The Mystery Of The 0th Root

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This video explores the tricky concept of the zeroth root. Looking at limits, we find the zeroth root seems to lead to contradictions and inconsistencies. As we approach zero in the denominator, the quantity goes to infinity or zero depending on the number we start with. With no clear intuitive definition, the zeroth root remains an elusive mathematical idea that is traditionally left undefined.
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

[@BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

[@arcaltoby5772]

Can you please tell me how Aleph-Null differs from Infinity? Like, why use Infinity as the limit? Could Aleph-Null be the limit because it's the first number larger than any finite number.

[@penguincute3564]

@@arcaltoby5772 aleph of null contains infinity but not infinity times infinity

[@kyokajiro1808]

in my opinion the ⁰√x is just simultaneously 0 and ∞ for any positive number that isn't 1

[@dinh1cam36]

If x

[@kyokajiro1808]

@@dinh1cam36 the way i see it, for x=1 it's 1, for x=-1 it's the unit circle on the complex plane, otherwise it's simultaneously 0 and infinity

[@marcusscience23]

an iconic trio: division by 0, 0th root, log base 1

[@talhochberg5062]

They are all division by 0 in disguise. ⁰√x=x^(1/0) log_1(x)=ln(x)/ln(1)=ln(x)/0

[@rangerdanger572]

I imagine this as one of those "tell us a little bit about yourself" segments but instead 1/0 keeps appearing with fake mustaches and wigs

[@brandoncukrow4033]

More iconic than Lebron D Wade and Bosh

[@NStripleseven]

Div0, div0 in a trench coat, div0 with a fake mustache

[@oceanb0rn05]

Well log base 1 has complex solutions. Log base 0 on the other hand...

[@OfficialUKGov]

Schrödinger’s root

[@FantyPegasus]

😂😂😂

[@usipussi6647]

Lmao

[@AnuragForJEE]

Shouldn't comment about something you know about 0.064% about.

[@OzOMega]

​@@AnuragForJEEshouldnt be on the internet if you cant take simple jokes

[@ManyamGupta]

​@@OzOMega well said and destroyed

[@Rhovanion85]

Zeroth also sounds like the final boss of a video game

[@FlatTopRob]

And there's no strategy guide to beat it.

[@cbaha4985]

⁠@@FlatTopRoband You don’t win the battle…you merely convince Zeroth that fighting is meaningless by dragging the fight out for several hours of trying to endure.

[@tft6182]

and now im reading “zeroth” not as zero-th but zer-oth

[@donotdoit8428]

Zeroth hand of golden buddha

[@user-jz7vf5iq7h]

Sephizeroth?

[@ianmathwiz7]

We could just have the zeroth root be a multifunction, like with other roots or with the complex logarithm. The zeroth root would have two branches if |z| is not 1, evaluating to either zero or infinity on each branch, and infinitely many branches if |z|=1. Granted, the resulting Riemann surface would be highly pathological, but I think that's to be expected in a case like this.

[@sumdumbmick]

this idiot doesn't know about multifunctions

[@deltapi8859]

I just wanted to say. This only screams for an analogue to (a little bit more complex, no pun intended) complex numbers. I see no issue here.

[@sakesaurus]

no. Square root is not a multi function. Why should 0th root be?

[@ianmathwiz7]

@@sakesaurus It most certainly is a multifunction, because it's defined as f(z) = exp(1/2*log(z)), and log(z) is a multifunction and the 1/2 means you get two branches of the complex logarithm before you run into the periodicity of the exponential function.

[@joejoemyo]

The only word I understood was "if"

[@MusicEngineeer]

Yes - as others pointed out, maybe a multifunction could come to the rescue. At 1:20, you ask, how can the 0-th root of 1 be 2 and 3 at the same time. That made me immediately think: well, the square-root of 4 is 2 and -2 at the same time - so having multiple solutions is not such a crazy thing actually. The somewhat weird thing is only that now our set of solutions would become infinite - I guess even uncountably so.

[@Guillau213]

The square-root of 4 is only 2. You can have -2 if resolving an equation, for example, x^2=4. Then, x=2 or x=-2. (Since sqrt(x^2)=|x|). But the square root of a number returns the positive output only, and not the negative one. Take the square-root function f(x)=sqrt(x). If f(4)=sqrt(4) would be equal to 2 and -2, then f would not be a properly defined function. I hope this clear things out!

[@ianmathwiz7]

@@Guillau213 What you said is correct for the square root function as it's conventionally defined for real-number inputs. However, the complex square root is typically defined to be a multivalued function, or a function defined on a Riemann surface. In this case, both 2 and -2 are valid square roots of 4. It's also possible to only take the principal branch of the complex square root function, in which case only 2 is a valid square root. But IMO this is not only arbitrary, but it destroys the mathematical properties of the Riemann surface.

