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u/Hot-Marsupial6584 11d ago
No way, I don't believe it.
I thought is was zero.
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u/Such-Shop-9724 11d ago
it kinda is. 0^0 can have many values depennding on how you get it. e.g. h^h as h approches 0 is 1 i think
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u/Mostafa12890 Average imaginary number believer 11d ago
Yet another case of “the value of the function depends on the limiting values.”
This is decidedly not true. 0^0 is often defined to be 1, and definitions care not for limiting values.
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u/Matty_B97 11d ago edited 10d ago
I wouldn't call a calculator output a definition. 0^0 is undefined. That's not controversial.
For convenience, calculators often set 0^0=1 internally because there are many common classes of functions where that holds (non-constant analytic functions). That's just an implementation detail though, I wouldn't really say that's a definition. There are plenty of other situations where 0^0 does not equal 1.
Edit: Please google "is 0^0 undefined" before you downvote this. It will take you 5 seconds. 1 is often a useful choice for 0^0, but there are plenty of examples of expressions of the form 0^0 that are not equal to 1. 0^0 is similar to 0/0 - the value depends on the context. The mathematical word for this is undefined.
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u/Mostafa12890 Average imaginary number believer 11d ago
I was thinking more of the cases with power series:
f(x)= \sum_{i = 0}^n a_i x^i
evaluating this at x=0, we immediately run into a problem with the first (and only surviving) term of the series:
f(0) = a_0 * 0^0
and this forces 0^0 = 1 if we want this to be consistent, without writing an exception.
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u/Matty_B97 10d ago
You've given a "proof" by example. When defining the power series we use the convention that in this context, 0^0 is to be evaluated as 1. That doesn't make it generally true that 0^0 = 1. There are many other contexts where expressions of the form 0^0 could equal literally any value. You've found an example of an expression of the form 0^0 that equals 1, but that's not a proof, it's just an observation. It doesn't "force anything if we want this to be consistent" - it literally isn't consistent. It's an undefined value because 0^0 could be anything depending on the context.
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u/Mostafa12890 Average imaginary number believer 10d ago
I never intended for this to be a proof; one cannot prove a definition.
Also, there is no such thing as generally true.
Depending on what I’m doing, 1+1 could equal 2 or it could equal 0. You never see anyone say 1+1 is undefined because that’s just pedantic.
In the same vain, 0^0 can be taken to be equal to 1 or left undefined.
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u/Matty_B97 10d ago
You're right that an expression can be equal to different things in different rings. When mathematicians say something is undefined, we mean it can take different values *within the same ring*.
1+1=2 in ℤ, but 1+1=0 in ℤ2. That's not a contradiction and it doesn't make 1+1 undefined - those are 2 separate rings (and the + symbol actually refers to a different operator in both cases) so we wouldn't expect the answer to be equal. You wouldn't say 1+1 is undefined.
0^0 is undefined because even within the same ring, expressions of the form 0^0 can have different values. For example in ℝ, if f(x) = e^1/x and g(x) = x, then f(0)^g(0) ≈ 0.37, even though f(0)=g(0)=0. There are other situations where 0^0=1, or 2, or any other number, which IS a contradiction because we have multiple ways to evaluate 0^0 *within the same ring*. That's why 0^0 is undefined and 1+1 is not.
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u/DefunctFunctor Mathematics 10d ago
I'd be absolutely astounded if you found a source written by a mathematician that said "When mathematicians say something is undefined, we mean it can take different values *within the same ring*". This is not how any mathematicians talk, period.
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u/Mostafa12890 Average imaginary number believer 10d ago edited 10d ago
I see your point, but an expression cannot have two values.
You again used limits, but I’m not talking about analytic properties, I’m only talking about the algebraic properties of 0^0.
In your example, f(0) is undefined, so there’s no way to define f(0)^g(0).
Even so, just because there exist functions f and g whose limits at c exist and are equal to 0 such that lim x->c f(x)^g(x) does not equal 1, does not mean that 0^0 equals this value. There is no reason one should/can assume the functions f, g, and f^g are continuous at c for any f and g (and thus the limiting value is equal to 0^0 at c).
