@@kevalsavla1727 in a limit situation, im pretty sure it'd be determined by the side from which you approach zero. If you approach zero from the left (from the negative half of the x axis, so x -> 0-, the limit would approach plus infinity. If you approach zero from the right(from the positive part of the x axis, so x -> 0+, the limit would approach minus infinity. All that assuming the expression you provided is something like -1/x as x approaches zero. Please correct me if im wrong.
1:20 I remember we had a angry teacher in school who sounded funny when he was angrier. There was an incident where a student did something similar to this but it wasn't obvious because the equation was complex. The teacher got angry af and the way he scolded became a meme in the school.
I had a spanish teacher who made seating plans so we wouldn't talk One guy never sat in his seat and one day the teacher slammed his desk and shouted RESPECT THE SEATING PLAN and that became a meme lol
0/0 = 0 really makes sense on practice at least in physics engine programming and similar things, because zero vectors might arise in some cases but common equations require to compute size of the vector
@@irhzuf no. if you take the limit of 1/x as x approaches 0 from the positive integers you get positive infinity, but if you approach 0 from the negative integers you get negative infinity. going to the same place, 0, you get results that are incredibly different. this is why we say it's undefined.
I would argue that 0/0 = 0 makes perfect sense. Division: How many times a number must be added together to get to another (In other words, how many times a number must be multiplied by to get to another) So, a/b = How many times you must add b to get to a You might say that 0/0 could then equal 1 because, for example, 6/6 = 1 because you must add 6 up one time. However, if you rephrase the above a little: a/b = How many times you must add b **to 0** to get to a This is exactly how you first learn to do multiplication. So, by the above definition for division, 0/0 would equal 0, you already start at 0, you must add 0 to itself no times. n/0 where n is non 0 would still be undefined, you would keep adding 0 and adding 0 but you will never ever reach n no matter how many times you add zero to itself. (Unless you add 0 to itself a non real number of times, if you assume there is a valid non real number that will get you there)
Nah clearly the answer is this: "Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends.”
He already has made a few videos about 0^0 before, and they are all relatively ba- erm, I mean, those videos are not very good and they, unintentionally, spread a misconception that BPRP should instead by trying to actively combat as a mathematics education channel.
My teacher: "so dear students today we'll study Schrödinger's equation. Well, this strange symbol which you see right here is called- Me: "Fork. It's called fork."
@Tejas It's discussed more often in chemistry here, when they teach atomic structure and then take it to mad levels. But yes it's modern physics basically. I edited my comment and made it 'subject neutral' lol.
I think the k not equal to 0 situation would have the same problem you get from say squaring numbers. You always have to be careful when you're dealing with multivalued operations.
@@MrKnivan Indeed, which fundamentally comes down to the same issue as multiplying both sides by 0. In fact, if we can't do one, why should we be allowed to do the inverse operation instead without proving that the inverse operation is good in the firstplace, (as n/0 is undefined, and one can prove this easily, we already know the inverse is not good).
There’s another issue with the first example you gave as well. When we multiply both sides by zero, we have zero over zero on the left side which we are trying to define but in the example you just cross it out. We can’t cross it out because when you do that, you assume it even has an answer so #1 uses circular reasoning to define 0/0
There is no neat closed form formula for this sum if you use elementary functions. However if you also include special functions, there is actually a formula: math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this
What if you integrated 0/0 under an arbitrary function f(x)? I’ve got another random thought: let the numerator and denominator denote the magnitude of vectors A and B in the z direction respectively. Let vector A extend exclusively in the x dimension, and vector exclusively in the y direction. Hmm, this probably won’t work because you end up with: f(z)=0(x units)/0(y units) but this was a fun thought to trail. p.s great job with all your content man. Crazy work ethic!
What i feel is that division is just comparing the magnitude of one thing with the magnitude of another. If the magnitudes are equal, (0=0) then the result of division is one.
My logic is "division is the amount of times 'b' needs to be taken out of 'a' until it reaches zero (in a/b)" E.g. "15/5" is "15 -5 -5 -5 = 0", five is repeated three times, therefore 15/5 = 3 Using that logic with 0/0, we get... zero. We don't have to subtract anything because we already reached our objective. AKA the amount of times '0' needs to be taken out of '0' until it reaches zero is... zero times. Bonus: And using that logic with 1/0 we get: "1 -0 -0 -0 -0 -0... = unreachable" therefore it's undefinable because it never ends
I really like your way of defining zero by zero and division in general. I also want to share some of my thoughts. You may know that zero is a number which denotes nothing, a concept which has been introduced in mathematics later and at first people didn't even use that number. But when they started using it people ran into a problem. How could they define zero divided by zero? I mean how can something consist of zero pieces? And yes, reasoning that way, division by zero is simply undefined. However I have an idea. If we have a number for nothing, why can't we have number for everything? I believe this way division by zero will be defined. Anything multiplied by zero yields zero and anything divided by zero yields everything. Everything can be number like infinity - a number which can't be reached but a number which wraps all other numbers. What I mean by that is you start from zero, make your way through the positive numbers, get past everything, then through the negative numbers and back at zero. This concept also applies to complex numbers. You start from zero, pick a direction, travel in it infinitely and finally arrive from the opposite direction. Hope I made myself clear.
But you also can subtract 0 of 0 indefinitely so: 0-0-0-0-0-0-0-...-0 = 0. Technically since any amount of time you subtract 0 is still true you can argue it is undefined or if you dont see a reason to stop since it is still true going into infinity you could say it can be infinity... which is questionable since it is not a number.
@@olivers.7821 Exactly what I have thought. But maybe we should look at infinity in a different way. Maybe we can treat it like a number just like how I explained in my previous comment. I know you might say that you can't measure the difference between infinity and any other finite number, but think about it - there are infinitely many numbers between 0 and 1, yet we still can calculate the difference which is one. Maybe we have to learn to work with such numbers which have unique properties like x = 2x. The only numbers that satisfy this equation are 0 (nothing) and infinity (everything).
@@veselinborisov369 the problem with infinity is that not every infinity is the same for example if you count normally you would count infinite numbers but if you counted every possible decimal place aswell you would have at least an infinite amout of numbers between an infinite amount of numbers which makes the second infinite way bigger and therefore different so anything is not equal to anything so you can not really define it
I disagree with the fact that you have to assume 0/0 a constant. (0/0)×1 = (0/0)×17 means (Something × 1) = (another something × 17) That means you can assume that the first 0/0 and the second 0/0 has different values and they can take "any" value.
A way to think of rational numbers of p/q is how many times do I need to subtract q from p to get to 0. For example in 15/3 you have to subtract 3 from 15 5 times to get to 0 and thus we say 15/3=5. Therefore, in 0/0, for any number r you pick if you subtract 0 r times from 0 you’ll get 0-0(r)=0-0=0 so any number r would satisfy 0/0 under that way of thinking. In theory we can choose to define it as 0 like in this video but to stay consistent with how fractions work in other situations we say it’s undefined.
Good approach at the start. I noticed the same problem where people would multiply both sides by 0 at the start even though that obviously never works to begin with.
In Lean, a/0 is defined to be 0 because - it has to be something (Lean only supports total functions) - 0 is already defined in our type (as opposed to 1 which we may never define) - it makes some results like a/c + b/c = (a + b)/c unconditional (you do not have to assume c != 0)
Division by zero is undefined precisely because the multiplicative inverse of zero does not exist. Why then when you multiply nothing by something nonexistent should you get anything at all? It makes no sense.
Sir, I have always thought 0/0 should be equal to 0. This is why. Correct me if I am wrong. We can consider the quotient when doing division as the number of times the dividend has to subtracted by the divisor for it to become 0. So in the case of 0/0 since the dividend is already 0 there is nothing to do hence the quotient should be 0 right. We can consider division again as the number of times the divisor has to added by itself to be equal to the dividend. When we consider 0/0 in this perspective we can see that since the divisor is 0 there is nothing to do to make it 0 hence making the quotient 0 again. Sir, please reply to this has been bugging me up for some time. And every others who are reading this if you have any opinion reply or please try to like this so that he can see this.
0 is an answer, but any number also fits that. Taking 0 away from 0 5 times still "gets" you to 0, it's just that 0 as a quotient is the smallest answer.