[@lih3391]

⁠​⁠​⁠@@Guillau213the complex squareroot is always multivalued, having 2 values that are negatives of other. The real squareroot only has 1 positive output

[@Guillau213]

@@ianmathwiz7 You are right for complex roots. I was referring to the problem using real values only, since I thought the question was to define the function for real values. But at this point, if we are using the complex root, we are no longer in the real analysis world, so the rules are different ;)

[@adamuhaddadi5332]

yeah just like the complex log :)

[@EdKolis]

The zeroth root is basically the infinite power, or what you'd get by applying compound interest for an infinite amount of time. Either zero, infinity, or one.

[@user-zz3sn8ky7z]

That point is addressed in the video, the left and right limits to those values don't "meet up"

[@ptrkmr]

Yes, but no. It’s +♾️ when going + to zero, but -♾️ going - to zero. Plot y=1/x to get a better idea. However, I’m an engineer, so for all intents and purposes 1/0 is +♾️ and -1/0 is -♾️

[@msq7041]

There are cases where it is none of those 3. ie (-1)^∞ does not converge, but is bounded. Its also not just x^∞, its technically x^(e^iθ*∞) for whatever thats worth

[@msq7041]

​@@ptrkmras a mathematician, you cannot imagine how cursed that is...

[@chrisjfox8715]

​@@msq7041I think OP's point is a practical use of it. There are applications of mathematics that inherently place conditions around what's feasible. For instance, a real world optimization problem in which the optimal value has units that can't possibly be negative in the real world despite the math alone concluding so.

[@abhidababy6746]

There is another problem AFTER negative numbers: complex numbers: ___---___ I clarify Zeroeth root of Negative/Complex Numbers

[@bp_iyarea]

-1√X = y where y^-1=X y^-1 = 1/y -1√2 = 0.5

[@bp_iyarea]

Xth root of y = y^(1÷x) and for I, that would be y^(-i) so i√2 = 0.7692389-0.6389613i

[@cycrothelargeplanet]

​@@bp_iyareaI'm never reading math video comments ever again, my brain hurts

[@ajilialyran]

I love the image of the same attitude applied to complex numbers 'cause it limits to a circle at infinity. It's such a pretty spiral!

[@bp_iyarea]

Fancy a McDonald's

[@ssaamil]

3:46 nice! I like this way of showing the "process" of taking limits. :D

[@denki2558]

The zeroth root actually appears on the "generalized mean". The formula is `(sum(x_i^p)/n) root p` where n is the number of data points, and p is a parameter to control what type of mean to compute. p=-1,1,0.5,2 correspond to the harmonic mean, arithmetic mean, RMS, and SMR respectively. The mean should be undefined when p=0 (zeroth root), but if we used the limit, the parameter surprisingly correspond to the geometric mean, whose traditional formula has no additions nor divisions.

[@santerisatama5409]

Youp, just explained in another comment how the thingie can be takens as the root seed of Stern-Brocot style concatenation of mediants.

[@viliml2763]

That's because this isn't just the zeroth root, it's the zeroth root of something to the zeroth power. Anything to the zeroth power is 1 and the zeroth root of 1 is 1 so it checks out, the geometric mean is recovered in the infinitesimal neighborhood of 1

[@santerisatama5409]

@@viliml2763 If we write the top of the exponent tower as A, and logarithms from that as A-1, A-2, A-3 etc, then it's not such a big deal to change A to 0 and add instead of negate. Stern-Brocot algorithm generates inverse fractions which are pairs of fractions which have versions of 1/1 as their mediants, and unitary operators manifest IIRC also in other ways.

[@santerisatama5409]

@@viliml2763 Other fascinating unitary operator contained in Stern-Brocot tree is that if we modify the fractions a/b to "simplicities" 1/ab, then each new row of simplicities adds up to 1.

[@deltapi8859]

I really like your style of doing math. Reasoning about which approaches you could take. Following them through and reflecting the merit of each approach. And one of those is going to be more suitable in the context of the field. This is totally not what you do in schools, because in schools you don't reason or try dead end paths for enquire. In schools everything has a "right" and "wrong" drawer things need to put in as fast as possible. Neither thinking nor science works this way.

[@MuffinsAPlenty]

Excellent point! Thinking many options through is such an important process in math!

[@RexxSchneider]

That's because teaching maths in schools is circumscribed by the need to maximise the performance of as many pupils as possible in a terminal exam on an arbitrary curriculum. If you want to change that, then vote out the politicians who think that they are the best people to decide what (and how) our children should learn.

[@deltapi8859]

@@RexxSchneider Don't even know how to answer this comment other than: perfect.

[@MichaelRothwell1]

Nice topic the zeroth root - I hadn't thought of this before! Interesting that in the first part you point out that ⁰√1 can take any value, then in the second part you say we had better define ⁰√1 to be 1... For me it's like 0/0 - it can take any value, so is undefined.

[@spretzel]

That's because he was using different definitions of the zeroth root in different sections of the video. Also, he only made these statements to show that they lead to absurd conclusions, not because he actually believes both statements.

