This Video Will Frustrate Your Teacher 

How is this comment two days old?

@@Bourbon_Biscuit because it was unlisted and only his channel members could see it

Yo thanks for the free one month sub

Please make a video on Sierpiński triangle ...∞

Oh no you didn't...

IKR

"other things that can grow arbitrarily large is your di-" "dice. Imagine a dice with infinite sides..."

"Other things that can grow arbitrarily large are your tendencies to think erotically

Edit: Wow this comment actually blew up?? Thx so much😎

That's probably why you aren't allowed to subtract infinity from itself as well, it's indeterminate

Thanks, that makes are lot more sense intuitively

Infinity is a concept, not a number. Both Σ^infinity_i=0 (i) and Σ^infinity_i=0 (2i) are the exact same size. In this case, what would "infinity - infinity" equal to if not "infinity"? So long as you'll think of infinity as a number, it'll be really hard for you to understand certain concepts because they'll seem contradictory when in reality they're not.

@@louisrobitaille5810 this isn't true unfortunately. There is a field of maths about different types of infinities

@Shimmy Shai 0/0 is the essence of calculus. Limit of ((f(x)-f(x-n))/n) when n approaches 0 is just a derivative. But... If you plug in n=0, you get (f(x)-f(x))/0=0/0. But you still define derivatives (0/0)! So this is not undefined, but indeterminate.

He probably took up mathematics out of spite

Haha, I'm a math teacher and this video frustrated me...because I fell for the clickbait title!

@@MultivectorAnalysis I feel you. Same here.

Right? I WISH I had students who cared enough to ask about stuff like this

More a maths teacher's delight, I would say. A number of fascinating tidbits most of which are beyond a high school maths course.

Bri has uploaded other videos that would indeed irritate math teachers. In those, he claims that he can divide by zero, and has solved-once and for all-indeterminate forms like 0/0, 0^0, 0.∞, ∞/∞ etc. Not surprisingly, he titles those videos "Don't show this to your math teacher." Before watching this video, I fully expected it to be a compilation of those videos!

i agree

A complex number with no imaginary part lies on the real axis. There's a name for such numbers...

They are sums, not proper sums though, as they contain infinite entries. And everyone knows as soon as infinity is involved anything that deals with finite systems doesn't hold. But who cares, math is about as rigorous as a lunatic mumbling about the endtimes, to determine the weather.

@@livedandletdie They are called "sums" but my point is that they are not obtained just from +. If you only used + then you could change order without problems. However it also involves the operation of taking a limit, and thats the one where order matters.

@@livedandletdie ​ ​ *They are sums, not proper sums though,...* This is utter nonsense. No mathematician speaks of a distinction between "proper" sums and "improper" sums. Besides, the structure of the operations in question are fundamentally different from the way addition works. *math is about as rigorous as a lunatic mumbling about the endtimes...* You sound like one of those right now. This is an extremely ignorant statement to make. Mathematical rigor is something that is extremely important. There are no disciplines of study with more rigor than mathematics.

Carla from the math discord server?

@@sriyansh1729 Yes and who are you and whats your favourite Zappa album?

Because it doesn't belong. The whole numbers include zero. The natural numbers do not. The natural numbers are the set of successors to zero. Because zero is not the successor of any number, but defined a priori, it is not natural.

@@ObjectsInMotion It's just a matter of convention, depending on your country. In France, the "entiers naturels" *N* include 0 and *N** does not. I find it better like that but I was mainly making a joke, as I said it depends where you're from.

@@ObjectsInMotion Purely convention. In my classes it depended on the professor’s preference. I’m in America, so most of my professors didn’t include zero. I’ve had a few that do though, and I’ve seen many mathematicians that include it. Personally I like it, because if you want the set without 0 you can write Z^+, aka the set of all positive integers, so N can be the set of all non-negative integers

Why? The natural numbers are the numbers used to count physical objects. Of course they don't include zero.

@@mrosskne But there sometimes exist zero of a given physical object. Of course the natural numbers include zero. At the end of the day, the only thing related to the natural numbers that annoys me is that there is, as far as I'm aware, no universally unambiguous way to write "the set of all integers greater than one" as "the set of natural numbers." ("the set of natural numbers" which includes zero can be unambiguously written as ℕ₀)

Yep, that's what he's saying, and that's the correct result by applying rigorous set theory.

that's because rational numbers can't have infinite digits

@@jacemandt but the rational numbers set contains all of the numbers between 0 and 1 so it has more numbers then in the natural numbers set.