As a fellow programmer, I disagree. If you're dividing by 0 often it means either "a really large number" (ex time left in a download) or "I should have divided by 1" (used a index instead of index+1)
@@gasparsigma You're raising up the point why it's not defined. It depends on the use case. My personnal experience differs, to which my original comment really only apply. Good to see the common default be different for others.
If 0/0 = k, then k must be 0. That's proven by this. However we can't assume that 0/0 actually is equal to any number, and since there are functions, which when you take the limit on them they get a 0/0 but still has a value that isn't zero, I think it should be undefined
@@nilsastrup8907 Are you incapable of doing basic arithmetic operations without appealing to the concept of limits? I am tired of repeating this in every comments section on RUclips, but for the 10083718304018384717940193772th time: lim x/y (x -> 0, y -> 0) not existing does not imply 0/0 is undefined, just like how lim floor(x) (x -> 1) not existing does not imply floor(1) is undefined. The logic you are using is borderline ridiculous.
i have a conter-argument. that proof is well made and hold for most situations, for example ln(1) because its multivalued and one of his answer is 0, but it does not hold for some specific situations? consider ln(0) = a, everybody knows 0^2 = 0^3, loging both sides and we have 2*ln(0) = 3*ln(0), 2a = 3a. if a is real and a=/= 0, then we can divide both sides by a and we have 2=3, but 2 is not 3, so a should be defined as 0 but ln(0) cannot be 0 if we consider ln(0) = 0 and elevate both sides by e, we have e^ln(0) = e^0 and 0 = 1, but 0 is not one, so ln(0) cannot be defined as 0. with this double proof by contradiction, we can agree that this proof needs more information to hold, without this we cannot be sure if it 0/0 can or cannot be defined as 0. again, nice video ☺
i have a conter-argument. that proof is well made and hold for most situations, for example ln(1) = 0, because its multivalued and one of his answer is 0, but it does not hold for some specific situations. consider ln(0) = n, everybody knows 0^2 = 0^3, loging both sides and we have 2*ln(0) = 3*ln(0), 2n = 3n. if n is real and n =/= 0, then we can divide both sides by n, and we have 2 = 3, but 2 is not equal 3, so n should be defined as 0. but ln(0) cannot be 0 too if we consider ln(0) = 0 and elevate both sides by e, we have e^ln(0) = e^0 and 0 = 1, but 0 is not equal 1, and then ln(0) cannot be defined as 0. with this double proof by contradiction, we can agree that it needs more information to hold, without information we cannot be sure if it 0/0 can or cannot be defined as 0. again, nice video ☺ (editing because im an ape and don't know how to write)
@@farklegriffen2624 That doesn’t quite work because writing 1/(0/0) = ... implicitly assumes that 1/0 is defined. But the only actual assumption made was that 0/0 exists and equals 0, there are no assumptions about the existence of any other number of the form x/0. So we can’t assume that just because 1/(a/b) exists for most a and b that it exists for the specific case of a=b=0. Likewise you can’t make any assumptions about the existence of, say, ln(0) existing. So just because ln(a/b) exists for most a and b doesn’t mean it exists for ln(0/0). Therefore if you want a contradiction that stems only from defining 0/0 = 0 then you have to make no other assumptions about any other as yet undefined expressions in the process.
@@angelmendez-rivera351 No, 1/0 doesn't need to be well defined just because 0/0 is defined. If you look at division as a binary operator called Div(x,y) where Div(x,y) = x/y for all x and y except 0, then what this video is doing is creating a new extended operator called maybe Divz(x,y) which is specifically 0 for x=y=0 and is undefined for all other y=0 parameters. There's no requirement in that extended system that 1/0 be defined.
@@Bodyknock Except that is false, because the video not only creates the operator, it also assumed certain properties of the operator, such as Div(x·y, z) = D(x, z)·y = x·D(y, z), properties that the video used in its proof. And the existence of those properties very much does require 1/0 to be defined. This is the problem with the video. Also, your argument that writing 1/(0/0) = ... implicitly assumes that 1/0 is defined is incorrect. It does not. It only assume that 1/(0/0) is defined.
I am from bangladesh .. your classes are awesome ..very much helpful for me .at first i am scared of differentiation .but now i just love it .❤️ Thanks a lot .
Maybe not from a math perspective but the only time I've ever had to divide by 0 was within programming and within the requirements that I was working with, it actually worked for exactly what we needed to have dividing by 0 equal to 0 as if you had multiplied, because it essentially allowed to turn a value into a boolean within the constraints of that language that we were using. The result was always either 0 or 1. the value was 0 or it was anything other than 0, and we counted it as 1 towards a sum of the total non 0 values.
My Intersection Formulas result in 0 ÷ 0 for overlapping equations, meaning that the 0 ÷ 0 might indeed be(Corrected from "by" on 12/25/2021 at 9:17 AM EST) all real numbers.
A division operation expresses the idea that the numerator is being segmented (denominator - 1) times resulting in denominator portions. If the denominator is zero and one refuses to accept this as undefined then it seems to me this could be interpreted as the segmenting operation being performed -1 times, resulting in zero equal portions. We may either interpret this as a negation of the division operation itself, leaving us with the original value, or interpret the negative of an operation as it's inverse, thus x/0 becomes another way of writing x*0. So, depending on which interpretation is applied, x/0 is either x or zero. Operation transformation must be accounted for before processing can begin.
I think that’s why he framed the proof as 0/0 not being equal to a non-zero number, rather than saying it’s always equal to 0. There are specific situations where 0/0 can be defined as 0, but you can’t define 0/0 as a non-zero number.
@@miguelwariskovih 1/0 = 1/(0/0) = 0/0 = 0. If you let 1/0 be an undefined operation, 0 must be undefined. But if you let 1/0 = 0, then you get into a whole mess of other contradictions :)
Asume that 0/0=0 for calcutations with limes: 0=0/0=(0/0)^(-1)=0^(-1) =>0=0^(-1)=1/0=Lim x→+0 1/x→Infinity =>0→infinity This a contradiction . So If you calculate with limes, then 0/0≠0. So don't define 0/0=0 for calcutations limes.
4:34 This proof you have heavily relies on the commutative property of multiplying with a fraction. I.e how (x*m)/n = x*(m/n) because (0*1)/0 = (0*17)/0 if you indeed multiply before you divide. But you've incidentally shown that this cannot be true if you define 0/0. because then if you let m and n both be 0 you get: (x*0)/0 = x*(0/0) k = k*x which is not true for all values k and x. But that fact in of itself is interesting because it means if you allow 0/0 math is no longer commutative in this certain way, unless you define it to be 0. So what you have really proven is that if k != 0 then (x*n)/m != x*(n/m)
He can because he introduce an axiom where 0/0=k. This is how imaginary number was invented. Mathematicians declared that there is a number i where i²=-1. He is doing the same thing declare that there is a number k so that 0/0=k. But doing that leads to contradiction. So the axiom 0/0=k cannot be true.
Amazing. Could you plz also give some STEP 3 questions a go. They are quite a lot more difficult and are more beautiful than the STEP 2 question than you attempted as well. There are some very beautiful one such as proving the irrationality of e etc.
Doesn’t this only apply if k is a complex number? We could alternatively define k as some non-zero non-complex number which has the property that 1/k = k and k != 1. After all, the inverse of 0/0 is still 0/0
If k is a non-complex number, what are the 0 and the 1 you talking about ? If you see the 0 and the 1 (and the division) as the 0 and the 1 (and the division) of a field K, and if you defined 0/0 as the solution in K/{1} to the equation 1/k=k, then why not ! But I think the point here is to solve it in R
@@julien31415 not entirely sure what you were trying to say for most of this, but yeah, the point was to define k inside R, though I just thought it might be something to point out
0/0 cannot be equal to 0 (or something else) in any case. If you are in the integers then quotient q and remainder r of the division n / m are defined by the equality n = q m + r where 0
Thats if you consider 0/0 a 1-1 functional output when it really has an infinite amount of outputs. I always liked to think of it as the symbol for the set of all numbers.
0/0 should not be defined at all, because 0 has infinite multiplicity. IE: 0/0 = 0^2 / 0^1 = 0^1 = 0, but also 0/0 = 0^1 / 0^1 = 0^0 = 1, and also 0/0 = 0^1 / 0^2 = 0^-1 = ??? Clearly, 0 = 1 is a contradiction, so 0/0 cannot be defined ever, right?
