0^0 = 1? Exploring 0^0 in 5 Levels from Exponents to Math Major 

He is the master of us all

Exactly🙏😊

yes

Rocco Si Fredi defined sex. this argument is enough for me. (cock shorter than 19cm is a nono) was with humor, but there is many ways to consider a problem.

a^0=a*0+1 a^1=a*1+0

What about 1*0?

I think they meant "any product that contains at least one 0 is equal to 0" lol

@@swenji9113 sorry, yeah. Got myself confused there with all the zeros and ones 😄 It's fixed now though, thanks!

Obviously math is not directly connected to reality in the way physics is, but sometimes a good way to think about math is to use a tangible example. If we take 0! We know that that equals 1. How is that possible when we usually do factorials by multiplying every positive number down to 1 from the n! ? So, 3!=3*2*1=6. 2!=2*1=2. 1!=1*1=1...but 0! Still equals one...how is that possible? Well, if you use a tangible example, imagine that you have three unique items and you want to arrange them in a certain way. There are six options. If you have two items, there are two ways to do so. If you have one item, there is only one way to do so. If you have no items, there is still one way to arrange them...by not arranging anything. Likewise with 0^0=1. If we consider the presenter's point again, there is still one way to organize a set with zero elements: as an empty set. You can think of this almost as a riddle in real life. How do you make $0 selling cupcakes? There's one way to do it: don't sell any cupcakes. How many ways are there to guarantee that you never, ever under any circumstances lose a game of chess? One: never play chess. It's a little unsatisfactory because it sounds more like a trick question than a valid analysis, but I hope that analogy helps. My apologies if it's confusing, annoying, both or TMI. If you disagree, I also understand. Philosophically, it's unappealing, even if it is logically consistent on some level.

@@AliceObscura I'm not entirely sure why you wrote an entire essay here but yes, the empty product argument was also in the 0! video. Only in that case, there's no deeper argument that contradicts it.

Thanks for the idea! I added it to my list

Humph. Did you invoke category theory for this? Just to generalize the problem, right? Because the conventions about the empty set are the set analogous of the conventions about zero. People insist on making this a big deal. It makes no sense.

How does one extend x^y from natural numbers to other number systems? For example, how would (1/2)^(1/2) be defined?

@@mesplin3 you don't need to do that, this is pre-calculus, foundational. Just choose an exponential definition that aligns on the naturals and obeys the additive rule e.g. a^b * a^c = a^(b+c). In the reals we have that a^b = e^ln(a)b = e^(floor(ln(a)b)) * e^{ln(a)b} where {.} is the fractional part, so you just need a definition for e^x for x in (0,1) and multiply it by e^n for integer n, such a definition can be obtained from a limit or from an infinite series. It's more complicated in other algebras like the complex numbers, or for matrices etc.

@@samueldeandrade8535 it's the same when talking about sets yes, since these are all 'small' categories. I suggest this should be the "Math Major" level though.

@@mesplin3 do you know how we usually define (1/2)^(1/2)? Or any power a^x.

This video makes a very common error: There is absolutely no requirement that functions must be continuous. It's nice if they are, but if you cannot extend a function *continuously* it doesn't mean you cannot extend it at all, nor that it is necessarily undesirable to do it. You just need to remember that for discontinuous functions, the limit (if it exists) does not necessarily equal the function value.

Excellent

Thanks so much! Definitely, I'll have to remember that

They really don't. At the moment it is agreed that 0^0 is undefined.

He has a video on that

> turn around > turn around again > wtph im facing the same direction

That's a fairly easy one, fortunately.

•(-a)(-b)=(-a)(-b)+a(0) •=(-a)(-b)+a(b-b) •=(-a)(-b)+ab-ab, (Common Fact: -b •=-b(a-a)+ab=ab, (a-a)is equal to 0 Therefore: (-a)(-b)=ab

@@master_of_blinchiki Now all that’s left to prove is that 180° rotations in 3D space around a fixed axis are bijective with the subset of the multiplicative group of the real numbers (or integers) that is comprised of all negative multiplicands. Step 2: Profit..?