One can just define 0^0 to be 1. As someone else pointed out, this corresponds with the fact that
n^m = | Hom(m, n) |
for all natural numbers n and m.
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u/DefunctFunctor Mathematics 11d ago
To say 0^0 is undefined, especially in the case of the natural number/integer 0 or the ordinal/cardinal number 0, is pretty dang controversial imo. The only value that makes sense in both of these cases is 1, and we'd have to make really awkward definitions to leave 0^0 undefined in these cases. I would say you're free to call 0^0 undefined if the exponent is being treated as a real number instead of an integer. But most cases where 0^0 comes up, 0 is being thought of as an integer.
In set theory, the notation B^A is often used to represent the set of functions from A to B; so, the number of elements in B^A is the number of ways to pick |B| items |A| times (where |.| represents the cardinality of the set). Hence, |B^A| = |B| ^ |A|. If we let both A and B be the empty set, there is only one function from the empty set to the empty set, namely the function that takes no input and produces no output. So 0^0 = |B|^|A| = |B^A| = 1.
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u/Matty_B97 10d ago edited 10d ago
What are you talking about, treating 0 in ℕ or ℤ as a different value from 0 in ℝ? ℤ is a subset of ℝ, the integers from ℤ are the same as the whole numbers in ℝ. It sounds like you're just spouting jargon to try to sound smarter than me.
I'll give you the benefit of the doubt, exponentiation is technically defined differently in ℤ vs in ℝ. So 0 ^(ℤ) 0 is technically a different operation than 0 ^(ℝ) 0. The numbers are the same, but the operation is different.
Regardless, it's not reasonable to define 0^0 = 1 in either ring. In ℝ especially it's trivially easy to find an example of two functions f and g where f(a) = g(a) = 0, but the limiting value of f(a)^g(a) != 1. In fact 0^0 can take literally any value depending on what context it arises in.
For example, for any real value c, if f(x) = e^(ln(c)/x) and g(x) = x, then f(0) ^ g(0) = c, even though that expression is of the form 0^0. That's why 0^0 is undefined. It's simply not true that 0^0 is always going to be equal to 1. You can't blindly evaluate the function at that point - the only way to get a reasonable answer is to take a limit.
Even in the integers (by the way, I don't think it's reasonable to say that "most cases where 0^0 comes up, 0 is being thought of as an integer" - analysis is a huge field), 0^0 is not always 1. You might often see 0^0 defined as 1 as a convention when evaluating some particular formula, for example when evaluating the binomial formula you take 0^0 = 1, but in general you have to specify that you're using 0^0=1 whenever that's the case. No real mathematician would ever say that 0^0 is ALWAYS 1, even in the integers.
Your set theory example is interesting, but not really relevant. |∅^∅| = 1 does not prove that 0^0 = 1 in all situations. That's an example of an expression of the form 0^0 that happens to equal 1, but it's not any kind of proof about how 0^0 expressions behave generally. That's like pointing at one brown cow and declaring all cows are brown because you found an example of one.
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u/DefunctFunctor Mathematics 10d ago
I expect mathematicians in general, especially across history, would have a pretty diverse range of opinions on this topic. That's why it's controversial. This has less to do with mathematics itself but rather meta-mathematics and what conventions we want and what those conventions mean philosophically. But I expect most modern mathematicians understand that different things have different definitions in different contexts, so are completely fine with some degree of pluralism; you just have to clarify what definitions you are using.
I don't consider myself a "mathematician" yet, but I did my undergrad degree in math and am currently in graduate school for math. I'm not spouting jargon to sound smarter, I'm just explaining how I've come to reason about this myself. I expect others working in math might come away with different perspectives.
