a good place for a floor equation.

[@goodplacetostop2973]

0:02 Good Place To Stop 10:27 Michael’s homework 10:33 Good Place To Stop

[@goodplacetostart9099]

You Should Include 0:02

[@goodplacetostop2973]

@@goodplacetostart9099 People were asking for you during the livestream, where were you?

[@goodplacetostart9099]

@@goodplacetostop2973 I had been hospitalised for few weeks and got discharged recently . I had some food infection , nothing related to Covid thankfully

[@goodplacetostop2973]

@@goodplacetostart9099 Oh wow, I didn't know that, I'm sorry to hear that.

[@pardeepgarg2640]

@@goodplacetostop2973 bruh you are good place to stop and one upper than you is good place to start So Is there is anyone who is Good place to live,die,eat,rest,..etc

[@MrRyanroberson1]

1:55 with 0^0 being undefined, you can't hope for it to be a solution even with 0 in the naturals

[@Bodyknock]

Actually, sometimes 0^0 is undefined but other times authors define 0^0 to be 1. For example in combinatorics it's common to define 0^0 = 1 since 1 is the number of permutations of the empty set. (Obviously you can't "prove" that 0^0 =1, it would always be by definition, but when it's useful to give it a defined value that value is normally 1 because it's the most natural value to assign.)

[@jimbolambo103]

@@Bodyknock Even then, you don't get a solution as 0^0 - 0 has equal 0.

[@Bodyknock]

@@jimbolambo103 Right, still not a solution, just pointing out that sometimes there's a defined value.

[@davidepierrat9072]

@@Bodyknock are you sure you are not talking about “0!” ? That’s the formula for the number of permutations of a set. 0!=1 by definition and no arguing with that

[@Bodyknock]

@@davidepierrat9072 You're right, I mispoke. n^n is the number of combinations of selecting objects n times from a set of n objects with replacement. In the case of the empty set and selecting 0 times, that's 0^0 = 1 (i.e. there's one way to make no selections with replacement out of the empty set.) P.S. Another isomorphic example is when counting the number of functions from a finite set to itself (e.g. the number of functions from {1,2,3} -> {1,2,3}). The number of such functions for a set of size n is n^n, and tn the case of the empty set, there's exactly one such empty function so it makes sense to define 0^0 = 1 here.

[@Grundini91]

1>=0 is not a contradiction. It is a tautology.

[@blackkk07]

(1) 1:40 That's an interest issue about what is 0^0. Maybe restrict x >0 is better. (2) In 6:40 there is no 'contradiction' for 1 >= 0. In fact if n = 1, the original condition x - [x] = n = 1 fails by the fact that 0 = 1 and n = 0). In fact, n = 0 and x \in [1, 2) are solutions, since x^n - [x] = 1 - 1 = 0 = n. Conclusion: I think a better way to discuss is to separate cases by n as follows: * If n = 0, get x \in [1, 2) are solutions (as (3) above). * If n = 1, no solution (as (2) above). * If n >= 2, as the video and get x = n = 2 and x = \sqrt[n]{n + 1}.

[@oida10000]

(1) he normally excludes 0 from the natrual numbers, so for him that is not an issue. Also if we consider consider it to be one, than x=1 is the only the value that would work here. Why? Because 0^0 is undefined and more often than not also considered to be 1 and not 0.

[@Triple_Trouble739]

@@oida10000 if n=0, we have the equation x^0 = floor(x) + 0. Provided x /= 0, we can simplify to 1=floor(x) so the whole interval of [1,2) is valid as that results in floor(x) = 1, no where in that interval was x = 0.

[@pablomartinsantamaria8689]

Since when 1≥0 is a contradiction?

[@samwalko]

The actual contradiction isn't hard to see, but yeah that one was pretty bad.

[@CHector1997]

It is an annoying mistake, the correct way is that x-floor(x)=1 is impossible because floor(x)+1>x (definition of floor function).

[@Horinius]

LOL Since Michael has made a silly mistake and didn't spot it 🤣

[@JM-us3fr]

Sometimes I think he puts these mistakes in there on purpose

[@Horinius]

@@JM-us3fr There is purely not true. If you are following Michael's videos enough, you would have seen that he has made quite a lot of mistakes and most of the time, he even didn't spot them at all.