You were a brilliant child if you figured out how 1 is the empty product. Also one of the reasons we define 0! as 1

haha this is exactly the reasoning behind empty operations, so there's a lot of truth to it. for the exact same reason, we usually define: • the sum of no numbers to be 0 • the product of no numbers to be 1 • the union of no sets to be the empty set • the intersection of no sets to be the entire domain (these last two also allow you to get rid of one of the axioms for a topology on a set)

Than you can just define 0 as a ghost in additive. So 67x is Same as 67x+0. 27= 27+0. You are only doing the 1 times 27 because by convention we leave out the symbol for multiplication but it’s still there. There is no difference between why the ghost property of 1 should be limited to only multiple and division.

@@Nosirt the reason 0 is not defined as ghost is 27 + 0 = 0, and we need extra plus symbol here. And what i mean is when you can't see, 27 you can also write it is (27) and now you can't see 1 but it can be there as 1(27)

@@malemsana_only 27+0= 27, not 0. Also the plus sign does not matter as the reason why you don’t put multiple sign is by convention. That convention can simply change. 1(27) has a hidden 1*27 there. It is not a fundamental thing in math, just convention of how we write multiple. With this, once again, when you write (27) you can insert a ghost +0 anywhere there too.

We may stick to the definition of exponentiation with a natural exponent: a⁰ = 1, aⁿ⁺¹ = aⁿ·a, so for any base a we have a⁰ = 1 by definition (as an empty product with no factors, which is equal to the multiplicative identity). This way we no longer get into the trap of dividing by zero, though the property aᵐ⁺ⁿ = aᵐ·aⁿ for all natural m, n still holds.

@@allozovsky Yes, that does get around the issue in the video where he’s dividing by zero. Although as a small caveat note that if we instead defined 0⁰ = 0 then exponent addition still works. (I.e. if a=0 then aⁿaᵐ = aᵐ⁺ⁿ for all m and n including if either or both are 0.) It’s not as pretty as simply defining it to be 1 for purposes of factoring but it’s just as consistent in this context.

Yeah, there are much better justifications for the empty product being equal to 1 than that argument.

In a sense, but i is actually a very well behaved number to add. It keeps lots of the properties of the real numbers like 0*x=0, ab=ba, and 1/x always being well-defined, in contrast to adding a number defined as 1/0, which would break the first property (and I think maybe the third? Anyway losing the first is already sufficiently annoying)

@@makhnoboi1996 what does i have to do with what I was saying? It's perfectly fine. I'm comparing the multi-valued equation x^2 = 1 to the multi-valued equation 0*x = 0. I'm saying we shouldn't treat only one of them seriously and dismiss the other one.

Although i'd like to agree with you, I'm not sure what your point is. Do you think that indeterminate forms are often ignored? What do you mean by that? That they are not studied or discussed enough? I absolutely agree that indeterminate forms are very interesting, I just feel like there are way too many discussions around indeterminate forms where noone defines what they're talking about...

@@swenji9113 yes, the indeterminate forms that can equate to the entire extended complex plane, i.e. 0/0 and its cousins, are always ignored and thrown away. And yes, they are not studied and discussed enough. I understand Wheel Algebra studies them, but I'm not sure that's the only way to make sense of them. Some indeterminate forms are different from others. For instance, 0/0 ∈ ℂ∪{∞}, whereas 1^∞ ∈ ℝ≥0. I believe we should be looking into the structures that come about when graphing functions that reveal these indeterminate forms, and we should especially incorporate the Riemann Sphere in order to capture the true essence of all of this.

@@samuelyigzaw I see, thank you for your answer!