After thinking some more about this and other fractions, I have another argument as to why 0/0 could/should be identified with 0. If you look at a modular integer, a, with modulus m, then there is an inverse integer, b = 1/a mod m if and only if gcd(a,m) = 1. But, if you want to generalize the concept of inverse, there is a way to do this to give every modular integer an inverse. Say gcd(a,m) = d. Thus there exists a subgroup of the integers mod m, the integers mod (m/d). You can pull your calculation of the inverse into this subgroup by calculating 1/(a/d) mod (m/d), and now you are guaranteed that gcd(a/d, m/d) =1. Then you can 1 to 1 pushback into your larger group of integers mod m. If 1/(a/d) mod (m/d) = b, then your pseudoinverse of a mod m should be (b×d). Something strange but fully consistent happens when you consider the case a=0, which is consistent for any integer m and can be generalized to any number m. Your pull function takes you from your initial group m to the trivial group {0}. Your one to one function has to take your initial number a=0 to a number in the trivial group (which there is only one value to choose from. In this group, 1=0, so your operation 1/0 is equivalent to 0/0 and results in the value 0. When you pushback to the original group, m, you have to use the inverse of the same 1 to 1 function that brought you to the trivial group in the first place, leaving you with your original value a=0. Thus, there is only one possible consistent value for 0/0 in your group m, and that value is 0.
0:20 - 0:33 *0/0 = x 0 = 0·x* Right off the bat, this is incorrect. Division a/b is actually defined as a·b^(-1), where b^(-1) denotes the unique solution to b·x = 1 IF it exists. IF it does exist, then you can take the equation a·b^(-1) = c, multiply both sides of the equation by b, to get [a·b^(-1)]·b = a·[b·b^(-1)] = a·1 = a = c·b. So a/b = c a = c·b necessarily requires that b^(-1) exists. Since 0^(-1) does not exist, it is not true that 0/0 = x 0 = 0·x. 0/0 is not undefined because it is equal to every number, it is undefined because it is equal to no number at all, since 0/0, by definition, is equal to 0·0^(-1), and 0^(-1) does not exist. If it did exist without contradicting associativity and distributivity, then 0/0 = 1 would be true, though. Also, if 0/0 were equal to every number, then that would make it "multivalued", at least in the naïve sense, not undefined. 4:06 - 4:25 *and now, because we have 0/0 = k, we can legitimately divide both sides by 0, and then when we see 0/0, well, by our assumption, that's equal to k, so on the left hand side, we have k·1, and on the right hand side, we have k·17* Ignoring the fact that you have not actually a provided an operational definition of division to even make sense of the claim 0/0 = k, and of something like (0·1)/0, or any other manipulations the problem here is that you are assuming a new form associativity, now. You are claiming that you know for sure that (0·17)/0 = (0/0)·17 and not 0·(17/0), and both of these claims are just assumptions themselves. 4:26 - 4:38 *and now, because we said that k is not equal to 0, and again, just imagine that we are saying that 0/0 is equal to 2, right? So if you have 2·1 = 2·17, of course you can divide that on both sides* But this assumes that k is a real or complex number, and is thus invertible, which again, is just an assumption. Though I can still grant this, since it is still relevant to make this assumption if you want to prove that, if 0/0 is real (complex), then 0/0 = 0, rather than just proving 0/0 = 0. 4:39 - 4:47 *so here, we can just divide by k on both sides* This runs into the same problem as before. You take (k·1)/k = (k·17)/k without even having given a definition for division by nonzero quantities in the first place, but even if I ignore this, you are still assuming new-associativity, which you are most certainly not warranted to have in this new arithmetic system. 4:52 - 5:07 *1 = 17, after we divide both sides by k. But of course, this is wrong, so I just say, "but you see that 1 is not equal to 17", so this right here is the end of the proof.* But this only proves that if we assume 0/0 is nonzero AND still a complex number, AND if we assume (a·b)/c = (a/c)·b = a·(b/c) for every triple of elements, then there is a contradiction. It does not prove 0/0 cannot be nonzero complex if you give up the above equality, nor does it prove it cannot just be some other new noncomplex quantity. Which is the crux of my objection. The rest of the video is just explaining how, therefore, if you ever wanted 0/0 to be defined, then 0/0 = 0 is the only option, because it does not cause any contradictions. But it likely does cause contradictions, and this is because it fails to account for the fact that this does not define what 1/0 is. In fact, since you feel comfortable in claiming that / as a binary operation satisfies (a·b)/c = (a/b)·c, I can say (1·0)/0 = (1/0)·0, but since 0/0 = (1·0)/0, because 0 = 1·0, this means 0/0 = (1/0)·0, which is impossible unless 1/0 is also defined. This is why one cannot actually ignore that / was not defined as an operation in the video to make sufficient sense of 0/0 = k as a definition and of the manipulations being made in the video. In this regard, 0/0 = 0 is just as problematic.
Interestingly I saw a different version of the 1=17 contradiction a while ago. It stumped me at first: a=b a^2 - b^2 = ab - b^2 (a+b)(a-b) = b(a-b) a+b = b 2b=b 2=1 Usually this is quite rightly refuted by arguing that step between lines 3 and 4, a division by zero, is a nonsense operation and everything after is invalid. But if we define 0/0 = 0, line 4 would instead be written 0=0, which ‘fixes’ this.
3/3=1 2/2=1 1/1=1 -1/-1=1 0.5/0.5=1 So 0/0=1 no? To add when we have some equations with x/x inside , we simplify it by 1 without making the case that x=0; so we already define 0/0 as 1 ,no? We have : x=0 x+1=1 x+x/x=1 but x=0 so 0+0/0=1 0/0=1
From what I've been told- anything divided by 0 is + or - infinity (because as x decreases, the value of 17/x increases and the value of -17/x decreases) but 0/0 is undefined. Why can't 0/0 be defined as + or - infinity, instead of 0?
OK so I was searching for about an hour on your channel for a video on limits from a/0 to 0/0 and found nothing. I told my teacher it was possible but don't know how to prove it or an equation that proves it. Can you help me? I also said that it was a real number but I guess that was wrong
What about the inverse of 0/0? 1/(0/0) = 0/0 1/k = k Which implies k = 1 or -1 But there is nothing special about 1 so let’s call a number n. n/(0/0) = n*0/0 = 0/0 n/k = k
What I feel is division is like a comparison like 4/2 equals 2 which says 4 is 2 times 2 (a comparison) and if we divide 0/0 which is again a comparison and we know both are equal then what I think is it should be equal to one ! Since equality is denoted by 1 not by zero ?
I figured it was like a fifth root. A fifth root has 5 potential answers, but that doesn't mean they're equal, just that they hold an equal property when based off of that function. √5≠-√5, but (√5)^2=(-√5)^2. This means that 0/0 should have a principle value, namely 0, just like √2 means 1.414... not -1.414..., but if one is to divide zero by zero, it can be anything within (-∞,∞) or i (-∞,∞).
i'm not sure about this but i've heard that proof that contradiction only proves if a statement is false OR that it is undecidable, normally (for example when deciding if a number is irrational or not) i think this shouldn't matter (maybe?) but this 0/0 problem is precisely what you'd want to avoid
There is a sense in 0/0 being any number. You get into the contradiction because you try to devide by k which is any number. But those any numbers are not equal, just like two infinities are not equal. What we can derive then is that the ratio of two k's. k1/k2 =. 17/.1 =1.7 So k1 is not equal to k2, which will be true to the fact that k can be any number. Whaat do you think sensei?
If we assume that 0/0 is any number it is not a single umber of our choice but all universe of numbers at the same time. In case of equation 1*0/0=17*0/0 as 0/0 can be any numer at the same time we just need to prove that there exist 2 numbers let's say p and q which satisfies the eqation 1*p=17*q. Solving for this is easy. Solution is p=17*r and q=1*r where r is any number.
i believe if we were to define 0/0, it should not be 0, but a number that acts identical to 0. x=0/0, x≠0, y is any number, xy=x as for how 0*x should behave, idk this would of course require a new number, but we did that with sqrt(-1) and i
The result of division is always unique. Suppose 10 is divided by 2 we get the result 5 which is unique. If we divide 0 by 5 we get the quotient 0 which is also unit. But when we divide 0 by 0 then we get 1 or 2 or 3 ..... which is not unique. Can I not say that this is Indeterminate?
