined to be TRUE. This, however, starts to make sense when you see that the taking the conjunction of a set X can be interpreted as "forall x in X, x is TRUE". If X is empty then this statement is trivially true.

bakhy 13y ago

best thing i read in quite a while!

vog 13y ago

I find it a bit disappointing that the article considers only limits that would justify "0^0 = 0" and "0^0 = 1". In fact, the termin "0^0" as a limit can reach any value - the same way as "0/0" can reach any value. Im stressing this because back in school, I really thought that although 0^0 is undetermined, it can only reach exactly 0 or 1, but nothing else. This is of course wrong, but no teacher was able to tell me why. For some time I even thought I found a new theorem and tried to prove it. Later, I told some math guru about this, he thought about a minute, and told me two functions f(x) and g(x) whose limit is each 0, but for which the limit of f(x)/g(x) is 2 (or any other value, if you adjust f(x) and g(x) accordingly). Having said that, in most cases "0^0 = 1" is a useful convention, especially in a purely algebraic context when polynomials are involved.

mxms 13y ago

Interesting article. I ran across a problem yesterday that was similar to P(x)^Q(x) = 1, which then asked me to find the sum of the solutions. I noticed that both P(x) and Q(x) share a root at some a. But I realized, 0^0 is most often defined as 1, and carried on. Checking the answer key later on showed that they chose to neglect that a, and call 0^0 undefined. I'm not sure how I really felt about it. Note, I also forgot to check when P(x) = -1, assuming that Q(x) is even there ;P