[@JuhaKona]

your videos are a beacon of clarity and inspiration!

[@profdc9501]

Another way to look at it is to examine the limit for y = exp(x) = lim n->infinity (1+x/n)^n You can get a parameterized version of the log function with n: lim n->infinity y^(1/n) = 1 + (1/n)log y changing parameter a = 1/n y^a = 1 + a log y for a->0 so for power a tends to zero, the result is linear with the log function.

[@pressfinchat]

The problem with limits is that it describes how the function behaves at it gets closer and closer to 64^(1/0), but NOT at 64^(1/0) itself. For example, you can have a limit x-> 1 equal to 1 but f(1) can be something completely different, like f(1) = 2, even though it should be 1 if we trust what the limit is telling us. So, even with regular numbers, the argument that “for number a, the limit of this function approaches this value so the function must be this value” is already completely invalid, let alone when using it to explain a number like 0.

[@7lllll]

the zeroth root is not a function, but a relation. zeroth root of 1 is the set of all real numbers, and the zeroth root of anything else is the empty set

[@santerisatama5409]

Not real numbers, but something much better which includes also 1/0. ;)

[@jacksonstarky8288]

The 0th root has been one of my greatest sources of entertainment in mathematics... and this video brilliantly highlights why. I obviously completely agree with the conclusions reached... and with the conclusion of the linked video. 🙂

[@felixstelis6666]

The reason why it does matter from which side is the limit is because the power is negative so the limit of n to the 0- of 64 is like saying the limit of n to 0+ of 1/64 and the opposite with the 0.5

[@periodictable118]

The zeroth root of 1 can be literally any number since any number^0 = 1 (Even 0^0=1 according to many sources). What this really means is that 1^infinity can equal anything, which becomes painfully obvious when you first learn limits.

[@omp199]

Your claim that 1^infinity can equal anything is not "painfully obvious". It's not even meaningful.

[@masonwheeler6536]

3:10: "So the zeroth root of 64 should be 64^1/0. And herein lies another problem with the zeroth root: we're dividing by zero." Is it really another problem, or is it exactly the same problem stated more clearly?

[@SJGster]

With division by 0, approaching 1/x from the left and right results in negative infinity and infinity, and it makes sense to define a number system where those are the same. Makes less sense to try to make 0 and infinity equal I guess.

[@ptrkmr]

Saying +/- inf is correct tho

[@vindi167]

i agree, i think +∞ and -∞ are equal. everything i've seen points to the idea of the numberline looping at ∞

[@SJGster]

@@vindi167 it's not really a question of objective truth, it's a question of what you want to do. In some cases, having them be the same and the number line being circular is useful, so use that. (E.g. dividing by 0). In other cases, having separate positive and negative infinity may be more useful. They're different number systems, just be clear which one you're using and don't mix them up.

[@RSLT]

Very good and unconventional explanation of limits. Great job!

[@tyronorxy5646]

I like the idea of it being equal to multiple values. It looks similar to superposition. What is if you try defining it with an "any of" prefix?

[@jesusbermudez6775]

Thanks for the info on the 0th root of a number. I've learned something knew.

[@TheLukeLsd]

Finalmente alguém fez um vídeo sobre esse assunto que eu já passei dezenas de horas fazendo inúmeras abordagens usando álgebra, comparação com exponenciação e até limites no expoente e no índice de das raízes.

[@gblargg]

1:30 Square roots in general have two values, e.g. sqrt(4) is +/-2. So having more than one value isn't immediately a problem.

[@giabao576]

what

[@AdamBomb5794]

Square roots do not have two values, otherwise it would not be a well defined function. The solution to x^2 = y will have two solutions for x, but that's a quirk of ^2, not the square root.

[@gblargg]

@@giabao576 SQUARE ROOTS IN GENERAL HAVE TWO VALUES. What you might be thinking of is the *principal* square root, the non-negative of the two.

[@gblargg]

@@AdamBomb5794 I looked it up more and it's a subtle linguistic distinction. The number 4 has two square roots, 2 and -2. However the symbol √ denotes only the non-negative (principal) square root.

[@steka68]

@@AdamBomb5794 Rather, it DOES have two values, thus is NOT a well defined function. You may CHOOSE to define it so it is well defined (for example so that the square root of a positive number is a positive number), but that's a rather arbitrary choice in general. Powers are generally well defined only if the exponent is an integer.

[@banrtv9446]

Glad to see you back buddy

[@LostSendHelp_YT]

what happens when you put it on a graph, will it be something like a quadratic where its big with in one set of rules and small in another?

[@freedom_aint_free]

Very interesting video and It was as interesting as it was underwhelming: a lot !

[@jaypaint4855]

The zeroth root is n^1/0, so it would inherently be undefined. I’m interested to see how you made a video out of it though.

