That's true. Let's take e^0. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. If you plug in x=0 and starting n=0, you get 0^0/0!+0^1/1!+0^2/2!+... All you have left is 0^0/0! By convention, 0^0 equals to 1 (this is an exception otherwise 0^0 is indeterminate) by convention and 0!=1 by definition. Therefore, e^0=1.
That's right because lets take 0 to the power of 2. That's the same thing as multiplying 1 by 2 times. 0 to the 1st power is multiplying 1 by 0 one time. So 0 to the power of 0 is the same thing as multiplying 1 by 0 zero times. Which means we're not multiplying 1 by 0 at all. We're not multiplying 1 by anything, so we get 1! You're right! How is it still undefined?!
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.
Yep, x^y is discontinuous at exactly (x, y) = (0, 0), but still many and many people try to tell each other a blind claim that 0^0 = lim_{(x, y) → (0, 0)} x^y which is undefined, which it is but they failed at their equality. Maybe there’s an unconscious belief that arithmetic operations + − × / ^ are all continuous everywhere, though a rare bird would claim / is continuous at (x, y) = (0, 0). And yet they would behave like ^ is. 🤦♂
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
when i tried to solve it gone like since we all know that when we dividing same base with different powers would result in subtraction of both powers and the bases would be same like 2^1÷2^1 = 2^(1-1)=2^0 = 1 and when we convert the exponents it would be 2÷2= 1 and for 0 everyone knows 0^2= 0 so 0^3÷0^1=0^(3-1) = 0^2=0 when converted 0÷0=0 and in that logic 0^2÷0^2=0^(2-2)=0^0 which is 0÷0 = 0
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.
Exponentiation is NOT a repeated multiplication. For example, matrix multiplication. Even to a set, it's still a quite questionable explanation, as A^B is a set of mapping from B to A
Your definition of exponentiation just doesn't work for anything other than positive intergers and for zero you are just *defining* it. Thus, it's not good to answer the question.
@@androkguz No, that is false. You were already corrected on this in another thread. 0 is a natural number, and it is cardinal number. It makes sense to talk about having 0 copies of x together, and multiplying them. And when you do, you get 1. This is not a matter of arbitrary definition.
The way I work it out is by how you work out exponentiating something by 0.5, which is the square root of the number. when its 0.25, its the 4th root of the number, etc. as the power gets closer to 0, the n in the nth root is going to be higher, since the n is determined by the formula 1/the power. as the n gets higher, the actual nth root of the number is going to get lower and lower or higher and higher, depending on if the number is higher or lower than 1 respectively, since the root of the number is a number powered by the n in the root. considering this, when the power is 0 (or in mathematician terms, approaching 0, since 0 is never actually used as a number, rather used the same way infinity is) then the n in the root is going to be equal to 1/0, which is approaching infinity. the infinite-th root of a number approaches 1 no matter if the number being rooted is higher or lower than 1. considering all of this which I probably got wrong since I just yapped based on reusing what I learned this year in my gcses, the root of 0 is undefined because it entirely depends on if the 0 is a true 0 or if the 0 is simply approaching 0, which are mathematically basically the exact same thing. you see when its approaching 0, there's still a tiny tiny value that is being powered, so therefore that value will approach 1 when rooted by infinity, however when the number is actually 0, then root will always be 0, considering that 0*0 is always 0. so you're left with 2 values of 1 and 0, so it is undefined. this call just be summed up to a single formula as well. x^y = (1/y)th root of x.
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.
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.
@@androkguz The empty product is 1 by definition; however, it is a highly motivated definition in that it is the _only possible_ value which preserves the associative property of multiplication on finite sequences. Since a _lot_ of our theorems and formulas involving products/exponents are based on the associative property, choosing the unique value for the empty product which preserves the associative property means it will work well with our theorems/formulas. I'm going to copy and paste an explanation I've given in the past: What is multiplication? That seems to be the fundamental question here. At its heart, multiplication is a binary operation, meaning that you have exactly two inputs. No more and no fewer. But obviously, this is too restrictive for many purposes. 30 cannot be written as a product of two primes. You need 30 = 2·3·5. But how does this work if multiplication is a binary operation? The answer is that you extend the meaning of multiplication. In particular, you look at iterated multiplication. For a·b·c, you have (a·b)·c. And for a·b·c·d, you have ((a·b)·c)·d. A priori, this is not in any sort of way canonical or unique. For a·b·c, you could have chosen a·(b·c), and for a·b·c·d, you could have chosen (a·(b·c))·d or a·((b·c)·d) or a·(b·(c·d)) or (a·b)·(c·d), etc. Luckily, multiplication is associative, so we don't have to worry about arbitrary choices. We can simply extend the meaning of multiplication to more than two factors. Although it's less obvious, you can also use the associative property of multiplication to extend the meaning of multiplication to fewer than two factors. The next few paragraphs explain how. Let S be a finite sequence of numbers. We will let Prod(S) denote the product of all of the elements in the sequence. If S and T are two sequences, we will denote the concatenation of S and T by S↔T. The concatenation of two sequences is a single sequence obtained by listing the entries of S followed by listing the entries of T. For some examples of this notation, consider S = {1, 2, 3} and T = {4, 7, 9, 10}. Prod(S) = 1·2·3 = 6. Also, S↔T = {1, 2, 3, 4, 7, 9, 10}. The associative property of multiplication implies that if S and T are finite sequences of numbers, then Prod(S)·Prod(T) = Prod(S↔T). In our example above of S and T, this means that (1·2·3)·(4·7·9·10) = 1·2·3·4·7·9·10. So what happens if you have a 1-term sequence? Let's say that U = {u} is a one-term sequence. What would Prod(U) be? *_If the associative property is expected to hold,_* then for every sequence T, Prod(U)·Prod(T) = Prod(U↔T). Now, U↔T is the sequence which starts with u and then lists every entry of T. Let's say T = {t1, t2, ..., tn}. Then Prod(U↔T) = u·t1·t2·...·tn. Of course, from the associative property, this is the same thing as u·(t1·t2·...·tn), which is equal to u·Prod(T). So Prod(U)·Prod(T) = u·Prod(T) for every sequence T. Therefore, the only possible conclusion is that Prod(U) = u. The product with one factor is the factor itself. Now what about the "empty" sequence? That is, what about a sequence with no terms? Let E be the empty sequence. What would Prod(E) be? Well, *_if the associative property is expected to hold,_* then for every sequence T, Prod(E)·Prod(T) = Prod(E↔T). Now, E↔T is the sequence which starts with the entries of E and then lists every entry of T. But there are no entries of E, so it just lists the entries of T. In other words, E↔T = T. So Prod(E)·Prod(T) = Prod(E↔T) = Prod(T) for every sequence T. In other words, for any sequence T, Prod(E)·Prod(T) = Prod(T). If you take T to be a 1-term sequence, then this states that Prod(E)·t = t for all numbers t. Therefore, the only possible conclusion is that Prod(E) = 1. The product with zero factors is 1.
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.
What you think and what the video discusses are two different things, and with you being wrong. If 0^1 equals 0, then how would 0^0 equal 1? What you’re saying makes no sense whatsoever.
Actually when we take a number N raised to the power M. It means we have to multiply 1 by N, M number of times. For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore 0 raised to the power 0 is 1 (1 not being multiplied with anything).
That's true, but sometimes 0^0 can be any value other than 0 and 1. Try the function y=x^(1/ln(x)). What is happening at x=0 and x=1? Does it have a value for both for these? You can tell both of these have indeterminate forms 0^0 and 1^infinity. As x goes to 0 from the right and x going to 1, both of these will go to e as a limit.
yee3816547290 0^0=1 by convention only works when you do power series. Let’s for example, we know that e^0=1. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. When you plug in x=0 and n=0 and all positive integers, you get 0^0/0!+0^1/1!+0^2/2!+…The rest of the sun equals to 0, and all you have left is 0^0/0! As you can see, by definition 0!=1 and by convention, 0^0=1. That’s why e^0=1. In general, 0^0 is indeterminate.
Shabbymannen 0^0 is the same as 0/0 using properties of exponents. 0^0=0^(4-4)=0^4/0^4=(0*0*0*0)/(0*0*0*0)=0/0, which is indeterminate. Think about this. 0/0=x then 0x=0. The thing here is x can have infinite answers to this (e.g. 1, sqrt(2), pi, arctan(3), etc.), but we want it to have a single value for this. That’s why the answer for 0^0 and 0/0 is indeterminate.
They got it wrong. a^b represents the number of functions from a set of b elements to a set of a elements. Essentially, for each element of the exponent set, you have base-many choices for where to send it. So 2^3 is the number of functions from {0, 1, 2} to {0, 1} For each element in {0, 1, 2} you have 2 choices where to send it. f(0) = 0 or 1 f(1) = 0 or 1 f(2) = 0 or 1 So the possible functions are 1. 0 ↦ 0, 1 ↦ 0, 2 ↦ 0 2. 0 ↦ 0, 1 ↦ 0, 2 ↦ 1 3. 0 ↦ 0, 1 ↦ 1, 2 ↦ 0 4. 0 ↦ 0, 1 ↦ 1, 2 ↦ 1 5. 0 ↦ 1, 1 ↦ 0, 2 ↦ 0 6. 0 ↦ 1, 1 ↦ 0, 2 ↦ 1 7. 0 ↦ 1, 1 ↦ 1, 2 ↦ 0 8. 0 ↦ 1, 1 ↦ 1, 2 ↦ 1 So there are 8 functions, making 2^3 = 8. When the exponent is 0, you have the functions from the empty set (set with no elements at all). This is difficult to imagine, but based on a logical technicality (vacuous truths), there is precisely one function from the empty set to any set (including from the empty set to itself). So in the set theory context, 0^0 = 1.
Thank you for your response! Might you mind explaining a bit more? I am unfamiliar with this topic and not sure what you mean by the number of "functions" or what the {0, 1, 2}/{0, 1}/etc sets mean exactly... @@MuffinsAPlenty
@@ninifeb14 In simple terms, you can think of a function as a black box with a hole for the input and a hole for the output. You might have a function that takes in whole numbers and gives whole numbers as output. You might put the number 1 into the input hole and get 2 out of the output hole, for example. It behaves consistently, so if you put a 1 into the input hole, you will always get a 2 out the other end. If you are given a black box, and you are given the numbers 0, 1, and 2 as inputs, and you are told that this particular black box can't give you any output other than 0 or 1, how many distinct behaviours can you imagine for this black box, before you try it out?
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.
well if we follow that pattern, 0 wouldnt exist because 0^1 would be the same as 0^(2-1) and that equals (0^2)/0 wchich is undefined. I was always taught that the "division=subtracting of exponents" rule does not apply to 0
Actually when we take a number N raised to the power M. It means we have to multiply 1 by N, M number of times. For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore 0 raised to the power 0 is 1 (1 not being multiplied with anything).
Yes but 0 isn't just the number you're multiplying by, it's also represents the number of times which you should be performing the multiplication. So it's telling you NOT to multiply by 0. So if you have 0 and do nothing to it, you still have 0. That's partially why it's undefined. Because there are multiple answers and both are true but neither can be proven.
@@dreddscott3873 No, that is a load of nonsense. If you multiply 0 times by 0, that is the same as not performing any multiplication at all, which is the same as multiplying by 1. Therefore, 0^0 = 1. It works the same as with any other base. 0^0 is not undefined. People have lied to you, and now you are lying to people. Brilliant is lying to people now too.
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.
@@angelmendez-rivera351 but any number to the power of 0 can also be written like 2^1÷2^1 which is 2^(1-1)= 2^0 = 1 and that disproves 0^0 = 1 since it would be 0^1÷0^1 which is 0÷0
@@randomguy-dky *but any number to the power of 0 can also be written like 2^1÷2^1 which is 2^(1-1)= 2^0 = 1 and that disproves 0^0 = 1 since it would be 0^1÷0^1 which is 0÷0.* No, this is incorrect, and even school teachers get this wrong. x^0 is _defined_ as the product of 0 copies of x. This is 1, and this is true for _all_ x. There are no exceptions, and thus, x^0 = 1 for all x. Now, IF x is invertible, THEN, then, the property x^(m + n) = x^m·x^n implies 1 = x^0 = x^1·x^(-1) = x·x^(-1), so in that case, x^(-1) denotes the inverse of x. But, 0 is not invertible, so that _does not apply_ here, and it is incorrect to say that 0^0 = 0·0^(-1), because it is false. So, no, that does not disprove 0^0 = 1. If it did disprove it, then it would also disprove 0^1 = 0. How? Because, by your logic 0^1 = 0^(2 + (-1)) = 0^2·0^(-1), and 0^(-1) is undefined, since 0 is not invertible. Also, for the love of Siesta, never ever use the symbol ÷. It is obsolete, and it is too much of a hassle. Just be a sensible human being and use / to denote division next time, please. We are all adults here, we are not in grade school.