[@b1akk-b7p]

Interesting problem: Show that for any natural n there exist natural a and b such that n = floor(3^a/2^b)

[@nanamacapagal8342]

My mind went straight to vectors and span ang logs and idk why it did, it just did The plan was to use î = [log3; 0] and ĵ = [0, log(1/2)] and somehow prove that all the natural numbers can be pinned to a lattice point in quadrant 1 this does not feel like the correct way to approach this

[@roberttelarket4934]

Mike I'd damn think after your marathon there wouldn't be a video until June, 2022!

[@Bodyknock]

Regarding x=0 and n=0, normally either 0^0 is left undefined or it's defined to be equal to 1. (For instance, in combinatorics, 0^0 = 1 is a useful definition because it corresponds to the number of permutations of the empty set.) I can't think of any contexts off-hand where 0^0 would be defined to be any number other than 1, when it's defined at all. So assuming you define 0^0 = 1, then 0^0 - Floor(0) = 1 which would mean x=0 and n=0 is not a solution.

[@Lucaazade]

In every context where 1 and 0 are symbols referring to different objects, for every object a, a^0 = 1 ≠ 0 is completely obvious. There is a concept entirely unrelated to the value a^0 = 1, in calculus where “0^0 is an indeterminate form” means limits lim f(x) → 0 and lim g(x) → 0 is not enough information to evaluate the limit lim f(x)^g(x).

[@Bodyknock]

@@LucaazadeYes, and of course for all non-zero a you have 0^a = 0. So on the one hand you have a set of functions of the form Fa(x) = x^a and also another set of functions Ga(x) = a^x. F and G are both well defined sets of functions for all non-zero a and x, and for all non-zero x you have Fa(0) = 0 and Ga(0) = 1. So the question then is what is F0(0) and G0(0) ? Note that if they have a value it is the same one because F0(0) = 0^0 and G0(0) = 0^0. It’s ok to define 0^0 as a number if you wanted to, but whatever choice you make is going to make either F0 or G0 discontinuous. It’s just that choosing 1 if you choose anything at all happens to work out nicely in practice in a lot of practical contexts.

[@sharmilasharma3227]

I challenge every one to solve the problem of floor function on the floor.

[@aayushsharma1345]

Thank God the problem didn't include the ceiling function😂.

[@goodplacetostop2973]

@@aayushsharma1345 Floating Michael Penn then 😂

[@aahaanchawla5393]

Round function would involve jumping from one place to another depending on the case

[@pardeepgarg2640]

Lmao , I lose 😹😹😹

[@Maya-yy4ye]

@Aayush Sharma 🤣🤣

[@tonyhaddad1394]

6:40 thank u mst michael for teach me that 0 > 1 all these years i was wrong !!!! 😥

[@cicik57]

so i guess it can be done shorter, just to note 0^0 -> 1 so x=0 and n= 0 does not match, n = 1 is evuivalent to {x} >0 means x is not natural, then you will gen difference in n then if n = 2 ,x = 2 or x > 3 is already too big , and if n = 3, x >= 2 is alreaty too big

[@shivam5105]

Great video, would never have thought of using those seemingly trivial inequalities

[@ronbannon]

Rewrite the problem as x^n - n = floor(x), graph the floor(x), and you see that the x^n-n intersects the floor on the interval between 1 and 2. Only two exceptions occur y = x -1 never equals the floor(x), and y = x^2 -2 equals the floor in two spots. All solutions are of the form x=(n+1)^(1/n), with only one exception, x=n=2.

[@goodplacetostop2973]

HOMEWORK : Find all real values of x such that 2 sin(x) = x² -4x + 6. SOURCE : 2015 W.J. Blundon Contest

[@arundhatimukherjee]

x²-4x+6 = (x-2)²+2 which is always greater than or equal to 2, whereas 2 sinx is always less than or equal to 2. Therefore, for some real x, if exists, they both must be equal to 2 (x-2)²+2=2 implies x=2; however 2sin2 is not equal to 2. So no real solution exists.🙂

[@goodplacetostop2973]

SOLUTION *SIKE, there's no solution* Completing the square we see that x² −4x + 6 = (x−2)²+ 2 ≥ 2. So the right hand side of the equation is greater than or equal to 2 for all values of x. It is equal to 2 only when x = 2 (and greater than 2 for all other values of x). The left hand side of the equation 2 sin(x) is less than or equal to 2 for all values of x. The left hand is equal to 2 for x = ..., −3π/2, π/2, 5π/2, .... The equation could only be true if there are values of x which simultaneously make both sides equal 2. But this is not possible.

[@pardeepgarg2640]

@@goodplacetostop2973 thanks for problem solvable for me But real challenge must for Imaginary solutions :?