@@mrubikscube7047 you are confusing rational numbers with real numbers here. When talking about numbers between 0 and 1 we are referring to real numbers. The rationals is a subset of the reals.

@@doge_69 actually they can, that’s not why. Take 1/3=0.333333333333... infinite digits

you would just use 0 where there was a 9 (with no carry), but really it doesn't matter. All that is required is that each digit is changed from the digit you borrowed from the corresponding number on the list.

@@phiefer3 That still doesn't resolve the issue that some rational numbers have multiple decimal representations, and it is possible for this algorithm to pick one of the representations while the other representation is in the list. The "add 1" algorithm has that fundamental flaw (regardless of which way you go when you encounter a 9), and insisting on using the "add 1" algorithm and then try to wrestle with this issue is much more of a hassle than just picking an algorithm that never writes a 9 or a 0. It is especially egregious that an educational video on the matter falls into this trap and doesn't even acknowledge it. If you're making a math video, and you make some simplifications in order to make it more accessible, the least you can do is to own up to the fact that simplifications are being made.

As stated in this video

This wouldn't actually work. Take the fractions 1/12 and 11/2 concatenating them would both give 112, so you do not get a unique integer for each fraction.

Define sum

@@lc1777 Combine two quantities to produce a resulting quantity.

Studying Symmetric pfaffian differential equation systems, you not only accept division by zero, you actually actively look for zero denominators since they simplify the job of solving them. Symmetric pfaffian systems look like this: dx/P = dy/Q = dz/R, for example. (P,Q,R) is a continuous vector field.

It would never even be possible to divide by zero if multiplication is an inverse of division. For example, "12/0" makes no sense because there is no number that when multiplied by zero will give you 12 as a result. Of course, any number multiplied by zero will result in 0. Moreover, I don't even think division by zero makes any logical sense. In 12/0, you'd be asking how many groups of zero are in 12, which is nonsense. On the other hand, if we had 15/3, we could say there are 5 groups of 3 in 15 and that 5 x 3 is 15.

@@JLKeener77 You are right, if we are talking about numbers as in "real numbers", or "complex numbers". But Math is the realm of the what ifs. So, what if we are not talking about those numbers? What about "surreal numbers"? In pfaffian systems such as the ones I proposed, numerators are infinitesimals, and division by zero is defined via parallelism of vectors. It is not an operation on numbers, but a relation on vectors. Division by zero as an operation is valid in certain branches of mathematics...

Infinite surface area enclosing finite volume is Gabriel's horn. Don't know what you mean by "cooler".

​@@mrosskne Dude! If you can't see why fractals are cooler-way, way cooler-than Gabriel's Horn, then either you don't know fractals, or you don't know the meaning of 'cooler'. Also, fractals can have infinite surface area _while being confined to a bounded region of space._ Gabriel's Horn is unbounded-so its infinite surface area isn't exactly surprising.

Can you guide me or suggest me some books so that i can be good at maths

​@@wardog697 Hmm, that's a tough one. It would depend on so many things-how much education have you had; are you a student; how much do you know already; what learning style do you prefer; what country are you in; how much time and money can you afford etc. You know what, I suggest you watch math videos by other channels as well: - "The Math Sorcerer" reviews all kinds of math textbooks. (Undergrad and above; not high school.) Math is a very broad subject, and no one book covers it all. - "Michael Penn" solves high-school, undergrad, and Math Olympiad problems. He also teaches full courses for undergrads. - "Dr Peyam" solves simpler problems. - "3Blue1Brown" is a legend for his animation software Manim. - "Zach Star", "Mathologer", "Numberphile", and "Stand-up Maths" cover interesting topics. - "Dr. Trefor Bazett", "Steve Brunton", and "Another Roof" cover advanced topics. Those are just off the top of my head. There are dozens more noteworthy channels which, hopefully, you'll discover. Plus, many universities have put entire courses online-videos of professors teaching those courses in classrooms. Once you subscribe to and start watching a few of the above videos, more will start popping up in your feed.