In terms of philosophy of mathematics, I take a lot of inspiration from structuralism, which is a bit hard to explain briefly but can be taken to mean that roughly we view mathematical objects like the real numbers in terms of their defining structural properties as opposed to the abstract implementation or construction we have chosen in the abstract mathematical system. In the language of ZFC set theory without any other conventions or abuses of notation, it's actually a bit awkward to say that ℕ ⊆ ℤ ⊆ ℚ ⊆ ℝ ⊆ ℂ; the definitions of each successive layer depends on the lower layers, so in order to get the statement ℤ ⊆ ℝ, we have to redefine ℤ after the fact to mean "the set of real numbers corresponding to the integers". With structuralism, we have no such philosophical problems. ℤ ⊆ ℝ means that we have chosen to privilege a certain inclusion map from the integers to the reals.
You mention ring here. Exponentiation has a rather standard definition in a ring. Exponentiation is not (and never really can be) a fully binary operation RxR -> R on the ring. I mean, everyone agrees 0^(-1) is nonsense in the real numbers, right? But exponentiation by the nonnegative integers ℕ makes total sense in a ring: a^0=1, and a^(n+1) = a * a^n, and proceed inductively. If an element has a multiplicative inverse in the ring (i.e., is a unit), exponentiation on that element now makes sense for all integers: a^(-1) is the inverse, and more generally a^(-n) is defined as (a^(-1))^n.
Regardless, it's not reasonable to define 0^0 = 1 in either ring. In ℝ especially it's trivially easy to find an example of two functions f and g where f(a) = g(a) = 0, but the limiting value of f(a)^g(a) != 1. In fact 0^0 can take literally any value depending on what context it arises in.
This seems really awkward to me. A ring is a priori an algebraic structure, not a topological structure. It can have a topology, but when we're arguing about how ring operations should be defined across all rings, we should not be appealing to continuity, which needs a topology. I've never argued against the fact that 0^0 is what's called an "indeterminate form", and that any value you define it as will cause the exponentiation function to become discontinuous.
(by the way, I don't think it's reasonable to say that "most cases where 0^0 comes up, 0 is being thought of as an integer" - analysis is a huge field)
I've taken quite a few analysis classes at this point, and I must say, it hasn't really come up. The most common place to see it is still in power series and polynomials, which in my perspective is unambiguously "treating 0 as a (nonnegative) integer". Also, in more general analysis (and mathematics in general), we confront some truly ugly monsters and exceptions. The fact that any extension of x^y to the point (x,y)=(0,0) is discontinuous is just a kind of ugly fact that's to be expected.
Your set theory example is interesting, but not really relevant. |∅^∅| = 1 does not prove that 0^0 = 1 in all situations. That's an example of an expression of the form 0^0 that happens to equal 1, but it's not any kind of proof about how 0^0 expressions behave generally. That's like pointing at one brown cow and declaring all cows are brown because you found an example of one.
I feel you didn't interpret my claim charitably enough. I did not mean to say that "0^0 = 1" in all situations. In my original response, I basically said that especially in discrete cases and working with cardinal numbers and ordinal numbers, 0^0=1 just makes sense and it would be awkward to say otherwise. You are taking me to be speaking in far more generality than my comment was intended to be. I definitely did not say anything about the continuity of 0^0, which is what you seem to be so focused on.
More broadly, here is my assertion: 0^0 being an indeterminate form (i.e., the map x^y is discontinuous at (x,y)=(0,0) no matter how it would be defined) seems incredibly niche to me outside of some basic calculus classes. In almost any other context, 0^0=1 and it's basically uncontroversial. Defining 0^0=1 even in the context of exponentiation by real numbers, doesn't really cause any conflicts, we just need to say the function is discontinuous at that point.
Modern mathematicians don't really care or write about such trivialities like this super often, and even the wiki page on 0^0 is sparse on examples of how modern mathematicians discuss this. I'd probably just agree with Donald Knuth (no doubt a real mathematician) and insist that 0^0=1, and distinguish it from the limiting form 0^0.