[@a_minor]

In my opinion, the function at 5:07 is the best way to define 0th root of a real number. The very last argument you made about the limiting values interchanging when taking x

[@olixx1213]

No it cannot be dismissed Even when written as 1/64^(1/x) , it also tends to 0 64^(1/x) CANNOT be assigned a value at x = 0 because it doesn't approch a single value when x approches 0

[@a_minor]

​@@olixx1213i think you might have misunderstood my point. To put it in simpler terms, my point is, that the case {x < 0 and a > 1} can be rewritten as the case {x > 0 and 0 < a < 1}, and vice versa. Hence making the problem at hand a bit less complex and making the function at 5:07 the closest way we can represent the 0th root.

[@olixx1213]

@@a_minor No you cannot , because we need x to be 0 since its the 0th root of a

[@levistepanian5341]

My understanding of calculus allows me to agree with your take: That a number (other than 0) "a" raised to the (1/0) power (infinity) (from the left or the right) should remain undefined, it is as if one is trying to define a number which has a 0 factor in the denominator, and while limits allow for this, each equation should be treated with the respect it deserves, every problem is different. I'll note that 0^(1/0) or 0^(infinity) is equal to 0 in all cases when using a limit, such that it can and has been defined properly, but any other number cannot be used in such a manner, it breaks.

[@saidfarid6382]

Hello dear professor Your lessons are really interesting and crucial,i do appreciate your job,i wish you peace and happiness under the sky of prosperity,all the best. Take care and have a good time. Your Student from Algeria

[@IloveRumania]

You know stuff gets crazy when even Bri leaves the expression undefined

[@skilz8098]

When it comes to mathematical computations and their limits such as a/0, tan(PI/2 + PI*N), vertical slope, etc... I personally rather list them or determine them to be indeterminate as opposed to "undefined". This for me pertains more to the context of the language than anything else. Take vertical slope for example which is basically the same as a/0 where a != 0, and tan(PI/2). They are basically the same thing. Vertical slope approaches either +/- infinity. For me this isn't undefined. It is however indeterminate because this is a many to one solution. Undefined to me means something that doesn't have a definition. And for things such as division by 0, vertical slope, tan(PI/2), etc... they are actually well defined. It's just that their results don't pass the vertical line test as they have more than one output. Take the general equation of a circle: (x-h)^2 + (y-k)^2 = r^2. Where the point (x,y) lies on its circumference, the point (h,k) is its center and r is its radius. This also doesn't pass the vertical line test because there are two outputs for most of its inputs. Yet, this equation or expression although isn't a function is well defined and isn't considered undefined. In fact, the equation of the circle for all tense and purposes is the same thing as the Pythagorean Theorem: A^2 + B^2 = C^2. It's just that one is relative to circles, where the other is relative to right triangles. Also, the slope of a given line from within the slope-intercept form y = mx+b is defined as (y2-y1)/(x2-x1) = dy/dx = sin(t)/cost(t) = tan(t) where t, theta is the angle that is between the line y = mx+b and the +x-axis. So for me I don't care for this idea that we've been taught that division by 0, and other phenomenon within mathematics is undefined. To me that means it doesn't have a definition, that it's not defined. Now, I can completely agree with indeterminate or ambiguous. As there is a many to one solution, or its output jumps all over the place based on specific ranges within its domain. This is just my take though. What if I were to tell you that that tangent function is actually continuous... Consider the fact that tan(t) = sin(t)/cos(t). We know that both sin and cos are continuous for all of its domain and that its domain is at least All Real Values since they are continuous wave functions that are periodic, oscillatory, rotational and transcendental. And the tan function can be composed of their ratios! Thus, for me, the tangent function also shares those same properties. Yeah, many will argue that the limit of the tangent function isn't the same when taking the left hand from the right hand limits... and to this argument I still wouldn't claim that it is undefined. I would claim that it is either indeterminate or ambiguous. The reason for this, is that tan(PI/2) approaches both + and - infinity and from the left and right hand limits it also approaches 0 besides just +/-infinity. This leads to the conclusion that it isn't undefined as this is no different than vertical slope or division by 0, it's just that we can not determine its output based on that single input as it is a many to one solution. This is what makes it indeterminate or ambiguous.

[@DB88888]

Hey, just wanted to say I only recently found your channel and I love what you are doing here. I do mathematical modeling in biotech R&D for living, and believe it or not, for most of my life i struggled with math, wspwcially as a kid: the way math was taught to me was very dogmatic, with little room for creativity or debates. It was only in uni when I literally started to build my own math out of necessity that I started to ubderstand what math really is. Now I have a newborn and I am often thinking how would I teach math to him: your approach is what I am going to follow with my kid. You are demistifying math as an a-priori god-given universal truth, and showing instead what math really is all about: a set of logically coherent rules which have consequences and limits. These rules can be whatever you want them to be, as long as you build them in a way that makes sense, but of course the limits, consequences, range of applicability and usefulness of your mathematics will very much vary based on which rules you decide to use. Thanks gor putting your content out there and keep u lp the good work!