Just ask Haskell for 0^0 or 0**0. The limits don't tell pretty much anything about values at discontinuities.

the error would be to assume that maths get satisfied from discreet calculations. maths don't care about computers limitations. A theorem is absolute, whatever the number of decimals. If it requires comps, then it's not a satisfying theorem. (and whatever the language, Euler would have loved the comp...and in my opinion he would have prefered Lisp, Scheme, or Prolog languages)

Well n^0 = n/n is not true for n=0 precisely. This "definition" of n^0 comes from an understanding of powers starting at 1 and possibly going back, but in general, we can define an empty product to be the the neutral element for multiplication regardless of what it is the empty product of. It is like saying that for n>0, n = n²/n and therefore 0 = 0/0 is indeterminate form

The function inside the parentheses approaches 0, as does the exponent outside. So if you wrote this as f(x)^g(x), with f(x)=exp(-1/t) being the term inside the parentheses and g(x)=t being the exponent, you get the indeterminate form 0^0 if you 'plug' the limit in. This shows that if f and g change at different rates, the indeterminate form 0^0 can result in different answers. An even simpler example might be lim(x->0) 0^x = 0, but lim(x->0+) x^x = 1.

He did, go check it out

Man, people LOOOOVE an useless discussion, huh?

@@samueldeandrade8535it’s not useless. It teaches you generalisable methods of how to deal with operator extensions and how to prove if they are well-defined or not.

​@@fullfungo eh, still seems pretty useless, cool trivia, but there are better ways to teach that...

@@fullfungo listen, it is. Most of mathematics is useless, just like most of art, just like most of philosophy. And that's awesome, because this is the biggest reason why we like them. So my "useless" accusation is already pretty fair. Now, 0/0 is NOT something really interesting. It is very easy to understand it, but people insist on treating it as something mystical. About your "generalisable methods of how to deal with operator extensions ... well-defined or not", man, I studied Math, do you think you are fooling me with all that? With some big words? Please, try again. Your statement is as deceitful as "we invented comolex numbers so quadratic equations would have solutions urrr durrr".

I added it to my list, thanks!

it's the empty product. Wasn't that mentioned in the video?

For any positive integer n, 0^n = 0. Now for calculus purposes, one can extend (x,y) -> x^y continuously at (0,a) to be 0 for any a>0. Such an extension isn't possible at (0,0). So there is a canonic way to define 0^x for x>0, but it's way more complicated than integer powers, for which we define x^0 to be 1 for any number x. This doesn't mean you can't study 0^x. However, there is a very good reason to think about x^0 first when you see any number raised to the 0

limit of 0^x as x approaches 0 from the right is 0 It's kinda mentioned in part 5 where x^y is discussed.

WolframAlpha disagrees, only real solution to x^x=1 is x=1

@andrewpang5111 wolfram alpha disagrees because 0^0 is controversial. However, if you do the algebra: x^x=1 Take the natural log of both sides: xlnx=0 therefore x=0 and lnx=0 meaning the solutions are 0 and 1 this does however require the assumption that 0ln0=0 and ln0 is undefined

​@@stelisc9326yeah lol this proof essentially supplants one indeterminate form for another; taking x to 0 makes ln(x) approach -infinity, so we're left with 0*infinity. interestingly enough, though, this is often defined in measure theory to be 0, so in that context i suppose it would work.

very true

Nope, the term on the left isn't defined. If you stated it with a sequence, then the sequence diverges to infinity for the standard definition of limits

No, the analytic continuation of the Riemann Zeta function evaluated at -1 is -1/12

How can it be -1?

@@alephnull8377 negative numbers ^0 = -1

because its not equal to 1/2

indeterminacy is a statement about the limits of operations, not the operations themselves. if you think about it, any justification for saying 0^0 = 0 (or anything other than 1) relies on thinking of what a function gets close to. however, there are plenty of justifications for 0^0 = 1 which only talk about fundamental definitions of operations (eg the tuple description and the empty product in this video). that simply can't be said for 0/0; if x = 0/0 then 0x=0, so x can be any number. it's not that we're throwing the rules out. the math "rules" (ugh there aren't rules in math but whatever) say that you should define things according to what feels most natural or convenient. it's inconvenient and unnatural to define 0/0 (or most other indeterminate forms), but it can be very convenient and very natural to define 0^0=1.

0^0 must be 1 as long as we accept the set theoretical or category theoretical foundations of mathematics. There is exactly one map from 0 ::= {} to itself, hence the exponential set 0^0 has exactly one element. Hence 0^0=1. Limts don't define function values of discontinuous functions. sgn(0) = 0 = | sgn(0) |, while | sgn(x) | = 1, for all other x.