I am a 11th grader in a cbse school India. Currently preparing for JEE. I mostly read books to understand the concepts. When I cant understand something I search about that topic on youtube or google. As for the time, I get 6 hours after school and coaching

@@nHans also how can you add scalars to vectors that seems impossible to me!! (0.0)

​@@wardog697 Well, since you're preparing for engineering entrance exams, I suggest you focus on those for now. I'm sure you have the recommended study books. In addition to the various entrance exams, you still have to get good grades in your 11th and 12th. That's already a lot to study. So my suggestion is, at this point, don't get distracted trying to learn advanced topics that are not in the syllabus. Adding scalars to vectors is one such topic. It's easy to explain. Remember how you add a real number to an imaginary number and get a complex number z = a+bi as a result? In the same way, you add a scalar and a vector and get a hybrid object (a+v), where a is your scalar and v is your vector. In component form: (a+bi+cj+dk), where i, j, and k are your unit vectors. Then you define addition, multiplication etc. in an analogous way. To be clear, the hybrid object that I created is not recognized by mathematicians. But instead, they use a very closely-related object called 'quaternion'. It comes in many flavors: the regular quaternion, split-quaternion, hyperbolic quaternion etc. But don't worry about all those for now. You'll have plenty of opportunities to study them after you get into engineering college. I'm an engineer. As engineers, we do study a lot of math. But for us, math is a tool for solving engineering problems. This is different from what mathematicians do-they study math for its own sake. Mathematicians also study far more math topics than engineers. They study axiomatic set theory, number theory, abstract algebra, topology etc., which we don't. Instead, most of our math is focused on calculus (including complex analysis, vector analysis, PDEs, and transforms), linear algebra, numerical methods, and probability and statistics. So if you're interested in math for its own sake, then instead of engineering, you might want to do B.Sc., M.Sc. and Ph.D. in math. But-I'm sure you're aware-career opportunities are far fewer for a math graduate. On the other hand, engineering is a well-paying, fulfilling career choice; you won't go wrong with it. And you can always study more math on your own, like I did. As for the extra time that you have, I strongly suggest getting physical exercise and playing sports with friends-not video games. Don't ignore your health! Remember: a healthy mind in a healthy body!

My school don’t have a subject called neither myath nor mythology

Bro just had a stroke

@@TrainingCuber emgish

@@darthmaul197 Durth Maul

This is automatically fixed because your list only includes 0.49999... and other numbers formed by 0 and 9, which, ofcourse, is *not* all numbers Interesting point nonetheless!

@@36sufchan the point is that due to the definition of real numbers 0.49999… is equal to 0.5

Either this, or require that the number you create never has a zero or a nine in its expansion. This way, it can't possibly have two distinct expansions and the problem is solved.

The way Cantor gets round this problem (which of course he was well aware of) is described in this article en.m.wikipedia.org/wiki/Cantor%27s_diagonal_argument in the section Uncountable set. Summary: Cantor's argument goes in three stages: first he uses the diagonal argument to show that the set T of all infinite binary strings is uncountable, then creates a bijection between T and the interval (0,1) [it is here that numbers with two binary expansions are dealt with], and finally a bijection between (0,1) and R. With this argument, he proves not only that R is uncountable, but that it has the same cardinality as T.

What are you talking about? 0.123 doesn't appear on your list. This doesn't contradict the proof.

All horses that are brown is in the set of all horses and therefore because of bijection we can say that all horses are brown because there's a 1 to 1 correspondence between the two sets. - Math in a nutshell. I hate Bijection and I hate the Transfinite number aleph null, and I hate how non-absolute math is today. Where they say dividing by 0 is a no no, but saying that proper subsets are equal to the set they're a minute fraction of is a ok.

@@livedandletdie, you might be thinking of something else? The "all horses are brown" paradox arises through a tricky mistake in an induction proof. I can't post links, but it's called "all horses are the same color" on Wikipedia if you're interested :)

@@livedandletdie I think that's better than "I'm stupid therefore mathematics is nonsense" argument

@@livedandletdie What do you mean "because of bijection"? You haven't shown that a bijection exists.

What are you confused about?

"take the biggest of them and add 1" No such biggest one exists. An important part of Cantor's diagonal argument that is usually glossed over is that the numbers we are making approach something. The edits we make get further from the decimal point, our successive numbers get closer and closer. The limit of the sequence is the number we claim is not in the list. Your attempt to pair up the naturals and the reals from 0 to 1 is a good idea, but unfortunately doesn't work. You have indeed included every terminating decimal in the list, but many numbers, such as 1/3, are missing. 1/3 is 0.33333... but ...33333 is not a natural number.