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.
hold on i might be missing something here from 7 to 7x⁰ and letting 0⁰=0 when x=0 it seemed that u were already claiming that 0⁰=1 at the beginning implicitly and then u continued to try give 0⁰ a different value other than 1 it's like 7 = 7×1^a let's say we don't know what 1^a when a=25 If you let 1^25=0, 7=0 while in reality, we implicitly made 1^a=1 by saying 7=7×1^a so once again i might be missing something but that explanation does not actually explain why 0⁰=1 because 0⁰=1 was used already by itself to show 0⁰=1 which is basically going in circles
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)
I don't know why people say this stuff about calculus. 0^0 is not an expression that has anything to do with limits. You can tell by the facts that it doesn't contain the symbol "lim" and that it doesn't have a variable name underneath it, followed by that funny little arrow, followed by another symbol after the arrow. How are people not spotting these pretty major clues?
Aditya Shankar it’s supposed to say as x approaches 0 from the right. The limit will go to 1. The answer will go have to get into calculus. This is one of the indeterminate forms using l’Hopital’s rule. To get the answer to be 1, y=x^x is the same as y=e^(xln(x)). Look at the exponent for xln(x). Change this into a fraction as ln(x)/(1/x). Take the derivative of top and bottom separately, which is (1/x)/(-1/x^2). Using algebra and simplification, you get -x. When x goes to 0 from the right, you get 0. All together, you get e^0, which is 1. That’s the answer to the limit.
@@justabunga1 Actually what's cool about the function x^x is that the limit as it approaches 0 is 1 from the left as well. The real part approaches 1 and the imaginary part approaches zero.
Logically, 1 seems like the best answer. Raising something to a power means multiplying it against itself that number of times. When multiplying something against itself 0 times, the base would be irrelevant in all cases, including 0
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.
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?
What you are missing is that this is the result of defining exponentiation that way. You could have defined it as is usually done without the 1 and the positive intergers work fine. Plus your definition makes it really hard to extrapolate to what the hell is exponentiation of a non-interger or even a complex number
To get from 0³ -> 0², you'd have to divide by 0. Since we established that 0^n where n is a non-zero positive integer = 0, 0³ = 0 We can write 0 as 0¹, therefore: 0² = 0³/0¹, but dividing be 0 is undefined, so what's the logic behind exponentiating 0? Correct if I made an error but I'm just not seeing this
Power means how many times digit is multiple so 0^0 mean so there we can write as 0^0×1 But according to what according to exponent definition zero repeat zero time so zero does not present there that's what zero means so the answer is 1
Power means how many times digit is multiple so 0^0 mean so there we can write as 0^0×1 But according to what according to exponent definition zero repeat zero time so zero does not present there that's what zero means so the answer is 1
The way I work it out is by how you work out exponentiating something by 0.5, which is the square root of the number. when its 0.25, its the 4th root of the number, etc. as the power gets closer to 0, the n in the nth root is going to be higher, since the n is determined by the formula 1/the power. as the n gets higher, the actual nth root of the number is going to get lower and lower or higher and higher, depending on if the number is higher or lower than 1 respectively, since the root of the number is a number powered by the n in the root. considering this, when the power is 0 (or in mathematician terms, approaching 0, since 0 is never actually used as a number, rather used the same way infinity is) then the n in the root is going to be equal to 1/0, which is approaching infinity. the infinite-th root of a number approaches 1 no matter if the number being rooted is higher or lower than 1. considering all of this which I probably got wrong since I just yapped based on reusing what I learned this year in my gcses, the root of 0 is undefined because it entirely depends on if the 0 is a true 0 or if the 0 is simply approaching 0, which are mathematically basically the exact same thing. you see when its approaching 0, there's still a tiny tiny value that is being powered, so therefore that value will approach 1 when rooted by infinity, however when the number is actually 0, then root will always be 0, considering that 0*0 is always 0. so you're left with 2 values of 1 and 0, so it is undefined. this can all just be summed up to a single formula as well. x^y = (1/y)th root of x.
I don't know what 0^0 is, but I know that 0^1 + 0^1 = 0^2 satisfies an equation, namely x^n + x^n = x^(n+1) for x in {0,2} and n in the natural numbers with 0. Could it be the case that 0^0 + 0^0 = 0^1 without actually calculating it? I think not even as an algebraic expression.
x^n + x^n = x^(n + 1) only works for powers of 2. That’s literally how powers of 2 are defined. The next power is twice the last. 2^n + 2^n -> 2(2^n) -> 2^(n + 1)
@@AvatarBowler You seem tol have much Maths Knowledge. Im only 15 so the question may seem dumb but cant you add *multiplied by one* to an equation which would lead through basic thinking to 1=0^0
@@AvatarBowler Yeah but logic thought(Not mathemathic) 1*0^0 shouls be the same as 1 cause 0^0 should be nothing and only 1 is left. That was my thought
If you visualise the rationale behind n^0=1, you would realise that our math is designed in such a way that n^0 should be the multiplicative identity, hence 1. Naturally, the same definition can be used for 0^0, and should again, be 1. Hence the most logical and apt definition for 0^0 is 1.
Actually when we take a number N raised to the power M, it means we have to multiply 1 by N (M number of times). For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore if we take N=0 ,then 0 raised to the power 0 is 1 (1 not being multiplied with anything).
x^n = x^(n-1) * x is the equation you mean. Lets fill in x=1 and n=1: 1^1 = 1^0 * 1. In this case 1^0 has to equal 1 in order to complete the equation. But when we fill in x=0 and n=1 we get: 0^1 = 0^0 * 0. In this case 0^0 can mean both 1 and 0.
@@bartjesper Nope. It has nothing to do with that. The idea is very simple. If you multiply by x exactly 0 times, then this is the same as not multiplying at all. Not multiplying at all is the same as multiplying by 1. Hence, x^0 = 1. Notice how this is true for ALL x, it does not matter whether x is 0 or nonzero. If you want to get rigorous, you define x^n not in terms of an equation, but as the product of the constant n-tuple where every term is x.
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
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...
It is because you still have one thing to choose and that is 'nothing'! For eg, if you are told that you can get maximum marks 10 in a test (without negative marking), it means you can have 11 scores- 0,1,2....10 Similarly you are told that you'll only get zero marks if you dont attempt the test, it means you'll have only one type of score- 0. I hope it clears up your doubt!
I know it's probably wrong, but I just think as exponents as a reference to a sequence where the base is the growth rate, and the exponent is the number you look at 2^4 = 1, 2, 4 , 8, 16, 32 Count to the fourth one and that's the exponent
@@annaluiseblume nah this exponential functional trick wont work as A^ -x =1/A (it only works when A is not equal to 0) So this wont work if this works the 0 itself cant exist as 0^1 = 0^(2-1)= 0^2 × 0^-1 ? So not possible
O^0 can never be any sort of value , simply 0 itself is merely a marking of the starting point of any value or all values, so it does not represent any specific value like infinity. It means that no matter whatever sort of mathematic manipulating tricks you can use to support your fallacious arguments, not any sort of value or such undefined values can never give you any sort of value either
If you take a sample space (continuous example) omega { x , y } modulus x < 0 > 1 there exists an infinite number of real numbers in a graph of finite space x = 1, y = 1. A square.
Im pretty sure that 0^0 =1 makes more since, because if you do lim(x=0)x^x you will find the limit to approach 1 from the positive side. But yes, restricting x to only be base or exponent and not both would make 0^0 both 1 and 0
Some college math classes actually define 0^0 as 1. I believe it. I mean… Look at e^0. We know that e^0=1. The sum of e^x is the sum of [(x^n)/n!] from n=0 to n=∞. (0^0)/0!=1, and 0!=1; therefore 0^0=1.
@@MuffinsAPlenty Yes, it's standard and so WRONG. It's an arbitrary and useless convention that contradicts the exponent definition and any kind of formal logic, provided you have some...
@@pelasgeuspelasgeus4634 "Yes, it's standard and so WRONG." Ah, you're a mathematics conspiracy theorist. I see there is no reason to take you seriously. I hope that you are one day able to get some help.
Let x=0^0 For any n, (n is not infinity) x^n=(0^0)^n=0^(0*n)=0^0=x Take n=2, x=0,1. Take n=-1, x=1,-1 The intersection is 1. 1^n=1 is always true. Therefore 1 is the only value of 0^0 that satisfies index laws. Index laws are invalid when 0 is denominator or 0 to negative power is encountered.
The idea of using the number 0 is a problem in math. One of the problems would be division by 0, the infinity symbol, and several indeterminate forms. In basic math, our math teacher said that 0 divided by any non-zero number is always zero. A value itself by adding or subtracting 0 is the value by itself. Any number multiplied by 0 is always 0. Any number divided by 0 is always undefined. The log base of a positive number other than 1 of 0 is always undefined.
0^0 could, depending on the context, be defined as 1. As does every other number raised to the power of 0. Assuming this is true we can raise every number to the power of 0 and multiply them. This will give us: 0^0 x 0^0 x … = 1 1^0 x 1^0 x … = 1 2^0 x 2^0 x … = 1 Etc.
I understand zero to be the absence of a value, therefore the equations that use approximations are incorrect because you are still substituting a value into an equation meant to be moot by definition.
@@05degreesI'm inclined to agree, but I still think there's an interesting discussion to be had of the idea. If you define a base as the number of single digits it uses, for example base 10 ( decimal) uses the 10 digits 0 1 2 3 4 5 6 7 8 9, then presumably there can be no base that uses 0 digits. But if you define b in a base b simply as the number that is raised to successive powers beginning with 0 from right to left by which to multiply each digit in a string and add the results, for example 359 in decimal is 3x10^2 + 5x10^1 + 9x10^0, then some sense can be made of the idea. Granted, what you may also have to allow is that the digits can represent quantities exceeding b. For example D in hex represents 13 in decimal. So you can have a number 3D9 in decimal that represents 3x10^2 + Dx10^1 + 9x10^0 = 439. Now in base 0, all digits other than 0 are going to be greater than 0. 3D9 in base 0 is 3x0^2 + Dx0^1 + 9x0^0, which of course equals 9, the rightmost digit alone. All the others are just noise. So effectively what you have here is a representation of every new quantity with just a single new digit. Once you've run out of those Indo-Arabic numerals you probably go on to uppercase letters, lowercase, emojis, etc etc. Granted this isn't a base either, if by base you also mean a system of representing new quantities by recycling digits already used. I think there are some further interesting implications, but that'll do for now.
A simple and easy way is to have x^x and have x = 0.9 , 0.8 , 0.7 , ... 0.1 and then 0.01 , 0.001 , 0.0001 , .... When you calculate these, from 0.9 to 0.5, the number becomes smaller. At 0.4, the number becomes higher. The lower we go, the higher the number gets. So, as x comes closer and closer to 0, the closer and closer x^x comes to 1.
The empty set cannot be chosen from the empty set because the empty set is not an element of the empty set. Set theory can't be used to prove that 0**0 == 1.
0⁰ should be considered as an operation between two determined, equal numbers, not to be confused as the limit of two different functions f(x)^g(x) which tend to 0 with different speeds. In this case x^x represents all numbers raised to themselves. So the correct result is 1. The same goes for 0/0 = 1, but the properties need to be reviewed or redefined to make everything coherent (in this case we need to find a justification that 0/0 ≠ 3 as an example).
Yes, I know :), but the function y = x/x appears to have the value 1, when x - > 0. It might be more correct to represent x/x with two perpendicular lines.
Hm, that actually makes perfect sense, since x·y = x is equivalent to x·(y−1) = 0, so either x = 0 (and y is any number) or y−1 = 0 (and y = 1 while x is any number). Yeah, that's an interesting observation.
0^0 doesn’t always approach to 1. It can be 0, e, or any other limit value. You can check out blackpenredpen if you want to know if the value approaches something other than 1.
@@seroujghazarian6343 take for example e^0 using sum notation. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. Let x=0 and n=0,1,2,... You get 0^0/0!+0^1/1!+0^2/2!+0^3/3!+... All you have left is 0^0/0! By convention, 0^0=1 (this is an exception where this happens when doing Taylor/Maclaurin series otherwise it's left to be indeterminate) and 0! is always 1. That's why e^0=1.
@@Rohan-nf5hc 1:10 "a^b can be viewed as the number of sets of b elements that can be chosen from a set of a elements". This is not even close to being correct. In fact, it's obviously incorrect. You can see that in the expression a^b, b can be greater than a, and yet there are no sets of b elements that can be chosen from a set of a elements if b is greater than a. So if the video was correct, a^b would be zero for all b > a. This is so far from being true that it is staggering that nobody making this video spotted the mistake. The correct statement is that a^b can be viewed as the number of functions from a set of b elements to a set of a elements. 2:17 "We're interested in limits of the form 0^0 when x = 0." This is a meaningless statement. 0^0 is not a statement of a limit. A statement of a limit would involve the symbol "lim" with a little arrow beneath it.