[@goodplacetostop2973]

@@pardeepgarg2640 You can go ahead and try to find solutions for complex numbers if you want to. I wont bring any solution for that though, this is left for the viewer like 😉

[@titassamanta6885]

Min value of the quadratic expression is 2. But sin x is less that or equal to 1. So, no solutions.

[@hoanglaikhanh1383]

I bring a much more easier solution: x^n= floor(x) + n then x^n is an integer, so x is an integer. If x is negative, then x^n - floor(x) smaller than zero, but n is not, a contradiction. If x is zero, ez. If x is positive then by induction x^n - floor(x) > x^n - x, which is bigger than n with every x>3, n>2. P/s: This solution is just for speacial case when from the equation we can get x is integer.In another case, i think michael will do it better :)

[@piotrbrzezinski243]

your first sentence is false, consider (sqrt(2))^2

[@cheerymoya]

There is another set of solutions for n=0 and for any x in the interval (1,2)

[@DavidSavinainen]

Isn’t the “single solution” technically an extension of the left solution, the n = 1 case, so that x = (n+1)^(1/n) for n >= 1?

[@Jimbob1337]

If x=1, n=0. (1^0=1, 1-1=0) If x=2, n=2. (2^2=4, 4-2=2) n cannot be equal to 1 because if it were, the index is ignored and x-x=0, which is clearly not the case. If n is greater than 2, then x cannot be resolved because even 1^3 - 1 = 0. 8-2, 27-3, 64-4 etc. Likewise x cannot be greater than 2 because the reduction does not equal n. 3^0=1, 1-3=-2. 3^1=3, 3-3=0. 3^2=9, 9-3=6 etc.

[@mikegallegos7]

Another nice ride ! Thanks, Prof.

[@goodplacetostart9099]

Good Place To Start At 0:02

[@CM63_France]

Hi, 4:15 : you didn't mistake, this time. x:xx : good place to continue For fun: 1 "ok, great". I am looking for a function that would equal the infinite product: f(x) = (1-x³)(1-x³/8)(1-x³/27)...(1-x³/n³)... I suppose you guess what I am looking for. Any Idea? Thanks you very much for that marathon last sunday.

[@lisandro73]

Wait, why 1 bigger or equal to 0 is a contradiction?

[@CM63_France]

I noticed this, but I am too lazy to watch the video again.

[@benlovesrainie]

I am wondering if 0^0 is undefined rather than equal to 0......

[@Bodyknock]

0^0 is normally either left undefined or defined as 1 depending on the context. For example, in combinatorics it's useful to define 0^0 = 1 since 1 is the number of permutations of the empty set. I'm not aware of any situations where 0^0 is defined to be 0, I think there's too many scenarios where the natural value for it would be 1 for it to be assigned a different value.

[@hydra147147]

Regarding (n+1)^(1/n) as a family of solutions, we should prove that the floor of this number is indeed 1.

[@pratikmaity4315]

So our claim is that (n+1)≤2^n for all n€N. We will use induction to prove this. n=1 is true. Assume that the statement is true for some n=k€N. So k+1≤2^k. Now k+2=(k+1)+1≤2^k+1

[@hydra147147]

@@pratikmaity4315 Yes, this is also the Bernoulli inequality for x=1 (which is similarly proven by induction).

[@Triple_Trouble739]

@@pratikmaity4315 The claim has to be strictly less than, otherwise we can get the floor(x) = 2, but otherwise, you are correct.

[@pratikmaity4315]

@@Triple_Trouble739 I didn't write the claim keeping the problem in mind....I mean in general for any natural n (n+1)≤2^n for n=1 equality holds basically

[@UltraMaXAtAXX]

Maybe a video on the Sylow subgroups for groups of order 69 and 420?

[@stevenwilson5556]

6:42 1 ≥ 0 is a contradiction… hmm I guess I need to refresh my math skills, I was not aware of this fact…

[@seroujghazarian6343]

1:53 Because 1 totally equals 0. Totally.

[@schweinmachtbree1013]

it does in the field with one element (:

[@seroujghazarian6343]

@@schweinmachtbree1013 ik. I'm just talking about /R. ;)

[@nanamacapagal8342]

This is mostly just from experimentation and I haven't formally proved this, but I figured X = - (N - 2)^(1/N) for even N might work.

[@not_vinkami]

4:44 I love how he spelled "eaeh"

[@zealot2147]

I figured out n=2 x=sqrt(3) on my own but didn't realize it was part of a whole family of solutions on (1,inf)

[@koenth2359]

1:40 regardless whether you consider 0 a natural number, 0^0 is an indeterminate form, for that reason it is not correct to say that (x=0, n=0) is a solution.