Hey! I have a somewhat different opinion on 0. I personally think of it as less of a number, and more of a concept. The closest analogy I have to 0 is infinity. For most practical purposes, they function the same way. Ex: multiply both sides of an equation by either 0 or infinity, it ruins the equation. “Divide” by either, you are doing sketchy math. Neither really fit in most mathematical groups. Maybe this is a dumb take, but it makes sense to me.
I would like to clarify. I don’t just think of 0 as similar to infinity. I would consider them inverses of each other - as much as non-numbers can be inverses. The math behind this goes something like this. *insert proof that .999 repeating equals 1. Therefore, 0 = infinitesimal (by subtraction). Infinitesimals are widely considered to be the opposite of infinity, and I believe they are the same thing.
@@greenytoaster They consider 0 and infinity to be inverses. They also believe that infinity and infinitesimal are the same thing. Their reasoning leads to the conclusion that infinitesimal is the inverse of 0. This contradicts another statement they made that 0=infinitesimal, because if n/infinity = infinitesimal, then infinity/n = 0 (which is not true when n is all reals except 0). This is why we say n/infinity approaches 0, not equal to
Look it up in the contradictionary. Although 0/0 should be left undefined, n mod 0 should be defined as n. I get annoyed when I'm programming something that mods something by 0 and the program aborts.
Why? In most programming languages % means remainder after division, and not modulo, so in that case it makes sense to be consistent with division for n % 0 and make the implicit n/0 undefined by default for any n And even when properly using modulus, mod m must give an element in the set {0, ..., m-1} by definition. And so, taking any number n mod 0 means it's an element in the set {} (I.e. doesn't even include 0), which is a contradiction. Therefore, to be consistent with mod m with m > 0, it returns undefined. n mod 0 returning n makes no sense, mathematically or even intuitively incompleteness is better than inconsistency in this case (mathematically, you can't have both be perfect in any system), since undefined tells you something "special" and means you can handle exceptions without leading to other (much more confusing and inconsistent) effects down the road. it's annoying, I agree, but aborting (which you as the programmer can catch) the better alternative. And honestly if it just aborts without telling you something so simple as attempted division/remainder by 0, or carries on with undefined behavior, that's the language's and/or compiler's fault (looking at you C)
There was some math teacher a few years ago who attempted to define 0/0 as "nullity" and put it among the extended real numbers...it leads to all sorts of contradictions
@@stlemur Oh, I don't find it _useful._ However, not being useful is different from leading to contradictions. For the record, I wrote a blog post about Anderson's idea in 2015. In it, I criticized the idea as being useless. So I'm not a defender of his.
If you define the result of division a / b = c as such number c that b * c = a, then any is quite a good answer, yo don't really have to multiply both sides by zero. But it should be noted, that in practical situation, that "any" does not mean "it can be anything you want", but rather "it could be anything, but we don't know what will work here just from seeing 0/0. We need more information". You should see every zero as a result of some limit or function and then you can divide it better. If you ever come to a parametric formula which can result in a division by zero, or even division of zero by zero, it kind of seems to me like trying to go to the next room through a wall, because it's the shortest path. Maybe you can just go around and not divide by zero. Maybe you can say what you want in a nicer way. It's more like 0/0 is not as much undefined, but rather impolite.
I'm not too sure about this demonstration. I think it does demonstrate that 0/0 , if we were to define it, couldn't belong to C* because the only k in C which has the property k*any=k*any is 0 but aren't there other numbers out of C that respect that rule? I'm mainly thinking about infinity: I see how +infinity and -infinity wouldn't cut it but if we just take infinity without a sign, I think that it could work as a definition of 0/0 according to this demonstration. Also, thanks for sharing interesting math concepts, like this, it really helps me develop some kind of a general culture on that field.
Personally i think 0/0 is undefined but i saw what wheel algebra says and "for me" if we should define 0/0 it should be the nullity, but i think it should remain undefined, at least if we talk about normal arithmetic, if we talk about limits, wheel algebra or anything else 0/0 changes
Dave!!!!!!!!!! 7:44
Dave!
I really miss the clean shaved blackpenredpen
What will be -1/0 infinity or -infinity
X^4 /1-x^3 integration plz solve it if one x has even power and one x has odd power but even power is greater than odd
@@kevalsavla1727 in a limit situation, im pretty sure it'd be determined by the side from which you approach zero. If you approach zero from the left (from the negative half of the x axis, so x -> 0-, the limit would approach plus infinity. If you approach zero from the right(from the positive part of the x axis, so x -> 0+, the limit would approach minus infinity. All that assuming the expression you provided is something like -1/x as x approaches zero. Please correct me if im wrong.
e=3, pi=3, 0/0=0
Engineers: *Finally I have them all*
lol im dead
don't forget g = 10 m/s²!
pi^2 = 10
e=pi=10 for astrophsycist buddies
Dont forget sinx=x
How did you hear me say 17 was my favourite number?
17 likes
He didn’t
@@davidepierrat9072 of course he did,
that should be the e p i c number
Mine is 07
"0 times Pikachu is equal to 0"
Pikachu: "So.....um...uh..."
But the Pikachu remains there 🤣
Shouldn't we just define 0/0 equal to Pikachu
*GONE! REDUCED TO ATOMS!*
@@swapnamoy6134 finally, 0/0 = pikachu
Gotta Catch Em All 🤣
bprp: _divides by zero_
Universe: **black hole**
@William Carter why are you talking about the creation of black holes
What the heck everyone is comment here , I don't know about too Much physics :D
@@pardeepgarg2640 wdym everyone, there's only one guy
@@Noname-67 yeah , but he comment 3 times :////
@William Carter But thanks to Pauli, atoms formed happily ever after
1:20 I remember we had a angry teacher in school who sounded funny when he was angrier. There was an incident where a student did something similar to this but it wasn't obvious because the equation was complex. The teacher got angry af and the way he scolded became a meme in the school.
I had a spanish teacher who made seating plans so we wouldn't talk
One guy never sat in his seat and one day the teacher slammed his desk and shouted
RESPECT THE SEATING PLAN
and that became a meme lol
2x = x because 2x * 0 = x * 0, you did say we can do any operation as long as we do it to both sides, teach
@@AirNeat2x * 0 = x * 0
2x = x
2x - x = 0
x = 0
nothing's wrong
@@greenytoaster 2x * 0 = x * 0
0 = 0
@@greenytoaster sorry you just multiplied by 0 but didnt
0/0 = 0 really makes sense on practice at least in physics engine programming and similar things, because zero vectors might arise in some cases but common equations require to compute size of the vector
Not to mention it keeps with 0/x=0
Tho we do loose x/x=1 I guess
@@Jade_TheCat as well as x/0 = undefined
@@greenytoaster It can be tho. It is hard to do for all numbers, but for most there are systems for that.
@@irhzuf interesting
@@irhzuf no. if you take the limit of 1/x as x approaches 0 from the positive integers you get positive infinity, but if you approach 0 from the negative integers you get negative infinity. going to the same place, 0, you get results that are incredibly different. this is why we say it's undefined.
Check out this video on why 0^0=1 is seriesly useful The convention 0^0=1 is seriesly useful ruclips.net/video/rJil85GHEyc/видео.html
Wow so 0 to the 0th power is 0 and 0÷0 is 0.
Well if 0/0 can't be 1 then 0^0 can't be 1 since 0^0 = (0^1)/0 = 0/0
I would argue that 0/0 = 0 makes perfect sense.
Division: How many times a number must be added together to get to another (In other words, how many times a number must be multiplied by to get to another)
So, a/b = How many times you must add b to get to a
You might say that 0/0 could then equal 1 because, for example, 6/6 = 1 because you must add 6 up one time.
However, if you rephrase the above a little:
a/b = How many times you must add b **to 0** to get to a
This is exactly how you first learn to do multiplication.
So, by the above definition for division, 0/0 would equal 0, you already start at 0, you must add 0 to itself no times.
n/0 where n is non 0 would still be undefined, you would keep adding 0 and adding 0 but you will never ever reach n no matter how many times you add zero to itself. (Unless you add 0 to itself a non real number of times, if you assume there is a valid non real number that will get you there)
Nah clearly the answer is this:
"Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends.”
Non-exitsting friends can have any number of cookies, just logic.
Remember, every moon unicorn is green.
