5 ms·

This is not true. A very natural way to define a^b is as the number of functions for a set with "b" elements to a set with "a" elements. In this case there is e

This is not true. A very natural way to define a^b is as the number of functions for a set with "b" elements to a set with "a" elements. In this case there is exactly one function from the empty set to the empty set, and we have proven 0^0 = 1. This is no different than defining addition and then proving 1+1 = 2.

alecbenzer 13y ago
I think this definition is somewhat non-standard for laypeople. Most people think of x^y to be defined as "x multiplied by itself y times" or something similar. By this definition, 0^0 is a convention/shorthand, not a truth. I think the "functions from b to a" definition is useful to note too, though
j2kun 13y ago
You're just pushing the arbitrariness of defining things one step further to the definition of a number. It makes it no less arbitrary that you've defined it and force us to accept the definition to get to your conclusion.
- mathgrad 13y ago
  There is a very natural definition of numbers as sets. We define 0 to be the empty set and we define the successor function by S(x) = {x} union x. Then the natural numbers are the smallest set containing 0 and closed under the successor operation. This is the standard way to define the natural numbers within ZFC set theory. This is admittedly very formal and not how the lay person thinks of natural numbers. However, it gets to the point of that although we may lie and say 0^0 = 1 is just a convention it is in fact a theorem within the system we work.
  - gohrt 13y ago
    > within the system we work. within a system we work. Which is the point. The system you are working in determines which value you use. You are more an algebraist than analyst.
  - j2kun 13y ago
    Levying the "authority" of ZFC doesn't change the fact that your definition is still arbitrary. I'm certain you could pick a different definition of natural numbers within ZFC and get 0^0 = 0.
    - Chattered 13y ago
      The given definition only tells you what natural numbers are; it doesn't directly tell you what 0^0 is. For that, the set theoretic definition of exponentiation was provided. That definition is not arbitrary, but is an instance of the very general definition of exponentiation given in category theory. Accordingly, A^B is the set of maps from B to A, and for non-empty finite sets, the number of such maps is the number of elements in A raised to the number of elements in B. If you extend this to cover empty and infinite sets, you get the full definition of cardinal exponentiation, which has 0^0 = 1 and has other cute things like 2^(the cardinal of the natural numbers) = the cardinal of the reals. There are a few obviously good properties about the given definition of natural numbers (the Von-Neumann ordinals), and one of them is that the size of each natural number is, well, that natural number. 0 = {} and contains 0 elements. 1 = {0} and contains 1 element. 2 = {0,1} and contains 2 elements. 3 = {0,1,2} and contains 3 elements. And so on. For any definition of natural numbers which has this feature, you find that when you consider their category where morphisms are all the functions between them, then again, the exponential (in the category theoretic sense) is precisely the one expected in arithmetic, and 0^0 = 1.
      - j2kun 13y ago
        A lecture on category theory and ZFC ignores my point. To run with your example: why must a definition of natural numbers have to involve sets of certain cardinality? You're telling me that because some people came up with a more general definition for natural numbers after the fact, that makes one definition of natural numbers more 'natural' than another. I claim you're just reinforcing my point: the definitions were chosen to make natural numbers a specific instance. Everything here is chosen in some way to further some goal (usually mathematical aesthetics), and eventually you get down to the bottom and what do you have? A bunch of definitions that one may choose to use or not. So whether I want to impose 0^0 = 1 by fiat is equivalent to whether I want to assume enough foundations of category theory to prove 0^0 = 1 in that system. You (and mathgrad) are just under the spell that because there is a whole lot more mathematics floating around (and big words and important people working in those fields) it somehow makes the latter less arbitrarily motivated by a desire to make theorems work out nicely. It's certainly pleasing that it does, but that doesn't make it somehow deeper or more natural than a different formalization in which 0^0 = 0. It just serves a different purpose. This is the entire point of the OP and the original comment: you can generalize the meaning of 0 and 1 and a^b to make it suit your need to express certain theorems and patterns, and that is a key part of the power of mathematics. Whether it's a "truth" or an "axiom" just depends on how far down the rabbit hole you're willing to go, and the distinction is irrelevant because as far as deciding what 0^0 should be they're equivalent.
Twisol 13y ago
I prefer thinking about 0^0 = 1 as an empty product (https://en.wikipedia.org/wiki/Empty_product https://en.wikipedia.org/wiki/Empty_product), since it generalizes nicely to any operation with an identity element. That is, if you apply any operation zero times, the result is that operation's identity. It's interesting that the analogous empty sum, 0*1 = 0, is a complete non-issue.
- einhverfr 13y ago
