6 ms·

Technically, 0^0 is an indeterminate form and has no specific solution. Accurate but unhelpful. Practically, 0^0 highlights the issue that most of us don't hav

• yaks_hairbrush 15y ago
  > How would you explain to a 10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers work out". Actually, that's exactly the reason 3^0=1: it was the definition that preserved the most identities. Agreed that this explanation doesn't really help intuition.
  • qntm 15y ago
    What is true in mathematics is whatever leads to no logical contradictions.
    • xyzzyz 15y ago
      There are cases when both a claim and its negation do not lead to contradiction, but them both being true obviously does.
      • qntm 15y ago
        Thus, mathematics unexpectedly turns into a Choose Your Own Adventure novel!
        • xyzzyz 15y ago
          It's an amusing way to put this, but yes, it's true. An example of such situation is the case of continuum hypothesis. Both it and its negation has been proven not to lead to contradiction, so while most mathematicians ignore it, effectively treating it as being true, many set and model theorists play with things like Martin's axiom, which makes sense only if continuum hypothesis is false.
  • kalid 15y ago
    Yeah -- that may have been the original motivation, but repeating it as an "explanation" reinforces the notion that math is a bunch of rules (vs. models you can construct and manipulate in your head). 0 probably started as a placeholder symbol for "naught", i.e. nothing to write, and the first scribes were taught "Just write a circle when you have nothing to report". But, with greater understanding of numbers 0 evolved into its own entity and we saw numbers on a "line", a powerful mental model (why not 2d numbers? N-dimensional numbers? etc.) Re-teaching that 3^0 = 1 "because the math is convenient" doesn't help us build a mental model of what exponents could be (I know you don't agree with this, just stating it again because the lack of intuitive explanations for math is a major pet peeve of mine).
    • tokenadult 15y ago
      the lack of intuitive explanations for math is a major pet peeve of mine I'm a very visual thinker, and that is one reason I enjoy the new Art of Problem Solving textbook Prealgebra by Richard Rusczyk, David Patrick, and Ravi Boppana-- https://www.artofproblemsolving.com/Store/viewitem.php?item=prealgebra https://www.artofproblemsolving.com/Store/viewitem.php?item=... it is full of interesting visual "explanations" and substitute for proofs in a book intended for a young audience. That said, I finally realized that I was limiting my mathematical development by insisting that every mathematical idea must appeal to my visual intuition. Some mathematical ideas are proven even if they don't appeal to visual intuition. In the words attributed to John von Neumann, "in mathematics you don't understand things. You just get used to them." http://en.wikiquote.org/wiki/John_von_Neumann http://en.wikiquote.org/wiki/John_von_Neumann That point of view makes a lot of sense to many of the best mathematicians. One more example of really interesting visual explanations of mathematical concepts is Visual Complex Analysis http://usf.usfca.edu/vca/ http://usf.usfca.edu/vca/ by Tristan Needham. The book is delightful, and well reviewed, but it is not the sole path toward getting used to complex analysis.
      • kalid 15y ago
        Thanks for the pointer! I found Needham's book awesome, I've barely made a dent in it but love the visualizations. I don't think visualization is the only intuitive method -- you can have a general "sense", not sure how to put it more specifically -- I have a "sense" about growth of e without a specific diagram. Agreed that not every concept can be understood... yet. There's a quote I love to rail on, in reference to Euler's formula: "It is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth." (Benjamin Peirce, 19th-century Mathematician) Really? Yes, it may be baffling at first, but we can _never_ understand it? Only if that's our attitude :).
  • Almaviva 15y ago
    You add a bunch of stuff and you get a total. You add nothing and 0 is the total because it's the identity and starting point for addition. You multiply a bunch of stuff and you get a product. You multiply nothing and you get 1 because that's the identity for multiplication. (That's the same as saying "the rules of algebra work out" but there's maybe something intuitive about multiplying nothing and getting back the thing that doesn't change the result of multiplication?)
• jkent 15y ago
  I successfully managed to explain 3^0 to 10 year olds (as a recovering high school math teacher) as: 3^2 = 9 3^1 = 3 (divide 9 by 3) 3^0 = 1 (divide 3 by 3) 3^-1 = 1/3 (divide 1 by 3) etc This can logically be explained as n^0=1 for all real numbers. Unfortunately this doesn't really handle 0^0 but fortunately 10 year olds are rarely that difficult.
  • pshapiro 15y ago
    Thanks, I was gonna say the same thing. :)
  • amalcon 15y ago
    This explanation works pretty well on adults: Just include the multiplicative identity (1) in the expansion. 3^3 = 3*3*3*1 = 27 3^2 = 3*3*1 = 9 3^1 = 3*1 = 3 3^0 = 1 = 1 and likewise: 0^3 = 0*0*0*1 = 0 0^2 = 0*0*1 = 0 0^1 = 0*1 = 0 0^0 = 1 = 1 I haven't yet tried this on an actual 10 year old, though.
    • jkent 15y ago
      Also good, but the last line doesn't work (in my opinion): to get from 0x1 to 1 you need to undo the x0, i.e. dividing by 0. Which is undefined ... Zero, confusing 10 year olds for centuries!
      • amalcon 15y ago
        True. If you approach it from a standpoint of extrapolation from known values, you have a division by zero one way or the other. What I'd intended was that the value of 1 was reached by applying the same algorithm that was applied to arrive at the other values: start with 1, multiply by the base once per instance of the exponent. No division involved. It's still incorrect if we want to be strict, of course. That algorithm is not quite the definition of exponentiation, because that algorithm can't really be extended to work outside rational exponents. Exponentiation is defined across complex numbers (ignoring 0^0 for the moment). I think this is acceptable because I'm only shooting for an explanation, which doesn't need to be strict.
