What is 0 to the power of 0?

Is zero to the power of zero 0, 1, or undefined? The answer reveals an important idea regarding the nature of math.

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[ganondorfchampin]

The thing about limits though is it doesn’t matter what 0^0 equals as it’s the limit of the value, not the value itself. The limit only equals the value if the function is continuous at the point, but in this case it’s clearly not. Thus defining 0^0 as one is completely consistent with limits. However, with the combinatorics formulas it actually relies on how 0^0 is defined, so defining it as 1 is correct.

[angelmendez-rivera]

0:10 - 0:21 No, this is a fallacious argument. Consider f(n) = lim sin(π·x)/(π·x) (x —> n). f(3) = 0. f(2) = 0. f(1) = 0. In fact, for every integer n > 0, f(n) = 0. Yet f(0) = 1. So no, you cannot conclude that 0^0 = 0 from the premise that, for every n > 0, 0^n = 0. That is just not how functions and arithmetic work.

0:21 - 0:32 Again, this is fallacious. In the context of real analysis, sqrt(x) is undefined for x < 0, but sqrt(0) = 0. So no, you cannot conclude 0^0 is undefined from 0^n for n < 0 being undefined. Again, this is not how arithmetic works.

Also, why is the video trying to use values of 0^n for n not equal to 0 in order to learn what is 0^0, instead of simply trying to compute 0^0 directly like any reasonable human being? What the video is doing is very much akin to saying "what is 2 + 2? Well, to learn the answer, we must first analyze 2 + x for every integer x not equal 0". No! Just compute 2 + 2 directly. This is not difficult, and computing 0^0 directly is not difficult either. We have a definition for x^n for every integer x and every integer n. Just use the damn definition to compute 0^0. Intuitively, the definition of x^n can be understood as simply multiplying n copies of x together. In the case that n = 0, we have x^0, which is simply the product of multiplying 0 copies of x together. 0 copies of x in a product is the same thing as the empty product, which is equal to 1. Hence 0^0 = 1. Was that so difficult? Or was it a computation that took less than 32 seconds? Seriously, why is this video promoting invalid argumentation?

0:37 - 0:45 No! For the third time, you cannot compute the output of a function f(0) by analyzing the outputs f(x) for nonzero x. That is not how functions work. And yes, it is true that for every nonzero integer x, x^0 = 1. But do you actually know why this is the case? Or are you just taking x^0 == 1 on faith? Because the reason is very simple: as I explained, 0 copies of x multiplied together is equal to the empty product, which is equal to 1. Since it does not matter what x is for this to be true, x^0 = 1 is true for every nonzero x, AND it is also true for x = 0, precisely because it does not matter what x is. So yes, it is true that 0^0 = 1, but not for the reasons the video suggests. The video is way off base here.

0:46 - 0:51 No, it is not. The only reason it seems there is no unique correct answer is because the scriptwriter of the video does not seem to understand the basics of how functions and arithmetic work. To determine what f(a) is for some a in the domain of f, you have to look at how f itself is defined, not at what f(x) is for x not equal to a. To determine what 2 + 2 is, you do not look what 2 + x is for x not equal to 2. No, you look at how + itself is defined, and you compute 2 + 2 directly. That is how definitions work. That is how arithmetic, and functions in general, work. If you try using a nonsensical method of deduction for computation, then of course you will get a nonsensical answer, like the video did. So here is a suggestion: maybe, just maybe, actually try to compute arithmetic expressions with a valid method of deduction.

0:51 - 0:56 I do not know of a single context in which 0^0 = 0 or 0^0 being undefined is regarded as better than 0^0 = 1 by the mathematical consensus. And neither does anyone else, apparently, because I always ask for examples, and no one is able to provide an answer that cannot be easily demonstrated to be wrong.

0:57 - 1:01 I agree, and only 0^0 = 1 satisfies these criteria. 0^0 = 0 and 0^0 being undefined do not. So I suppose we have the answer then.

2:10 - 2:15 The fact that the video says this about the limit of a function without actually providing a rigorous definition is problematic, because it obscures the topic, and it makes it easier to pass off blatantly incorrect statements about limits as being correct because they _sound_ correct based on the nonrigorous description the video provides, especially since the video is clearly not meant for people who have studied calculus or real analysis. As such, this is misleading. Later in the argument, the fact that this description of limits is not actually a definition will be very much relevant, which is why I bring it up now.

