Amazing idea to use the fact that a rational sequence converging to an irrational number must eventually have as complicated of a denominator as you want!!!!
All these discontinuities are “removable”, meaning that limits exist at all points, the function just isn’t equal to the limit at the rationals. Fun theorem: it’s impossible to have a limit at every point but an uncountable number of removable discontinuities. So this is the worst we can do and still have limits everywhere.
8:20 If for any irrational number x you can choose a d such that there is no integer inside (x-d), (x+d). Then it must also be that for any integer n there isn't an irrational number inside (n-d), (n+d). (since if there is an irrational number inside (n-d), (n+d), then the integer must be inside that irrational numbers interval). This doesn't sound correct to me, since we already established that n-e/N is irrational in the limit of N to infinity. So we can only assume that this process doesn't hold for all irrational numbers. Unless I'm missing something
That does not imply, no. If for any specific irrational number x you can choose a 'd' such that there is no integer inside ((x-d),(x+d)), then that only means that, for any integer k, there exists some small h such that *that specific* x is not in ((k-h),(k+h)). That is, for any *specific* irrational number, you can make an interval small enough around any integer so as not to include that number. However, because there are uncountably infinite irrational numbers, you can't just list off all of the irrationals and repeat this process to exclude all of them - by Cantor's Diagonalization proof, if you exclude a countably infinite number of irrational numbers from the interval, you will always be able to still find an example of one you 'missed', even after taking that into account. Your conclusion does not hold, because the choice of value of 'd' does not have to be the same 'd' for every possible irrational 'x', essentially. It doesn't matter if there are irrational numbers in every nonzero size interval around an integer, it only matters if *that specific* irrational number 'x' is not in chosen size interval around any integer.
The function is cool and all but it's quite amazing to know that you can limit the size of the denominator of the rationals between any 2 irrationals to any number that you want!
(That is to say, I think the proof of the limit at irrational values is easier to see if you prove it at rational ones as a lemma, and then make a few adjustments to make sure you're not too close to some simple fraction that isn't x_0)
delta epsilon proof for discontinuity on rationals. Let x0 = p/q in Q, x0 in [0, 1]. Then f(x0) = 1/q. Let epsilon < 1/q. Let delta > 0. By the density of irrational numbers in R, B_delta(x0) contains an irrational number y0. We have that |x0-y0| < delta, but |f(x0) - f(y0)| = |1/q - 0| = 1/q > epsilon, so f is discontinuous for rational numbers
i did some googling after watching this --- apparently there's also a funciton which is continuous everywhere but differentiable nowhere --- "Karl Weierstrass when he constructed a function that is continuous everywhere but differentiable nowhere" thatsmaths.com/2012/10/25/the-popcorn-function/ en.wikipedia.org/wiki/Weierstrass_function
wow, this is quite beautiful. I think of it as a bunch of frogs that hop 1/2, 1/3, 1/4, ..., 1/N starting from 0, and they all hop over this very tiny range (x_0 - delta, x_0 + delta).
It is actually not that hard to prove continuity at irrationals using just sequences, given a certain result about rational approximations to irrationals. The "tricky" case here is a sequence of rattionals p_n/q_n approaching x_0. The result we need is that for any such sequence, the sequence of denominators q_n tends to infinity (I believe this is most easily shown by contradiction). From this it then follows that 1/q_n tends to 0 and we are done.
For the selection of $\delta$, you only need to show that the set $\mathbb{Q}_N = \{p/q \mid 0 \le q \leq N \}$ is closed where $p$ and $q$ are integers. (Since it's a finite union of the closed sets $\mathbb{N}/n$, this is easy.) You then take $delta = d(x, \mathbb{Q}_N)$ where $d$ is the distance function. Since $\mathbb{Q}_N$ is closed and does not contain $x$, we know $\delta$ is not zero.
I stumbled across that very function on one of my pupil's test sheet yesterday at about the same moment you posted your video... It was just printed there by their teacher for curiosity and made me think of Dirichlet's function and lebesgue integration.. Nice coincidence!
It’s also Riemann integrable as it only has countably many discontinuities as it’s only discontinuous are the rationals, yet it has no anti derivative on any [a,b] cause it’s not continuous for every x an element of [a,b] for any [a,b]. I honest to god think there was something wrong going on in the heads of the original analysts, it’s truly beautiful what a collection of such perverse minds can think up.