@@omp199 "The correct statement is that a^b can be viewed as the number of functions from a set of b elements to a set of a elements." Yes. This is correct. Brilliant has said in some comment somewhere that they intended to say: "a^b can be viewed as the number of _multisets_ (i.e., sets where repetition of elements is allowed) of b elements that can be chosen from a set of a elements" which has the same cardinality as functions from b to a. Still, though, what they actually said was wrong.
But 3^2 = 3×3 But 0^0 = 0 raised to how many times? Im in class 9 and I asked my teacher about it today when she was teaching degree of polynomial But she said how many times are you going to multiply 0 with... I wasn't satisfied with her answer so found you video.. But still I'm confused
@@justabunga1 nope. Algebra takes it 1. Eg there is 1 and only one function from the empty set to itself and that is the necessary bijection. So the cardinality of (empty set)^(empty set) (set of functions from the empty set to itself) is 1. And we all know that card(F^E)=Card(F)^Card(E) And Card(empty set)=0 0^0=1
it should be 0 because 0 to the power of 0 basically means 0 repeating itself 0 times, which is 0. the same way 2 to the power of 3 is 2 repeating itself 3 times. if a number is repeated 0 times doesnt that mean its 0??
No, 0^0 should be equal to 1. Many arguments tend to use this in order to prove that 0^0 = undefined, 0^0 = 0^1-1 = 0^1/0^1 = undefined, though, what many misunderstands is that this equation is ONLY true if X is not equal to 0, it follows this pattern (x^a-b = x^a/x^b) which doesn't apply to a variable with 0 as it's value. Another argument is by using a limit (lim x > 0+ for x^0) = 1, and use the opposite limit (lim x > 0+ for 0^x) = 0, People argue that since this is is multivalued then it is undefined, though keep in mind that these are not ABSOLUTE values, limits is a function where a variable 'approaches' a value and not AT a specific value, meaning that this is be neglectable, and 0^0 itself isn't a limit. Another thing, 0^0 is used in many different mathematical functions, for example, the binomial theroem which was mentioned in the video requires 0^0 to be defined as the simplest calculated integer for the equation, without it being defined as 1 this theorem would need to undergo many complex configurations in order to be totally accurate. You are stating that 0^0 should be equal to 0, but that is not how exponents work. Exponents is an operation which determines how many times a value should be multiplied by itself, and not "repeating" itself (relates more to multiplication), consider exponents to be how many times you can arrange a certain number, if we take this definition and use x^0 for it , then we can ask the question "how many times can I arrange x for a list that doesn't exist"? The answer is 1 as there is only one way you can arrange them, and that is the initial state.
Why not zero then? Multiplication is a short form of addition, and exponents/indices/powers are a short form of multiplication. 0 x 0 = 0. Anything else appears to be accepting a falsidical paradox. ~~~~~ Read below for further explanation ~~~~~ The other patterns with _actual values to the power of zero that equal one_ explain this result with a break down either back track to exponential pattern like 2^4 = 16 then 2^3 = 8 to reach 2^0 =1 accepting the pattern presumption as proof... *OR* by using equation proofs that have the second last step showing the number or variable to one exponential value subtracting the same exponential value as proof to anything to power of zero equals one. Seems legit since two proofs by reversing an equation say so. Or is it really? We are expected to accept these explanations because so far it has proven mathematically useful to do so. But all these equations only seem to make sense through these double check proofs. Yet fails the common sense part of the equation. Zero times Zero done zero amount of times still leaves you at zero. I mentioned the falsidical paradox because that's exactly what the Achilles and Tortoise Paradox is. Scholars could never prove mathematically that Achilles could actually catch the Tortoise despite real life examples. Not until Convergent series were discovered with calculus that we saw how math proved what we already knew in real life.
consider: (0+1)^0 which is equal to: 0C0 • 0^0 • 1^(0-0) In this case (0+1)^0= (1)^0 = 1 So 0C0 • 0^0 • 1^(0-0) has to be equal to 1. And it can only be if 0^0=1. So it is not undefined, unless we claim and prove that 1^0 is not equal to 1.
It is not indeterminate....in fact you can use a direct limit to calculate the correct answer of 1 0^0 = lim_{x->0} x^x = lim_{x->0} exp(x*ln(x)) = ..... (apply De L'Hospital rule) ..... = 1
But if 0^0 is the number of sets of 0 elements that can be chosen from a set of 0 elements, then wouldn’t that be equal to 0? Because there wouldn’t be any elements to choose from, so there wouldn’t be any way to choose them
It's ome: the own empty set. The set with zero elements, the only set that is contained into the empty set is the empty set itself, so that's why its one. If it was zero, then the empty set wouldnt exist. But it does.
0^0 is undefined. Because a number raised to the zero es one due to the exponent property. For example 3^0 = 1 because 3^0= 3^3 / 3^3 Exponents substraction 3^0 = 27/27 3^0= 1 Now. If we put 0 0^0= 0^3 / 0^3 0^0= 0/0 0/0 = 1? No it doesn't. If 0^0 was 1, then 0/0 would also be one, which doesn't make any sense. So 0^0 is undefined
Not the same. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
1 is a consistent answer. You _cannot_ find a valid contradiction to having 0^0 = 1. All of the arguments which are typically used to say that 0^0 must be undefined either use invalid reasoning (such as applying mathematical formulas where they do not apply) or confusing the arithmetic expression 0^0 with the limiting form (→0)^(→0).
If we could find what 1/0 is equal to we can solve this mathematically inquiry. This is because n^0 is equal to n × 1/n. E.g., 2^0 is 1 because 2 × 1/2 is equal to 1, 1,000,000^0 is 1 because 1,000,000 × 1/1,000,000 is equal to 1. Even works for negatives. I.e., to find 0^0 we just need to calculate 1/0 and then multiply that result by 0 and then boom, that's the real answer for 0^0.
If you guys want my opinion into he answer to 0^0. I thinks it is to an extent 0, and here's why. Like I previously explained, n × 1/n is equal to n^0. So for 0^0, if we want to find that we need to do 0 × 1/0, so what's 1/0? Well because 0 cannot go into 1, we place a 0, 1-0=1, 0 cannot go into 1, so we get 0.0/still 0. Do it even you're go to get an infinite amount of zeros without any _subsequential_ digits. So we get 0 simply, then we 0 × 0, which equals 0. So we can say 0^0=0. But 0 × 1/0 is equal to 0/0 which equals 0? That defeats the rules of division. So really unless math is drastically changed, there seems to not be a valid formula for particularly 0^0.
Wrong, 0^0 writing is itself crime in maths. There is no such thing as it depends! Q: is irrational + irrational = irrational? A: False (eg π + 5-π) and not it depends So 0^0 is undefined.
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
International Conference on Differential & Difference Equations and Applications ICDDEA 2017: Differential and Difference Equations with Applications pp 293-305 | Cite as log0=log∞=0 and Applications Authors Authors and affiliations
But you people, when you 'invented' 0, your definition of 0/0 was WRONG. 0/0 is not equal to 0, as was stated in the 'invention'. It (0/0) is now considered undefined. 0 is also NOT equal to ∞, how in the world is that possible?
You say 2^1 = 2 because it's the number of sets of 1 element that can be chosen from 2 elements. Then you say that means 0^0 = 1 because it's the number of sets of 0 elements that can be chosen from a set of 0 elements. That does not follow. You can't make any sets from the null set except null. If you count that, then 2^1 should be two sets plus the null set so the answer should be 3.
"You can't make any sets from the null set except null." The null set is a set of size 0. So it is 1 set of size 0 you can make from any set. "If you count that, then 2^1 should be two sets plus the null set" But the null set is size 0, not size 1, so it shouldn't be counted in the collection of sets with 1 element.
Take log0 = y and first doing deffierentaite and then integrate we get value of logo and then z = 0 power 0 and take log an both side and put the value of log0 and we get 0 power 0 is 1
"0^(1-1) = 0¹/0¹" This step is invalid. The exponentiation rules you are using here do _not_ apply when 0 is the base. For example, would it be valid to say 0^1 = 0^(2-1) = 0^2/0^1 = 0/0? No, it wouldn't. But that argument is just as valid as the argument you presented for 0^0 = 0/0.
Многие думают что на ноль можно умножать а делить "почти" совсем ни как нельзя... Типа X×0 = 0 это нормально лишь потому что 0/X = 0...? Но из этого же следует что сам X = 0/0...? Х=0⁰...? ну и где логика... Давайте рассмотрим один из вариантов как обычно происходит действие деления... 6:2=6/2=(2+4)/2=2/2+4/2=1+(2+2)/2= =1+2/2+2/2=1+1+2/2=3 (без остатка...) 7:2=7/2=(2+5)/2=2/2+5/2= =1+(2+3)/2=1+2/2+3/2= =1+1+(2+1)/2=1+1+2/2+1/2= =1+1+1+1/2=3+1/2=3 с остатком 1... И это также можно с помощью принятых форм математических записей выразить как 3½ или 3.5... А что же происходит когда якобы производят деление на ноль... многие говорят что это будет равно какой то бесконечности... 15:0=15/0=(0+15)/0=0/0+(0+15)/0= =0/0+0/0+(0+15)/0=0/0+...+0/0+15/0... и при дальнейших действиях всегда такое деление будет c постоянным остатком в виде того что "делилось" изначально... в данном случае остаток 15... и почему то вот об этом остатке или забывают или неосознанно замалчивают считая только бесполезные бесконечные действия не приводящие ни к какому результату деления... Если быть немного логичным то видно что даже при бесконечном количестве таких действий деления (а точнее бездействий) вся сумма таких действий равна нулю с постоянным остатком того что было изначально делимым... То есть само такое деление не происходит... сколько было изначально столько и остаётся в остатке неделимо... X:0=X/0=(0/0)×N+X/0=N×(0/0) с неразделённым остатком X где N×(0/0)=0 и N число мнимых манипуляций не производящих деления... поэтому N=0... а не бесконечность... отсюда и получается два ответа при делении на ноль... относительный ответ равен 0... но именно ноль бессмысленных манипуляций... а безотносительный ответ равен самому значению делимого X... В примере 15/0 = 0 целых 15 нулевых... или же 0 целых и 15 в остатке... именно умножая это число на ноль можно получить первоначальное данное значение... Но об этом как правило неумышленно умалчивают... ведь этому не научили...
Полное непонимание современной математики темы умножения на ноль и тем более деления на ноль... При записи умножения числового значения X на ноль получаем ----- X×0=0 X /\ 0/0 Перенос ноля через знак равно превращает равенство в качельное неравенство типа ---- 100% /\ 0% Сам знак процентов кстати пишется как 0/0... То же самое и с делением на ноль... ---- X/0=0 Х /\ 0×0 Перенос ноля через знак равно превращает равенство в качельное неравенство... Я называю это нулёвыми значениями (не нулевыми а именно нулёвыми... "ни разу взятыми" или "ни разу трачеными" то есть "нерастрачеными" если это об умножении на ноль... "ни разу делёнными" или "неразделёнными" если это о делении на ноль)... И непонимание до сих пор этого простого меня очень удивляет... 220 вольт делить на ток 0 Ампер это сопротивление = 0 Ом... но это просто не потраченое напряжение... 1 торт не взятый кусками ни разу (деленный на ноль) 1/0 это 0 кусков взятых но это все тот же 1 неразделённый торт... 5 монет ни разу не взятых 5×0 это 0 взятых монет но вопрос как правило звучит "сколько будет" а не сколько взято... так вот будет все те же 5 невзятых "нулёвых" монет... И это всего лишь маленькая вершина айсберга действий с нолем... Сложение и вычитание нуля не меняет первоначального значения... а почему? да потому что на самом деле не происходит самого математического действия как такового... ничего не прибавляется и не убавляется при этом... С умножением и делением на ноль происходит примерно тоже самое... Ничего не происходит при этих действиях... всего лишь описывается что первоначальные значения не изменяются... хотя ответов получается два относительный = 0 безотносительный = 100% = 1×Х(нулёвое) в зависимости от поставленного вопроса... И безотносительный ответ имеет гораздо больше смысла... 5 метров × 5 метров × 0 метров = 25 метров² × 0 метров = ? Относительно нуля ответ 0 метров³ Безотносительно нуля = 25 метров² ---- 25м²(=100%) /\ 0м³ : 0м (= 0м²) Перенос нуля (при умножении на ноль или делении на ноль) через знак равно превращает равенство в качельное неравенство ----- 1×X(нулёвое)(=100%) /\ 0(=0%) Безотносительный ответ при действиях умножения и деления с нулем не учитывает как само действие с нулем так и его измерение... 5 яблок : 0 корзин = ? Относительный ответ 0 яблок на корзину... Безотносительный ответ 5 неразделенных яблок (без корзин)... Убираем ноль и его измерение из вычисления и получаем нетронутые первоначальные данные и можем дальше с ними что то вычислять... ---- 5 яблок нулёвых (= 100%) /\ 0 корзин × 0 яблок/на корзину (= 0) Чисто качельное 100% неравенство... Откуда здесь могут взяться какие то бесконечности? Или черные дыры? Кстати 0(нулёвый) / 0(нулёвый) = 1 Впрочем как и любое другое число раз делённое на само себя... Это всего лишь малая часть моего личного взгляда на действия с нулем и он не ограничивается только этими действиями...