[@theking2000]

awesome intuitive approach

[@SJrad]

then when x < -1 and we’re talking the limit from the right, it explodes into infinity but half of the time it crosses into the complex world depending on if there’s an even-th root

[@ofekshochat9920]

Great video! But I dont agree with your final conclusion (of it being too hard to explain). Negative roots are just 1/|the root|, so this makes total sense...

[@user-my4yv9zb3p]

that's why i love math. Sometimes things so complicated that they make you laugh.

[@Psylent]

For some reason, I can’t add this video to my watch later list. And I can’t add it to my playlist. It was hard to physically watch this video. WTF youtube? I don’t think I’ve ever had this problem. I like this video

[@PanduPoluan]

I always immediately think of roots as the reciprocal exponentiation. So square root = x ^ (1/2) Cube root = x ^ (1/3) Zeroth root = x ^ (1/0) Define 1/0 first , _then_ you can calculate the 0th root.

[@Tom-vb6fk]

This logics fails however for x^(1/x), as when x approaches 0+, x^(1/x) approaches 0, and when x approaches 0-, x^(1/x) approaches 0. Thus x at 0 is defined here.

[@PanduPoluan]

@@Tom-vb6fk No. If x at 0 is defined, then you just put in x=0 in the formula... which you can't. Rather, the *_limit_* of the formula as x approaches 0 from both sides is defined. But a limit is just that: a limit. It's not the actual result of x at 0.

[@Tom-vb6fk]

@@PanduPoluan thats wrong, this shows that x is continuos at 0, and it converges to 0. Matter of fact x^(1/0) is defined for x is an element from (-1,1). You should learn what makes 1/0 undefine first

[@Rudxain]

IEEE-754 numbers can solve this: 1/+0 = +∞, 1/-0 = -∞ . This is because +0 and -0 aren't "exactly 0", *they represent the limit as we approach 0 from the corresponding side.* This explains most of the weird quirks of FP math

[@deltalima6703]

I like this video. Would have liked to see some consideration of complex numbers and also the gamma function. I doubt it matters with something this ill bahaved though.

[@qazmatron]

This COULD make an interesting GRAPH. Plot y = "xth root of k" by plotting y = k ^ (1/x), for some selected values of k. • With k=1, the plot is a horizontal line at y=1. • With k=1.25, the plot has y=1.25 at x=+∞, exploding to y=+∞ as x approaches 0 from the right, AND y=0.8 at x=-∞, collapsing to y=0 as x approaches 0 from the left. • With k=0.8, the plot is a mirror image of the previous graph. The plot has y=0.8 at x=+∞, collapsing to y=0 as x approaches 0 from the right, AND y=1.25 at x=-∞, exploding to y=+∞ as x approaches 0 from the left.

[@JJ_TheGreat]

5:11 I was just starting to think about that... But what about where x < 0?

[@DavidRTribble]

You could have added a graph showing the limit trends (0 or ∞) on either side of -1 and +1.

[@Arycke]

You should make a video with the graph.

[@SeasOfCheese929]

f(x) = x^(1/n) is a continuous function for all n. We could look at the behavior of this seqence of functions as n goes to infinity?

[@josephshaff5194]

Nice ! Allright ya earned a sub !

[@GhostEmblem]

The cards for you videos dont appear for me I have a lot of youtube addons so maybe that has something to do with it ?

[@lambdadelta3105]

what if we *do* define it though? As in, what if we do a similar thing to what we do to give -1 a square root, in that we take some 0th root and give it a value thats on its own terms (such as one we could call q) and then try to make this work?

[@andunyaa]

Amazing!!!

[@anglewinder]

Self evident I’m sure, and bit more work, but I’m confident that a simple graph could better illustrate the the limits and the resulting disconnects. Still a concise explanation.

[@yunamadsimackomen881]

You can rewrite zeroth root of x as (e^(1/0))^lnx. Rewrite the expression , now you take the limit as n approaches 0 on, and you can see that the positive limit expression is undefined (+inf) at 1, and 0 for all the negative values, and the negative limit expression is undefined at 1 (+inf as well) and 0 for all the positive values. That means that the function has no real value anywhere, because the left side contradicts the right side of the limit, and the one point where both limits agree is undefined.

[@BrattyBrett]

Would you say that replacing the zero in the maths system with a square and the square works just like 0, however the square represents space as a zero. Would be a easier way to represent zero being added to maths to represent space.

[@tylerbakeman]

It’s one of those situations where something is undefined, so redefine it (or use category theory to see why we can/ cant)

[@bullpup1337]

around 0:40 - is this a generally accepted definition? The one I know defines the root as just the positive solution, otherwise the result is not a number, but a set of possible values (or the root loses its functionality).

[@user-ky5dy5hl4d]

What about if you start using not 0th root but go below zero like negative twoth root. What about let's say, negative second root; -n written on the radical sign? Like -2 root of 5?

