Dirichlet Invented this Function to Prove a Point
ฝัง
- เผยแพร่เมื่อ 29 มี.ค. 2024
- In 1829, Dirichlet invented the first nowhere continuous function.
FAQ : How do you make these animations?
Animations are mostly made in Apple Keynote which has lots of functionality for animating shapes, lines, curves and text (as well as really good LaTeX). For some of the more complex animations, I use the Manim library. Editing and voiceover work in DaVinci Resolve.
Supporting the Channel.
If you would like to support me in making free mathematics tutorials then you can make a small donation over at
www.buymeacoffee.com/DrWillWood
Thank you so much, I hope you find the content useful.
Life has much improved since I stopped worrying about analytic monstrosities and decided to live C_infinity
Welcome to physics
@@_Xeto There's a derivative for that
What the? I don’t understand the last part
C_infinity refers to functions that have infinitely many derivatives. In other words, the commenter means they decided to exclusively work with very smooth functions (a nice place indeed, but not as nice as functions that are equal to their Taylor series)
@@whatisrokosbasilisk80 certainly they are not. Some of the numerical approximating methods which are PRECISELY what physicists do when describing the real world is densely packed with many very nice very smooth functions, say gauss curves for example. Smoothness is nice and convenient, thus it is used.
"On sets of measure zero, always bet on Lebesgue... and his overwhelming integrability."
bear witness to the one who left it all behind, the one who is free
hey where did u study lebesgue integrals on i ve seen quite an amount of this but i gotta see more of it
💀💀💀 Fraudmat about to get the airport treatment with this one
Found the jujutsu kaisen fan
As the king of calculus, Bernhard Riemann faced the Dirichlet function.
Interesting to think that these jumps occur at infinite scales too. Whether you zoom in or zoom out, the density of the discontinuities is going to be just as thick!
Yeah, you can't really visualize it perfectly ... a line riddled with infinitely tiny holes packed infinitely dense, or two complementary lines where one has holes exactly where the other is filled.
Irônicamente elas tendem a ter densidade "completa" porque a quantidade de números irracionais é um infinito além do infinito dos números racionais e pode-se dizer que a quantidade de irracionais entre dois racionais é maior que de todos os racionais. E a integral dessa função tende ao valor da constante do número irracional. Mas ao mesmo tempo entre quaisquer dois irracionais definidos haverá pelo menos um racional e entre quaisquer dois racionais definidos haverá pelo menos um irracional. Uma coisa louca lidar com infinidades, infinitos e infinitesimais.
that's a pretty good description of the concept of dense sets in calculus and topology
My Calculus Professor dropped this on us at the end of a Friday lecture to give something to discuss at Happy Hour.
To prove "a point" 😂
OP has proved his point, twice.
A slightly more fun example: consider a function that is equal to 0 for all irrational arguments and 1/q for any rational number p/q, where p and q are coprime and q is positive. It's not hard to show that such a function is discontinuous for all rational values and continuous for irrational ones.
Thomae's function
The modified Dirichlet function is fun
It is also neat to show that the Dirichlet function is not Riemann integrable, but the modified one is
Nitpick police coming in
That function is also continuous for 0;
@@matheusjahnke8643 The nitpick police is wrong. Zero is a rational number of the form 0/1 (0 and 1 are coprime since gcd(0,1) = 1, and 1 is positive), so the value of my function there is 1/1 = 1. But there are irrational numbers in any neighborhood of zero, and those are sent to zero, so the function is discontinuous at zero too.
@@pavlosurzhenko4048 so i guess you are... the nitpick fbi? heahaeheha
In the description you wrote "In 1729" and I believe you meant to write "In 1829" since this would be roughly around Fourier's time..
ramanujans typo strikes again....
What a smooth explanations of proofs! Great job !
But what happens to the FT of that function??? Also converges to 1/2 ?
Because the function is nowhere continuous it doesn't meet the criteria to have its Fourier transform taken. The point of this function was to essentially construct a function for which the Fourier transform couldn't be taken
I believe it will be constantly 1. Observe that if f and g are Lebesgue integrable functions and the measure of {x|f(x) != g(x)} is zero, then the Fourier series of g equals the Fourier series of f. The result follows by letting f=1 for all x and g be the dirichlet function.
@@hansyuan4116 this will depend on how we are defining the Fourier Transform. If we use the Riemann Integral, then, as the video is discussing, the lack of piecewise continuity will make it so we can’t take the Fourier Transform.
If we use the Lebesgue Integral then, as you suggest, we can extend the functions for which we can apply the Fourier Transform to.
I was about to comment that your pronunciation of Dirichlet may have been wrong as all teacher I've had pronounced "dirikley".
They were all wrong, my life was a lie !
