This Video Will Frustrate Your Teacher
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Corrections:
7:03 Have non-zero imaginary component
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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 ...∞
2:39
"other things that can grow arbitrarily large..."
why did my mind go places?
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😎
The way I think about the Rearrangement Theorem is that conditionally convergent series are essentially subtracting one infinity from another. So it makes sense that literally any number can result.
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.
As a teacher, the most frustrating thing about this video is the sponsored section in the middle of the video.
i agree
Kinda pity you, Bri... Seems like you were unlucky to get bad teachers
He probably took up mathematics out of spite
You won't "frustrate" (or trick) any maths teacher with those misconceptions. They went to university... to study maths...
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!
Great visual with the cylinder. Perfect for HS geometry students to understand the relationship between surface area and volume.
The one that breaks my brain the most is the Banach-Tarski paradox. If I break a solid object into finitely many parts, then move those parts around and put them back together, I should have the same volume, right? No, we can split a sphere into five pieces and use them to reassemble two spheres the same size as the original.
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?
Changing order of addition doesn't matter: infinite sums are not sums, you are not changing order of addition you are taking the limit of a different series. I also don't get the title: any teacher will surely be aware of all of the situations presented in the video.
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?
Good stuff. Man, I was dead certain this was going to be another -1/12 video.
Really liked the phrase "Your knowledge can grow arbitrarily large". 😂
I always cry when I see people not including 0 into the natural numbers.
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)
The thing that frustrates me the most is the title 🤦🏾
6:59 Actually that's WRONG! 0.207879 is a complex number with no imaginary part.
A complex number with no imaginary part lies on the real axis. There's a name for such numbers...
Nice sum up of the first three lectures of my Computer Science analysis 1 course.
For the fractions example, the way I would show that they are the same size is that you can concatenate the numerator and denominator and get a unique integer for each fraction. The only integers that wouldn't show up are digits that end with 0, as well as 1-9. The nine integers can be inserted anywhere, with them just scotching all number assignments to the right, and the integers ending with zero can be inserted in some regular way, such as putting them into the position with their own number.
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.
I don't really see which teacher would be frustrated by a student knowing that thes are "misconceptions" about maths....
Why would this frustrate your teachers? These are some of the first results you're taught in the related courses.
Me: For once please try to be reasonable !
Math: Haha ! In your dreams fool !
It really did frustrate my English teacher. She couldn't understand a word.
I only know of the cantor diagonalization in binary (base 2).
Things like this will only frustrate math teachers who never really studied math in the first place beyond first year calculus. Any math teacher with an active interest in mathematics will be able to talk about these "facts" and show students how they work. As at least one other commenter has said.... It's a pity that you only had weak math teachers.
02:40 missed opportunity to sneak in a wrong.
I think better than the fact that the rationals and naturals have the same cardinality is that the integers and naturals have the same cardinality!
I knew this!!! Still... you explain so well
also the like button throws confetti
From the thumbnail, I assumed this video would be about math teachers being frustrated by the blackboard bold N with the double line on the left instead of the diagonal.
1:43 "That's obvious, right?"
*immediately thinks of Zeno's paradox of infinitely dividing the distance between point A and point B by 2*
Nothing is obvious when infinity's sitting at the table with mathematics 🥲.
Please spare my brain
Can you make a video with log base inf of inf?
4:56 I genuinely can't wrap my head around this one, even if you were to ignore duplicates and if you were to pair it up diagonally like you have here. If the first column is all of the natural numbers then how could you pair a natural number to everything in the grid if there's more numbers in the entire grid than in the first column alone?
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.
why the diagonalization cant be applied to the naturals too? 5:00
So you suppose people have bad teachers?
Shearing a shape preserves the n-dimensional volume, but increases the n-dimensional surface area.
so youre saying that there are MORE numbers between 0 and 1 but rational numbers and natural numbers are the same size?
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
"Just add 1 to each digit" doesn't work because you may encounter 9. "Add 1 to each digit except 9 where you subtract 1" might work, but might run afoul of 0.999...=1 related shenanigans. Use 5 if the number on the list doesn't have a 5, and use 6 otherwise. Much safer.
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.
0,9 repeating isn't less than one, it's equal to one
As stated in this video
I'm in middle school if math is like this then I quit trying to make it my best subject
That's great ☺️👍👍
About the natural vs Rational numbers thing, if n(N) = n(Q) then how come that N is a subset of Q? At the very least 0 also comes under Q but it does not come under N, while all elements of N come under Q
"Except thats not always true... CUT! -1/12 get outta here-"
7:07 i^i is still a complex number, as the complex numbers are a superset of the real numbers.