@@mattermonkey5204 "no such biggest one exists" that problem also stands for cantor's diagonal argument you'd need to go through an infinitely long number which just like the biggest number of an infinity can't be found because infinity. So in order to use infinite Numbers generalization is necessary to show what makes sense and what doesn't "The limit of the sequence is the number we claim is not in the list" that's exactly what the proof also shows for my list of all natural numbers. That's why i think such a proof can't be used on infinities n333...=n∞=0.333... just because it's impossible without using infinity doesn't mean it's impossible with using infinity

@@per2632 An "infinitely long number" is not a problem. 1/3 is an example. The problem with the proof for naturals is the the sequence doesn't have a limit, it doesn't converge. Meanwhile for the reals, later steps only change the number a very, very small amount, since we're deep into the decimal places. That one does converge.

Lim x->∞ f(X)=ln(X)=∞ should ease both of these concerns. It converges to infinity and ln(X)gets ever growingly smaller compared to y but still goes to infinity just the same so 10th square root of X isn't a problem

I've just found out that even though integers can go towards infinity and can create specific infinities like an infinite amount of 1's ,infinite integers somehow don't count as natural numbers like why?? apparently the defining factors of the natural numbers are divisibility and primes but that's also possible for defined infinities like an infinite series of 1's

Here's a related idea, which shows that there are as many even whole numbers as there are odd numbers. (Intuitively, there should be fewer even numbers, since only 50% are even?) Well, create the bijection which maps an even number 2n to the number n. Since every eve number 2n has a matching whole number n, there are equally many even numbers as there are whole numbers. The idea is that you basically create an algorithm such that for any number n, you eventually reach it. Going back to the fractions, for any specific fraction m/n, you eventually reach it, and you can find an upper bound for how many steps you will have to take.

No you can't.

This is class 9 maths

Nothing big we did this sort of stuff in 8th lmao

@@darthmaul197 what is your point?

@@fungamingwithdhairya yep it is Coz even i am in 9th 😋

Technically you should define what you mean by 0.9999... and show it actually converges and is a number, then you can do that

Yes, some infinities can be larger than others. The Integers, Naturals, and Rationals are the same size, while the Reals are bigger. (Actually, R, R^2, and R^3 are all the same size as well). The way these things are defined is that two sets have the same size (or "cardinality") if there can be a one-to-one correspondence between them. Of course, the integers are bigger than the naturals in the sense that one is a proper subset of the other, but in terms of what kind of infinity they are, they're the same. It depends what you mean by "bigger".

There isn't any such thing as "further". Since both sets are infinite, you are the same percentage along each set : zero. There is a bijection between the sets, therefore they are the same size. For a rigorous proof, refer to the video you just watched.

~2.665. What was the point of asking this?

Yes the actual proof doesn't have duplicates, he simplified the argument

True, and there are two options to fix the proof. One is to just skip duplicates in the list, but formalising that is a bit messy. The other is to say that this proof shows that there are at least as many natural numbers as fractions, but there are also at least as many fractions as natural numbers, so there must be equally as many of them.

Good enough. That means there are at least as many natural numbers as rational numbers. The inequality in the other direction is obvious, so we have equality. (some technical details aside)

Not really well defined as stated. You could define a function as: f(x) = lim_{n ->\infty} x^n. It's not a very interesting function though. It's only well-defined on (-1,1] and becomes f(x) = 0 for x < 1 f(1) = 1. If you allow for extended real values then f(x) = \infty for x > 1. But it is not defined for x

For X>1 the function is just infinity For X

You might as well ask for the graph of X to the power of purple.

@@mrosskne seems logical

Yes, there are multiple branches of complex exponentiation. Given z^w, you can define it as exp(w∙log(z)). Since log(z) has multiple values, so does z^w. But you can define log(z) = |z|+i∙arg(z). And arg(z) has multiple values because there are infinitely many angles which are coterminal to each other. That all being said, if we fix a principal branch of the argument function, then this choice then gives us a principal branch of complex exponentiation. Typically, (−π,π] is used as the outputs of the principal argument function. This would then make e^(−π/2) be the principal value of i^i. But even if you consider all other branches, regardless of your branch, i^i has a purely real value since: i^i = exp(i∙log(i)) = exp(i∙(ln|i|+i∙arg(i))) = exp(−arg(i)), and arg(i) is real.