[@PlutoTheSecond]

It's an indeterminate form for use with limits in calculus. Other areas of math, especially combinatorics, define 0^0 as being equal to 1. There is an argument that the value of an expression is distinct from the use of that expression as a limiting form. You can find it in a Wikipedia article specifically on 0^0.

[@koenth2359]

@@PlutoTheSecond Even if you would define 0^0=1, (x=0, n=0) is not a solution, regardless whether 0 is a natural number. Michael would at least have spent some words to the conundrum, so I think he was not aware of it in the moment and just assumed 0^0=0.

[@Basedgwad]

graphical method make it surprisingly easy

[@wyboo2019]

x=0 & n=0 gives 0^0 on the LHS and that's VERY iffy to say 0^0=0

[@wyboo2019]

at 6:40 you say "1 >= 0 which is a contradiction," but 1 IS >= 0

[@RandomBurfness]

1:47 0^0 is undefined!

[@Sxilder1k]

I thought it was well accepted that 0**0 should be 1, not 0…

[@Sxilder1k]

@Andrew Layton in any case, it should not be considered as 0, and hence be a solution, should it?

[@schweinmachtbree1013]

@Andrew Layton 0^0 is only undefined in the context of limits of functions. In the context of arithmetic of numbers, which is the context in this video (as there is no limits/calculus in the video), 0^0 is 1

[@gustavinho1986]

There is a simple way to show that there is no solution for x>=2 and n>2. For that, n=x^n-[x]>=[x]([x]^(n-1)-1)>=2(2^(n-1)-1). This implies that n+2>=2^n, which is only true for n={1,2}. Thus, when n>3 the solutions must have [x]=1, what give us x=(n+1)^(1/n), (n>2). We are still left to consider the cases where n={1,2} and x>=1. We can see that there is no solution for n=1, since 1=x-[x]

[@WhattheHectogon]

6:40, how is 1 >= 0 a contradiction? Great video!

[@warmpianist]

You can check in the original equation x - floor(x) = 1, which obviously has no solutions.

[@ChrisVenus]

@@warmpianist If you look at n= 1 in the original equation you are right that it doesn't work. I don' think that means that 1 >= 0 is a contradiction though. Like, putting n = 1 into the equation works. The equation correctly evaluates to 1 >= 0. That doesn't contradict anything itself...

[@warmpianist]

@@ChrisVenus actually on the factorization step floor(x)^n-1 won't work for n = 1 anyways.

[@leiferikson850]

1 >= 0 means 1 is either bigger OR equal to 0. However, 1 is NOT equal to 0.

[@MushookieMan]

@@leiferikson850 There's no contradiction. 1 is indeed greater than or equal to zero.

[@roboto12345]

Try the APMO 2021 P1

[@nocontextmuhtar1750]

Hey i want to suggest a problem how do i do that?

[@MathElite]

Awesome suggestion Good Place To Stop! I really like the title too

[@goodplacetostop2973]

😎

[@pardeepgarg2640]

@@goodplacetostop2973 😎

[@HagenvonEitzen]

But if n=0 and 0 n

[@utkarshsharma9563]

Hey I heard today that the Riemann hypothesis has been solved by a mathematician in Hyderabad, India. Does anyone know anything about it? I haven't found much on the internet

[@utkarshsharma9563]

@@last3239 but I read that he's a physicist or something. Did you read the paper? I can't understand all that stuff yet so I couldn't read it

[@TrimutiusToo]

X=0 n=0 cannot be a solution, because 0^0 is undefined

[@xiaoshou6752]

Yeah, I was thinking the same...

[@schweinmachtbree1013]

actually 0^0 = 1, but it is true that we still don't get a solution

[@xiaoshou6752]

@@schweinmachtbree1013 That's absolutely untrue, it can be proven that defining 0^0 is nonsense. It is however true that the limit of x^x as x -> 0+ equals 1.

[@TrimutiusToo]

@@xiaoshou6752 it is also true that limit of (0)^x as x -> 0 equals to 0 That is why it is undefined, because depending on how you approach 0 you will get different answers

[@schweinmachtbree1013]

@@xiaoshou6752 0^0 is only undefined in the context of limits/calculus, and there is no calculus in this video. In algebra, anything to the power of 0 is an empty product, which is equal to 1 (more abstractly, in any monoid, anything to the power of 0 is the identity element). You can find discussion of 0^0 = 1 on wikipedia, for example.