There is exactly one map from the empty set 0 to the empty set 0. Hence the set 0^0 of all maps from 0 to 0 has exactly one element. Hence 0^0 = 1. Set theory is the standard foundation of mathematics. If you use category theory to found mathematics, you get the same conclusion.

No

​@@yuseifudo6075he has a point though

Idk man, probably you could explain all of them?

Because I found (1/100000)^(1/100000) is close to zero, if u make the denominator smaller on each parentheses it will get even closer to 1...

Wasn't that the point of the video? No matter how you explain it 0⁰ never really makes sense, but it is still defined as 1 sometimes since it avoids a lot of unnecessary complications lol

There is no natural algebraic interpretation of 0/0 that makes it 1. If you think in terms of limits, then it's not true in general either

@swenji9113 any number divided by itself is 1. I'm not serious about math but I think it's ridiculous that 0!=1 and 0^0=1 but saying 0/0=1 is going too far

@@denverrsouthers5531 lol I agree, it's funny

@@denverrsouthers5531 it depends on the 'order' of the 0, when we're talking about meromorphic functions (aka laurent series, laurent polynomials, rational functions etc.). It's only 1 when the order of both the zeroes are the same. This shows up when we want to factorise or cancel factors out from a rational function and we need to consider multiplicities of roots of both the numerator and denominator. Notice that when we cancel we are doing it purely algebraically, not using calculus or limits (though the justifications are derived from complex analysis, the algebra of these polynomials is consistent). Basically for some field F (usually the complex numbers) then we extend the field to contain zeroes and isolated singularities of each positive integer order (say this is G) and consider the algebra of its rational functions over some formal variables G(x,y,...). As an example, you can say unequivocally that x/x = 1, but x^2/x = x, this is correct in the algebra. However, plugging in x = 0 we get 0/0 = 1 and 0/0 = 0 which doesn't work, so you have to remember where the zeroes come from rather than assigning 0/0 a particular value. Instead we would have to write it 0/0 = 1 and 0^2/0 = 0, but this only works since we have 0^2 =/= 0 which is not generally something you want in all situations. In contrast, 0^0 doesn't have any problems algebraically when assigned 1, you just have to remember that the value under the limit is not necessarily the value of the original function at that point under its domain of definition (so functions that hold values of 0^0 are not necessarily continuous, where before they wouldn't even be defined there without holding a value for 0^0). And 0^0 = 1 is the only valid choice if you were to assign it a value, so that it aligns with the natural construction of exponential categories A^B is the category of all functors from category B to category A, and |A|^|B| represents the size of this category, and 0 is the size of the category of no elements, so 0^0 is the size of the category of all functors from the empty category to the empty category, and there is only one such functor, the identity functor, so 0^0 = 1 (if you must give 0^0 a value). 0! = 1 is actually just a simple, noncontroversial result: permutations must be between 1 and 2^n for the size of the set, so between 1 and 2^0 for empty sets, so empty sets have 1 (between 1 and 1) permutation. This is because permutations are actions on a set and produce a subset of the set (part of the powerset which has 2^n elements), and there is always the identity permutation (does nothing to the set). You can also think of both 0^0 and 0! as empty products, which are 1 since 1 is the multiplicative identity (the empty sequential operation if it exists must always be an identity of that operation, e.g. the empty sum is 0 since 0 is the additive identity). But 0/0 cannot be thought of this way as it's not formed by a product.

@@denverrsouthers5531bro u know that division is repeated subtraction right, so if u divide 0 by 0 u can subtract it both infinitely and 0 times so there is nog definition

You are wrong

No

Let him cook 😅...continue good sir

how about instead of just saying "uh you're wrong actually" you provide your actual argument? otherwise just don't even leave a comment, let alone two

Ma man, why do you have 2 comments and none of them actually explain or even mean anything? At least give good arguments lol. If thousands of mathematicians struggle to define 0⁰ without any flaws, why would you know how to do it. Literally the whole video is all about why there is no definitive answer to what 0⁰ equals💀