@@pierro281279He was German with French-speaking ancestors (from what is now the French-speaking part of Belgium). In Germany, his name is usually pronounced "Dirikley" nowadays, but the French way, i.e. "Dirishley", isn't wrong either, imo.
Whatta rebel my man Dirichlet!
Great video!🎉
Whoa, yeah it becomes continuous. That raises moe questions… Great video! I hope you’re going to do a follow-up to this.
Lovely as always. 💜
The last point really drives home what a monstrosity this function is
Mindblowing. Great explanation!
Dirac's delta function has entered the chat
Dirac delta is continuous when you define it properly, as a linear functional on a Schwartz space (a distribution) since it is bounded (as an operator)
Also Fourier transform is defined differently for distributions i.e.
Fourier transform (Dirac delta(f))=Dirac delta(- Fourier transform(f))
dirac delta has nothing to do with today’s topic
Very cool!
Great video! At 0:30, that infinite sum of trig functions would also be a weierstrass function right? Would love to see you do a video on that topic, a similarly weird function to the one you talked about this time
Fourier series are especially useful for solving PDEs such as the heat equation. such a series would certainly be differentiable whereas the point of the wierstrass function is that it is differentiable nowhere.
This just makes me realize that the intuition I had developed to explain why the cardinality of the reals is larger than the cardinality of the rationals is not correct
Well that's fun. Just gotta relearn my fundamental understanding of infinity again
Intuition is kinda cursed when thinking about these things. Notice that we're really close to stating the continuum hypothesis, and that can't be proven or disproven.
@@jordanrodrigues1279Indeed, it was proven to be independent in the 90s
@@jordanrodrigues1279 you mean those things are cursed while intuition fails because it isn't;
Hey this isnt the Normal Album!
Was searching for a Will Wood (the artist) comment
This might be the greatest math-pun I've ever read
Why does it become continuous if you take only the rational or irrational parts? Wouldn't there be an infinitely dense forest of Removable points of discontinuity if you do that?
Continuity means different things depending on the domain. In fact, calculus books that call a function with points excluded from its domain "discontinuous" at non-domain points are kinda misleading because such a thing is technically not a function at all on the "full" reals (rather it's what is called a "partial function"), since the definition of a function requires that each point of the domain be associated with _some_ point of the codomain. If you restrict the domain to, say, only rational points, then what happens is that in effect this domain cannot "see" where what you are calling "discontinuities" are, and thus it "thinks" the function is continuous. (Think about what'd happen if the reals just didn't exist, because you did not define them yet!)
Well imagine if you only have rational numbers, for each x in Q, f(x) = 1, you are continuous on your domain
@@shimrrashai-rc8fqbut then 1/x must be continuous too?
If you exclude 0 from the domain, yes, 1/x is continuous. If you do include 0, then no matter what you define 1/0, the function will be discontinuous.
@@skallos_ but since everything divided by 0 doesn't exist can't you just say that these functions then are contiuous. Or if they are just contiuous on a given interval aren't they just partly contiuous?
This is so cool
always coming across this function in physics
What would be the integral of this function ? (between 0 and 1 for example)
Undefined ? 1/2 ? some other ratio ?
The c for rationao and d for irrational idea was the first thought I had when I saw the thumbnail. Kinda proud of having basic understanding of maths.
There is a definition of phi as the most irrational number, where it’s written as a continuous fraction of ones all the way down.
If two takes the place of the last digit in this continuous fraction construction for phi, then those two numbers are not separated by a rational number.
I think this runs into some methodological issues. However, you can do this digit substitution with finite values of phi, for which it is possible to find a rational number. It is only problematic to find it when the digits occur at the end. Still, any continuous fraction can be flipped inside out, though you may run into problems with infinity, though an equality sign by definition makes an equation finite if one side has finite value.
Is this a reupload? I remember seeing this long ago
I didn't know that there's a rational number between any two irrationals. Because there's also an irrational between any two rationals, does that mean you can't go from one irrational to another without going through a rational, so the number line alternates in some sense between rationals and irrationals?
Would be interesting to know about the significance of this. What has this function been used for?
For proving a point.
The most epic pun
this is for 2:21
)
also have you heard of the musician will wood? also good stuff
Thank you. We'll need another at 3:57 if you've got one. It was driving me more crazy than it had any right to.
will wood fan spotted
how does the function become continuous if restricted to just one of the domains?
OK ngl, that last bit killed me ☠
Isn't that there are more irrational numbers than rational numbers? Won't this affect the density of the lower line? What would the Fourier transformation for this function look like?
Wait, how is one part or the other, contiuous, at the end of video?
how does it become continuous if you restrict the domain just to rational or irrational? won't you have holes?