Siri Easter egg
0/0 = sadness
But what about -3 cookies distributed equally to -3 friends and each friend gets 1 cookie. I,e -3/-3=1
@@smsarfaz1213 but that's because -3 * 1 = -3
0/0 should be defined with the symbol %
INDEED
clever lol
bruh
This was really good! Could u do a video on 0⁰ once? I'd love to see you talk about it!
Eddie woo has done a video on it
@@lc1777 ah yes I've seen that :) thanks for the recommendation though!
For most uses it's = 1
He already has made a few videos about 0^0 before, and they are all relatively ba- erm, I mean, those videos are not very good and they, unintentionally, spread a misconception that BPRP should instead by trying to actively combat as a mathematics education channel.
@@angelmendez-rivera351 oh okay I'm sorry, im new to the channel 😅
You knowing I said 17 really caught me off guard omg
My teacher: "so dear students today we'll study Schrödinger's equation. Well, this strange symbol which you see right here is called-
Me: "Fork. It's called fork."
@Tejas It's discussed more often in chemistry here, when they teach atomic structure and then take it to mad levels. But yes it's modern physics basically.
I edited my comment and made it 'subject neutral' lol.
I love his explanations and videos so much, always informative and attentive, nice video!
I think the k not equal to 0 situation would have the same problem you get from say squaring numbers. You always have to be careful when you're dealing with multivalued operations.
Explain
@@luffnis well because -1 is not equal to 1, but (-1)² = (1)² = 1, so we may naively assume that -1 = 1 since their squares are equal.
@@MrKnivan naively indeed, since this is a reason why most quadratic equations have two different solutions.
@@MrKnivan Indeed, which fundamentally comes down to the same issue as multiplying both sides by 0. In fact, if we can't do one, why should we be allowed to do the inverse operation instead without proving that the inverse operation is good in the firstplace, (as n/0 is undefined, and one can prove this easily, we already know the inverse is not good).
There’s another issue with the first example you gave as well. When we multiply both sides by zero, we have zero over zero on the left side which we are trying to define but in the example you just cross it out. We can’t cross it out because when you do that, you assume it even has an answer so #1 uses circular reasoning to define 0/0
How to do summation of a series of factorial ?
Like 1! +2! +3! +.....n!
Wolfram alpha i assume
I don't know, I would maybe try different values of n and checked if I see a pattern, then I'd prove by induction or something.
@@nilsastrup8907 no there is no pattern , its factorial so the next term is just the previous term multiplied with which number term it is
@@leonhardeuler675 never heard of that :)
There is no neat closed form formula for this sum if you use elementary functions. However if you also include special functions, there is actually a formula: math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this
What if you integrated 0/0 under an arbitrary function f(x)?
I’ve got another random thought: let the numerator and denominator denote the magnitude of vectors A and B in the z direction respectively. Let vector A extend exclusively in the x dimension, and vector exclusively in the y direction. Hmm, this probably won’t work because you end up with: f(z)=0(x units)/0(y units) but this was a fun thought to trail.
p.s great job with all your content man. Crazy work ethic!
the equivalence relation for rationals is defined as (a,b)~(c,d) iff ad=bc, so the first case does hold
What i feel is that division is just comparing the magnitude of one thing with the magnitude of another. If the magnitudes are equal, (0=0) then the result of division is one.
My logic is "division is the amount of times 'b' needs to be taken out of 'a' until it reaches zero (in a/b)"
E.g. "15/5" is "15 -5 -5 -5 = 0", five is repeated three times, therefore 15/5 = 3
Using that logic with 0/0, we get... zero. We don't have to subtract anything because we already reached our objective. AKA the amount of times '0' needs to be taken out of '0' until it reaches zero is... zero times.
Bonus: And using that logic with 1/0 we get:
"1 -0 -0 -0 -0 -0... = unreachable" therefore it's undefinable because it never ends
nice
I really like your way of defining zero by zero and division in general. I also want to share some of my thoughts. You may know that zero is a number which denotes nothing, a concept which has been introduced in mathematics later and at first people didn't even use that number. But when they started using it people ran into a problem. How could they define zero divided by zero? I mean how can something consist of zero pieces? And yes, reasoning that way, division by zero is simply undefined. However I have an idea. If we have a number for nothing, why can't we have number for everything? I believe this way division by zero will be defined. Anything multiplied by zero yields zero and anything divided by zero yields everything. Everything can be number like infinity - a number which can't be reached but a number which wraps all other numbers. What I mean by that is you start from zero, make your way through the positive numbers, get past everything, then through the negative numbers and back at zero. This concept also applies to complex numbers. You start from zero, pick a direction, travel in it infinitely and finally arrive from the opposite direction. Hope I made myself clear.
But you also can subtract 0 of 0 indefinitely so: 0-0-0-0-0-0-0-...-0 = 0. Technically since any amount of time you subtract 0 is still true you can argue it is undefined or if you dont see a reason to stop since it is still true going into infinity you could say it can be infinity... which is questionable since it is not a number.
@@olivers.7821 Exactly what I have thought. But maybe we should look at infinity in a different way. Maybe we can treat it like a number just like how I explained in my previous comment. I know you might say that you can't measure the difference between infinity and any other finite number, but think about it - there are infinitely many numbers between 0 and 1, yet we still can calculate the difference which is one. Maybe we have to learn to work with such numbers which have unique properties like x = 2x. The only numbers that satisfy this equation are 0 (nothing) and infinity (everything).
@@veselinborisov369 the problem with infinity is that not every infinity is the same for example if you count normally you would count infinite numbers but if you counted every possible decimal place aswell you would have at least an infinite amout of numbers between an infinite amount of numbers which makes the second infinite way bigger and therefore different so anything is not equal to anything so you can not really define it
I disagree with the fact that you have to assume 0/0 a constant.
(0/0)×1 = (0/0)×17 means
(Something × 1) = (another something × 17)
That means you can assume that the first 0/0 and the second 0/0 has different values and they can take "any" value.
Yes but he was seeing that if it would be defined as a constant, what constant it would be. And in that case the best option would be 0.
@@Hesselaer As a constant, 0/0 has to equal to 0, but in general, it's basically a variable.
AKA it’s undefined
The assumption wasn't applied there.
A way to think of rational numbers of p/q is how many times do I need to subtract q from p to get to 0. For example in 15/3 you have to subtract 3 from 15 5 times to get to 0 and thus we say 15/3=5. Therefore, in 0/0, for any number r you pick if you subtract 0 r times from 0 you’ll get 0-0(r)=0-0=0 so any number r would satisfy 0/0 under that way of thinking. In theory we can choose to define it as 0 like in this video but to stay consistent with how fractions work in other situations we say it’s undefined.
You could also say its 0 because it how many times do you NEED and you dont need to do it(if you have to define for whatever reason)
@@tadeogaldo4346 that’s a good point!
0÷0=∞. Ноль может состоять из бесконечного кол-ва нулей
Amazing video man! Loved it!
Good approach at the start. I noticed the same problem where people would multiply both sides by 0 at the start even though that obviously never works to begin with.
In Lean, a/0 is defined to be 0 because
- it has to be something (Lean only supports total functions)
- 0 is already defined in our type (as opposed to 1 which we may never define)
- it makes some results like a/c + b/c = (a + b)/c unconditional (you do not have to assume c != 0)
I LOVE LEAN
2:55 : bprp forgot the c in contradiction.
Me: happy also bprp forgets to do "+ C" sometimes
In indefinite integration?
@@advaykumar9726 yes
contradition + c
Division by zero is undefined precisely because the multiplicative inverse of zero does not exist. Why then when you multiply nothing by something nonexistent should you get anything at all? It makes no sense.
The "fork" psi is like the second letter you use after phi when talking about homomorphisms of groups, rings etc., it's not that uncommong
@ཏྦཱལ་ག་པོ། nope, he used lower case psi and that's what I meant
And the time-independent Schrodinger Equation
Edit: I didn’t see the bonus part when I wrote this haha my b
Sir, I have always thought 0/0 should be equal to 0.
This is why. Correct me if I am wrong.
We can consider the quotient when doing division as the number of times the dividend has to subtracted by the divisor for it to become 0. So in the case of 0/0 since the dividend is already 0 there is nothing to do hence the quotient should be 0 right.
We can consider division again as the number of times the divisor has to added by itself to be equal to the dividend. When we consider 0/0 in this perspective we can see that since the divisor is 0 there is nothing to do to make it 0 hence making the quotient 0 again.