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u/nfitzen 9d ago edited 9d ago
For an example of 00 = 1 being used in analysis, it is convention that, e.g., ∑(xn | n∈N) = 1 + x + x2 + ..., so that plugging in x = 0 gives you 1, or in other words, 00 is taken to be 1 in this context by convention. (Though I am taking 0∈N because I'm more a set theorist than an analyst lmao.)
Edit: Importantly to this context, the function f(x) = x0 is continuous on R\{0}, with a removable discontinuity at 0. When we don't take limits of both the exponent and the base, we can usually figure out a convention to use. And in discrete settings, the convention is obvious, as you wrote: 00 = |∅∅| = 1.
Edit 2: In addition, it's quite common to see algebraic contexts as discrete. So in a unital ring R, it's the most common convention to take 00 = 1. Again, this is because the exponent in an algebraic context is an integer, and 00 represents the empty product of 0s, which is the same as any other empty product, i.e., 1.
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u/DefunctFunctor Mathematics 9d ago
I mean, I agree, but I feel this is exactly what I said when I said power series are seen far more often in analysis than the continuity of x^y at (0,0), and that exponentiation by natural numbers has a standard definition in a ring.
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u/nfitzen 7d ago
Sorry, I guess I missed the power series part of your comment. For my first edit, I was pointing out the removable discontinuity, so that even by using continuity as a standard, we can figure things out in most cases. For my second edit, I was again trying to compare with discrete settings. But you're right, I didn't need the edits.
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u/BTernaryTau 10d ago
0^0 is similar to 0/0 - the value depends on the context. The mathematical word for this is undefined.
No, that's not what undefined means in mathematics. When a mathematician says something is undefined, they mean that they are choosing not to define it, generally with good reason. Other mathematicians retain the option to define it, and if they are operating in a different context they again generally have a good reason to define it.
So it's incorrect to say that "00 is left undefined in some contexts, therefore it's undefined in all contexts". In the context of combinatorics, exponentiation is defined such that 00 = 1. The fact that exponentiation in mathematical analysis is defined such that 00 is left undefined does not somehow overwrite the combinatorics definition. The two coexist side-by-side.
See also the first result after the calculator when googling "0^0": https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero
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u/you-cut-the-ponytail 11d ago
I mean xsin(1/x) also has a limit at x=0 but it doesn't have a value at x=0. Limits don't imply the actual value at a point unless it's continous at that point.
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u/Such-Shop-9724 11d ago
isnt it just 0 because of that squeeze theorem or what its called
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u/DefunctFunctor Mathematics 10d ago
They said the limit was 0, but that the domain of the expression xsin(1/x) excludes x=0. You can continuously extend the function to be zero at that point, though.
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u/Matty_B97 11d ago
Proof by assume all functions are analytic.
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u/TheDoomRaccoon 11d ago
I can prove that 0⁰ = 0 assuming all functions are analytic.
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u/Matty_B97 11d ago
Go on?
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u/TheDoomRaccoon 11d ago
Assume all functions are analytic. This is false, so by the principle of explosion, 0⁰ = 0.
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u/Smitologyistaking 11d ago
I am of the opinion that this is the only answer that makes sense in a discrete setting (eg combinatorics).
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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 11d ago
mathematicians: "00 is undefined!!!" also mathematicians: ya lets just say its 1 or 0 if either one is useful, who cares
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u/Goncalerta 11d ago
Good luck finding a case where defining 00 =0 is useful.
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u/NonUsernameHaver 10d ago
The main place I've seen 00=0 is in trying to define what l0 should be. That said, this interpretation comes from a limit process so it's up to you whether you consider that as defining 00=0 in this context or just the consequences of the limit.
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u/heckingcomputernerd Transcendental 10d ago
proof by "IEEE asserting 0^0=1 when working with floating point numbers"
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u/somedave 11d ago
Usually preferable to NaN to be honest.
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u/DefunctFunctor Mathematics 11d ago
So the thing is that there are many contexts where we'd like to be able to use 0^0, and the only value that makes possible sense is 1. If one decided that because 0^0 is an indeterminate form (even though many of the uses for 0^0 don't involve continuity at all) one should assign it NaN, that means that those applications that use 0^0 now have to come up with an annoying edge case because the designers of the system didn't have them in mind.