2:15 - 2:20 What is a limit "of the form 0^0 when x = 0"? This is gibberish. Limits do not "take forms", and limits are not evaluated _at_ a point, they are evaluated _near_ a point, which is why, in the notation, we write x —> a, rather than x = a, so talking about letting x = 0 here is just wrong. But people would not notice this, because the only way the would notice it if is they actually knew how limits are rigorously defined. By putting familiar-sounding words together to form a sentence-sounding blob of nonsense, people come off with the impression that what you said here makes sense, but it does not.

2:20 - 2:26 No. We want to know the value of 0^0. So why would we try to calculate lim x^0 (x —> 0) instead? You do know that lim f(x) (x —> a) and f(a) are not the same thing, right? In general, they are not equal. In fact, they are equal if and only if f is continuous at a. So by assuming that, with f(x) = x^0, lim f(x) (x —> 0) = f(0), you are assuming that f is continuous at 0, even though you have not proven it, and if you make this assumption, yet it turns out the assumption is false, then you are going to produce a contradiction. The issue is that, rather than blaming the contradiction on the false assumption you are making, you will simply blame it on 0^0 being undefined. So you start with a false assumption to arrive at a false conclusion. Now, since 0^0 = 1, it does so happen that f is continuous, and so lim x^0 (x —> 0) = 0^0. But the video fails to acknowledge this, because...

2:40 - 2:43 With f(x) = 0^x, we have that lim f(x) (x > 0, x —> 0) = 0, but f(0) = 1. This is fine, but the problem is that here, the video chooses to assume that lim 0^x (x > 0, x —> 0) = 0^0, which is an incorrect assumption. This assumption is equivalent to lim f(x) (x > 0, x —> 0) = f(0), and so of course, a contradiction has been produced. The video blames this contradiction on the expression 0^0, rather than blaming it on the incorrect assumption that lim 0^x (x > 0, x —> 0) = 0^0. This is such a basic mistake, this is actually one of those mistakes calculus textbooks teach you to avoid during chapter 1 or 2! Yet this video is making precisely that mistake.

2:47 - 2:53 Yes, lim x^[1/ln(x)] (x > 0, x —> 0) = e, but this does not prove 0^0 = e. The video is pretending that you can replace x and 1/ln(x) with 0 to obtain 0^0. That is not how limits work. Notice how the limit has the condition (x > 0, x —> 0), or x —> 0+, as the video writes it, not x = 0. x^[1/ln(x)] is not defined when x = 0, because ln(0) is not defined. Since we have x —> 0+, rather than x = 0, we have that x > 0, and so x is not equal to 0, and neither is 1/ln(x). So by evaluating the limit, you are not actually evaluating 0^0. Stop confusing the two. Again, this is a basic mistake that calculus students are typically taught to avoid early on. I have no idea why this video's scriptwriter was not able to avoid it.

2:54 - 3:00 If these conflicts were actually the consequences of valid rigorous proofs, then I would agree with the video. But since these conflicts are a product of the video's mistakes rather than being mathematical facts, the video is just blatantly wrong about this point. Even within calculus and analysis, 0^0 = 1 is still true. In fact, we use 0^0 = 1 implicitly when working with power series.

3:00 - 3:05 No, they are not. These arguments are incredibly inconsistent with how mathematicians actually define limits of functions, and even more inconsistent with the limit theorems that you learn in every real analysis course. Indeterminate forms are not a valid concept, and 0^0 = 1 is the only conclusion with real analysis, once you take mathematical rigor and the definition of exponentiation into account.

3:15 - 3:18 Hold it right there. The usage of mathematical terminology and notational conventions can change from one time period to the next, from one location to another, and from one discipline of study to the other. Mathematical logic and theorems, though, do not change. Mathematical facts do not change. The addition of two integers is an integer, regardless of what maths discipline you are working in. 0 is the additive identity, regardless of what maths discipline you are working in. The empty product is 1, regardless of what discipline you are working. The Pythagorean theorem does not suddenly become false when you start doing set theory rather than geoemetry. 0^0 = 1, regardless of what discipline of mathematics you are working in.

[stefanjoeres]

Exponentiation is repeated multiplication. If you raise something to the power of 0, it's an empty product which is defined as 1. That's why everything to the power of 0 is 1. I'm studying math and we have to use expressions where 0^0 could occurr almost every week. Usually you say 0^0=1 because otherwise you could easily disprove many important mathematical laws.
If you are talking about limits, 0^0 obviously isn't defined and depends on the limit you arr looking at. But that goes for many expressions you don't have the limit of, that is the nature of limits.