[@byronwatkins2565]

I know this is picky, but 1 is not in the half-closed interval [0, 1)...

[@schweinmachtbree1013]

that's not picky :) it is good to point "technical" things like this out because they can often cause part of a proof to be invalid (which most of the time is easily fixed by dealing with a particular case on its own; for example with intervals you might have to make a separate little argument for the endpoints than for the interior).

[@byronwatkins2565]

@@schweinmachtbree1013 In this case making both ends open, [0, 1], fixes the problem.

[@ezequielangelucci1263]

hi michael, last days i was doing a problem but i dont know how to solve it x-floor(x)=1/x

[@Mathcambo]

Good teacher

[@Legerine]

don't changing it from X to Floor of X change the equation?

[@matthieumoussiegt]

6:42 1>0 a contradiction 🤔

[@roberttelarket4934]

If you're on the top of the Empire State Building elevator going down you should be doing a floor equation!

[@sinecurve9999]

6:48 but 1>0...

[@mrrashedali]

Nice shirt

[@MicheleCaine]

Here is my challenge for you. Find f(x) such that : ceil(f(floor(x))) = floor(f(ceil(x))) for x natural in [0,1) if there exists

[@michaelcampbell6922]

A: If you don’t consider 0 to be a natural number, there are no solutions B: If you consider 0 to be a natural number, then 0 is the only one, so floor(x)=ceil(x)=0, and so ceil(f(0))=floor(f(0)). This means that f(0)=n with n in integers. This means that f(x)=ax^b +n for a in nonnegative integers and b in positive integers C: If you meant integers in that range, same solution as B D: If you meant rational or real numbers in that range, then the only solutions are f(x)=n for n in integers

[@siddharthbharadwaj3221]

0⁰ is an indeterminate value, right? 1:54

[@PlutoTheSecond]

There are examples in math (e.g. combinatorics) where 0^0 is defined as 1. It still does not yield a solution here though because 1 ≠ 0.

[@roberttelarket4934]

If it's a hot summer day and you're at the beach without paper and pencil it's great place to do a floor equation on the sand with your finger!

[@Alex_Deam]

Not if you're Archimedes being approached by a Roman soldier (supposedly)

[@roberttelarket4934]

@@Alex_Deam: I didn't expect anyone to know that about Archimedes.

[@Alex_Deam]

@@roberttelarket4934 Pretty sure it was in a kids' pop maths book that got me into maths and hence years later watching channels like this lol

[@guilhermefranco2949]

And to think there's a chance of we being alive when they prove Riemann Hypothesis, what about it, Hilbert?

[@emanuellandeholm5657]

RH could be undecidable.

[@zeranderman4170]

The first solution i can think of is x=1 and n=0

[@nastrimarcello]

I think something is wrong in this part here: X^n=n , x=0, n=0. Because 0^0 is undefined (and not 0, right?). Then it follows that X^n is undefined.

[@Bodyknock]

Sometimes authors define 0^0 = 1 depending on the context. In combinatorics for instance defining 0^0 = 1 is useful because it corresponds to the numbers of permutations of the empty set. Whether or not it's defined therefore is a matter of what's most convenient for the subject at hand.

[@nastrimarcello]

@@Bodyknock thanks for the input, but does 0^0 equals to 1 or is undefined? I don't see this in the video

[@Bodyknock]

@@nastrimarcello In the video he said he was leaving it up to the viewer if you wanted to define 0^0 or not. In fact if you define 0^0 = 1 it's still not a solution to the video's problem with x=0 and n=0 because 0^0 - Floor(0) = 1 which doesn't equal n=0. So really it doesn't matter much in this case, you get the same result either way.

[@itays7774]

bruh you're presenting 0^0=0 as a possible solution just like that

[@brsram]

X= 0, n = 0? What is 0^0????

[@noumanegaou3227]

But 0^0 is not define

[@mcwulf25]

Nice

[@sgdufbaoaah8692]

f(x)=[ax^2 + b]^3 and f[g(x)]=g[f(x)] . Find g(x) ? help me please.

[@MicheleCaine]

consider g(x)=f^-1 (x) then solve for x

[@sgdufbaoaah8692]

@@MicheleCaine thank you brother

[@kruksog]

0 is not a natural! Rahhhh! If zero had to be invented/discovered, it's not natural.

[@helo3827]

mistake at 1:38 and 2:22, anything to the power of 0 is 1.