  But here's your problem: infinity * n = infinity, right? 0 * n = 0, right? 0 * infinity = ? Ok, this is relevant here particularly because: Lim 1/x as x -> 0 from the positive side is infinity, right? So 0 * that is..... lim 1/x as x -> from the negative side is negative infinity, right? So 0 * that is..... That's why 0/0 doesn't work as such. You don't know how 0 is derived or what it means. If we have x^2/x, and take the limit as x -> 0, we get 0. If we take x/x^2 and take the limit as x -> 0 we get +/- infinity. But that doesn't mean that x/x or x^2/x are not continuous functions, any more than 1 or x are not continuous functions.
  - Twisol 13y ago
    You are conflating the limiting behavior of a function with the value of a function. In (standard) analysis, there is no actual value called infinity - it's just used to describe how a function behaves arbitrarily close to a given value. When you have a 0^0 limiting form (or 0/0, or 0 * infinity), the function's behavior is indeterminate. The "0" and "infinity" you're looking at aren't precisely 0 or infinity, but only arbitrarily close. The actual behavior of the function depends on the expressions that approach 0 and infinity, hence `x/(x+1)` and `x/e^x` having different limiting behaviors as x increases without bound. But if you are literally considering the function at a specific point which produces a so-called indeterminate form, the answer is simpler. In some cases, the function is undefined (e.g. 1/x where x=0). In the case of 0^0, there is a precise value we can assign: 1. And this doesn't conflict with the 0^0 limiting form: the limit of a function at a point can be different from the actual value of the function at said point. A quote of Knuth, taken from Wikipedia (http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_power_of_zero http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_powe...): > Knuth (1992) contends strongly that 0^0 "has to be 1", drawing a distinction between the value 0^0, which should equal 1 as advocated by Libri, and the limiting form 0^0 (an abbreviation for a limit of f(x)^g(x) where f(x), g(x) -> 0), which is necessarily an indeterminate form as listed by Cauchy: "Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side."
Bahamut 13y ago
There is a bit of conflation going on here - the notation a^b being discussed is exponentiation of real numbers (or complex if you'd like to extend it that far). Of course, that notation is what inspired a^b being used to denote the number of maps from a set with cardinality b to a set with cardinality a (and the notation A^B to denote the set of all maps from B to A), but set theory was developed centuries after the likes of calculus and other important developments that touch on the reasoning used in the article linked in the OP.
chacham15 13y ago
Actually this gave me the idea: a^b is the number of mappings from set a to set b. Now, if set a is empty and set b is too, it is natural to believe that there are no mappings because there is nothing to map from or to.
- mathgrad 13y ago
  A function is a relation for elements in a to elements in b such that for every element in a there is a unique element in b. This is vacuously true of the empty relation when a is empty.
  - chacham15 13y ago
    I didnt discuss a function, I discussed a mapping. They are different constructs. Imagine the question like this: you have two groups of people, students and teachers. How many possible ways are there to assign students to teachers? To put it another way, how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment? Now, if there are no students and no teachers, you have no arrows, not one.
    - mathgrad 13y ago
      I am a mathematician I don't need examples. What is your formal definition of a mapping. In mathematic it is normally used as a synonym for a morphism in a given category and in the category of sets this would be a function. So in our context a mapping is a function.
      - chacham15 13y ago
        > I am a mathematician I don't need examples. I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve->Bob, and Eve->Carl. How many arrows are there if there are no teachers and no students?
impendia 13y ago
I think it's a little bit different. There are many natural ways of defining the expression a^b, which coincide when a and b are (say) positive integers, but there is basically only one way of defining 2 (namely, 1 + 1).
- mathgrad 13y ago
  Look up Peano arithmetic to see how one proves 1+1 = 2 within the standard formal system.
rickhanlonii 13y ago
That's a great point! Under that definition of a^b, 0^0=1 follows from the definition and is not itself defined. It's exactly the same as 1+1=2. However, what I said is absolutely true for the definition of a^b that the author of the article is using. In that system, 0^0 does not follow from the definition, and must be defined itself (or not at all a la Cauchy). I should have made that more clear.
doetoe 13y ago
I agree that your definition is a very natural one, but I think your conclusion is mistaken. Defining a function in the same, set-theoretical terms (http://en.wikipedia.org/wiki/Implementation_of_mathematics_in_set_theory#Functions http://en.wikipedia.org/wiki/Implementation_of_mathematics_i...), a function is a subset of the cartesian product, and the Cartesian product of the empty set with itself is empty. EDIT: I realize that you are right. The Cartesian product of the empty set with itself is empty, but it does have a subset, namely the empty set, which happens to be a function. I'm not so sure this settles anything though. There is exactly one function 0 -> a for every non-empty set a and there is no function b -> 0 for any non-empty set b. Now you have to decide how to define a function to treat the case 0 -> 0. I'm no expert in set theory, but I would think this can be defined to be either without getting into any kind of trouble.