Ноль не имеет численного значения... он лишь описывает отсутствие чего либо... Практически все действия с нолем на самом деле не происходят... Многие пытаются ноль "всунуть" в основные математические действия... при этом абсолютно не понимая смысла самой записи таких действий... но с нулем есть только математические "бездействия" и чаще всего действия "умножения" и "деления" связанные с нулем говорят что есть что то безотносительное до той поры пока вместо нуля в таких выражениях не появится числовое значение... Лишь после этого выражение становится относительным... Если напряжение = 0 и сила тока = 0 то это не значит что при этом всегда нет сопротивления... Если скорость = 0 и время = 0 то расстояние при этом может быть каким угодно (в том числе и отсутствовать)... Поймите главное перенос нуля через знак равно изменяет смысл равенства на качельное неравенство 100% того что было изначально и есть до сих пор и с другой стороны 0% того что якобы "взято"... И никаких бесконечностей и всяких "черных дыр" при явном нуле в таких действиях никогда не будет... Чисто математически ЛЮБОЕ "действие" когда Х не равное нулю "умножается" на ноль или "делится" будет равно ВСЕГДА нулю... то есть отсутствию таких отношений... Но смысл совершенно не в этом... любое такое "действие" описывает лишь неизменность самого стопроцентно имеющегося значения X при этих нулевых "операциях" с ним...
Что касается "деления" 0/0... (или по другому выражения типа 0⁰...) Если вы делите два различных нуля один на другой (с различными мерами измерения) то "относительный" математический ответ этого будет 0 = 0% того что использовано... Но безотносительный ответ будет равен 100% того что было дано изначально и не было использовано в ходе бездейственного "деления" отсутствия одной величины на отсутствие другой... Если делить один ноль сам на себя (с одной и той же мерой измерения) то ответ равен 1 раз... И никаких 2 раза... 3 раза... и т.п. у отсутствия величины в виде ноля не будет... Интересно как можно объяснить 0/0 = 0⁰ с точки зрения "практических" равенств... Многие считают что 0⁰ = 1... Напряжение U = 0 вольт... Сила тока I = 0 ампер... Сопротивление R = U/I = 0/0 = 0⁰ = 1...? Ом...? Весело... Дистанция S = 0 километров... Время t = 0 часов... Cкорость V = S/t = 0/0 = 0⁰ = 1...? километров/час...? Смешно... Объем V = 0 метров³... Ширина W = 0 метров... Высота H = 0 метров... Длина L = V/(W×H) = 0/(0×0) = 0/0² = 0‐¹ = 1/0...? = ...? Сколько будет...? метров? Интересно сможет хоть кто то это объяснить хоть как то математически...?
Многие математики почему то считают что у нуля нет обратной величины... Другие свято верят что величина обратная нулю это "бесконечность"... К сожалению (ну или к счастью) у "бесконечности" есть обратная величина равная 1/бесконечность... И как бы она ни была мала она НИКОГДА не будет равна нулю... И уж точно она не имеет безотносительного значения... К тому же она имеет знак плюс или минус в зависимости от того с каким знаком берется сама "бесконечность"... (если конечно она хоть как то вообще может быть "взята"...) У полного отсутствия в виде нуля есть обратная величина... это полное присутствие... и для нуля это равно единице... то есть 100% присутствие чего либо... 1/0 "относительный" ответ математически равен нулю... но именно он не имеет смысла а вот безотносительный ответ как раз равен "ни разу делённой" то есть нераздельной (нулёвой) единице... Никакой бесконечности при делении на ноль не бывает... если только сама бесконечность не делится на ноль... 1/0 равна 0 целых и 1 в остатке... полностью неделённая единица... Умножьте обратно 0×0 целых и прибавьте остаток 1... получите изначальное имеющееся число якобы "делённое" на ноль... Деление на целые части заканчивается когда вы не можете больше "отсоединить" от делимого количества записанного в делитель... При нуле находящимся в делителе вы не сможете "отсоединить" вообще ничего от делимого числа пытаясь вычесть ноль... даже при "бесконечных" таких попытках... поэтому для действия деления на ноль это равно всегда ноль целых... а остальное неделимый остаток...
Много есть искусственных точек нулевого отсчета в различных измерениях различных величин... Но в большинстве своем они не имеют никакого отношения к делению на ноль... Я лишь изложил некоторые видения своей теории математических "действий" с нулевыми значениями... На сегодняшний день никто не смог переубедить меня в этом... и даже наоборот после дискуссий на эту тему мое личное убеждение в моей правоте возрастает все больше... Вам же желаю всех благ в деле поиска и распространения знаний...
This is a commonly repeated argument, and I can understand why it seems convincing. The exponential "rules" that most people learn in elementary school actually have restrictions on them, but most people aren't used to seeing these restrictions. These rules work when you have a _strictly positive_ base; however, some of these rules do not work when the base is 0 or negative. For example, (a^b)^c = a^(bc) is not, in general, a valid rule if the base is negative and if the exponents aren't positive integers. ((−1)^2)^(1/2) = 1^(1/2) = 1 but (−1)^(2*1/2) = (−1)^1 = −1 So these are not the same. The rule you're using here, that a^(b−c) = a^b/a^c is not valid *at all* when a = 0. Let's suppose that it is valid: that 0^(a−b) = 0^a/0^b (I think already writing it this way, you may start to see a problem.) So if this is valid, then we could say 0^2 = 0^(4−2) = 0^4/0^2 = 0/0 The step in the first equality just uses the fact that 4−2 = 2. I think we agree with this. The step in the third equality just uses that 0^4 = 0 and 0^2 = 0. I think we agree with that as well. So the only possible error in this argument is 0^(4−2) = 0^4/0^2. Of course, using 0^2 here is pretty irrelevant. You could use any number instead of 2 and get the same issue. So the problem is that 0^(a−b) is not equal to 0^a/0^b. This then shows that your argument that 0^0 = 0/0 is faulty as well.
Not the same. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
I once said that 0^0 is 1 in class and everyone started laughing at me.
That's true. Let's take e^0. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. If you plug in x=0 and starting n=0, you get 0^0/0!+0^1/1!+0^2/2!+... All you have left is 0^0/0! By convention, 0^0 equals to 1 (this is an exception otherwise 0^0 is indeterminate) by convention and 0!=1 by definition. Therefore, e^0=1.
@@justabunga1 WOW great explintion
Happened to me yesterday, even the teacher told me "that s impossibile" :/
@@TheWayOfTheOtaku tell them the exponential function says hi!
That's right because lets take 0 to the power of 2. That's the same thing as multiplying 1 by 2 times. 0 to the 1st power is multiplying 1 by 0 one time. So 0 to the power of 0 is the same thing as multiplying 1 by 0 zero times. Which means we're not multiplying 1 by 0 at all. We're not multiplying 1 by anything, so we get 1! You're right! How is it still undefined?!
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.
Exactly.
Wrong
Yep, x^y is discontinuous at exactly (x, y) = (0, 0), but still many and many people try to tell each other a blind claim that 0^0 = lim_{(x, y) → (0, 0)} x^y which is undefined, which it is but they failed at their equality. Maybe there’s an unconscious belief that arithmetic operations + − × / ^ are all continuous everywhere, though a rare bird would claim / is continuous at (x, y) = (0, 0). And yet they would behave like ^ is. 🤦♂
@@gigachadkartik Projection much
At least provide reason as to why you believe that, otherwise you sound like an old granny yapping her mouth.@@gigachadkartik
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
Yes! Exactly!
when i tried to solve it gone like
since we all know that when we dividing same base with different powers would result in subtraction of both powers and the bases would be same
like
2^1÷2^1 = 2^(1-1)=2^0 = 1
and when we convert the exponents it would be
2÷2= 1
and for 0
everyone knows 0^2= 0
so
0^3÷0^1=0^(3-1) = 0^2=0
when converted
0÷0=0
and in that logic
0^2÷0^2=0^(2-2)=0^0
which is
0÷0 = 0
@@randomguy-dky That exponents property is not applicable when the base is zero.
@@programmer4047 o
@@randomguy-dky 0÷0=1 not 0
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
Correct. Life is unfair. High-quality educational vids has less views compare for those who just discuss 1/2 + 3/4 in 8 minutes got million of views..
what Video
@@yasyasmarangoz3577 Eddie Woo channel video
The first result result u search "zero power zero"
@@alice_in_wonderland42 No, I want him to show me which he means.
@@alice_in_wonderland42 Eddie Woo does not get it wrong, though. This video by Brilliant does get it wrong, though.
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.
Exponentiation is NOT a repeated multiplication. For example, matrix multiplication. Even to a set, it's still a quite questionable explanation, as A^B is a set of mapping from B to A
Your definition of exponentiation just doesn't work for anything other than positive intergers and for zero you are just *defining* it. Thus, it's not good to answer the question.
@@androkguz No, that is false. You were already corrected on this in another thread. 0 is a natural number, and it is cardinal number. It makes sense to talk about having 0 copies of x together, and multiplying them. And when you do, you get 1. This is not a matter of arbitrary definition.
The way I work it out is by how you work out exponentiating something by 0.5, which is the square root of the number. when its 0.25, its the 4th root of the number, etc. as the power gets closer to 0, the n in the nth root is going to be higher, since the n is determined by the formula 1/the power. as the n gets higher, the actual nth root of the number is going to get lower and lower or higher and higher, depending on if the number is higher or lower than 1 respectively, since the root of the number is a number powered by the n in the root. considering this, when the power is 0 (or in mathematician terms, approaching 0, since 0 is never actually used as a number, rather used the same way infinity is) then the n in the root is going to be equal to 1/0, which is approaching infinity. the infinite-th root of a number approaches 1 no matter if the number being rooted is higher or lower than 1.
considering all of this which I probably got wrong since I just yapped based on reusing what I learned this year in my gcses, the root of 0 is undefined because it entirely depends on if the 0 is a true 0 or if the 0 is simply approaching 0, which are mathematically basically the exact same thing. you see when its approaching 0, there's still a tiny tiny value that is being powered, so therefore that value will approach 1 when rooted by infinity, however when the number is actually 0, then root will always be 0, considering that 0*0 is always 0. so you're left with 2 values of 1 and 0, so it is undefined.
this call just be summed up to a single formula as well. x^y = (1/y)th root of x.
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.
Sorry about that. We should have clarified that we're using multi-sets, where you're allowed to pick the same element multiple times.
Let us try 2^0, the number of sets of 0 elements that can be chosen from a set of 2 elements ??
@@Rashidamin there is only one set which satisfies this condition : the empty set
@@micrapop_6390 and how many sets of 0 elements can be chosen from a set of 0 elements? The set itself!
@@seroujghazarian6343 yeah, that is why the binomial coefficient (0 0) is 1 :)
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.
😊
Wow
Biggest /woooosh I've seen in a while.
"The empty product is equal to 1"
Everything you say is based on this. Prove it please
@@androkguz The empty product is 1 by definition; however, it is a highly motivated definition in that it is the _only possible_ value which preserves the associative property of multiplication on finite sequences. Since a _lot_ of our theorems and formulas involving products/exponents are based on the associative property, choosing the unique value for the empty product which preserves the associative property means it will work well with our theorems/formulas. I'm going to copy and paste an explanation I've given in the past:
What is multiplication? That seems to be the fundamental question here.
At its heart, multiplication is a binary operation, meaning that you have exactly two inputs. No more and no fewer. But obviously, this is too restrictive for many purposes. 30 cannot be written as a product of two primes. You need 30 = 2·3·5. But how does this work if multiplication is a binary operation? The answer is that you extend the meaning of multiplication. In particular, you look at iterated multiplication. For a·b·c, you have (a·b)·c. And for a·b·c·d, you have ((a·b)·c)·d. A priori, this is not in any sort of way canonical or unique. For a·b·c, you could have chosen a·(b·c), and for a·b·c·d, you could have chosen (a·(b·c))·d or a·((b·c)·d) or a·(b·(c·d)) or (a·b)·(c·d), etc. Luckily, multiplication is associative, so we don't have to worry about arbitrary choices. We can simply extend the meaning of multiplication to more than two factors.