Ooh, I'd like to see your take on "all horses are brown" - now that causes arguments 😆
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.
I'm sending this to my teacher...
for the -1/12, only if you supersum. There's a good breakdown by Mathologer
I don't need this video to frustrate my teacher, I already have bad marks!
Hey Brian! Curious about the sum from n=1 to infinity of 1/(n^n). Does it converge to anything interesting? Does it converge at all? I’m only in year 10 so I have no idea of how to test these things but I’d really like to know!
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)
Please make a video on Sierpiński triangle ...∞
Today I learned that the set of functions f : N -> N isn't countable, but the set of functions f : N -> N that are eventually zero is.
"Other things that can grow arbitrarily large"
😉
hey Bri how will you solve 2^root2
~2.665. What was the point of asking this?
A simpler counterexample to the statement "Combining Imaginary Numbers Produces Imaginary Numbers" is i⋅i=-1.
Showing this to myath teacher
My school don’t have a subject called neither myath nor mythology
Bro just had a stroke
@@TrainingCuber emgish
@@darthmaul197 Durth Maul
Doesnt i^i have multiple solutions?
Is there even a principle branch with complex exponents?
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.
I've been thinking about the natural numbers vs decimals between 0 and 1. What if you just reversed the digits of the decimal and assigned it to that number? We ignore leading 0s with natural numbers and trailing 0s with decimals, so would this work?
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.
yeah I just LOVE math...
you can do the same thing with Natural Numbers as you did with the numbers between 0 and 1 both are uncountable
No you can't.
I Think Finite Volume means Finite Surface Area Because The Volume is Infinite and The Area is also infinite Notice pi is Infinite
"Textbook knowledge aren't always true" said my teacher.
Please please reply,which software do you use to make these clean videos?
This video would not frustrate any good math teacher. You can't actually sum an infinite list of numbers. By definition of "infinite" there is no last number so the sum will never be complete. What you can do, however, is find a limit to a sequence of partial sums.
Define sum
@@lc1777 Combine two quantities to produce a resulting quantity.
What have most got in common.. they involve "infinity"... the hole into which common sense and usefulness vanishes. 1,2,3, ... where the "..." stands in for something you can't write, or imagine let alone do.
Thanks for a delightful video full of fascinating mathematical tidbits.
I'm afraid I wasn't convinced by your Cantor's diagonal proof, because of the non-uniqueness of the decimal representation of real numbers.
Consider the following list of real numbers:
0.49999999...
0.0900000...
0.0090000...
0.0009000...
etc.
The diagonal argument gives you 0.5000..., the same as the first number.
Of course, this is easily fixed by requiring the numbers to be written in "standard" decimal form (should end an an infinite sequence of 0s rather than 9s), but thought it was worth pointing out...
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.
What the hell is going on
My brother in christ, shes a geometry teacher, nothing frustrates a geometry teacher
They frustrate you
Good luck.
If stuff like that can frustrate Bri's math teacher, then I've got some things to drive them stark raving mad: 🤪
- Fractals:
• shapes with non-integer (fractional) dimensions
• infinitely long perimeter enclosing a finite area
• infinite surface area enclosing a finite volume (kinda like Gabriel's horn, but cooler)
• lengths and areas that vary depending on the scale you use to measure them
- Think complex numbers are weird? Wait till you hear about dual, split-complex, quaternions, octonions ... For example:
• How many square roots does -1 have? Only +i and -i? Wrong-there are more. Many more.
• How many square roots does +1 have? Only +1 and -1? Wrong again!
• How many square roots does 0 have? Only 0? Guess what-still wrong. Even 0 has many distinct square roots.
- And my favorite paradox: The real number line is continuous and has an infinite number of points. There are no gaps. And yet, no two points are adjacent to each other. Same with 2D plane, 3D volume etc.
Sadly, despite all the advances we've made in math, we still can't divide by zero. 😢
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.
That supposed size equivalence between rational and natural sets is extremely fishy: some infinities can be larger than others, like how R² is bigger than R and smaller than R³. You could do the same thing between Naturals and Integers: Count it in the order of 0, 1, -1, 2, -2, and so on, and every integer would have a respective natural... but if you stop at any point, you can clearly see that you would be "further" down through the naturals compared to integers.