Sir, please reply to this has been bugging me up for some time. And every others who are reading this if you have any opinion reply or please try to like this so that he can see this.
0 is an answer, but any number also fits that. Taking 0 away from 0 5 times still "gets" you to 0, it's just that 0 as a quotient is the smallest answer.
@@VollkinSea i agree with you
As a programmer. *Most* of the time when a division by 0 occurs and we don't want to throw an exception, we actually meant 0 as the result.
but we use terminary operation and not divide by 0 directly?
As a fellow programmer, I disagree. If you're dividing by 0 often it means either "a really large number" (ex time left in a download) or "I should have divided by 1" (used a index instead of index+1)
@@novamc7945 of course you don't directly. But business logic doesn't really care about that.
@@gasparsigma You're raising up the point why it's not defined. It depends on the use case. My personnal experience differs, to which my original comment really only apply. Good to see the common default be different for others.
@@gasparsigma for floats I agree with big number but for integer division I'm okay with 0
If 0/0 = k, then k must be 0. That's proven by this. However we can't assume that 0/0 actually is equal to any number, and since there are functions, which when you take the limit on them they get a 0/0 but still has a value that isn't zero, I think it should be undefined
He ignored limits in this video, but that doesn't need we have to and that you even should ignore them
@@nilsastrup8907 Are you incapable of doing basic arithmetic operations without appealing to the concept of limits? I am tired of repeating this in every comments section on RUclips, but for the 10083718304018384717940193772th time: lim x/y (x -> 0, y -> 0) not existing does not imply 0/0 is undefined, just like how lim floor(x) (x -> 1) not existing does not imply floor(1) is undefined. The logic you are using is borderline ridiculous.
Pls bring back the black pen red pen Yay!!! intro . It's really nice
Outstanding video!!! Thank you for producing and posting!! :) :) :)
i have a conter-argument.
that proof is well made and hold for most situations, for example ln(1) because its multivalued and one of his answer is 0, but it does not hold for some specific situations?
consider ln(0) = a, everybody knows 0^2 = 0^3, loging both sides and we have 2*ln(0) = 3*ln(0), 2a = 3a.
if a is real and a=/= 0, then we can divide both sides by a and we have 2=3, but 2 is not 3, so a should be defined as 0
but ln(0) cannot be 0
if we consider ln(0) = 0 and elevate both sides by e, we have e^ln(0) = e^0 and 0 = 1, but 0 is not one, so ln(0) cannot be defined as 0.
with this double proof by contradiction, we can agree that this proof needs more information to hold, without this we cannot be sure if it 0/0 can or cannot be defined as 0.
again, nice video ☺
Yes......You guess little bit right...but my favourite number is 1729....
Taxicab number
MY FAVOURITE NUMBER IS 0 CUZ THATS ALWAYS MY MARKS
i have a conter-argument.
that proof is well made and hold for most situations, for example ln(1) = 0, because its multivalued and one of his answer is 0, but it does not hold for some specific situations.
consider ln(0) = n, everybody knows 0^2 = 0^3, loging both sides and we have 2*ln(0) = 3*ln(0), 2n = 3n.
if n is real and n =/= 0, then we can divide both sides by n, and we have 2 = 3, but 2 is not equal 3, so n should be defined as 0.
but ln(0) cannot be 0 too
if we consider ln(0) = 0 and elevate both sides by e, we have e^ln(0) = e^0 and 0 = 1, but 0 is not equal 1, and then ln(0) cannot be defined as 0.
with this double proof by contradiction, we can agree that it needs more information to hold, without information we cannot be sure if it 0/0 can or cannot be defined as 0.
again, nice video ☺
(editing because im an ape and don't know how to write)
- Are you sober now?
- Ozzy: **divides by 0**
Nice done. Can you show us a contradiction if we assume 0/0=0?
1/(0/0) = 0/0
⇒ 1/0 = 0
@@farklegriffen2624 That doesn’t quite work because writing 1/(0/0) = ... implicitly assumes that 1/0 is defined. But the only actual assumption made was that 0/0 exists and equals 0, there are no assumptions about the existence of any other number of the form x/0. So we can’t assume that just because 1/(a/b) exists for most a and b that it exists for the specific case of a=b=0.
Likewise you can’t make any assumptions about the existence of, say, ln(0) existing. So just because ln(a/b) exists for most a and b doesn’t mean it exists for ln(0/0). Therefore if you want a contradiction that stems only from defining 0/0 = 0 then you have to make no other assumptions about any other as yet undefined expressions in the process.
@@Bodyknock The assumption that 0/0 = 0 necessarily requires that 1/0 be well-defined.
@@angelmendez-rivera351 No, 1/0 doesn't need to be well defined just because 0/0 is defined. If you look at division as a binary operator called Div(x,y) where Div(x,y) = x/y for all x and y except 0, then what this video is doing is creating a new extended operator called maybe Divz(x,y) which is specifically 0 for x=y=0 and is undefined for all other y=0 parameters. There's no requirement in that extended system that 1/0 be defined.
@@Bodyknock Except that is false, because the video not only creates the operator, it also assumed certain properties of the operator, such as Div(x·y, z) = D(x, z)·y = x·D(y, z), properties that the video used in its proof. And the existence of those properties very much does require 1/0 to be defined. This is the problem with the video.
Also, your argument that writing 1/(0/0) = ... implicitly assumes that 1/0 is defined is incorrect. It does not. It only assume that 1/(0/0) is defined.
Can you make a Video about the "Wheel-theory"? This Topic is really intresting and also about the 0/0 Problem.
I am from bangladesh .. your classes are awesome ..very much helpful for me .at first i am scared of differentiation .but now i just love it .❤️ Thanks a lot .
Please do a video about Quaternions
3:50 wtf that is my favourite number
Maybe not from a math perspective but the only time I've ever had to divide by 0 was within programming and within the requirements that I was working with, it actually worked for exactly what we needed to have dividing by 0 equal to 0 as if you had multiplied, because it essentially allowed to turn a value into a boolean within the constraints of that language that we were using. The result was always either 0 or 1. the value was 0 or it was anything other than 0, and we counted it as 1 towards a sum of the total non 0 values.
Yes, because 0/0 is a whole number that is why you got 1 or 0.
You are great RUclipsr and I am your biggest fan
My Intersection Formulas result in 0 ÷ 0 for overlapping equations, meaning that the 0 ÷ 0 might indeed be(Corrected from "by" on 12/25/2021 at 9:17 AM EST) all real numbers.
2:17 this allows for 0/0 resulting in NaN
Is that symbol used as End of Contraction (oppositely pointed arrows) ?
Yes.
A division operation expresses the idea that the numerator is being segmented (denominator - 1) times resulting in denominator portions. If the denominator is zero and one refuses to accept this as undefined then it seems to me this could be interpreted as the segmenting operation being performed -1 times, resulting in zero equal portions. We may either interpret this as a negation of the division operation itself, leaving us with the original value, or interpret the negative of an operation as it's inverse, thus x/0 becomes another way of writing x*0. So, depending on which interpretation is applied, x/0 is either x or zero. Operation transformation must be accounted for before processing can begin.
The problem about dividing by 0 is that we already assume it is equal to 1 when we cross it out when dividing.
Facts
But wouldn't there also be some equation that'd lead to contradiction if you assume 0/0=0?
can you think of a example?
I think that’s why he framed the proof as 0/0 not being equal to a non-zero number, rather than saying it’s always equal to 0. There are specific situations where 0/0 can be defined as 0, but you can’t define 0/0 as a non-zero number.
@@miguelwariskovih 1/0 = 1/(0/0) = 0/0 = 0. If you let 1/0 be an undefined operation, 0 must be undefined. But if you let 1/0 = 0, then you get into a whole mess of other contradictions :)
Asume that 0/0=0 for calcutations with limes:
0=0/0=(0/0)^(-1)=0^(-1)
=>0=0^(-1)=1/0=Lim x→+0 1/x→Infinity
=>0→infinity
This a contradiction .
So If you calculate with limes, then 0/0≠0. So don't define 0/0=0 for calcutations limes.
Could we define 0/0 rigurously if we use the set R + {∞}?
4:34 This proof you have heavily relies on the commutative property of multiplying with a fraction.
I.e how (x*m)/n = x*(m/n)
because (0*1)/0 = (0*17)/0 if you indeed multiply before you divide.