Although it's less obvious, you can also use the associative property of multiplication to extend the meaning of multiplication to fewer than two factors. The next few paragraphs explain how.
Let S be a finite sequence of numbers. We will let Prod(S) denote the product of all of the elements in the sequence. If S and T are two sequences, we will denote the concatenation of S and T by S↔T. The concatenation of two sequences is a single sequence obtained by listing the entries of S followed by listing the entries of T. For some examples of this notation, consider S = {1, 2, 3} and T = {4, 7, 9, 10}. Prod(S) = 1·2·3 = 6. Also, S↔T = {1, 2, 3, 4, 7, 9, 10}.
The associative property of multiplication implies that if S and T are finite sequences of numbers, then Prod(S)·Prod(T) = Prod(S↔T). In our example above of S and T, this means that (1·2·3)·(4·7·9·10) = 1·2·3·4·7·9·10.
So what happens if you have a 1-term sequence? Let's say that U = {u} is a one-term sequence. What would Prod(U) be? *_If the associative property is expected to hold,_* then for every sequence T, Prod(U)·Prod(T) = Prod(U↔T). Now, U↔T is the sequence which starts with u and then lists every entry of T. Let's say T = {t1, t2, ..., tn}. Then Prod(U↔T) = u·t1·t2·...·tn. Of course, from the associative property, this is the same thing as u·(t1·t2·...·tn), which is equal to u·Prod(T). So Prod(U)·Prod(T) = u·Prod(T) for every sequence T. Therefore, the only possible conclusion is that Prod(U) = u. The product with one factor is the factor itself.
Now what about the "empty" sequence? That is, what about a sequence with no terms? Let E be the empty sequence. What would Prod(E) be? Well, *_if the associative property is expected to hold,_* then for every sequence T, Prod(E)·Prod(T) = Prod(E↔T). Now, E↔T is the sequence which starts with the entries of E and then lists every entry of T. But there are no entries of E, so it just lists the entries of T. In other words, E↔T = T. So Prod(E)·Prod(T) = Prod(E↔T) = Prod(T) for every sequence T. In other words, for any sequence T, Prod(E)·Prod(T) = Prod(T). If you take T to be a 1-term sequence, then this states that Prod(E)·t = t for all numbers t. Therefore, the only possible conclusion is that Prod(E) = 1. The product with zero factors is 1.
@@androkguz wrong usage
bro wrote down an entire high school semester
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.
What you think and what the video discusses are two different things, and with you being wrong. If 0^1 equals 0, then how would 0^0 equal 1? What you’re saying makes no sense whatsoever.
Actually when we take a number N raised to the power M. It means we have to multiply 1 by N, M number of times. For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore 0 raised to the power 0 is 1 (1 not being multiplied with anything).
@@BEASTMAN992 exactly
@@BEASTMAN992: (0ˣ = 0) _only_ (| x > 0)
no matter how many times we mutiply we get the value = 0 (0x0=0 or 0^anything = 0
Please don't add music, it's distracting
The background music is distracting
It’s not some kind of rock music of something it’s just chill music trying to help you focus
0^0=1 is always the best definition.
Function value is different from limit.
The two need not to be equal.
That's true, but sometimes 0^0 can be any value other than 0 and 1. Try the function y=x^(1/ln(x)). What is happening at x=0 and x=1? Does it have a value for both for these? You can tell both of these have indeterminate forms 0^0 and 1^infinity. As x goes to 0 from the right and x going to 1, both of these will go to e as a limit.
As for 0^0 is concerned,
limit is not a necessary consideration.
Infinity is not a specific number,
it is a topic of limit.
yee3816547290 0^0=1 by convention only works when you do power series. Let’s for example, we know that e^0=1. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. When you plug in x=0 and n=0 and all positive integers, you get 0^0/0!+0^1/1!+0^2/2!+…The rest of the sun equals to 0, and all you have left is 0^0/0! As you can see, by definition 0!=1 and by convention, 0^0=1. That’s why e^0=1. In general, 0^0 is indeterminate.
@@justabunga1 : Can you give any example that does not relate to limits, where 0^0 could be shown to be indeterminate, or anything other than 1?
Shabbymannen 0^0 is the same as 0/0 using properties of exponents. 0^0=0^(4-4)=0^4/0^4=(0*0*0*0)/(0*0*0*0)=0/0, which is indeterminate. Think about this. 0/0=x then 0x=0. The thing here is x can have infinite answers to this (e.g. 1, sqrt(2), pi, arctan(3), etc.), but we want it to have a single value for this. That’s why the answer for 0^0 and 0/0 is indeterminate.
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.
They got it wrong. a^b represents the number of functions from a set of b elements to a set of a elements.
Essentially, for each element of the exponent set, you have base-many choices for where to send it.
So 2^3 is the number of functions from {0, 1, 2} to {0, 1}
For each element in {0, 1, 2} you have 2 choices where to send it.
f(0) = 0 or 1
f(1) = 0 or 1
f(2) = 0 or 1
So the possible functions are
1. 0 ↦ 0, 1 ↦ 0, 2 ↦ 0
2. 0 ↦ 0, 1 ↦ 0, 2 ↦ 1
3. 0 ↦ 0, 1 ↦ 1, 2 ↦ 0
4. 0 ↦ 0, 1 ↦ 1, 2 ↦ 1
5. 0 ↦ 1, 1 ↦ 0, 2 ↦ 0
6. 0 ↦ 1, 1 ↦ 0, 2 ↦ 1
7. 0 ↦ 1, 1 ↦ 1, 2 ↦ 0
8. 0 ↦ 1, 1 ↦ 1, 2 ↦ 1
So there are 8 functions, making 2^3 = 8.
When the exponent is 0, you have the functions from the empty set (set with no elements at all). This is difficult to imagine, but based on a logical technicality (vacuous truths), there is precisely one function from the empty set to any set (including from the empty set to itself). So in the set theory context, 0^0 = 1.
Thank you for your response! Might you mind explaining a bit more? I am unfamiliar with this topic and not sure what you mean by the number of "functions" or what the {0, 1, 2}/{0, 1}/etc sets mean exactly... @@MuffinsAPlenty
@@ninifeb14 In simple terms, you can think of a function as a black box with a hole for the input and a hole for the output. You might have a function that takes in whole numbers and gives whole numbers as output. You might put the number 1 into the input hole and get 2 out of the output hole, for example. It behaves consistently, so if you put a 1 into the input hole, you will always get a 2 out the other end.
If you are given a black box, and you are given the numbers 0, 1, and 2 as inputs, and you are told that this particular black box can't give you any output other than 0 or 1, how many distinct behaviours can you imagine for this black box, before you try it out?
@@omp199 Thank you for your response
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.
well if we follow that pattern, 0 wouldnt exist because 0^1 would be the same as 0^(2-1) and that equals (0^2)/0 wchich is undefined. I was always taught that the "division=subtracting of exponents" rule does not apply to 0
Actually when we take a number N raised to the power M. It means we have to multiply 1 by N, M number of times. For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore 0 raised to the power 0 is 1 (1 not being multiplied with anything).
Yes but 0 isn't just the number you're multiplying by, it's also represents the number of times which you should be performing the multiplication. So it's telling you NOT to multiply by 0. So if you have 0 and do nothing to it, you still have 0. That's partially why it's undefined. Because there are multiple answers and both are true but neither can be proven.
@@dreddscott3873 No, that is a load of nonsense. If you multiply 0 times by 0, that is the same as not performing any multiplication at all, which is the same as multiplying by 1. Therefore, 0^0 = 1. It works the same as with any other base. 0^0 is not undefined. People have lied to you, and now you are lying to people. Brilliant is lying to people now too.
maxw3[[
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.
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.
Exactly. I find it unbelievable that so many people do not understand such a simple concept.
@@angelmendez-rivera351 but any number to the power of 0 can also be written like
2^1÷2^1 which is 2^(1-1)= 2^0 = 1
and that disproves 0^0 = 1 since it would be 0^1÷0^1
which is 0÷0
@@randomguy-dky *but any number to the power of 0 can also be written like 2^1÷2^1 which is 2^(1-1)= 2^0 = 1 and that disproves 0^0 = 1 since it would be 0^1÷0^1 which is 0÷0.*
No, this is incorrect, and even school teachers get this wrong. x^0 is _defined_ as the product of 0 copies of x. This is 1, and this is true for _all_ x. There are no exceptions, and thus, x^0 = 1 for all x. Now, IF x is invertible, THEN, then, the property x^(m + n) = x^m·x^n implies 1 = x^0 = x^1·x^(-1) = x·x^(-1), so in that case, x^(-1) denotes the inverse of x. But, 0 is not invertible, so that _does not apply_ here, and it is incorrect to say that 0^0 = 0·0^(-1), because it is false. So, no, that does not disprove 0^0 = 1.
If it did disprove it, then it would also disprove 0^1 = 0. How? Because, by your logic 0^1 = 0^(2 + (-1)) = 0^2·0^(-1), and 0^(-1) is undefined, since 0 is not invertible.
Also, for the love of Siesta, never ever use the symbol ÷. It is obsolete, and it is too much of a hassle. Just be a sensible human being and use / to denote division next time, please. We are all adults here, we are not in grade school.
@@angelmendez-rivera351 ok
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.
Smart!
hold on i might be missing something here
from 7 to 7x⁰
and letting 0⁰=0 when x=0
it seemed that u were already claiming that 0⁰=1 at the beginning implicitly and then u continued to try give 0⁰ a different value other than 1
it's like
7 = 7×1^a
let's say we don't know what 1^a when a=25
If you let 1^25=0, 7=0 while in reality, we implicitly made 1^a=1 by saying 7=7×1^a
so once again i might be missing something but that explanation does not actually explain why 0⁰=1 because 0⁰=1 was used already by itself to show 0⁰=1 which is basically going in circles
errr no. all you did is say 7 = 7x^0.
@GDPlainA No. I didn’t. That isn’t what I did at all.
@GDPlainA correct. and it doesn't explain anything lol
This is one of those things that you never thought about until stumbling upon this video
No, i searched it.
@@AmanSingh-fh4zb I will touch my eyes ¡O¡
@@AmanSingh-fh4zb also I subbed to you
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)
I don't know why people say this stuff about calculus. 0^0 is not an expression that has anything to do with limits. You can tell by the facts that it doesn't contain the symbol "lim" and that it doesn't have a variable name underneath it, followed by that funny little arrow, followed by another symbol after the arrow. How are people not spotting these pretty major clues?
Hey, Brilliant! Why dont you take limit of x^x as x approaches 0
Aditya Shankar it’s supposed to say as x approaches 0 from the right. The limit will go to 1. The answer will go have to get into calculus. This is one of the indeterminate forms using l’Hopital’s rule. To get the answer to be 1, y=x^x is the same as y=e^(xln(x)). Look at the exponent for xln(x). Change this into a fraction as ln(x)/(1/x). Take the derivative of top and bottom separately, which is (1/x)/(-1/x^2). Using algebra and simplification, you get -x. When x goes to 0 from the right, you get 0. All together, you get e^0, which is 1. That’s the answer to the limit.
@@justabunga1 Actually what's cool about the function x^x is that the limit as it approaches 0 is 1 from the left as well. The real part approaches 1 and the imaginary part approaches zero.
@@Eliseo_M_Pthat only applies for real numbers only in calculus since we only discuss most of the parts.
The music is quite distracting
Logically, 1 seems like the best answer. Raising something to a power means multiplying it against itself that number of times. When multiplying something against itself 0 times, the base would be irrelevant in all cases, including 0
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.
Precisely.
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?
What you are missing is that this is the result of defining exponentiation that way. You could have defined it as is usually done without the 1 and the positive intergers work fine.
Plus your definition makes it really hard to extrapolate to what the hell is exponentiation of a non-interger or even a complex number
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
Anything raised to the power of Zero is 1
So If we follow that rule 0^0 is equal to one.
@@donandremikhaelibarra6421
But you need to see how powers work for that
@@donandremikhaelibarra6421 Also, 0 raised to anything is 0. So by that logic...