To be fair, I don't know how someone would build a rigorous, mathematical proof, but I'd imagine mathematician would know.
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.
TL;DR version of the .99…=1 refutation : [everything in brackets can be omitted, so the argument distills down to 2 yes-no questions and 1 four-word statement]
You admit 9/9=1? (1)
And 1/9 = 0.11…? (2)
Now [in (2)] multiply by 9. [Label this (3)]
[LHS of (3) = LHS of (1). RHS = 0.99… .]
The only thing that has to be given any real intellectual assent is scalar multiplication.
This stuff should teach the Uni teachers, for the normal teachers it is no matter
This thumbnail almost looks like new twitter logo
It’s the set of every natural numbers.
This video is 19.55% ads 😂
Who's "your teacher"? All my math teachers knew this at least from high school and up 🧐. (I didn't ask my earlier teachers because I didn't know about those concepts back then and I don't want to talk to them ever again.)
why isnt it n.(n+1)/2 so ∞.(∞+1)/2
Real numbers are complex numbers (they are in the set of complex numbers)
A real number is complex
0 is a natural number bro.
This won't trigger anyone mathematically inclined.
Actually, rearranging the terms gives you a different result
typo at 1:00 :)
Hi Brian, I love watching your vids, but i have a question and it feels interesting to me.
*what's the graph of x^infinity?*
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
for anyone who wants a proof for the rearrangement thing in the 1st part th-cam.com/video/U0w0f0PDdPA/w-d-xo.html
When I was in school, I found a way to irritate teachers. Not by giving wrong answers. Not by giving correct answers. But rather, by being ultra-pedantic and correcting _them_ frequently. Basically, by being a smartass.
(This was decades before _"I wish I was high on potenuse.")_
- The teacher said that the roots of a quadratic equation are complex only if the discriminant is negative; otherwise they're real. I interrupted, saying that the roots are _always_ complex, because even real numbers are complex.
- Naturally, I always protest when people say things like _"imaginary numbers are real"_ or _"complex numbers are real."_ In a math class, "real number" is a technical term and doesn't mean what it means in everyday English. Use it accordingly; don't talk like a philosopher or mystic!
- Trying to explain subsets, the teacher said that you can separate the integers into two disjoint groups: odd and even numbers. I objected, because clearly the odd numbers don't form a group! 😎
- Another favorite: _"You can't add scalars to vectors."_ Of course you can! In fact-if I may be allowed to boast-in high school, I independently rediscovered hyperbolic quaternions simply to prove this point.
- When asked to pick a number between 1 and 10, I pick-depending on my mood-π, _e,_ φ, or √69.
What can I say? I loved math and knew a lot more than my classmates at any time. So I got bored easily.
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!
2:39 -are your mom
wonderfull and great
Thank you! Cheers!
isn't the Cantor Diagonalization Argument(5:30) kinda stupid because the same argument can be used for the Natural numbers
n1=1
n2=2
n3=3
nx=x
so we just take the biggest of them and add 1
and thus we would show that there are more Natural numbers than Natural Numbers which is obviously false
and as he said 2 sets have the same size if we can pair up every element which is clearly possible here
n1=0.1
n2=0.2
n10=0.01
n485=0.584
nx=0.(x but mirrored)
so if we use that argument on this we just get the result that we haven't used all the Natural numbers because they are infinite obviously
"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
1+2+3... doesn't equal -1/12. thats a false proof that shows that making assumptions in infinite series leads to wierd answers like -1/12
these are just 3 different vsauce videos but not explained as well...
You sound like cgp grey
just use tau
3hlate
Instead of trying to prove there exists a number not on the real numbers list, why not use that proof of a number not on the list - to construct the list. Start with an arbitrary number, change the first digit by 1, you have a 2nd number, for the second number change the first number in the second number, and the second number in the first, that's the third number and so on. You might say aha, I've found a number that doesn't exist in your list! But no - you've just found the next number on it.
Why did I learn math in school? What’s up with this? We need to redo math itself this makes no sense and feels like loopholes. Like you can state any ‘rule’ in math then put an * after and still be right and wrong. The system we use to measure our universe is just a logistical fallacy!
What are you confused about?
N contains 0
5:28 always confused me. Can't we to it the other way around too?