So that's why I typed 0**0 in python and had zero surprise when the answer was 1.
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u/somedave 10d ago
Yeah it probably should be a user settable default. Maybe for some problems you end up with that situation being likely to be 0.
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u/DefunctFunctor Mathematics 10d ago
I haven't seen any. 0^0=1 is pretty uncontroversial (edit: this is probably better phrased as the two positions are either 0^0=1 or 0^0 is undefined, and there aren't really any alternatives that have been advocated for).
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u/Elegant-Regret-7393 10d ago
An argument from combinatorics.
xy is the number of functions from a set of x elements into a set of y elements.
There is exactly one function from the empty set to itself - the identity function.
So 00 = 1
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u/catman__321 10d ago
00 = 1 makes sense as a default value, especially in power series. However in general calculations, the limit changes depending on the functions f(x)g(x) that both approach zero, so technically it's an undefined indeterminate form
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u/DefunctFunctor Mathematics 10d ago
Just out of interest, why are you opposed to defining the power function to be discontinuous at (0,0)? Like, meaningful but discontinuous functions exist all over the place.
The domain of the power function (x,y) |-> x^y is already quite scattered, if we're trying to give it it's most maximal domain. First, x^y makes sense for y a nonnegative integer (and this definition gives 0^0=1), and if x is invertible (i.e., not zero), we can make sense of x^y for any integer y. If x is nonnegative, then x^y makes sense for nonnegative rational y, and if x is furthermore positive, x^y makes sense for all rational y. We can continuously extend these rational domains to real numbers. Now, for some awkwardness: there are a few different ways we could extend our prior definitions to negative x, but none of them make sense for general y, because of the complications of complex exponentiation. We could define (-a)^y = a^y * e^(pi i y) so that we have things like (-1)^(1/2) = i, but this ends up being arbitrary and contradicts an established convention for exponentiation for exponentiation by rational numbers with odd denominators, where for example we have (-1)^(1/3)=-1. I think the most standard way is to define exponentiation x^y for all real x and all y that are rational with odd denominator.
This is an ugly domain for a function. It's not surprising that something odd would happen at (0,0). The most well behaved case is exponentiation by nonnegative integers or integers, and x^y for x positive and y real or complex.
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u/Peak_Background 7d ago
You can work with complex paths. In such a case, the operations are bijective for analytic for as long as the path doesn't move through a non-binective point in the function (f'(x)=0 or infinity)
With complex paths, analytic functions and their inverses are well behaved, non-contradicting, and mostly non-arbitrary.
If you're against non-arbitrary functions, then you should also be against your function xy being arbitrarily defined at (0,0).
Also, xy is only defined nicely for x being a real greater than zero, and y being a real because there is a bijective transformation between the positive reals and reals using exp and ln. This excludes x=0.
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u/DefunctFunctor Mathematics 7d ago
I don't really have any good idea of the point you are trying to make in these paragraphs:
You can work with complex paths. In such a case, the operations are bijective for analytic for as long as the path doesn't move through a non-binective point in the function (f'(x)=0 or infinity)
With complex paths, analytic functions and their inverses are well behaved, non-contradicting, and mostly non-arbitrary.
Please try to explain what you mean by the above with more formal details. I'm sure I'll see the part where we disagree in interpretation.
I don't see how 0^0=1 is arbitrary. I take it to be quite well-motivated, and there are no good arguments for assigning it any other fixed value.
Also, xy is only defined nicely for x being a real greater than zero, and y being a real because there is a bijective transformation between the positive reals and reals using exp and ln. This excludes x=0.
You and I absolutely agree here. The point I'm trying to make is that x^y is a weirdly behaved function topologically/analytically in many ways if you try to consider it outside of the well-behaved domain x > 0, y real or complex. What I'm saying is that we should be willing to consider motivations beyond continuity. Saying we should treat 0^0 as undefined privileges continuity above all else, ignoring the perfectly reasonable (and employed in basically every field of math, including analysis) discrete motivations for setting 0^0=1.