• xyzzyz 15y ago
  What is "indeterminate form"? What does it mean for expression to "have a specific solution"? You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R -> R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems -- for instance, this is the source of the whole "indeterminate form" notion. If a function f is continuous in (a, b), then for every two sequences a_n, b_n, such that lim a_n = a, lim b_n = b, we have lim f(a_n, b_n) = f(a, b). That's why lim (a_n)^(b_n) = a^b if (a, b) != (0, 0), and this is "determinate form". But if (a, b) = (0, 0), then no matter how we define 0^0, it does not follow that lim (a^n)^(b^n) = a^b = 0^0, because in this case, lim (a_n)^(b_n) can be every positive value, and so mathematicians used to call it "indeterminate form" (it's not common today, though). So, since this problem is unsolvable in a consistent (continuous) way, we define 0^0 = 1, to be consistent with exponentiation rules, at least. I've never seen a need for an "intuitive" explanation of exponentiation -- the usual definition is as intuitive as one can get. The thing is, most people do not know, _why_ expressions like pi^e are supposed to make sense -- they just take exponentiation as given. Only then they need to make up some explanation why "exponentiation rules" are like this, and what exponentiation is about. Hell, people don't even know what real numbers are! How are they supposed to make sense of exponentiation with exponent other than natural number?
  • Dove 15y ago
    What is "indeterminate form"? http://en.wikipedia.org/wiki/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.
    • xyzzyz 15y ago
      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.
• dxbydt 15y ago
  >most of us don't have a good conceptual model for what exponents really do Instead of matching math to real world objects (1= one banana, 2 = two bananas, 1+2 = 3 bananas etc. ) and building up to exponentiation, multiplication etc. thereby introducing all sorts of paradoxes, group theory dodges all that and treats the whole thing as a very consistent rule-based system. Things fall into place quickly once the rules are laid out explicitly. Consider: finite abelian group with only 3 elements a,b,c. Given a+b=c, a+c=a, what's b+b ? Hmmm...okay, if a plus c is a, then c is acting like zero. So b+c must be b. since addition is commutative (abelian gp), b+a must be a+b which you said was c. So now we know b+a=c, b+c=b, so b+b better be a ! Students are easily convinced because you've laid out the rules very explicitly. In fact, they'll try to convince you that b plus b better be a because that's the only way to make the cayley table work out! (http://en.wikipedia.org/wiki/Cayley_table http://en.wikipedia.org/wiki/Cayley_table) There are several books that argue that the teaching of Abstract Algebra must precede Calculus for this very reason. With Calculus, the mapping of math to real-world objects leads to all sorts of messy realities. With group theory, you dodge that mess by simply stating rules upfront.
  • ubasu 15y ago
    Group theory is formed by abstracting out the observed properties of number systems. If you want to show that some collection is a group, you will have to do the computations to show that they follow the rules, in which case, it helps to have an understanding of the mechanics of the computation.
    • dxbydt 15y ago
      > it helps to have an understanding of the mechanics My claim is the exact opposite. I claim you don't need to understand the mechanics ( just blindly abide by the rules of the group or abelian group or finite simple group or whatever), which is why the approach is better. If you show a monkey red means stop and green means go and reinforce these rules by rewarding with a banana, eventually the monkey will stop when he sees the red. Not because he understands the mechanics of traffic management. Simply because he is abiding by the rules. Similarly, large portions of math can be approached by either the definitional route ( ie. rules ie. define propositions & theorems that logically follow if those props held ) or via trying to understand actual mechanics by mapping everything to real world phenomena ( x = distance, dx/dt = velocity, d/dt(dx/dt) = acceleration etc. ) which are problematic because the mapping breaks down due to the nature of physical reality ( like friction etc. ) How would one explain say Hilbert's 7th problem via the actual mechanics ? If a is algebraic and b is irrational show a^b is transcendental. What does that even mean when you map them to the real world ? Instead, the solution is to build upon theorems that logically follow from the axioms you start out with. Problem: http://en.wikipedia.org/wiki/Hilbert%27s_seventh_problem http://en.wikipedia.org/wiki/Hilbert%27s_seventh_problem Solution: http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/ http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-pr...
• MtrL 15y ago
  I just think of it like X to the 0.5 is square root, X to the 0.3 is cube root, X to the 0.25 is fourth root, etc Therefore X^0 is what you get if you're taking the 'infiniteth' root = 1
• lloeki 15y ago
  > Technically, 0^0 is an indeterminate form and has no specific solution. Precisely, as this is the true mathematician answer: "it depends where 0^0 comes from". As a f(x,y): RxR->R function, come from the top of the R² plane and 0^0 is 0 but come from the right side and it's 1. Limits and extension by continuity give us this easily enough for fh:x->x^0 and fv:y->0^y. Writing this I asked myself, what if we came from some funky other path, like the diagonal, or a curve? h: R->RxR, x->(x, 0) defines "coming from the top", and foh = fh v: R->RxR, x->(0, y) defines "coming from the top", and fov = fv d: R->RxR, x->(x, x) defines coming along the diagonal, where things could get interesting. s: R->RxR, t->(e^(at)sin(t), e^(at)cos(t)) defines coming along a log spiral whose tangent at t=0 is vertical, so fos looks like fun around t=0. Now what happens if we build a path function p: RxR->RxR, (t, z)->? that endlessly approaches v when z->0? the log spiral with z=1/a as a parameter is a possible one. With such a p function, what does lim fop(x) when x->0 (which is a function of z) look like when subsequently z->0? Damn. It was supposed to be a two-line comment.
• hbt 15y ago
  I've always thought the best explanation was 7^2=7 * 7=49 7^2/7^2=7 * 7/ 7 * 7=49/49=7^(2-2)=7^0=1 But 0^0 was never intuitive to me.