[@darryllmaybe3881]

I mean, it makes sense to leave it undefined. Anything to the 0th power is 1, so the 0th root of one could be anything, but you can't even take the 0th root of any number other than 1 because the 0th power of any real number could never be anything except for 1. It's kind of like the 0th root of 1 is undefined and the 0th root of any number other than 1 is a domain error.

[@berkaykokturk2671]

Is it actually non defined or do not exist? When we say right hand limit and left hand limit is not equal, limit of x as approaches a does not exist. Then the case we disguise in the videos has one property that i mentioned. The left limit and right limits of the integers between 1 and 0 and bigger than one differs. And also 0th root of 1 is not just equal to one but also equal every real number. So, it seems "DNE" fits more to this analysis.

[@ICQ200]

*NICE VIDEO, THANKS, LOVE IT*

[@mineonlydiamonds]

How about imaginary-th square root?

[@christopherneufelt8971]

You are absolutetly Blasphemous! I suscribe!

[@Tripleye]

I don’t get it, if you put a different input then you get a different output right? Correct me if i’m missing something

[@HopeRock425]

Did anyone else start laughing when seeing the thumbnail? This is the funniest thing I've seen all week. The concept is hilarious impossible.

[@globaltruth957]

In the third grade when we were exposed to multiplying and dividing for the first time, we learned that multiplication was just repeated addition and division was repeated subtraction. Therefore, division by zero means subtracting zero until you can't subtract anymore. Therefore division by zero equals infinity. That has nothing to do with this video, though, even though it was mentioned as being undefined.

[@CC-1.]

you can show √4 As 4^1/2 Hence to define the root 0 You'll need to get 1/0 and since 1/0 is infinity and not a number everything ends here it's like you have one end of a infinite line but since you don't have the other side You can not measure it. (summary)

[@DR-54]

I would just use a different numbering system entirely where 1/0 is equal to the number v and that 2/0 is thus 2v and 0/0 is 0v which reduces into 0. -1/0 is -v. -2/0 is -2v. infinity/0 is infinity(v) (and thus this is now entering the surreal number territory). i/v is iv, which in a way is just 4. I haven't really played with this nor really just checked if anyone else has done it but I think this is just an easy ass way to deal with the undefined 0s without breaking everything. Some calculators can't handle the square roots of negative values, yet some have imaginary numbers built into it. Some calculators can't handle division by 0, but a calculator that implemented v into it would be able to. In this system, things like the 0th root are no longer mysterious because it's like taking the imaginary root of the number. The general case is simply going to be undefined, or infinitely many solutions possibly I've not thought this out too deeply.. The solution is going to be unknowable, until it suddenly is knowable. anything to the 0th power is best defined as 1. This would mean that, in 0^x, all values of x which are either positive is equal to v, if x is 0 then it is equal to 1, and all negative values of x is equal to 0. It's a piecewise that is actually true here to my knowledge but I'm going to guess my 2am math skills are wrong here because of some random ass shit I didn't think about. v^v is the 0th root of v and is thus not possible as stated above. I would like to imagine this number system as being dimensional. The way these dimensions work are as long strings of every value on their respective number line who then intersect at 0 with all other number lines added to this dimensional approach.These dimensions can then be transformed using functions. This is like bending, pulling, compressing, or cutting the string of the dimension This can be done by application of the new variable w, with w effectively forming a z (real) and series of other axes to define the non-real strings's transformations upon the other strings, wherein the functions applied to dimensions are f(w), then by using f(w)=d^w, with d being the dimension transformed, upon two strings in which one is the imaginary number line and the other is the real number line, then the imaginary number line now forms a cylinder in which the radius is defined by f(w)=d^w and it intersects precisely with the z line where the imaginary number line now becomes real. When this concept is applied to v with the same formula you get a case wherein everything 0 and below is exactly within the real number line except for it being 2 gravitational points where all negative values are at 1 and only x=0 makes the z dimension equal to 0, and then everything else conforms to the 0 axis precisely. It's like a big ass horn. Now you can just keep adding dimensions to other kinds of numbers and you'll get this gigantic mess whenever you change the function for f(w) as it conforms it all into their special intersecting cases. My imagination may have gone wild here but this is literally how I visualize differing number systems when transformed by different functions. I don't know why my brain finds it easier to visualize things in higher dimensions than just the number itself.

[@wayneyadams]

The zeroth power of any number is one for a very good reason. Exponents are subtracted in division, so if the numerator and denominator have the same exponent meaning the numbers are equal the result is 1. n^x/n^x = n^(x-x) = n^0 = 1. It is not defined to be one, it IS one by calculation.

[@adayah2933]

It is defined to be 1...

[@wayneyadams]

@@adayah2933 Did you read and understand my post about why it is one, not defined as one?