[@viniciuscastro4957]

x=0, n=0 is not a solution. 0^0...

[@facr]

This video is a contradiction or greater than a contradiction.

[@amirb715]

at 1:54, 0^0=exp(0*ln(0)) is not equal to 0

[@МихаилУжов-е2й]

For real? 0^0, does it make sense? I've been taught that this is uncertainty.

[@schweinmachtbree1013]

0^0 is undefined in the context of limits of functions, but the number 0 to the power of 0 is 1. See the wikipedia article Zero to the Power of Zero if you're interested :)

[@DavidCodyPeppers.]

God Loves You and so do I. Peace! \o/

[@lisandro73]

I thought 0 to the power of 0 has no solution

[@juliopuerta5049]

4:49 if n-1, n-2, or n-3, or... is/are negative, the floor of x raised to it isnt bigger than 1, right??

[@roberttelarket4934]

Sorry I have no intention of doing any homework. I stopped that in 1973!

[@ourgoalisto6737]

No Goals here , sorry

[@chritophergaafele8922]

What can these problems be used for (Application), Computer science, biology, chemistry or in Physics

[@schweinmachtbree1013]

this is Pure Math channel I'm afraid xD

[@ahmadkalaoun3473]

0 to the 0 th power is undefined Great video :)

[@udic01]

6:45 1>=0 is not a contradiction. But if n=1 we get x-[x]=1 and that is impossible because {x} is in [0,1) 1:47 0^0 is undefined!!!!!

[@YaamFel]

0^0 is usually defined to be 1 in analysis

[@udic01]

@@YaamFel even so it isn't a solution to the equation...

[@YaamFel]

@@udic01 It's not, but your reasoning for why it's not is still wrong. I never claimed it was a solution either

[@thibaud5764]

Just I'm not sure but isn't 0⁰ not defined like it's either 0 or 1 ? Or am I wrong

[@schweinmachtbree1013]

The number zero to the power of the number zero is one (there are several ways of seeing this - you can read the wikipedia article Zero to the Power of Zero if you are interested), but people will often leave 0^0 undefined because in the context of limits it is indeterminate: if f and g are continuous functions which take the value 0 at some number b, i.e. f(b) = g(b) = 0 and lim_{x->b} f(x) = lim_{x->b} g(x) = 0, then the limit as x->b of f to the power of g is not determined; it can take many different values depending on what f and g are, e.g. we could have lim_{x->b} f(x)^(g(x)) = 1, or lim_{x->b} f(x)^(g(x)) = 0, and in fact this limit can take any value, or it may not exist at all. But to reiterate, although the expression 0^0 is indeterminate in the context of limits/calculus, there is no limits/calculus in this video, so here 0^0 equals 1 (and so the case n=0 does not actually give a solution to the equation in the video)

[@thibaud5764]

@@schweinmachtbree1013 thank you a lot I now understand the mistake I make ( talking about limit in an algebra problem). It helps me a lot !

[@schweinmachtbree1013]

@@thibaud5764 you’re welcome :)

[@michamazur2179]

How 0^0=0?

[@giovanicampos4120]

1:45 ok if I consider x = n = 0 as a solution, then I get 0⁰ = 0 and that's my favourite identity in math ❤️

[@schweinmachtbree1013]

en.wikipedia.org/wiki/Zero_to_the_power_of_zero

[@YTCalynk]

1:53 0^0 is undefined, so it cannot be a solution.

[@jimschneider799]

At 1:50 - no, x = n = 0 is not a solution, because 0^0 is undefined.

[@JalebJay]

n=0 shouldn't be a solution as 0^0 holds no value.

[@Bodyknock]

Actually in some contexts authors define 0^0 = 1. For instance, in combinatorics it's useful to define 0^0 = 1 since the number of permutations of the empty set is 1. So really it just depends on the situations and the author whether or not 0^0 is undefined or defined to be 1.

[@angelowentzler9961]

But… 0^0 is undefined

[@arimermelstein9167]

0^0 is undefined, no?

[@ezequielangelucci1263]

si

[@schweinmachtbree1013]

@@ezequielangelucci1263 no

[@ezequielangelucci1263]

@@schweinmachtbree1013 yes, it is

[@schweinmachtbree1013]

@@ezequielangelucci1263 no, it isn't. 0^0 = 1.

[@ezequielangelucci1263]

@@schweinmachtbree1013 demostrate it

[@luizgilbertooliveiramessia2217]

I think the same 0^0 is undefined. Is not a solution.