But you've incidentally shown that this cannot be true if you define 0/0.
because then if you let m and n both be 0 you get:
(x*0)/0 = x*(0/0)
k = k*x
which is not true for all values k and x.
But that fact in of itself is interesting because it means if you allow 0/0 math is no longer commutative in this certain way, unless you define it to be 0.
So what you have really proven is that if k != 0 then (x*n)/m != x*(n/m)
At 4:15 How can you divide both sides of an equation by 0. I don't think that's a correct way.
He can because he introduce an axiom where 0/0=k. This is how imaginary number was invented. Mathematicians declared that there is a number i where i²=-1. He is doing the same thing declare that there is a number k so that 0/0=k. But doing that leads to contradiction. So the axiom 0/0=k cannot be true.
Well, someone had to do it. IT'S INDETERMINATE
Amazing. Could you plz also give some STEP 3 questions a go. They are quite a lot more difficult and are more beautiful than the STEP 2 question than you attempted as well. There are some very beautiful one such as proving the irrationality of e etc.
Doesn’t this only apply if k is a complex number? We could alternatively define k as some non-zero non-complex number which has the property that 1/k = k and k != 1. After all, the inverse of 0/0 is still 0/0
If k is a non-complex number, what are the 0 and the 1 you talking about ?
If you see the 0 and the 1 (and the division) as the 0 and the 1 (and the division) of a field K, and if you defined 0/0 as the solution in K/{1} to the equation 1/k=k, then why not !
But I think the point here is to solve it in R
@@julien31415 not entirely sure what you were trying to say for most of this, but yeah, the point was to define k inside R, though I just thought it might be something to point out
I am glad you finally posted this. I am compelled ten times a day to define 0/0. I can finally get some sleep.
0/0 cannot be equal to 0 (or something else) in any case.
If you are in the integers then quotient q and remainder r of the division n / m are defined by the equality n = q m + r where 0
Thats if you consider 0/0 a 1-1 functional output when it really has an infinite amount of outputs.
I always liked to think of it as the symbol for the set of all numbers.
0/0 should not be defined at all, because 0 has infinite multiplicity.
IE:
0/0 = 0^2 / 0^1 = 0^1 = 0, but also
0/0 = 0^1 / 0^1 = 0^0 = 1, and also
0/0 = 0^1 / 0^2 = 0^-1 = ???
Clearly, 0 = 1 is a contradiction, so 0/0 cannot be defined ever, right?
But if you have any number n divided by itself, you can write it as n^0. And n^0=1. So 0/0 must equal 1.
yeah, it would be so easy, that is if 0^0 wasn't undefined as well.
After thinking some more about this and other fractions, I have another argument as to why 0/0 could/should be identified with 0.
If you look at a modular integer, a, with modulus m, then there is an inverse integer, b = 1/a mod m if and only if gcd(a,m) = 1. But, if you want to generalize the concept of inverse, there is a way to do this to give every modular integer an inverse. Say gcd(a,m) = d. Thus there exists a subgroup of the integers mod m, the integers mod (m/d). You can pull your calculation of the inverse into this subgroup by calculating 1/(a/d) mod (m/d), and now you are guaranteed that gcd(a/d, m/d) =1. Then you can 1 to 1 pushback into your larger group of integers mod m. If 1/(a/d) mod (m/d) = b, then your pseudoinverse of a mod m should be (b×d). Something strange but fully consistent happens when you consider the case a=0, which is consistent for any integer m and can be generalized to any number m. Your pull function takes you from your initial group m to the trivial group {0}. Your one to one function has to take your initial number a=0 to a number in the trivial group (which there is only one value to choose from. In this group, 1=0, so your operation 1/0 is equivalent to 0/0 and results in the value 0. When you pushback to the original group, m, you have to use the inverse of the same 1 to 1 function that brought you to the trivial group in the first place, leaving you with your original value a=0. Thus, there is only one possible consistent value for 0/0 in your group m, and that value is 0.
Nope!
Not reading that
It's a huge cultural achievement that university scientists are allowed to think about such questions.
It’s like dividing fractions;
Example:
10/35 = 10 * 1/35 = 10/35
Therefore
0/0 = 0 * 1/0 = 0
Idk if that’s true or not
ye
Doesn't that work with k=infinity as well ?
0:20 - 0:33 *0/0 = x 0 = 0·x*
Right off the bat, this is incorrect. Division a/b is actually defined as a·b^(-1), where b^(-1) denotes the unique solution to b·x = 1 IF it exists. IF it does exist, then you can take the equation a·b^(-1) = c, multiply both sides of the equation by b, to get [a·b^(-1)]·b = a·[b·b^(-1)] = a·1 = a = c·b. So a/b = c a = c·b necessarily requires that b^(-1) exists. Since 0^(-1) does not exist, it is not true that 0/0 = x 0 = 0·x. 0/0 is not undefined because it is equal to every number, it is undefined because it is equal to no number at all, since 0/0, by definition, is equal to 0·0^(-1), and 0^(-1) does not exist. If it did exist without contradicting associativity and distributivity, then 0/0 = 1 would be true, though. Also, if 0/0 were equal to every number, then that would make it "multivalued", at least in the naïve sense, not undefined.
4:06 - 4:25 *and now, because we have 0/0 = k, we can legitimately divide both sides by 0, and then when we see 0/0, well, by our assumption, that's equal to k, so on the left hand side, we have k·1, and on the right hand side, we have k·17*
Ignoring the fact that you have not actually a provided an operational definition of division to even make sense of the claim 0/0 = k, and of something like (0·1)/0, or any other manipulations the problem here is that you are assuming a new form associativity, now. You are claiming that you know for sure that (0·17)/0 = (0/0)·17 and not 0·(17/0), and both of these claims are just assumptions themselves.
4:26 - 4:38 *and now, because we said that k is not equal to 0, and again, just imagine that we are saying that 0/0 is equal to 2, right? So if you have 2·1 = 2·17, of course you can divide that on both sides*
But this assumes that k is a real or complex number, and is thus invertible, which again, is just an assumption. Though I can still grant this, since it is still relevant to make this assumption if you want to prove that, if 0/0 is real (complex), then 0/0 = 0, rather than just proving 0/0 = 0.
4:39 - 4:47 *so here, we can just divide by k on both sides*
This runs into the same problem as before. You take (k·1)/k = (k·17)/k without even having given a definition for division by nonzero quantities in the first place, but even if I ignore this, you are still assuming new-associativity, which you are most certainly not warranted to have in this new arithmetic system.
4:52 - 5:07 *1 = 17, after we divide both sides by k. But of course, this is wrong, so I just say, "but you see that 1 is not equal to 17", so this right here is the end of the proof.*
But this only proves that if we assume 0/0 is nonzero AND still a complex number, AND if we assume (a·b)/c = (a/c)·b = a·(b/c) for every triple of elements, then there is a contradiction. It does not prove 0/0 cannot be nonzero complex if you give up the above equality, nor does it prove it cannot just be some other new noncomplex quantity. Which is the crux of my objection.
The rest of the video is just explaining how, therefore, if you ever wanted 0/0 to be defined, then 0/0 = 0 is the only option, because it does not cause any contradictions. But it likely does cause contradictions, and this is because it fails to account for the fact that this does not define what 1/0 is. In fact, since you feel comfortable in claiming that / as a binary operation satisfies (a·b)/c = (a/b)·c, I can say (1·0)/0 = (1/0)·0, but since 0/0 = (1·0)/0, because 0 = 1·0, this means 0/0 = (1/0)·0, which is impossible unless 1/0 is also defined. This is why one cannot actually ignore that / was not defined as an operation in the video to make sufficient sense of 0/0 = k as a definition and of the manipulations being made in the video. In this regard, 0/0 = 0 is just as problematic.
Interestingly I saw a different version of the 1=17 contradiction a while ago. It stumped me at first:
a=b
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
2b=b
2=1
Usually this is quite rightly refuted by arguing that step between lines 3 and 4, a division by zero, is a nonsense operation and everything after is invalid. But if we define 0/0 = 0, line 4 would instead be written 0=0, which ‘fixes’ this.
3/3=1
2/2=1
1/1=1
-1/-1=1
0.5/0.5=1
So 0/0=1 no?
To add when we have some equations with x/x inside , we simplify it by 1 without making the case that x=0; so we already define 0/0 as 1 ,no?