alecbenzer 13y ago

I think this definition is somewhat non-standard for laypeople. Most people think of x^y to be defined as "x multiplied by itself y times" or something similar. By this definition, 0^0 is a convention/shorthand, not a truth. I think the "functions from b to a" definition is useful to note too, though

j2kun 13y ago

You're just pushing the arbitrariness of defining things one step further to the definition of a number. It makes it no less arbitrary that you've defined it and force us to accept the definition to get to your conclusion.

mathgrad 13y ago
There is a very natural definition of numbers as sets. We define 0 to be the empty set and we define the successor function by S(x) = {x} union x. Then the natural numbers are the smallest set containing 0 and closed under the successor operation. This is the standard way to define the natural numbers within ZFC set theory. This is admittedly very formal and not how the lay person thinks of natural numbers. However, it gets to the point of that although we may lie and say 0^0 = 1 is just a convention it is in fact a theorem within the system we work.
- gohrt 13y ago
  > within the system we work. within a system we work. Which is the point. The system you are working in determines which value you use. You are more an algebraist than analyst.
- j2kun 13y ago
  Levying the "authority" of ZFC doesn't change the fact that your definition is still arbitrary. I'm certain you could pick a different definition of natural numbers within ZFC and get 0^0 = 0.
  - Chattered 13y ago
    The given definition only tells you what natural numbers are; it doesn't directly tell you what 0^0 is. For that, the set theoretic definition of exponentiation was provided. That definition is not arbitrary, but is an instance of the very general definition of exponentiation given in category theory. Accordingly, A^B is the set of maps from B to A, and for non-empty finite sets, the number of such maps is the number of elements in A raised to the number of elements in B. If you extend this to cover empty and infinite sets, you get the full definition of cardinal exponentiation, which has 0^0 = 1 and has other cute things like 2^(the cardinal of the natural numbers) = the cardinal of the reals. There are a few obviously good properties about the given definition of natural numbers (the Von-Neumann ordinals), and one of them is that the size of each natural number is, well, that natural number. 0 = {} and contains 0 elements. 1 = {0} and contains 1 element. 2 = {0,1} and contains 2 elements. 3 = {0,1,2} and contains 3 elements. And so on. For any definition of natural numbers which has this feature, you find that when you consider their category where morphisms are all the functions between them, then again, the exponential (in the category theoretic sense) is precisely the one expected in arithmetic, and 0^0 = 1.
    - j2kun 13y ago
      A lecture on category theory and ZFC ignores my point. To run with your example: why must a definition of natural numbers have to involve sets of certain cardinality? You're telling me that because some people came up with a more general definition for natural numbers after the fact, that makes one definition of natural numbers more 'natural' than another. I claim you're just reinforcing my point: the definitions were chosen to make natural numbers a specific instance. Everything here is chosen in some way to further some goal (usually mathematical aesthetics), and eventually you get down to the bottom and what do you have? A bunch of definitions that one may choose to use or not. So whether I want to impose 0^0 = 1 by fiat is equivalent to whether I want to assume enough foundations of category theory to prove 0^0 = 1 in that system. You (and mathgrad) are just under the spell that because there is a whole lot more mathematics floating around (and big words and important people working in those fields) it somehow makes the latter less arbitrarily motivated by a desire to make theorems work out nicely. It's certainly pleasing that it does, but that doesn't make it somehow deeper or more natural than a different formalization in which 0^0 = 0. It just serves a different purpose. This is the entire point of the OP and the original comment: you can generalize the meaning of 0 and 1 and a^b to make it suit your need to express certain theorems and patterns, and that is a key part of the power of mathematics. Whether it's a "truth" or an "axiom" just depends on how far down the rabbit hole you're willing to go, and the distinction is irrelevant because as far as deciding what 0^0 should be they're equivalent.