It is interesting whether it is fractal and how to produce its graph an a computer while it uses just a subset of rational numbers event if using long math.
They don't really do anything noteworthy to this particular function since it can only pick out whether a number is rational or irrational. Perhaps one could define a function which is zero on trancendental numbers and positive on algebraic ones, with a value related to that numbers minimal polynomial over either Z or Q. There is probably some choice of values that would give such a function similar properties as Thomae's function, but "see" algebraic vs trancendental distinctions instead of rational vs irrational.
I am a little confused here how we can do that. If we don't control the p as well in p/q. Let's say we have 5 as the denominator and then we move on to the denominator being 1000. So in the former case we have 1/5 included and in the latter 201/1000 so if 201/1000 is included then 1/5 must be too in the inside the range
Upside down, it looks like a draped curtain. Right by 1/2 on either side, there's a huge notch, same by 1/3 etc. Not the way the Wikipedia pic is done, but if you leave no gaps.
I remember having to prove this exact same thing with epsilin delta as a problem in a calc 1 assignment. Probably the hardest problem I had in calculus. My solution was more than a page long though arguing by offering money.
x_0 was defined as p/q (such that q > 0, p and q are integers, and the fraction p/q is in its reduced form), and f(x) was defined to be equal to 1/q for a rational x (a rational x that is like x_0), so f(x_0) = 1/q.
Peyam: a function that is so crazy that it is continuous at every irrational number but discontinuous at every rational number. Me: Pffffft. No way. Impossible. Peyam: let f = ....... Me: hmmm. okayyy. Where are you going with this?? Peyam: let's evaluate a few points. Me: Sure why not. Peyam: the graph looks like this ... Me: if the graph is like that ... then that would mean on a neighborhood for an irrational point ... f would take on values close to .... no way .... OMGGGGGGG. IT IS TRUEEEE!!!! Peyam: mwuahahahaha 😎 Peyam needs an evil laugh xD
The real numbers are not real. They are a figment of man’s imagination. That makes it so strange that our human “creation” escapes our intuition and behaves differently… like also the Vitali set, it escapes our intuition, or even our intention (sets with no measure!). Or why pi is ubiquitous, in the Basel summation, and in the integral of the gaussian… So strange. Contemplating these mysteries is a pleasure reserved to mathematicians only 😊
This guys enthusiasm is contagious .
Merci :)
it is
04:33 today I came to know the meaning of wtf for mathematician is 'what to find' 😅😅
So, What's The Function?
I love this function because it's perverse. It's a great one to pull out for when students need a counterexample.
To be honest I was having popcorn and watching yt when your notification came. . . Such a coincidence
lol
Amazing idea to use the fact that a rational sequence converging to an irrational number must eventually have as complicated of a denominator as you want!!!!
This is so beautiful! Thank you very much, Dr Peyam, for such a detailed explanation. I finally got it!
All these discontinuities are “removable”, meaning that limits exist at all points, the function just isn’t equal to the limit at the rationals. Fun theorem: it’s impossible to have a limit at every point but an uncountable number of removable discontinuities. So this is the worst we can do and still have limits everywhere.
The reference to Stromae though
8:20 If for any irrational number x you can choose a d such that there is no integer inside (x-d), (x+d). Then it must also be that for any integer n there isn't an irrational number inside (n-d), (n+d). (since if there is an irrational number inside (n-d), (n+d), then the integer must be inside that irrational numbers interval).
This doesn't sound correct to me, since we already established that n-e/N is irrational in the limit of N to infinity.
So we can only assume that this process doesn't hold for all irrational numbers.
Unless I'm missing something
That does not imply, no. If for any specific irrational number x you can choose a 'd' such that there is no integer inside ((x-d),(x+d)), then that only means that, for any integer k, there exists some small h such that *that specific* x is not in ((k-h),(k+h)). That is, for any *specific* irrational number, you can make an interval small enough around any integer so as not to include that number. However, because there are uncountably infinite irrational numbers, you can't just list off all of the irrationals and repeat this process to exclude all of them - by Cantor's Diagonalization proof, if you exclude a countably infinite number of irrational numbers from the interval, you will always be able to still find an example of one you 'missed', even after taking that into account.
Your conclusion does not hold, because the choice of value of 'd' does not have to be the same 'd' for every possible irrational 'x', essentially. It doesn't matter if there are irrational numbers in every nonzero size interval around an integer, it only matters if *that specific* irrational number 'x' is not in chosen size interval around any integer.