We have :
x=0
x+1=1
x+x/x=1
but x=0 so
0+0/0=1
0/0=1
From what I've been told- anything divided by 0 is + or - infinity (because as x decreases, the value of 17/x increases and the value of -17/x decreases) but 0/0 is undefined.
Why can't 0/0 be defined as + or - infinity, instead of 0?
because infinity is not a number, it's a concept used to define unbound growth (in the positive or negative direction)
What if I make 0/0 = 0^0? I saw this done in a video on another channel in the past:
0/0 = 0^1/0^1 = 0^(1-1) = 0^0 and vice-versa.
please tell exact 0 ^ infinity
Great topic for discussion
OK so I was searching for about an hour on your channel for a video on limits from a/0 to 0/0 and found nothing. I told my teacher it was possible but don't know how to prove it or an equation that proves it. Can you help me? I also said that it was a real number but I guess that was wrong
What about the inverse of 0/0?
1/(0/0) = 0/0
1/k = k
Which implies k = 1 or -1
But there is nothing special about 1 so let’s call a number n.
n/(0/0) = n*0/0 = 0/0
n/k = k
Exactly, with this logic, you can deduce that assuming 0/0 = 0 implies that n/0= n•0 =0
What I feel is division is like a comparison like 4/2 equals 2 which says 4 is 2 times 2 (a comparison) and if we divide 0/0 which is again a comparison and we know both are equal then what I think is it should be equal to one ! Since equality is denoted by 1 not by zero ?
Seems appropriate to use Psi for both 0/0 and the wave function
I figured it was like a fifth root. A fifth root has 5 potential answers, but that doesn't mean they're equal, just that they hold an equal property when based off of that function. √5≠-√5, but (√5)^2=(-√5)^2.
This means that 0/0 should have a principle value, namely 0, just like √2 means 1.414... not -1.414..., but if one is to divide zero by zero, it can be anything within (-∞,∞) or i (-∞,∞).
i'm not sure about this but i've heard that proof that contradiction only proves if a statement is false OR that it is undecidable, normally (for example when deciding if a number is irrational or not) i think this shouldn't matter (maybe?) but this 0/0 problem is precisely what you'd want to avoid
I don't know why this was in my recommendations but I watched it and liked it
What if you were to define 0/0 to be +infinity or -infinity?
That’s not a number
wat if you do (e^ipi + 1)/(e^ipi + 1)
There is a sense in 0/0 being any number. You get into the contradiction because you try to devide by k which is any number. But those any numbers are not equal, just like two infinities are not equal. What we can derive then is that the ratio of two k's.
k1/k2 =. 17/.1 =1.7 So k1 is not equal to k2, which will be true to the fact that k can be any number.
Whaat do you think sensei?
If we assume that 0/0 is any number it is not a single umber of our choice but all universe of numbers at the same time. In case of equation 1*0/0=17*0/0 as 0/0 can be any numer at the same time we just need to prove that there exist 2 numbers let's say p and q which satisfies the eqation 1*p=17*q. Solving for this is easy. Solution is p=17*r and q=1*r where r is any number.
i believe if we were to define 0/0, it should not be 0, but a number that acts identical to 0. x=0/0, x≠0, y is any number, xy=x
as for how 0*x should behave, idk
this would of course require a new number, but we did that with sqrt(-1) and i
If act identical to 0, it's 0. Don't overdo it.
The result of division is always unique. Suppose 10 is divided by 2 we get the result 5 which is unique. If we divide 0 by 5 we get the quotient 0 which is also unit. But when we divide 0 by 0 then we get 1 or 2 or 3 ..... which is not unique. Can I not say that this is Indeterminate?
Hey! I have a somewhat different opinion on 0. I personally think of it as less of a number, and more of a concept. The closest analogy I have to 0 is infinity. For most practical purposes, they function the same way. Ex: multiply both sides of an equation by either 0 or infinity, it ruins the equation. “Divide” by either, you are doing sketchy math. Neither really fit in most mathematical groups. Maybe this is a dumb take, but it makes sense to me.
I would like to clarify. I don’t just think of 0 as similar to infinity. I would consider them inverses of each other - as much as non-numbers can be inverses. The math behind this goes something like this. *insert proof that .999 repeating equals 1. Therefore, 0 = infinitesimal (by subtraction). Infinitesimals are widely considered to be the opposite of infinity, and I believe they are the same thing.
Then how do you explain an infinitesimal being the inverse of zero. Zero is a value and infinity/infinitesimals are concepts.
@@ccbgaming6994...they just said that they think of 0 more as a concept...
@@greenytoaster They consider 0 and infinity to be inverses. They also believe that infinity and infinitesimal are the same thing. Their reasoning leads to the conclusion that infinitesimal is the inverse of 0. This contradicts another statement they made that 0=infinitesimal, because if n/infinity = infinitesimal, then infinity/n = 0 (which is not true when n is all reals except 0). This is why we say n/infinity approaches 0, not equal to
Can a positive number raised to any power be equal to a negative number
1=17, for some belief systems.
(1)⁴=(1)⁵ therefore 4=5
Here laws of indices fails.
Please explain
Look it up in the contradictionary.
Although 0/0 should be left undefined, n mod 0 should be defined as n. I get annoyed when I'm programming something that mods something by 0 and the program aborts.
Why? In most programming languages % means remainder after division, and not modulo, so in that case it makes sense to be consistent with division for n % 0 and make the implicit n/0 undefined by default for any n
And even when properly using modulus, mod m must give an element in the set {0, ..., m-1} by definition. And so, taking any number n mod 0 means it's an element in the set {} (I.e. doesn't even include 0), which is a contradiction. Therefore, to be consistent with mod m with m > 0, it returns undefined. n mod 0 returning n makes no sense, mathematically or even intuitively
incompleteness is better than inconsistency in this case (mathematically, you can't have both be perfect in any system), since undefined tells you something "special" and means you can handle exceptions without leading to other (much more confusing and inconsistent) effects down the road. it's annoying, I agree, but aborting (which you as the programmer can catch) the better alternative.
And honestly if it just aborts without telling you something so simple as attempted division/remainder by 0, or carries on with undefined behavior, that's the language's and/or compiler's fault (looking at you C)
There was some math teacher a few years ago who attempted to define 0/0 as "nullity" and put it among the extended real numbers...it leads to all sorts of contradictions
Are you talking about James Anderson from the University of Reading? If so, I'm not familiar with any contradictions introduced by his system.
@@MuffinsAPlenty Imagine you actually let a system use nullity as a number. You will end up with, effectively, NaNs propagating through everything.
@@stlemur Oh, I don't find it _useful._ However, not being useful is different from leading to contradictions.
For the record, I wrote a blog post about Anderson's idea in 2015. In it, I criticized the idea as being useless. So I'm not a defender of his.
If you define the result of division a / b = c as such number c that b * c = a, then any is quite a good answer, yo don't really have to multiply both sides by zero. But it should be noted, that in practical situation, that "any" does not mean "it can be anything you want", but rather "it could be anything, but we don't know what will work here just from seeing 0/0. We need more information". You should see every zero as a result of some limit or function and then you can divide it better. If you ever come to a parametric formula which can result in a division by zero, or even division of zero by zero, it kind of seems to me like trying to go to the next room through a wall, because it's the shortest path. Maybe you can just go around and not divide by zero. Maybe you can say what you want in a nicer way. It's more like 0/0 is not as much undefined, but rather impolite.
try it as limit of f(x)/g(x) as x approaches 0 = 0/0 its easier
I'm not too sure about this demonstration. I think it does demonstrate that 0/0 , if we were to define it, couldn't belong to C* because the only k in C which has the property k*any=k*any is 0 but aren't there other numbers out of C that respect that rule? I'm mainly thinking about infinity: I see how +infinity and -infinity wouldn't cut it but if we just take infinity without a sign, I think that it could work as a definition of 0/0 according to this demonstration.
Also, thanks for sharing interesting math concepts, like this, it really helps me develop some kind of a general culture on that field.
so what are 0*log(0) and 0^0 if 0/0=0?
Personally i think 0/0 is undefined but i saw what wheel algebra says and "for me" if we should define 0/0 it should be the nullity, but i think it should remain undefined, at least if we talk about normal arithmetic, if we talk about limits, wheel algebra or anything else 0/0 changes
But why lim x->0 (x/x) = 1?