[EAtheatreguy]

The first thing you learn about limits is that they aren't the same as the non-limit expression. Limits can give you a value different from function of the term the limit is approaching or give you a value when the function is undefined. Just because 0^0 is an indeterminate form for evaluating a limit, doesn't disprove that 0^0 is 1

[stephendonovan]

1:13 That's not even remotely true. 3^2 is 9, but there are only 6 ways to pick 2 elements out of a set of three. There's a completely different operator for that.

[andrewfinn]

At 1:13 can you please explain that definition more? I don’t see how it would work for when b > a or even tests like 3^2. Thanks.

[ucan]

I once said that 0^0 is 1 in class and everyone started laughing at me.

[frfrchopin]

I think it's 1 because when we define addition, it can be expressed as (set) + (times counted).
For multiplication, (times counted) × (times added). Here, we set the original number set (set) to 0 as default for multiplication, and when we do exponentation, (times added) ^ (times multiplied) is the basic form and we set the default (set) and (times counted) to each 0 and 1. If (times multiplied) is 0, we do the operation of 0 + 1 which is counting up once from (set), (set) + (times counted). Now, since we don't multiply anything to this value which is 1, it simply remains unoperated, resulting in
1. Same applies for 0.

[grudley]

Here we are with a video that gives a relatively clean answer to the question, when a video that gets it wrong has 5 million views

[yee]

0^0=1 is always the best definition.
Function value is different from limit.
The two need not to be equal.

[anonymoususer]

I’d also like to add this as to why 0^0 = 1:

A linear/quadratic/cubic equation such as y = 8x^2 + 4x + 7 could also be written as y = 8x^2 + 4x^1 + 7x^0. If x = 0 and we let 0^0 = 0 then the resulting equation becomes y = 8(0) + 4(0) + 7(0) = 0, but if we let 0^0 = 1 then the resulting equation becomes y = 8(0) + 4(0) + 7(1) = 7 which is the correct value for y.

[Ciofey]

0^0 should be consistent under the multiplicative zero, which is 1. Otherwise 1*x is not always x. For instance:

2^2 = 1*2*2 = 2*2
2^1 = 1*2 = 2
2^0 = 1 = 1

Unless 1 has special properties, you should be able to use it the same way for other bases:

0^2 = 1*0*0 = 0*0
0^1 = 1*0 = 0
0^0 = 1 = 1

And if you find it strange to just type a 1 and nothing else in a multiplication, you should also be able to write

0^2 = 1*1*...*1*0*0 = 0*0
0^1 = 1*1*...*1*0 = 0
0^0 = 1*1*...*1 = 1

So, basically, either 0^0=1, or 1 has to be redefined.

[oqlapsldim]

Please don't add music, it's distracting

[mahadahmedbaloch]

Also, exponents can be seen as numbers multiplying by 1
3^4=1*3*3*3*3
3^1=1*3
3^0=1
0^2=1*0*0
0^1=1*0
0^0=1

[MousMohamed]

I saw in another video that any number to the power of 0 equals 1 because it's like dividing two same numbers having both the power of one (ex. 3^1/3^1) which can be represented by the difference of power (3^(1-1) = 3^0 = 1). The problem with 0 is by doing this same process, you have to divide 0 by 0, which normally is considered undefined. So I guess the second reasoning is better overall.

[mhoeij]

The limit argument has no merit because the only thing that it demonstrates is that the x^y function is discontinuous at (0, 0).
But that has no impact on the value at (x, y) = (0, 0).
For example, the floor(x) function is discontinuous at x=0, but this has no impact on the value at x=0.

[amantiwari]

See 0^0 is always equal to 1 in algebra.

But in case of calculus, specially limits 0^0 is considered as indeterminate form which can have multiple answers. But here all zeroes are approaching zeroes or limiting zeroes not exact zero, that's why this is indeterminate form here.

Although in algebra that zero is exact zero. (Exact 0)^(Exact 0) = 1 (Always True & Valid)

[gayMath]

I dont get why this isn't the answer.
0^2=1x0x0
0^1=1x0
0^0=1
0^-1=1/(0)
0^-2=1/(0x0)
when you look at the negative numbers you can see 1 is necessarily introduced, so even using the same weird pattern and assumption system used in the video to argue for other answers, one can be argued for using the same logic?
is there something about this I'm not getting?

[willbagthegreat]

I think it's 1 because ^0 means the number is used as a factor 0 times, or never. So, there's no zero as a factor that would make the product 0.

[xxz]

1:37 if there's nothing in the box, then why is it one?


Also if it's like 1322^0 then u wouldn't write anything, therefore it is nothing. same with zero, so then why is it one? It just doesn't make sense...