@@HeavyMetalMouse thank you
The function is cool and all but it's quite amazing to know that you can limit the size of the denominator of the rationals between any 2 irrationals to any number that you want!
I see this comment is from 3 months back while the video was posted 8 minutes ago
Which seems crazy
Wtf??
What
Wtf..some 👽 stuff at work
How on earth do you make a video every single day?? Incredible!
I make them in batches haha
I thought it would be easiest to show that the limit is 0 everywhere. If x_0 is rational, choose delta to be 1/(N!) such that 1/N
(That is to say, I think the proof of the limit at irrational values is easier to see if you prove it at rational ones as a lemma, and then make a few adjustments to make sure you're not too close to some simple fraction that isn't x_0)
Thank you Dr Peyam, I've read a lot of explanations to this problem and yours is definitely the best one.
delta epsilon proof for discontinuity on rationals. Let x0 = p/q in Q, x0 in [0, 1]. Then f(x0) = 1/q. Let epsilon < 1/q. Let delta > 0. By the density of irrational numbers in R, B_delta(x0) contains an irrational number y0. We have that |x0-y0| < delta, but |f(x0) - f(y0)| = |1/q - 0| = 1/q > epsilon, so f is discontinuous for rational numbers
0:22 doc dropped fire beat tho
Thanks for explaining it!
That’s easy to understand！
But why we can restrict the delta, to exclude an infinite set of form p/n.
i did some googling after watching this --- apparently there's also a funciton which is continuous everywhere but differentiable nowhere --- "Karl Weierstrass when he constructed a function that is continuous everywhere but differentiable nowhere" thatsmaths.com/2012/10/25/the-popcorn-function/
en.wikipedia.org/wiki/Weierstrass_function
i was reading it the other day!
I love how "want to find" is written WTF. :D
Thanks for such a great content
This freaks me out. How can this be? I too happened to be eating popcorn while watching this and I almost never eat popcorn.
So basically it discontinuous in a countable set of points
Indeed the rationals are countable
wow, this is quite beautiful. I think of it as a bunch of frogs that hop 1/2, 1/3, 1/4, ..., 1/N starting from 0, and they all hop over this very tiny range (x_0 - delta, x_0 + delta).
It is actually not that hard to prove continuity at irrationals using just sequences, given a certain result about rational approximations to irrationals.
The "tricky" case here is a sequence of rattionals p_n/q_n approaching x_0. The result we need is that for any such sequence, the sequence of denominators q_n tends to infinity (I believe this is most easily shown by contradiction). From this it then follows that 1/q_n tends to 0 and we are done.
John Conway called this function "Stars Over Babylon".
fantastic explanation
Sir you are legend
Plz recommend more TH-cam channel of Math
Dr Peyam hahaha
@@drpeyam 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
why we can continue the process up to p/N ? why not beyond that?
wow this just might be my new favorite function! Love this video!!
For the selection of $\delta$, you only need to show that the set $\mathbb{Q}_N = \{p/q \mid 0 \le q \leq N \}$ is closed where $p$ and $q$ are integers. (Since it's a finite union of the closed sets $\mathbb{N}/n$, this is easy.) You then take $delta = d(x, \mathbb{Q}_N)$ where $d$ is the distance function. Since $\mathbb{Q}_N$ is closed and does not contain $x$, we know $\delta$ is not zero.
Stromae + an insanely cool function; what more could you ask for? Amazing stuff!!
That strange image of the pop-corn function reminds me of the Sierpinski gasket (Pascal's triangle mod2). Perhaps there is some connection!
I stumbled across that very function on one of my pupil's test sheet yesterday at about the same moment you posted your video... It was just printed there by their teacher for curiosity and made me think of Dirichlet's function and lebesgue integration.. Nice coincidence!
It’s also Riemann integrable as it only has countably many discontinuities as it’s only discontinuous are the rationals, yet it has no anti derivative on any [a,b] cause it’s not continuous for every x an element of [a,b] for any [a,b]. I honest to god think there was something wrong going on in the heads of the original analysts, it’s truly beautiful what a collection of such perverse minds can think up.
How do we show that it's indifferentiable?
It is interesting whether it is fractal and how to produce its graph an a computer while it uses just a subset of rational numbers event if using long math.