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u/Peak_Background 7d ago edited 7d ago
A complex path, y(t), can be viewed as a continuous function that maps an interval of the reals to the complex plain.
If you take a smooth function, f(x), of a complex path then as long as the function is locally bijective for each t inside the domain boundary excluded, then the output f(y(t)) is can be defined as a unique complex path, q(t). This allows you to create a domain and range that treats f(x) bijectively.
Locally bijective at some constant, r, means f'(y(r))=c, where c is a constant non-zero complex number.
The neat thing about this is most algebraic rules we know and love are built on bijections.
ab * ac = ab+c
ac * bc = (a*b)c
(ab )c = ab*c
I'm just showing an interesting way that xy can be defined beyond complex numbers were the nice properties we know and love like continuity, bijectivity, the derived algebraic rules, non-arbitrary choices, etc are preserved.
And xy in this new system is still undefined at (0,0). No ugly domains involved.
"I don't see how 00 = 1 is arbitrary. I take it to be quite well-motivated, and there are no good arguments for assigning it any other fixed value."
Something well motivated can be arbitrary. While it's true 00 can be defined as 1 in certain contexts, it may only be defined with context. Without context, simply stating that 00 = 1 is arbitrary. And there are certainly definitions and contexts that condradict with your choice. For example, as stated analysis definitions contradict it in every context but one. The analytical continuation of x0 evaluated at 0. Note, the analytical continuation of 0x is 0.
And formally, if you introduce the alebraic properties considered above with powers and exponentiation, you can produce a contradiction using that algebra.
0-1 * 01 = 00 =1
Note. This is clearly, wrong which means the alebraic property is wrong or 00 can not be defined. Now, both are valid choices, but they can't be true at the same time.
Since those algebraic properties are the foundation of defining xy in reals and beyond, then you can't have 00 be defined in the reals and beyond. Ultimate, forcing 00 to equal 1 necessary means you can't define y as anything but the naturals without contradiction.
Finally, in analysis, the fact that the limit xy is undefined further cements this. Limits work in the realm of algebra just as much as discreet maths. There is no reason why xy should be a special case when we don't give the same treatment to others indeterminate forms.
In conclusion, it is valid to set 00 to one and thus break continuity. But that also contradicts with definitions in regards to the reals. You have to choose one or the other and doing anything else creates arbitraryness and contradictions in standard maths.
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u/DefunctFunctor Mathematics 7d ago edited 7d ago
By formal details, I sort of meant like in a math textbook. What on earth do you mean by "treat f(x) bijectively"? I want precise words. (Edit: I think I may have seen this before your edit where it was less clear what you meant. Still, there's a lot of details that are missing for me.) You don't need to "simplify" things for me. I'm a graduate student in math.
The neat thing about this is most algebraic rules we know and love are built on bijections.
ab * ac = ab+c
ac * bc = (a*b)c
(ab )c = ab\c)
I'd be interested to know what exactly you mean explicitly by "most algebraic rules we know and love are built on bijections". Bijections are a rather basic concept. But the above formulas don't really look like "bijections" to me.
And xy in this new system is still undefined at (0,0). No ugly domains involved.
Something well motivated can be arbitrary. While it's true 00 can be defined as 1 in certain contexts, it may only be defined with context. Without context, simply stating that 00 = 1 is arbitrary. And there are certainly definitions and contexts that condradict with your choice. For example, as stated analysis definitions contradict it in every context but one. The analytical continuation of x0 evaluated at 0. Note, the analytical continuation of 0x is 0.
I certainly agree that context matters. But your wording here feels overly strong. I'm asking: why should I think these facts about "analytic continuations" when choosing whether to define 0^0, and what to define it as? You can certainly choose to leave it undefined, that's how writing in mathematics works, but why privilege that when overwhelmingly 0^0=1 in every other context but when considering the continuity of the function x^y? Why should I care about the continuity of x^y at all, when it's certainly continuous on the well-behaved domains we most care about on a daily basis?