[@adayah2933]

@@wayneyadams Yes, but it only makes sense on the surface. Given a > 0, the very standard approach is to *define* a^0 as 1, then a^n inductively for positive n, then a^-n as 1/a^n. (The definition can be further extended to rational and real exponents, but for the purposes of this discussion it is not necessary.) Next the identities a^(x+y) = a^x * a^y and (a^x)^y = a^(xy) are *proved* from these definitions. You seem to be doing something completely different. You *assume* the identity a^(x+y) = a^x * a^y (or similar) without saying how a^x is defined, and then you prove a^0 = 1 from that identity. For this approach to make sense, you would need to define a^x in such a way that the identity a^(x+y) = a^x * a^y becomes a more direct consequence of the definition then a^0 = 1. It's technically possible, but rather exotic and nobody really does that, unless for fun or education. Moreover, your approach fails to show that 0^0 = 1, which is an equally useful convention as a^0 = 1 for a > 0. That's why it is better to have that as a definition, not a consequence. If you disagree, let me ask: how do you define a^x?

[@wayneyadams]

@@adayah2933 I am using standard mathematical operations for exponents. There are no assumptions, or unsupported definitions. Exponentiation is shorthand notation for repetitive multiplication. For example, nxn = n^2 and nxnxn=n^3 (nxn)x(nxnxn) = nxnxnxnxn = n^5 n^2 x n^3 = n^5 In the same way division is accomplished by subtracting exponents. Let's divide. (nxn)/(nxnxn) = 1/n n^2/n^3 = n^(2-3) = n^-1 n^-1 = 1/n There is no reason n cannot be negative. n0. The only place it breaks down is when n=0 which is the only place we have to define the value.

[@yakirfrankoveig8094]

What is the problem with saying it means to the power of infinity according to the limit

[@alejrandom6592]

0:31 the arrow is in the wrong direction. Example: (-1)²=1 => -1=sqrt(1) which is nonsense.

[@mrblakeboy1420]

you could also treat it as n^0=x, which makes n be anything if x=1 and undefined if x≠1

[@signisot5264]

keeping in mind that this is the definitional inverse function of x^0, it would make sense that 1 maps to every positive and negative value simultaneously, and that 0 itself would be as hard to define in this equation as 0^0. When taking the limit of the function from the right we're effectively re-charting x^infinity, and from the left we're taking x^(-infinity) which is to say (1/x)^(infinity). The behavior of the zeroth root really makes sense for all numbers, with some special behavior at 1 and some undefined value at 0 as per the 0^0 reason.

[@syed3344]

I dont rlly comment on videos but ur videos r very interesting,thank you.

[@leonmelbin1326]

another way of finding that is it undefinied is for example. x^1/2 = the sqrt of 2 and x^1/3 = sqrt of 3 so if we were to do the zeroth root we will have to have x^1/0 which is underfined therefore the zeroth root is undefined

[@nick_fous]

In 1:07 you said that x to the 0th power is one for any positive real number. However, a negative number to the 0th power equals to 1, too.

[@user-on2dt7dv8n]

This also goes into the rabbit hole of x divided by 0. A log (or anything else) divided into 2 pieces is one of the ways people teach division to children. If you want to cut a log into 1 piece, you don't make a cut at all. If you want to divide a log into 0 pieces, you delete the log from existence. So x/0 might be 0.

[@michaelhunte743]

Given that between 0 and 1 it converges, it may be that the function is a representation of a distribution.

[@WillRennar]

x^(-y) is the same as (1/x)^y; this is why scientific notation for small decimals is generally written as x*10^(-y), such as the pressure of space being notated as 1x10^(-17) torr rather than 0.00000000000000001 torr. So in your examples at 5:25, 64^-(1/x) would equal (1/64)^(1/x) which would be between 0 and 1, while 0.5^-(1/x) would be the same as *1/0.5)^(1/x), which would just be 2^(1/x) and thus be greater than 1...thus the values for 64 coalescing to 0 and the values for 0.5 coalescing to infinity holds true to the earlier definition.

[@Arycke]

The values don't coalesce, though. Coalesce means to come together.

[@Arycke]

You're statement about those 2 things, x^(-y) and (1/x)^y, being the same in the first sentence, is not why scientific notation for small decimals is generally written the way it is. It is written that way out of convenience. In the way you have stated it, you claim a fact about exponents is the reason why and technically it's not. It is indeed a fact that what you said is true for x for real numbers except 0, but yeah.

[@LFTRnow]

What happens with the 0th root of i? (or j for our engineering friends), what about other complex numbers, etc 2, 3, 4... terms? What about 0th root of say 0+3i+4j+5k+6l?

[@emnot8982]

So hilarious.....but awesomely explained

[@SingleBoy-je5kp]

So according to the rule root of a number is equivalent to power of that number by 1/rootdigit since any number power zero is 1 then 1 root 0 is equivalent to 1 power infinity which may contain any values which is wrong as one multiplied by itself is 1 despite how many times it is carried out but taking limit will lead to positive or negative infinity

[@Bill_Woo]

You said the answer. x is the set of all nonnegative real numbers. It's just that you can only take the 0th root of the number 1. But x can be anything. Using undefined /0 to get something else - verboten. BTW what is (-1)^0? BTW what is (i)^0?