Next time, can you please feed your viewers popcorn? You could even brand it as Peyam's popcorn.
Hahaha
Thanks.
What, if anything noteworthy, do the transcendental numbers do to this function?
They don't really do anything noteworthy to this particular function since it can only pick out whether a number is rational or irrational.
Perhaps one could define a function which is zero on trancendental numbers and positive on algebraic ones, with a value related to that numbers minimal polynomial over either Z or Q. There is probably some choice of values that would give such a function similar properties as Thomae's function, but "see" algebraic vs trancendental distinctions instead of rational vs irrational.
I am a little confused here how we can do that. If we don't control the p as well in p/q. Let's say we have 5 as the denominator and then we move on to the denominator being 1000. So in the former case we have 1/5 included and in the latter 201/1000 so if 201/1000 is included then 1/5 must be too in the inside the range
That is some high level mathematics...I still always come around to see if something helps me...btw u are very well received by Indians..
Upside down, it looks like a draped curtain. Right by 1/2 on either side, there's a huge notch, same by 1/3 etc. Not the way the Wikipedia pic is done, but if you leave no gaps.
So if it is integrable, what is the integral? Just f(x) = 0, right?
0 indeed
@0:24 i dont think this is a reference to Junji Ito's Tomie, but i thought it was going to be haha
But what if x0 is so close to a rational number that there is no interval that doesn't contain both of them???
very nice video!
I remember having to prove this exact same thing with epsilin delta as a problem in a calc 1 assignment. Probably the hardest problem I had in calculus. My solution was more than a page long though arguing by offering money.
Omg Calc 1? 😱
@@drpeyam Ya our calc 1 was hard. It wasnt on a test luckily but part of a compulsary assignment
Such a beautiful proof!! But how am I going to solve it by myself in the exam 💀
Wow, that's awesome!
Hey what does stromae mean?
You should have ended the video with a clip from the song by Gershon Kingsley or Hot Butter.
Sir what is your qualification and which university
that is so crazy!!
Alors on danse tan tatan , LOOOOOL
The best
Very good sir
I don't get this: how is f(x_0) = 1/9 at 6:05?
x_0 was defined as p/q (such that q > 0, p and q are integers, and the fraction p/q is in its reduced form), and f(x) was defined to be equal to 1/q for a rational x (a rational x that is like x_0), so f(x_0) = 1/q.
@@redvel5042 Oh yes. I misread the 'q' as '9'. 🤦🏾♂️
@@redvel5042 Thank you.
@@MurshidIslam np
this function in funtastic
is
hello dr peyam!!!!!!
Popcorn is function .😂 Lol
سلام دکتر
چطوری میتونم با شما ارتباط برقرار کنم؟
Peyam: a function that is so crazy that it is continuous at every irrational number but discontinuous at every rational number.
Me: Pffffft. No way. Impossible.
Peyam: let f = .......
Me: hmmm. okayyy. Where are you going with this??
Peyam: let's evaluate a few points.
Me: Sure why not.
Peyam: the graph looks like this ...
Me: if the graph is like that ... then that would mean on a neighborhood for an irrational point ... f would take on values close to .... no way .... OMGGGGGGG. IT IS TRUEEEE!!!!
Peyam: mwuahahahaha 😎
Peyam needs an evil laugh xD
This is a proof but ,not a visual proof ,i am not able to visualize this
alors on danse 😂😂😂
this teacher is funny he's the best
The real numbers are not real. They are a figment of man’s imagination. That makes it so strange that our human “creation” escapes our intuition and behaves differently… like also the Vitali set, it escapes our intuition, or even our intention (sets with no measure!). Or why pi is ubiquitous, in the Basel summation, and in the integral of the gaussian… So strange. Contemplating these mysteries is a pleasure reserved to mathematicians only 😊
Want To Find
WTF!
WTF
I am hungry of pop-corn 😋🍚
A'lor on danse
Alors on danse 😂😂
Is it even possible to have consecutive irrational numbers?
Not possible
@@drpeyam how do we prove it?
@@BlokenArrow maybe archimede proprety
@@BlokenArrow density of irrational numbers
Here q is x for all the confused out here
Use excel, at least!
?
Never been so early
somehow the thirds are screwed up in the plot
Hahaha
@@drpeyam :)
Why your expressions is like a gay ??
🇷🇺🇷🇺