While illustrative, this only works for integer exponents
@@androkguz
What happens with non-integer exponents
To get from 0³ -> 0², you'd have to divide by 0. Since we established that 0^n where n is a non-zero positive integer = 0, 0³ = 0
We can write 0 as 0¹, therefore:
0² = 0³/0¹, but dividing be 0 is undefined, so what's the logic behind exponentiating 0? Correct if I made an error but I'm just not seeing this
Power means how many times digit is multiple so 0^0 mean so there we can write as 0^0×1
But according to what according to exponent definition zero repeat zero time so zero does not present there that's what zero means so the answer is 1
Power means how many times digit is multiple so 0^0 mean so there we can write as 0^0×1
But according to what according to exponent definition zero repeat zero time so zero does not present there that's what zero means so the answer is 1
I think it is undefined
0^0=0^1-1=0^1*0^-1
0^-1 is undefined so 0^1*0^-1 is undefined so 0^0 is also undefined
You know 0^3=0
Then 0^3 = 0^(-1+4) =0^(-1) × 0^4
0^(-1) is undefined
So 0^(-1) × 0^4 is undefined
Thus 0^3 is also undefined
Whenever I see 1/0 I treat it as infinity, as 1/∞ is infinitely small so I treat it as 0, substitute 1/(1/∞) = 1 * ∞ = ∞
No, the implication arrow points at the wrong side. It only means that if 0^0 is undefined, then 0^1 * 0^-1 is also undefined.
The way I work it out is by how you work out exponentiating something by 0.5, which is the square root of the number. when its 0.25, its the 4th root of the number, etc. as the power gets closer to 0, the n in the nth root is going to be higher, since the n is determined by the formula 1/the power. as the n gets higher, the actual nth root of the number is going to get lower and lower or higher and higher, depending on if the number is higher or lower than 1 respectively, since the root of the number is a number powered by the n in the root. considering this, when the power is 0 (or in mathematician terms, approaching 0, since 0 is never actually used as a number, rather used the same way infinity is) then the n in the root is going to be equal to 1/0, which is approaching infinity. the infinite-th root of a number approaches 1 no matter if the number being rooted is higher or lower than 1.
considering all of this which I probably got wrong since I just yapped based on reusing what I learned this year in my gcses, the root of 0 is undefined because it entirely depends on if the 0 is a true 0 or if the 0 is simply approaching 0, which are mathematically basically the exact same thing. you see when its approaching 0, there's still a tiny tiny value that is being powered, so therefore that value will approach 1 when rooted by infinity, however when the number is actually 0, then root will always be 0, considering that 0*0 is always 0. so you're left with 2 values of 1 and 0, so it is undefined.
this can all just be summed up to a single formula as well. x^y = (1/y)th root of x.
I don't know what 0^0 is, but I know that 0^1 + 0^1 = 0^2 satisfies an equation, namely x^n + x^n = x^(n+1) for x in {0,2} and n in the natural numbers with 0. Could it be the case that 0^0 + 0^0 = 0^1 without actually calculating it? I think not even as an algebraic expression.
x^n + x^n = x^(n + 1) only works for powers of 2. That’s literally how powers of 2 are defined. The next power is twice the last.
2^n + 2^n ->
2(2^n) ->
2^(n + 1)
Best explanation
0^x = 0, but x can't be a negative number. In the 1 case, x can be any number
It's right that it's undefined.
0^0 = 0^3 ÷ 0^3 =0÷0
Therefore, according to our definitions of the index laws, 0^0 should be undefined.
How do you come to 0^0=0^3÷0^3?
@@AvatarBowler You seem tol have much Maths Knowledge. Im only 15 so the question may seem dumb but cant you add *multiplied by one* to an equation which would lead through basic thinking to 1=0^0
@@AvatarBowler Yeah but logic thought(Not mathemathic) 1*0^0 shouls be the same as 1 cause 0^0 should be nothing and only 1 is left. That was my thought
your accusation is not correct from any angel
If you visualise the rationale behind n^0=1, you would realise that our math is designed in such a way that n^0 should be the multiplicative identity, hence 1. Naturally, the same definition can be used for 0^0, and should again, be 1. Hence the most logical and apt definition for 0^0 is 1.
Actually when we take a number N raised to the power M, it means we have to multiply 1 by N (M number of times). For example 3 raised to the power 2 means 1*3*3. Similarly a number N raised to the power 1 is 1*N. However, a number N raised to the power 0 means we are keeping 1 as it is and NOT multiplying it with N. Therefore if we take N=0 ,then 0 raised to the power 0 is 1 (1 not being multiplied with anything).
x^n = x^(n-1) * x is the equation you mean. Lets fill in x=1 and n=1: 1^1 = 1^0 * 1. In this case 1^0 has to equal 1 in order to complete the equation. But when we fill in x=0 and n=1 we get: 0^1 = 0^0 * 0. In this case 0^0 can mean both 1 and 0.
@@bartjesper Nope. It has nothing to do with that. The idea is very simple. If you multiply by x exactly 0 times, then this is the same as not multiplying at all. Not multiplying at all is the same as multiplying by 1. Hence, x^0 = 1. Notice how this is true for ALL x, it does not matter whether x is 0 or nonzero. If you want to get rigorous, you define x^n not in terms of an equation, but as the product of the constant n-tuple where every term is x.
@@angelmendez-rivera351 I never said x^0 is not 1
So they arbitrarily assign an answer to a zero equation depending on what's convenient for them. That's a stupid way to have a math system.
Undefined
Correct
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
@@AlbertTheGamer-gk7sn meow?
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...
1322^0 = 1
x^0 = 1
AkkiCat 21109 1 way of choosing set of 0 elements from a set of 0 elements.
It is because you still have one thing to choose and that is 'nothing'!
For eg, if you are told that you can get maximum marks 10 in a test (without negative marking), it means you can have 11 scores- 0,1,2....10
Similarly you are told that you'll only get zero marks if you dont attempt the test, it means you'll have only one type of score- 0.
I hope it clears up your doubt!
I know it's probably wrong, but I just think as exponents as a reference to a sequence where the base is the growth rate, and the exponent is the number you look at
2^4 = 1, 2, 4 , 8, 16, 32
Count to the fourth one and that's the exponent
Either 0^0 is equal to 0 or 1, it would be undefined, like 0/0
It would my dear friend it would
If you set a • 0^(-1) = 0 you extend the field of the real numbers to a meadow in which we also have 0^0 = 0.
@@annaluiseblume nope it could be any number so infinite gor sure
@@thunderpokemon2456 but then you lose 0^(a-b) = 0^a • 0^(-b)
@@annaluiseblume nah this exponential functional trick wont work as
A^ -x =1/A (it only works when A is not equal to 0)
So this wont work if this works the 0 itself cant exist as
0^1 = 0^(2-1)= 0^2 × 0^-1 ?
So not possible
O^0 can never be any sort of value , simply 0 itself is merely a marking of the starting point of any value or all values, so it does not represent any specific value like infinity. It means that no matter whatever sort of mathematic manipulating tricks you can use to support your fallacious arguments, not any sort of value or such undefined values can never give you any sort of value either
If you take a sample space (continuous example) omega { x , y } modulus x < 0 > 1 there exists an infinite number of real numbers in a graph of finite space x = 1, y = 1. A square.
I would like to identify errors in proofs of 0^0 being undefined.
Im pretty sure that 0^0 =1 makes more since, because if you do lim(x=0)x^x you will find the limit to approach 1 from the positive side. But yes, restricting x to only be base or exponent and not both would make 0^0 both 1 and 0
When you use this formula (n⁰=n/n) you have 0⁰=0/0 and n/0 is undefined
Some college math classes actually define 0^0 as 1.
I believe it.
I mean… Look at e^0. We know that e^0=1.
The sum of e^x is the sum of [(x^n)/n!] from n=0 to n=∞.
(0^0)/0!=1, and 0!=1; therefore 0^0=1.
I don't know where you were taught math but something very wrong was happening there...
@@pelasgeuspelasgeus4634 No, it's pretty standard that 0^0 = 1 in discrete mathematics.
@@MuffinsAPlenty Yes, it's standard and so WRONG. It's an arbitrary and useless convention that contradicts the exponent definition and any kind of formal logic, provided you have some...
@@pelasgeuspelasgeus4634 "Yes, it's standard and so WRONG."
Ah, you're a mathematics conspiracy theorist. I see there is no reason to take you seriously. I hope that you are one day able to get some help.
Thanks for the clarification
The limit of x^x as x approaches 0 is 1 also
Let x=0^0
For any n, (n is not infinity)
x^n=(0^0)^n=0^(0*n)=0^0=x
Take n=2, x=0,1.
Take n=-1, x=1,-1
The intersection is 1.
1^n=1 is always true.
Therefore 1 is the only value of 0^0 that satisfies index laws.
Index laws are invalid when 0 is denominator or 0 to negative power is encountered.
I don't actually get the idea of zero but its how our maths teacher show so I will go with it n get the marks
The idea of using the number 0 is a problem in math. One of the problems would be division by 0, the infinity symbol, and several indeterminate forms. In basic math, our math teacher said that 0 divided by any non-zero number is always zero. A value itself by adding or subtracting 0 is the value by itself. Any number multiplied by 0 is always 0. Any number divided by 0 is always undefined. The log base of a positive number other than 1 of 0 is always undefined.
0^0 could, depending on the context, be defined as 1. As does every other number raised to the power of 0. Assuming this is true we can raise every number to the power of 0 and multiply them. This will give us:
0^0 x 0^0 x … = 1
1^0 x 1^0 x … = 1
2^0 x 2^0 x … = 1
Etc.
we need two kinds of exponentiation, actually. One for calculus, one for probability.
I just realized brilliant has a channel.
And it's amazing.
I'd say it's brilliant
Amazing explanation
a ^ 1 : a^1 = a ^ 0= 1( because a^ 1-1 = a^ 0 and b : b = 1)
I understand zero to be the absence of a value, therefore the equations that use approximations are incorrect because you are still substituting a value into an equation meant to be moot by definition.
Zero is not tbe absence of value
@@angelmendez-rivera351 but it literally is though...
@@angelmendez-rivera351 Yes it is
@@kiteinthesky9324 Repeating a debunked claim doesn't undebunk it, y'know
In any base n, 11 = n^1 + n^0 and 11 = n + 1. So in base 0, 11 = 0^1 + 0^0 and 11 = 0 + 1 = 1. 0^1 = 0, so 0^0 = 1.
Unfortunately there is no base 0. (But 0^0 is still 1.)
@@05degreesI'm inclined to agree, but I still think there's an interesting discussion to be had of the idea.
If you define a base as the number of single digits it uses, for example base 10 ( decimal) uses the 10 digits 0 1 2 3 4 5 6 7 8 9, then presumably there can be no base that uses 0 digits.
But if you define b in a base b simply as the number that is raised to successive powers beginning with 0 from right to left by which to multiply each digit in a string and add the results, for example 359 in decimal is 3x10^2 + 5x10^1 + 9x10^0, then some sense can be made of the idea. Granted, what you may also have to allow is that the digits can represent quantities exceeding b. For example D in hex represents 13 in decimal. So you can have a number 3D9 in decimal that represents 3x10^2 + Dx10^1 + 9x10^0 = 439.
Now in base 0, all digits other than 0 are going to be greater than 0. 3D9 in base 0 is 3x0^2 + Dx0^1 + 9x0^0, which of course equals 9, the rightmost digit alone. All the others are just noise. So effectively what you have here is a representation of every new quantity with just a single new digit. Once you've run out of those Indo-Arabic numerals you probably go on to uppercase letters, lowercase, emojis, etc etc. Granted this isn't a base either, if by base you also mean a system of representing new quantities by recycling digits already used.
I think there are some further interesting implications, but that'll do for now.
X^0= x^1 * x^-1
=x * ¹/x
x^0= x/x = 1
Therefore...
0^0= 0/0 = undefined
So what do you guys think?
Index laws are invalid when 0 is denominator or 0 to negative power is encountered.
0=0^1=0^(2-1)=0^2/0^1=0/0
So what do you think?
Correct, answer is undefined and not it depends, stupid video
A simple and easy way is to have x^x and have x = 0.9 , 0.8 , 0.7 , ... 0.1 and then 0.01 , 0.001 , 0.0001 , ....
When you calculate these, from 0.9 to 0.5, the number becomes smaller. At 0.4, the number becomes higher. The lower we go, the higher the number gets. So, as x comes closer and closer to 0, the closer and closer x^x comes to 1.
You can't magically get something from nothing. And these high end math "experts" arguing this is far more mind boggling
The empty set cannot be chosen from the empty set because the empty set is not an element of the empty set. Set theory can't be used to prove that 0**0 == 1.
0⁰ should be considered as an operation between two determined, equal numbers, not to be confused as the limit of two different functions f(x)^g(x) which tend to 0 with different speeds. In this case x^x represents all numbers raised to themselves. So the correct result is 1. The same goes for 0/0 = 1, but the properties need to be reviewed or redefined to make everything coherent (in this case we need to find a justification that 0/0 ≠ 3 as an example).
But how is 0/0 = 1? Equation 0·x = 0 is true for _any_ value of x.
Yes, I know :), but the function y = x/x appears to have the value 1, when x - > 0. It might be more correct to represent x/x with two perpendicular lines.