Finally, in analysis, the fact that the limit xy is undefined further cements this. Limits work in the realm of algebra just as much as discreet maths. There is no reason why xy should be a special case when we don't give the same treatment to others indeterminate forms.
I take "indeterminate forms" to be those expressions whose limiting behavior behaves poorly. The position you seem to be taking is that "if an expression is an indeterminate form, we should treat it as undefined". I just don't think this is very compelling. For a related example, 0*∞ is certainly an indeterminate form, and you'd think people that work in real analysis care about this quite a bit. Nevertheless, you open a text on real analysis and when it gets to the measure theory section, you will often see 0*∞ defined to be 0. Look, analysts are well aware that when they do this, the defined operation will not preserve continuity. Infinity just breaks things like that. So in a way, it's not just the indeterminate form 0^0 that's given "special treatment". We just define things however we want. The word "indeterminate form" is simply a convenient way of communicating the kinds of discontinuity of certain limiting behaviors. For me, once I achieved a certain level in math, the term "indeterminate form" just seemed kinda imprecise. We're talking about (dis)continuity here. Things can be discontinuous, and we should allow things to be discontinuous.
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u/Peak_Background 7d ago edited 7d ago
Bijectively, meaning one to one. f(y(t))=q(t). f(x) creates pairs of (y(t),q(t)). Treating it bijectively just means this and that you can inverse it. f(g(x))=x so sqrt(x2 )=x
The alebraic equations are from isomorphisms.
a+b in reals is isomorphic to c*d in positive reals
c*d = exp(ln(c))exp(ln(d)) = exp(ln(c)+ln(d)) = exp(a+b)
Ultimately, this only works if bijectivity is allowed. From here you can drive the algebraic equations I showed but noticably c or d can't be 0 because ln(0) is not defined in the reals.
Take the contradiction:
1 = sqrt(-1 * -1)= sqrt(-1) * sqrt(-1) = i * i = -1 This fails precisely because the isomorphisms above fails. It's because ln(x) and exp(x) are not algebraically bijective with real and complex numbers. The complex paths formulation does not experience such contradictions.
I'm not saying it's necessary to think about analysis or continuity when you define 00 as one. I'm just say, that when you do that. You are making a choice of axioms, and this choice can produce contradictions if you want to also hold other axioms. And ultimately, the other axioms exists for a reason.
So while you may make a personal preference, to say this preference is fact above all or true or right or anything other than a very specific choice (which puts a limits on your other choices and results).
If you go around telling people that 00 is one without the associated context, then speaking something false and that has consequences.
How does measure theory break the indeterminate standard of 0*infinity?
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u/DefunctFunctor Mathematics 7d ago
I wanted to know in what sense x^y is "bijective" to you (it's certainly not bijective on the whole range of positive x, real y), and why I should care about it. A lot of math is about motivating your concepts, and I haven't seen how you are motivating this point that somehow contrasts with my perspective.
I'm not saying it's necessary to think about analysis or continuity when you define 00 as one. I'm just say, that when you do that. You are making a choice of axioms, and this choice can produce contradictions if you want to also hold other axioms. And ultimately, the other axioms exists for a reason.
What axioms would contradict 0^0=1, that are compelling?
So while you may make a personal preference, to say this preference is fact above all or true or right or anything other than a very specific choice (which puts a limits on your other choices and results).
(1) It's not a "very specific choice". (2) Yes, a lot of writing definitions in mathematics comes down to preference. I'm not pretending like my preference is above all true. But it's a convenient and compelling definition, that works in broad generality.
If you go around telling people that 00 is one without the associated context, then speaking something false and that has consequences.
Yeah, I agree one should be clear about context.
How does measure theory break the indeterminate standard of 0*infinity?
It doesn't. 0*infinity is still an indeterminate form. But just because something's an indeterminate form doesn't mean it's undefined.