No. It comes down to each element in the set of natural numbers having a finite length. This constraint is not placed on real numbers. Things like sqrt(2) or pi have a non-finite decimal expansion so any mapping of reals would not map them. In essence: Countable infinity = infinite number of finite elements Uncountably infinite = infinite number of non-finite elements

[0,1] actually has the same "size" as R. Actually any interval of real numbers has the same "size" as the entire set of real numbers.

Thank you! Cheers!

I had a stroke reading this

"a finite amount of paint can paint an infinite amount of area, no matter how limited the amount of paint is." This is the correct statement, assuming your paint can be spread arbitrarily thin.

Because infinity makes things weird. A bijection between sets means they are the same size. That’s why when talking about infinite sets, we use bijections to say they are the same size. It is not saying 1=2 however, not even close. It’s saying ∞=∞ in certain cases, even if they came from two different sets

Stay in school, moron.

"Why would there be a 1 to 1 correspondance between setA and setB when A is not even a measurable part of the set B." Why don't you try reading a proof for once, instead of complaining about the proper application of logic giving counterintuitive results? The answers to your questions would be easily found, if you'd only be willing to open your eyes and see the world around you for a few moments, instead of blindly proclaiming your sight to be superior to everyone elses'. Speaking less hyperbolically, the reason we say setA and setB have the same size if there is a bijection between them is because that is what we have defined the phrase "setA and setB have the same size" to mean. It turns out that this is a very generally useful definition, and slapping "except if one of the sets is a proper subset of the other" at the end of it makes it less useful. (I'll admit that for my part, I wish we'd stop using terminology which uses analogies to finite things for infinite things. I find "there exists a bijection between A and B" to be far easier to understand than "A and B are the same size," and it isn't signficantly harder to say or write. But my opinions don't change the fact that those sentences mean the same thing when you are talking about mathematics)

That’s crazy but who asked

Not five as far as I know. But some finite number, yes. That's why axiom of choice is often referred to as "incorrect"

​@@DanTheManTerritorial You're right in that Banach and Tarski did not mention a specific number when they published their paradox in 1924; they only said it was finite. But mathematicians have been assigning actual numbers ever since: In 1929, Von Neumann said without proof that 9 pieces were sufficient. In 1945, Sierpiński showed that 8 were sufficient. Finally, in 1947, Robinson proved that 5 were necessary and sufficient-no less, no more. As for the axiom of choice (AC): In 1991, Pawlikowski showed that you don't need the full AC; a weaker version-called the "ultrafilter lemma"-is sufficient. It's now 2022-you gotta keep up, buddy, or you'll get left behind!

The key to that paradox is at least one of those pieces must be constructed in a way that gives it extremely strange properties. Specifically, the set of points in that piece is non-measurable. This means the "size" of the set literally cannot be assigned a value, because no assignable value makes sense. And if the size of a set of points that compose a shape is undefined, then the volume of that shape is also undefined. Normally, moving and rotating a shape will always keep its volume the same. But what happens if you try to move and rotate an object that literally has no definable volume? Well in that case, it IS possible to gain or lose volume by just moving or rotating, and this is exactly what happens in the Banach-Tarski paradox. The construction of these non-measurable sets is only possible given the Axiom of Choice. The Axiom of Choice may look relatively innocent if you read the definition, but the problem is that it is non-constructive: it guarantees that certain objects exist even if it is literally impossible to construct them. So the objects exist because the Axiom of Choice says they do, but you have no way to describe them. And one consequence of this is non-measurable sets.

@@DanTheManTerritorial As far as I've heard (which admittedly isn't very), the axiom of choice is actually becoming more well-accepted over time. Just because a mathematical result is counterintuitive doesn't mean it's wrong.

@@nHans What's the catch? Clearly it's impossible to create 2 spheres' worth of matter from a single sphere, else you'd be able to make infinite matter. Is this just a hypothetical thing?

It’s the set of every natural numbers.

There's a convergence test which is not often used in calculus courses despite often being taught. It's call the "Root Test". The Root Test would be perfect for your series. I'll leave it at that, but if you would like more information or hints, I'd be happy to comment further.

@@MuffinsAPlenty thanks very much!

It converges since 1/(n^n) < 1/(2^n) for any natural n (except 0)