For the real numbers you have one rule which is a number cannot end with a zero (as in 0.482000 is not allowed, but 0.482 is), and natural numbers cannot start with a zero (as in 000284 is not allowed but 284 is). The examples in the brackets also show that every unique real number can be assigned to a unique natural number as follows: 0.abcdef -> fedcba.
Even if you ignore the fact that you can do what you showed in the video the other way around, every time you add a new real number you can assign it a new natural number. Saying "I can keep adding more real numbers" doesn't mean anything if you can still assign all of them natural numbers.
Just to clarify, obv if this has stuck for years, was discovered by a mathematician, and other mathematicians to this day don't dispute it, I must be missing something. I'm not claiming to have discovered something new or anything, I'm just trying to figure out what'd I miss
well, it does not work the other way around. Problem is that real numbers can have infinitely many digits after the decimal point without any pattern. Natural numbers cannot. While you can have a real number like 4.1951763...., its impossible to reverse this process and make a natural number out of it, we can't have something like ...36715914 with infinitely many digits to the left, that would not be a natural number per definition.
Also the way this proof works is that we first assume that our list is complete. For every natural number we WILL have exactly one corresponding real number and for every real number we WILL have exactly one corresponding natural number. What the proof then just does is that it constructs a new real number from our given "complete" list, and by construction it cannot be in our list, because it will ALWAYS be off by at least 1 digit. This shows that there is no corresponding natural number. We cannot just add this real number to our list because we already assumed it was complete, and if we have to add this number it shows that it wasnt complete after all right? This is just a typical proof by contradiction, we assume the opposite of what we want to show and then somehow derive a contradiction, in this case it's that our assumption is incorrect
@@hww3136 1. It talks about real numbers from 0-1, if it was all real numbers then I agree. Once you only take the reals from 0 to 1 it's basicall one group of numbers after the decimal point (reals from 0-1) against numbers before the decimal point (naturals), which are basically the same.
2. This process of adding a new number also works for the naturals, take the one's digit from the first number, than take the ten's digit from the second number, the 100's from the third and so on, then add one to each digit. You'll get a new number.
@@ozargaman6148 can real numbers have infinitely many digits after the decimal point? can natural numbers have infinitely many digits, excluding 0, "before the decimal point"? You just need to understand this difference. Real numbers can take infinitely many digits to represent, while every natural number always has finitely many digits.
Also the set of real numbers between 0 and 1 and the set of all real numbers actually have the same cardinality. I can even explicitely give a bijection as (2-1/x)/(1-x) or tan(x*π+π/2)
@@hww3136 you can't just decide decimals can go forever and naturals can't. Yes, you can't put zeros before a natural, but you also can't put zeros after a decimal
@@ozargaman6148 huh? thats just what follows from the definition of naturals and reals, and how their decimal representation is defined. So mathematicians already decided that long ago. By the way if you're not happy with Cantors diagonalization argument, there is also another way to prove that there is no bijection without having to rely on any decimal interpretation stuff, using nested intervals, although its a bit less intuitive, but gives the exact same results
7:30 This is how i would like to prove
Let x be = 0.99...
10x - x = 9x
= 9.99... - 0.99...
= 9
Since 9x = 9 we can divide both sides by 9 to get :-
x = 9/9 = 1
Therefore 0.99999..... = 1 😀
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
4:55 You are counting some rationals more than once. For ex. 1/1, 2/2, 3/3...
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)
asnwer= n1 isit 🤣🤣🤣🤣🙂🙂🙂
hi
I wish these interesting ideas and facts were known by today's mathematics teachers... I mean, if I didn't know about these, I wouldn't consider the title of 'math teacher' appropriate for myself.
Everything is getting more and more unqualified, ordinary and awful. Let's see how will we end up....
Good news then : they are known by today's maths teachers. It turns out that TH-cam video titles aren't actually a source of knowledge.
@@mrosskne Not everyone is as lucky as you, mister.
Gabriel's Horn, or how a mathematical object cannot even have the half of it that's containing a finite amount of paint, be painted with the paint.
Because a finite amount of paint cannot cover an infinite area.
But you can have every point of it covered in paint, and every point between it covered in paint, yet there's not enough paint to cover it.
Or as I call it, a useless mathematical garbage concept that is beyond ridiculous. It's either that or a finite amount of paint can paint an infinite amount of area, no matter how limited the amount of paint is. Gabriel's Horn, is dumb.
I have 0 paint and I can with 0 paint paint the whole plane with it.
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.
No, i^i is not *a* real number. It is *many* real numbers.
257th comment(Fermat prime)