Hm, that actually makes perfect sense, since x·y = x is equivalent to x·(y−1) = 0, so either x = 0 (and y is any number) or y−1 = 0 (and y = 1 while x is any number). Yeah, that's an interesting observation.
The indeterminate form "0^0" is actually (->0)^(->0), not (exact 0)^(exact 0), the latter is 1
0^0 doesn’t always approach to 1. It can be 0, e, or any other limit value. You can check out blackpenredpen if you want to know if the value approaches something other than 1.
@@justabunga1 A- that's (->0)^(->0)
B-Say that to the Taylor series of e^x and 1/x
Also, I'm talking about exact values, not limits
@@seroujghazarian6343 take for example e^0 using sum notation. y=e^x is the same as the sum from n=0 to infinity of x^n/n!. Let x=0 and n=0,1,2,... You get 0^0/0!+0^1/1!+0^2/2!+0^3/3!+... All you have left is 0^0/0! By convention, 0^0=1 (this is an exception where this happens when doing Taylor/Maclaurin series otherwise it's left to be indeterminate) and 0! is always 1. That's why e^0=1.
@@justabunga1 1/x's taylor series (Σ(n=0,infinity)(-1)ⁿ(x-1)ⁿ) also gives 0⁰=1
turn down the music
I think this is the only best explanation on yt
It contains some quite major errors, so no.
@@omp199 ohk thx
@@omp199 BTW can you explain ?
@@Rohan-nf5hc 1:10 "a^b can be viewed as the number of sets of b elements that can be chosen from a set of a elements". This is not even close to being correct. In fact, it's obviously incorrect. You can see that in the expression a^b, b can be greater than a, and yet there are no sets of b elements that can be chosen from a set of a elements if b is greater than a. So if the video was correct, a^b would be zero for all b > a. This is so far from being true that it is staggering that nobody making this video spotted the mistake.
The correct statement is that a^b can be viewed as the number of functions from a set of b elements to a set of a elements.
2:17 "We're interested in limits of the form 0^0 when x = 0." This is a meaningless statement. 0^0 is not a statement of a limit. A statement of a limit would involve the symbol "lim" with a little arrow beneath it.
@@omp199 "The correct statement is that a^b can be viewed as the number of functions from a set of b elements to a set of a elements."
Yes. This is correct. Brilliant has said in some comment somewhere that they intended to say: "a^b can be viewed as the number of _multisets_ (i.e., sets where repetition of elements is allowed) of b elements that can be chosen from a set of a elements" which has the same cardinality as functions from b to a. Still, though, what they actually said was wrong.
0:46 Actually, n/0 or 0^-n = Complex Infinity,
But 3^2 = 3×3
But 0^0 = 0 raised to how many times?
Im in class 9 and I asked my teacher about it today when she was teaching degree of polynomial
But she said how many times are you going to multiply 0 with... I wasn't satisfied with her answer so found you video.. But still I'm confused
Lim x^x as x->0 from the right is 1
Claw Jet that’s correct, but that’s only for calculus part. If you’re talking about algebra, then the answer is indeterminate.
@@justabunga1 nope. Algebra takes it 1.
Eg there is 1 and only one function from the empty set to itself and that is the necessary bijection. So the cardinality of (empty set)^(empty set) (set of functions from the empty set to itself) is 1.
And we all know that card(F^E)=Card(F)^Card(E)
And Card(empty set)=0
0^0=1
@@seroujghazarian6343 not always that only works for non-zero numbers. That just means it can be any number besides 0.
@@justabunga1 did you even read my reply?
0^(0) = undefined
it should be 0 because 0 to the power of 0 basically means 0 repeating itself 0 times, which is 0. the same way 2 to the power of 3 is 2 repeating itself 3 times. if a number is repeated 0 times doesnt that mean its 0??
No, 0^0 should be equal to 1. Many arguments tend to use this in order to prove that 0^0 = undefined, 0^0 = 0^1-1 = 0^1/0^1 = undefined, though, what many misunderstands is that this equation is ONLY true if X is not equal to 0, it follows this pattern (x^a-b = x^a/x^b) which doesn't apply to a variable with 0 as it's value. Another argument is by using a limit (lim x > 0+ for x^0) = 1, and use the opposite limit (lim x > 0+ for 0^x) = 0, People argue that since this is is multivalued then it is undefined, though keep in mind that these are not ABSOLUTE values, limits is a function where a variable 'approaches' a value and not AT a specific value, meaning that this is be neglectable, and 0^0 itself isn't a limit.
Another thing, 0^0 is used in many different mathematical functions, for example, the binomial theroem which was mentioned in the video requires 0^0 to be defined as the simplest calculated integer for the equation, without it being defined as 1 this theorem would need to undergo many complex configurations in order to be totally accurate.
You are stating that 0^0 should be equal to 0, but that is not how exponents work. Exponents is an operation which determines how many times a value should be multiplied by itself, and not "repeating" itself (relates more to multiplication), consider exponents to be how many times you can arrange a certain number, if we take this definition and use x^0 for it , then we can ask the question "how many times can I arrange x for a list that doesn't exist"? The answer is 1 as there is only one way you can arrange them, and that is the initial state.
Math does not change for on context to the next. 0^0 is always undefined.
Why not zero then? Multiplication is a short form of addition, and exponents/indices/powers are a short form of multiplication. 0 x 0 = 0. Anything else appears to be accepting a falsidical paradox.
~~~~~ Read below for further explanation ~~~~~
The other patterns with _actual values to the power of zero that equal one_ explain this result with a break down either back track to exponential pattern like 2^4 = 16 then 2^3 = 8 to reach 2^0 =1 accepting the pattern presumption as proof...
*OR* by using equation proofs that have the second last step showing the number or variable to one exponential value subtracting the same exponential value as proof to anything to power of zero equals one. Seems legit since two proofs by reversing an equation say so. Or is it really?
We are expected to accept these explanations because so far it has proven mathematically useful to do so. But all these equations only seem to make sense through these double check proofs. Yet fails the common sense part of the equation. Zero times Zero done zero amount of times still leaves you at zero. I mentioned the falsidical paradox because that's exactly what the Achilles and Tortoise Paradox is. Scholars could never prove mathematically that Achilles could actually catch the Tortoise despite real life examples. Not until Convergent series were discovered with calculus that we saw how math proved what we already knew in real life.
Obviously, you haven't studied much mathematics, Zachary.
consider: (0+1)^0
which is equal to:
0C0 • 0^0 • 1^(0-0)
In this case
(0+1)^0= (1)^0 = 1
So
0C0 • 0^0 • 1^(0-0) has to be equal to 1. And it can only be if 0^0=1.
So it is not undefined, unless we claim and prove that 1^0 is not equal to 1.
Rmemeber that 0C0 is already defined to be 1 by using Gamma Function as well, which extends the definition for factorials and also proves that 0!=1
Nope, it's 1.
Otherwise, the exponential function would not be continuous in 0, which is absurd
It is not indeterminate....in fact you can use a direct limit to calculate the correct answer of 1
0^0 = lim_{x->0} x^x = lim_{x->0} exp(x*ln(x)) = ..... (apply De L'Hospital rule) ..... = 1
Assuming that 0^0 = lim_{x->0} x^x. But why are you assuming that? On what basis are you making that assumption?
@@omp199continuity
But if 0^0 is the number of sets of 0 elements that can be chosen from a set of 0 elements, then wouldn’t that be equal to 0? Because there wouldn’t be any elements to choose from, so there wouldn’t be any way to choose them
It's ome: the own empty set. The set with zero elements, the only set that is contained into the empty set is the empty set itself, so that's why its one. If it was zero, then the empty set wouldnt exist. But it does.
0^0 is undefined.
Because a number raised to the zero es one due to the exponent property.
For example
3^0 = 1 because
3^0= 3^3 / 3^3
Exponents substraction
3^0 = 27/27
3^0= 1
Now. If we put 0
0^0= 0^3 / 0^3
0^0= 0/0
0/0 = 1?
No it doesn't.
If 0^0 was 1, then 0/0 would also be one, which doesn't make any sense.
So 0^0 is undefined
Index laws are invalid when 0 is denominator or 0 to negative power is encountered.
0=0^1=0^(2-1)=0^2/0^1=0/0
Correct
Not the same. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
0 / 0 =undefined
explained so well
It is wrong brother, watch this: ua-cam.com/video/lKmW-mGqXQU/v-deo.html
#undefined
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
If 0^0=1
But why 0 div 0 eq error
Same with 0^3
0^4/0^1 = 0/0
@@thexoxob9448 bro u have to understand me
Ansower me , What is 0/0 ??
@@Muslim-man1 undefined. Look at my reply above. Thus 0^3 should be undefined
Look
3^0 = 1
2^0 = 1
1^0 = 1
0^0 = 1 So, Thats means the same for
0⁰ × 0-¹ = 0/0 = 1
Plz bro just focus
@@Muslim-man1 Aleph-0.
The music on this video is really messing with my ability to listen to the explanation
0^0 means 1 multiplied by 0, 0 times.
I personally tried it and i think it is undefined for sure
Yeah 1 and 0 are sht answers
If it's 1, then explain to me this :
0⁰ =
0^1-1 =
0¹ × 0^-1 =
0 × 1 / 0 =
0 / 0
So if you're right, 0 / 0 must be 1 too, if not so it's undefined
If there isn't a consistent answer, it should be undefined.
1 is a consistent answer. You _cannot_ find a valid contradiction to having 0^0 = 1. All of the arguments which are typically used to say that 0^0 must be undefined either use invalid reasoning (such as applying mathematical formulas where they do not apply) or confusing the arithmetic expression 0^0 with the limiting form (→0)^(→0).
0⁰ is 1
A × 1/A = 1, so 0⁰ × 1/0⁰ should be 1
If 0⁰ was 0, 0 × 1/0 = 1, but 1/0 isn't a number.
So, 0⁰ is 1
(1 × 1/1 = 1
If we could find what 1/0 is equal to we can solve this mathematically inquiry. This is because n^0 is equal to n × 1/n. E.g., 2^0 is 1 because 2 × 1/2 is equal to 1, 1,000,000^0 is 1 because 1,000,000 × 1/1,000,000 is equal to 1. Even works for negatives. I.e., to find 0^0 we just need to calculate 1/0 and then multiply that result by 0 and then boom, that's the real answer for 0^0.
If you guys want my opinion into he answer to 0^0. I thinks it is to an extent 0, and here's why. Like I previously explained, n × 1/n is equal to n^0. So for 0^0, if we want to find that we need to do 0 × 1/0, so what's 1/0? Well because 0 cannot go into 1, we place a 0, 1-0=1, 0 cannot go into 1, so we get 0.0/still 0. Do it even you're go to get an infinite amount of zeros without any _subsequential_ digits. So we get 0 simply, then we 0 × 0, which equals 0. So we can say 0^0=0. But 0 × 1/0 is equal to 0/0 which equals 0? That defeats the rules of division. So really unless math is drastically changed, there seems to not be a valid formula for particularly 0^0.
I used the limit x raised to x, as x approaches to 0.. it's 1 .
🔥🔥🔥
Zero is not a number, which is why 0^0 is not 1 in every context.
...zero is not a number? man you crazy
Wrong, 0^0 writing is itself crime in maths. There is no such thing as it depends!
Q: is irrational + irrational = irrational?
A: False (eg π + 5-π) and not it depends
So 0^0 is undefined.
This comment makes no sense. What does adding irrational numbers have to do with 0^0?
@@MuffinsAPlenty It's an example ffs
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
Well, if we were living in Minecraft, drawing a circle will also be a "crime in math".
International Conference on Differential & Difference Equations and Applications
ICDDEA 2017: Differential and Difference Equations with Applications pp 293-305 | Cite as
log0=log∞=0 and Applications
Authors
Authors and affiliations
log0 = -∞ and log∞ = ∞
The mathematicians at this conference are evidently quite dumb :)
Who invented 0 😎 we indians 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
We know 0 is equal to infinity ♾️
I appreciate it
But you people, when you 'invented' 0, your definition of 0/0 was WRONG. 0/0 is not equal to 0, as was stated in the 'invention'. It (0/0) is now considered undefined. 0 is also NOT equal to ∞, how in the world is that possible?
0^0 is like 0/0 which is undefined though isn’t it?
Nope, not the same.
A box full of 0 has 1
filling.
Wait, so 0^0 could be anything?
Make it 2,763. No reason.
You say 2^1 = 2 because it's the number of sets of 1 element that can be chosen from 2 elements. Then you say that means 0^0 = 1 because it's the number of sets of 0 elements that can be chosen from a set of 0 elements. That does not follow. You can't make any sets from the null set except null. If you count that, then 2^1 should be two sets plus the null set so the answer should be 3.