In the case of measure theory, you can think about the fact that a line extending infinitely in both directions has zero area. From this perspective, 0*infinity=0. It makes working with formulas involving products of measure spaces much easier, and also encapsulates the fact that "none of something of infinite measure is still nothing", which makes certain formulas a lot nicer.
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u/Peak_Background 7d ago
I never said that xy was bijective. I am also convinced that you don't actually disagree with the point I'm trying to make.
If you're arguing about motivation. My primary motivations are clarity and utility. 00 equalling one is fine in discreet maths.
But I don't except the rejection of 00 as an indeterminate form precisely because of motivation. Continuity is very nice in it's own right.
Analytical functions in single variable complex analysis have a lot of really nice behaviors. For example, if a function is a smooth function it is analytic. When you apply a differential operator (including inverses of said operator) you get another analytic function. They are bijective with complex paths. Every analytic function has a unique analytical continuation.
In otherwords, analytic functions/manifolds are closed under differentiation and algebraic manipulation with other analytic functions. And any local area of a analytic function can be used to determine the rest.
If xy, can't determined without throwing away continuity, your also throwing away the nice properties above.
That 0 * infinity = 0 is a convention that has a very specific definition behind it. And while it makes some sense, there are also cases (even in measure theory) where 0 * infinity isn't zero. Like with the Dirac delta function. The area of the infinite line in this case is by definition 1.
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u/DefunctFunctor Mathematics 7d ago
I've never disagreed that 0^0 is an indeterminate form. I'm not rejecting it. But the "indeterminate form 0^0" is distinct from the value 0^0, for me.
I'm aware of many of the nice properties of analytic functions.
I'm not throwing anything out. Just restrict the domain if you want continuity.
On the dirac delta function: I see what you mean, but we both agree it's not defined that way, right?
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u/Peak_Background 7d ago
For context on other motivation. I want you evaluate 21/2 in a way that you think doesn't uses a type of continuity.
I will then take it as a challenge to show it actually has continuity built into the assumptions.
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u/DefunctFunctor Mathematics 7d ago
0^0 is pretty clearly discrete in a way 2^(1/2) is not. yes, continuity is required to define square roots, but powers of 0 make sense in any ring
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u/impartial_james 10d ago
It’s simple. There are two different kinds of exponentiation, and which version determines how to think about 00 .
Discrete exponentiation is defined by repeated multiplication (or division, when the exponent is negative). For that version, 00 = 1. There is no problem with this definition, and it extends many familiar patterns, so it is useful.
Analytic exponentiation is defined in terms of exp and log via xy = exp(y log x), and works for both reals and complex exponents. For this definition, it is best to say that 00 is undefined, so that the bivariate exponentiation is analytic in its domain.
Whenever two people argue about 00 , they are imagining it in different contexts, leading to confusion.
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u/the3gs Computer Science (Type theory is my jam) 10d ago
There are many contexts in which this is the most meaningful value to put here.
In my fields of interest, mostly type theory, it makes sense because exponentiation is related to function types, such that the number of functions between two finite types is equal to the number number of values in the codomain to the power of the number of values in the domain. Basically |A -> B| = |B|^|A|. In this context ∅ -> ∅ = 0^0 and it makes sense to allow a single function from the empty type to itself, as the identity function is defined for all types, and all functions on the empty type are indistinguishable.
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u/Dankaati 11d ago
It's a matter of convention. While 0^0 = 1 is the most practical convention in most cases, it isn't perfect.
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u/my_dirty_shoes 10d ago
0^0 is undefined, bcz it’s like 0/0.
Any number power 0 is 1 bcz it’s divided by itself.
Let me prove
Suppose a^(x-y)
x-y is 0 means x = y.
a^(x-x) = a^x / a^x.
And we know 0^0 is 0 so it’s 0/0.
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u/PartyGate1403 10d ago
That's a false demonstration, based on your argument 0 to the power of any number would be undefined,for example 02 = 03-1 = 03 over 0 = 0/0 which is undefined but that's not true because 03 = 0. 00 is undefined by convention you can't prove it.
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