Twisol 13y ago

I prefer thinking about 0^0 = 1 as an empty product (https://en.wikipedia.org/wiki/Empty_product https://en.wikipedia.org/wiki/Empty_product), since it generalizes nicely to any operation with an identity element. That is, if you apply any operation zero times, the result is that operation's identity. It's interesting that the analogous empty sum, 0*1 = 0, is a complete non-issue.

einhverfr 13y ago
But here's your problem: infinity * n = infinity, right? 0 * n = 0, right? 0 * infinity = ? Ok, this is relevant here particularly because: Lim 1/x as x -> 0 from the positive side is infinity, right? So 0 * that is..... lim 1/x as x -> from the negative side is negative infinity, right? So 0 * that is..... That's why 0/0 doesn't work as such. You don't know how 0 is derived or what it means. If we have x^2/x, and take the limit as x -> 0, we get 0. If we take x/x^2 and take the limit as x -> 0 we get +/- infinity. But that doesn't mean that x/x or x^2/x are not continuous functions, any more than 1 or x are not continuous functions.
- Twisol 13y ago
  You are conflating the limiting behavior of a function with the value of a function. In (standard) analysis, there is no actual value called infinity - it's just used to describe how a function behaves arbitrarily close to a given value. When you have a 0^0 limiting form (or 0/0, or 0 * infinity), the function's behavior is indeterminate. The "0" and "infinity" you're looking at aren't precisely 0 or infinity, but only arbitrarily close. The actual behavior of the function depends on the expressions that approach 0 and infinity, hence `x/(x+1)` and `x/e^x` having different limiting behaviors as x increases without bound. But if you are literally considering the function at a specific point which produces a so-called indeterminate form, the answer is simpler. In some cases, the function is undefined (e.g. 1/x where x=0). In the case of 0^0, there is a precise value we can assign: 1. And this doesn't conflict with the 0^0 limiting form: the limit of a function at a point can be different from the actual value of the function at said point. A quote of Knuth, taken from Wikipedia (http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_power_of_zero http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_powe...): > Knuth (1992) contends strongly that 0^0 "has to be 1", drawing a distinction between the value 0^0, which should equal 1 as advocated by Libri, and the limiting form 0^0 (an abbreviation for a limit of f(x)^g(x) where f(x), g(x) -> 0), which is necessarily an indeterminate form as listed by Cauchy: "Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side."

Bahamut 13y ago

There is a bit of conflation going on here - the notation a^b being discussed is exponentiation of real numbers (or complex if you'd like to extend it that far). Of course, that notation is what inspired a^b being used to denote the number of maps from a set with cardinality b to a set with cardinality a (and the notation A^B to denote the set of all maps from B to A), but set theory was developed centuries after the likes of calculus and other important developments that touch on the reasoning used in the article linked in the OP.

chacham15 13y ago

Actually this gave me the idea: a^b is the number of mappings from set a to set b. Now, if set a is empty and set b is too, it is natural to believe that there are no mappings because there is nothing to map from or to.

mathgrad 13y ago
A function is a relation for elements in a to elements in b such that for every element in a there is a unique element in b. This is vacuously true of the empty relation when a is empty.
- chacham15 13y ago
  I didnt discuss a function, I discussed a mapping. They are different constructs. Imagine the question like this: you have two groups of people, students and teachers. How many possible ways are there to assign students to teachers? To put it another way, how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment? Now, if there are no students and no teachers, you have no arrows, not one.
  - mathgrad 13y ago
    I am a mathematician I don't need examples. What is your formal definition of a mapping. In mathematic it is normally used as a synonym for a morphism in a given category and in the category of sets this would be a function. So in our context a mapping is a function.
    - chacham15 13y ago
      > I am a mathematician I don't need examples. I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve->Bob, and Eve->Carl. How many arrows are there if there are no teachers and no students?

impendia 13y ago

I think it's a little bit different. There are many natural ways of defining the expression a^b, which coincide when a and b are (say) positive integers, but there is basically only one way of defining 2 (namely, 1 + 1).

mathgrad 13y ago
Look up Peano arithmetic to see how one proves 1+1 = 2 within the standard formal system.

rickhanlonii 13y ago

That's a great point! Under that definition of a^b, 0^0=1 follows from the definition and is not itself defined. It's exactly the same as 1+1=2. However, what I said is absolutely true for the definition of a^b that the author of the article is using. In that system, 0^0 does not follow from the definition, and must be defined itself (or not at all a la Cauchy). I should have made that more clear.

doetoe 13y ago

I agree that your definition is a very natural one, but I think your conclusion is mistaken. Defining a function in the same, set-theoretical terms (http://en.wikipedia.org/wiki/Implementation_of_mathematics_in_set_theory#Functions http://en.wikipedia.org/wiki/Implementation_of_mathematics_i...), a function is a subset of the cartesian product, and the Cartesian product of the empty set with itself is empty. EDIT: I realize that you are right. The Cartesian product of the empty set with itself is empty, but it does have a subset, namely the empty set, which happens to be a function. I'm not so sure this settles anything though. There is exactly one function 0 -> a for every non-empty set a and there is no function b -> 0 for any non-empty set b. Now you have to decide how to define a function to treat the case 0 -> 0. I'm no expert in set theory, but I would think this can be defined to be either without getting into any kind of trouble.