"You can't make any sets from the null set except null."
The null set is a set of size 0. So it is 1 set of size 0 you can make from any set.
"If you count that, then 2^1 should be two sets plus the null set"
But the null set is size 0, not size 1, so it shouldn't be counted in the collection of sets with 1 element.
Take log0 = y and first doing deffierentaite and then integrate we get value of logo and then z = 0 power 0 and take log an both side and put the value of log0 and we get 0 power 0 is 1
guys imma bit baffled by his definition of exponentiation. does it rlly work?
someone pls explain.
isn't this the formula for WARP DRIVE and Star Trek?
*0^0=0/0*
because
0^0
= 0^(1-1)
= 0¹/0¹
= 0/0
and 0/0 can be *every number* because
x*0=0 is how to prove that 0/0=x
"0^(1-1)
= 0¹/0¹"
This step is invalid.
The exponentiation rules you are using here do _not_ apply when 0 is the base.
For example, would it be valid to say
0^1 = 0^(2-1) = 0^2/0^1 = 0/0?
No, it wouldn't. But that argument is just as valid as the argument you presented for 0^0 = 0/0.
Index laws are invalid when 0 is denominator or 0 to negative power is encountered.
0=0^1=0^(2-1)=0^2/0^1=0/0
Многие думают что на ноль можно умножать а делить "почти" совсем ни как нельзя...
Типа X×0 = 0
это нормально лишь потому что 0/X = 0...?
Но из этого же следует что
сам X = 0/0...? Х=0⁰...? ну и где логика...
Давайте рассмотрим один из вариантов как обычно происходит действие деления...
6:2=6/2=(2+4)/2=2/2+4/2=1+(2+2)/2=
=1+2/2+2/2=1+1+2/2=3 (без остатка...)
7:2=7/2=(2+5)/2=2/2+5/2=
=1+(2+3)/2=1+2/2+3/2=
=1+1+(2+1)/2=1+1+2/2+1/2=
=1+1+1+1/2=3+1/2=3 с остатком 1...
И это также можно с помощью принятых форм математических записей выразить как 3½ или 3.5...
А что же происходит когда якобы производят деление на ноль...
многие говорят что это будет равно какой то бесконечности...
15:0=15/0=(0+15)/0=0/0+(0+15)/0=
=0/0+0/0+(0+15)/0=0/0+...+0/0+15/0...
и при дальнейших действиях всегда такое деление будет c постоянным остатком в виде того что "делилось" изначально...
в данном случае остаток 15...
и почему то вот об этом остатке или забывают или неосознанно замалчивают считая только бесполезные бесконечные действия не приводящие ни к какому результату деления...
Если быть немного логичным то видно что даже при бесконечном количестве таких действий деления (а точнее бездействий) вся сумма таких действий равна нулю с постоянным остатком того что было изначально делимым...
То есть само такое деление не происходит...
сколько было изначально столько и остаётся в остатке неделимо...
X:0=X/0=(0/0)×N+X/0=N×(0/0) с неразделённым остатком X
где N×(0/0)=0 и N число мнимых манипуляций не производящих деления...
поэтому N=0... а не бесконечность...
отсюда и получается два ответа при делении на ноль...
относительный ответ равен 0...
но именно ноль бессмысленных манипуляций...
а безотносительный ответ равен самому значению делимого X...
В примере 15/0 = 0 целых 15 нулевых...
или же 0 целых и 15 в остатке... именно умножая это число на ноль можно получить первоначальное данное значение...
Но об этом как правило неумышленно умалчивают... ведь этому не научили...
Полное непонимание современной математики темы умножения на ноль и тем более деления на ноль...
При записи умножения числового значения X на ноль получаем
-----
X×0=0 X /\ 0/0
Перенос ноля через знак равно превращает равенство в качельное неравенство типа
----
100% /\ 0%
Сам знак процентов кстати пишется как 0/0...
То же самое и с делением на ноль...
----
X/0=0 Х /\ 0×0
Перенос ноля через знак равно превращает равенство в качельное неравенство...
Я называю это нулёвыми значениями (не нулевыми а именно нулёвыми...
"ни разу взятыми" или "ни разу трачеными" то есть "нерастрачеными" если это об умножении на ноль...
"ни разу делёнными" или "неразделёнными" если это о делении на ноль)...
И непонимание до сих пор этого простого меня очень удивляет...
220 вольт делить на ток 0 Ампер это сопротивление = 0 Ом... но это просто не потраченое напряжение...
1 торт не взятый кусками ни разу (деленный на ноль) 1/0 это 0 кусков взятых но это все тот же 1 неразделённый торт...
5 монет ни разу не взятых 5×0 это 0 взятых монет но вопрос как правило звучит "сколько будет" а не сколько взято... так вот будет все те же 5 невзятых "нулёвых" монет...
И это всего лишь маленькая вершина айсберга действий с нолем...
Сложение и вычитание нуля не меняет первоначального значения... а почему?
да потому что на самом деле не происходит самого математического действия как такового... ничего не прибавляется и не убавляется при этом...
С умножением и делением на ноль происходит примерно тоже самое...
Ничего не происходит при этих действиях... всего лишь описывается что первоначальные значения не изменяются...
хотя ответов получается два
относительный = 0
безотносительный = 100% = 1×Х(нулёвое)
в зависимости от поставленного вопроса...
И безотносительный ответ имеет гораздо больше смысла...
5 метров × 5 метров × 0 метров =
25 метров² × 0 метров = ?
Относительно нуля ответ 0 метров³
Безотносительно нуля = 25 метров²
----
25м²(=100%) /\ 0м³ : 0м (= 0м²)
Перенос нуля (при умножении на ноль или делении на ноль) через знак равно превращает равенство в качельное неравенство
-----
1×X(нулёвое)(=100%) /\ 0(=0%)
Безотносительный ответ при действиях умножения и деления с нулем не учитывает как само действие с нулем так и его измерение...
5 яблок : 0 корзин = ?
Относительный ответ 0 яблок на корзину...
Безотносительный ответ 5 неразделенных яблок (без корзин)...
Убираем ноль и его измерение из вычисления и получаем нетронутые первоначальные данные и можем дальше с ними что то вычислять...
----
5 яблок нулёвых (= 100%) /\ 0 корзин × 0 яблок/на корзину (= 0)
Чисто качельное 100% неравенство...
Откуда здесь могут взяться какие то бесконечности? Или черные дыры?
Кстати 0(нулёвый) / 0(нулёвый) = 1
Впрочем как и любое другое число раз делённое на само себя...
Это всего лишь малая часть моего личного взгляда на действия с нулем и он не ограничивается только этими действиями...
Ноль не имеет численного значения...
он лишь описывает отсутствие чего либо...
Практически все действия с нолем на самом деле не происходят...
Многие пытаются ноль "всунуть" в основные математические действия... при этом абсолютно не понимая смысла самой записи таких действий...
но с нулем есть только математические "бездействия" и чаще всего действия "умножения" и "деления" связанные с нулем говорят что есть что то безотносительное до той поры пока вместо нуля в таких выражениях не появится числовое значение...
Лишь после этого выражение становится относительным...
Если напряжение = 0 и сила тока = 0 то это не значит что при этом всегда нет сопротивления...
Если скорость = 0 и время = 0 то расстояние при этом может быть каким угодно (в том числе и отсутствовать)...
Поймите главное перенос нуля через знак равно изменяет смысл равенства на качельное неравенство 100% того что было изначально и есть до сих пор и с другой стороны 0% того что якобы "взято"...
И никаких бесконечностей и всяких "черных дыр" при явном нуле в таких действиях никогда не будет...
Чисто математически ЛЮБОЕ "действие" когда Х не равное нулю "умножается" на ноль или "делится" будет равно ВСЕГДА нулю...
то есть отсутствию таких отношений...
Но смысл совершенно не в этом...
любое такое "действие" описывает лишь неизменность самого стопроцентно имеющегося значения X при этих нулевых "операциях" с ним...
Что касается "деления" 0/0...
(или по другому выражения типа 0⁰...)
Если вы делите два различных нуля один на другой (с различными мерами измерения) то "относительный" математический ответ этого будет 0 = 0% того что использовано...
Но безотносительный ответ будет равен 100% того что было дано изначально и не было использовано в ходе бездейственного "деления" отсутствия одной величины на отсутствие другой...
Если делить один ноль сам на себя (с одной и той же мерой измерения) то ответ равен 1 раз...
И никаких 2 раза... 3 раза... и т.п. у отсутствия величины в виде ноля не будет...
Интересно как можно объяснить 0/0 = 0⁰ с точки зрения "практических" равенств...
Многие считают что 0⁰ = 1...
Напряжение U = 0 вольт...
Сила тока I = 0 ампер...
Сопротивление R = U/I = 0/0 = 0⁰ = 1...? Ом...?
Весело...
Дистанция S = 0 километров...
Время t = 0 часов...
Cкорость V = S/t = 0/0 = 0⁰ = 1...? километров/час...?
Смешно...
Объем V = 0 метров³...
Ширина W = 0 метров...
Высота H = 0 метров...
Длина L = V/(W×H) = 0/(0×0) = 0/0² = 0‐¹ = 1/0...? = ...?
Сколько будет...? метров?
Интересно сможет хоть кто то это объяснить хоть как то математически...?
Многие математики почему то считают что у нуля нет обратной величины...
Другие свято верят что величина обратная нулю это "бесконечность"...
К сожалению (ну или к счастью) у "бесконечности" есть обратная величина равная 1/бесконечность...
И как бы она ни была мала она НИКОГДА не будет равна нулю...
И уж точно она не имеет безотносительного значения...
К тому же она имеет знак плюс или минус в зависимости от того с каким знаком берется сама "бесконечность"... (если конечно она хоть как то вообще может быть "взята"...)
У полного отсутствия в виде нуля есть обратная величина... это полное присутствие... и для нуля это равно единице... то есть 100% присутствие чего либо...
1/0 "относительный" ответ математически равен нулю... но именно он не имеет смысла а вот безотносительный ответ как раз равен "ни разу делённой" то есть нераздельной (нулёвой) единице...
Никакой бесконечности при делении на ноль не бывает... если только сама бесконечность не делится на ноль...
1/0 равна 0 целых и 1 в остатке... полностью неделённая единица...
Умножьте обратно 0×0 целых и прибавьте остаток 1...
получите изначальное имеющееся число якобы "делённое" на ноль...
Деление на целые части заканчивается когда вы не можете больше "отсоединить" от делимого количества записанного в делитель...
При нуле находящимся в делителе вы не сможете "отсоединить" вообще ничего от делимого числа пытаясь вычесть ноль...
даже при "бесконечных" таких попытках...
поэтому для действия деления на ноль это равно всегда ноль целых...
а остальное неделимый остаток...
Много есть искусственных точек нулевого отсчета в различных измерениях различных величин...
Но в большинстве своем они не имеют никакого отношения к делению на ноль...
Я лишь изложил некоторые видения своей теории математических "действий" с нулевыми значениями...
На сегодняшний день никто не смог переубедить меня в этом... и даже наоборот после дискуссий на эту тему мое личное убеждение в моей правоте возрастает все больше...
Вам же желаю всех благ в деле поиска и распространения знаний...
1:43 *shows binomial theorem* Brain.exe has stopped working
0^0 = limit a->0 a^0 = 1
But lim(x→a) f(x) may not be equal to f(a).
0⁰=0^(1-1)=0¹/0¹=undefined.
This is a commonly repeated argument, and I can understand why it seems convincing. The exponential "rules" that most people learn in elementary school actually have restrictions on them, but most people aren't used to seeing these restrictions. These rules work when you have a _strictly positive_ base; however, some of these rules do not work when the base is 0 or negative.
For example, (a^b)^c = a^(bc) is not, in general, a valid rule if the base is negative and if the exponents aren't positive integers.
((−1)^2)^(1/2) = 1^(1/2) = 1
but (−1)^(2*1/2) = (−1)^1 = −1
So these are not the same.
The rule you're using here, that a^(b−c) = a^b/a^c is not valid *at all* when a = 0.
Let's suppose that it is valid: that 0^(a−b) = 0^a/0^b (I think already writing it this way, you may start to see a problem.)
So if this is valid, then we could say
0^2 = 0^(4−2) = 0^4/0^2 = 0/0
The step in the first equality just uses the fact that 4−2 = 2. I think we agree with this. The step in the third equality just uses that 0^4 = 0 and 0^2 = 0. I think we agree with that as well. So the only possible error in this argument is 0^(4−2) = 0^4/0^2.
Of course, using 0^2 here is pretty irrelevant. You could use any number instead of 2 and get the same issue. So the problem is that 0^(a−b) is not equal to 0^a/0^b.
This then shows that your argument that 0^0 = 0/0 is faulty as well.
Correct
Not the same. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.