[@dmimiguo]

- Daddy, why they say it is impossible to divide by zero? - C'mon son. Sit down. We need to talk about USS Yorktown.

[@rgbatom5145]

Can we please redefine the symbol to just mean the nth root and put a 2 next to the square root? It would look so much nicer with other roots

[@JonathonV]

A few questions: 1:07 You said that any positive real number raised to the exponent of zero equals 1, but negative reals raised to the exponent zero also equal 1, so you don’t need to make that distinction: any nonzero real number will work. In fact, more broadly, any nonzero complex number raised to the exponent zero equals 1. As for zero itself, that’s a bit more debatable. 😄 Those immersed in calculus would say zero to the exponent zero is undefined, but if you ask someone who deals mostly with combinatorics, they’d probably say it equals 1. (The Google calculator, for example, says it equals 1.) 5:16 I’m not sure it’s so cut-and-dried to define the zeroth root of 1 to be 1. The “note” says that 1 to any exponent equals 1, which is true, but any nonzero real raised to the exponent of 0 equals one, which is just as consistent of a pattern, so mathematicians could just as easily define it as ♾️, though technically you could define it as anything you want if you don’t want to call it undefined. (I suppose defining it as 1 gives it a nice symmetry.)

[@carlosrivas2012]

Bonito ejercicio. Gracias.

[@Fifacardscreator]

I’m an aspiring mathematician and I’m actually working on division by 0, on the exponential part I found that if x is positive then x^1/0 (basically the 0th root)=1/0 If x=1 then x^1/0=1 And if x is negative then x^1/0=+/-1/0, I can’t detail everything but if y’all really want maybe I’ll make a video on the subject

[@qy9MC]

The only case it where might work is the 0thRoot(1)={x, x∈R/{0}} because we are basically solving for x^0=1 and that’s defined for all real numbers except 0. And for non 1 numbers we have 0thRoot(x≠1)= ∅ basically no solution for anything else. So I would be implying that 1^(1/0)= x which doesn’t make any sense except if you look at it as x^0=1. So to go from that to the other expression I’ll need to take the 0th power then to evaluate 0/0=1!? Anyway, stuff is weird ignore this comment, it just breaks down.

[@josephshaff5194]

And Reload. Those are good too. Not many know about them!

[@StevenTAbell]

Take a hint from the IEEE-754 floating point math standard: NaN = Not-a-Number. This turns strange results like this into something at least vaguely computable. It has been a part of every FPU you have ever touched, but most people still don't know about it.

[@myomyat2216]

NaN happens because FPUs don't know the answer to something It isn't inherently related to mathematics

[@SIXSHAMAN]

On the complex plane, n^(1/z) has ESSENTIAL SINGULARITY at z = 0. So undefined is the only possible answer.

[@robfielding8566]

// just note that in geometry, there are objects that square to 1, 0, or -1. // these are square roots of 1. directions in space are such a thing. // these are exactly as weird as i*i=-1. u*u=1. v*v=1. // note that when you square a number, there are multiple possible choices // it is not defined which one it is, without other context x^2=4 implies (x=2 or x=-2 or x=2u or x=2v or ...). which can be converted into: x^2=4 implies x=2 or x^2=4 implies x=-2 or x^2=4 implies x=2u or ... We are not bothered by this, where a multiply by zero loses information about a: 0 a = b if you were to divide by 0, then you are saying that a could be anything. b/0 implies (a=0 or a=1 or a=1/2 or a=-2 or a=i or ...) // note that this is not a problem if a=0. // but by convention, we won't write b=0a as b/a=0 when a=0. 0a = b implies (b=0 and b/a=0) There is something we are not doing right in our notation when we track side-conditions.

[@reptilianinmateusz8533]

My brain is too smooth for this

[@whisper3856]

How to Scare Your Math Teacher and/or Math Class: 1. Write 0th root of any number 2. Write “Solve this” under it 3. Let chaos ensue

[@nangld]

You need non-standard analysis with infinitesimals and infinitely large entities to do that. Then it is perfectly defined, you just have different infinitesimals corresponding to different roots of 0.

[@MuffinsAPlenty]

Could you explain a bit more about how nonstandard analysis would help define a 0th root?

[@Questiala123]

Pretty close to the infinite root, simplifying as x^1/infinity, 1/infinity is approaching zero if not already. X^0 is 1 and therefore the infinite root of x is 1. But this works for any equation no matter the number, so they’re all 1. 1^infinity is undefined as well this being one of those reasons. This root also states every number is every other number and is therefore undefined. 0 is the opposite of infinity and negative infinity and as it involves 1/0 which approaches those exact values, and numbers approach two seperate values depending on the way we approach zero.

[@dannygjk]

iirc the zeroth root of x can be written as x to the power of zero which = 1. yes?

[@markusbanach-stb5892]

Try defining the inverse element of zero. That should do the trick.

[@haroldlake1005]

So I ask you when a=0, which the root zero is. Thank you