Well yes but actually no. Rationals and reals are both what we call "dense", meaning at any Intervall no matter how small you have an infinitely many numbers from both. So that makes the holes kinda disappear.
So you might say if we forget about all reals there would obviously be a "hole" at π to which I would say: well yes but I can find a rational thats as close to it as you like.
So basically the holes are infinitely small and everything still works just fine
So, can you Fourier transform the Dirichlet function?
I expected this to lead somewhere.
1:13 should this not converge to x = 1 at 0, because the original function is equal to x = 1 at 0?
If "Because FU, that's why" had a function.
how does restricting it to the rational inputs make it continuous?
The function is defined as f(x)=a if x is rational, and f(x)=b if x is irrational, therefore if we restrict it to rational inputs, the function cannot give the value of b, as b is output when x is irrational, therefore as a is the only value of the function when x is rational, and we only can input rational numbers, the function simplifies to f(x)=a, which is continuous.
@@magma90 I understand that as we restrict the domain to the rational inputs, f takes only the value a (which is 1 in pur case). But I don't understand how that makes it continuous. My understanding is that if a function is any close to being continuous on some interval I, it needs to be defined on all real numbers in I. If, for example, I define the function g(x) that equals x² when x is an integer, and 0 when it is not. If I restrict the domain to only integers, my functions would g(x) = x². But it is not continuous since it is defined only for integer values. Am I missing something here?
@@pizzarickk333you're missing the more general definition of continuity. The most general version of it is that between arbitrary topological spaces. You treat both the domain and the codomain as their own spaces (so for our purposes irrational inputs simply don't exist). If you can't detect any discontinuities, then the function is continuous. For example, if you define a function on rational values to be 0 for x < 0 and 1 for x >= 0, you can still find that the function is discontinuous at 0, but for constant functions there aren't any discontinuities at all.
Now the induced topology on natural numbers is a discrete one, which means that every function with natural numbers as its domain and some other topological space as a codomain is continuous. People generally don't talk about continuous functions on natural numbers because it's not very meaningful - they are _all_ continuous.
As for more precise definitions, I'm not going to give you the most general one, since it requires some intuition building first, but if you have two sets X and Y with a well defined concept of distance between two points (let's write it as d(x, y)) then you can say that a function f from X to Y is continuous at a point x iff for any epsilon > 0 there is such a delta > 0, such that for all x' if d(x', x) < delta then d(f(x'), f(x)) < epsilon. Hopefully you can see how it's basically the same definition as in your real analysis class.
@@pizzarickk333 I think some of your confusion may come from the (opaque) definition of continuous functions on weird sets. In your example on the integers, we have a function g:Z->Z with g(x) = x^2. The usual topology on the integers is the discrete topology; that is, ever subset of Z is open. By the definition of topological continuity, it is quite easy to see g(x) is continuous as every function from a discrete topology to an arbitrary topology is continuous (see note below)
If you haven't studied topology, sadly, this approach isn't too intuitive. The key idea, however, is that defining a function consists of three things: the domain, the co-domain, and the map. It's normally implied that the domain is the real numbers, but, for domains like Z, definitions that are formulated over the reals no longer make sense. Topology helps us deal with these cases. I hope this helped!
N.B. For two topological spaces (X,𝜏x), (Y,𝜏y), we say a function f:X→Y is continuous if for every open V⊆Y, its inverse image is open; that is f^-1(V)={x∈X|f(x)∈Y}∈𝜏x. A direct consequence of this definition is that every function from a discrete topology to an arbitrary topology is continuous. As the discrete topology is simply 𝜏x=P(X) (the power set of X), trivially for any V we have f^-1(V)∈𝜏x as the power set contains all subsets. Thus, the claim holds.
@@pizzarickk333 Continuous always implies "within the domain". If you evaluate continuity as "it needs to be defined on all real numbers in I" you haven't restricted the domain.
Its almost like the function is in a "super position" of sorts?
Wait!! That last sentence of the video felt like mike drop. Why dies the function become continuous when we restrict the domain to just the rational or just the irrational? Wouldn't this be a line with infinity many holes in it?!
Okay, but what's the Fourier transform of this function?
With the proof you just gave, could you say irrational numbers and rational numbers alternate in a certain way? Or not?
This is incredible. I have so many questions. If you can find a rational number between any two irritational numbers, and an irritational between any two rational numbers, doesn't that imply that there are an equal number of rational and irrational numbers, since you cannot have a continuous sequence of either? Then why are there infinitely more irrational numbers than rational numbers? The function is interesting, but the two small proofs used to construct it are way more interesting. They imply a lot of crazy things.
Can you make a follow up video covering a section of the number line, and why there must be more irrational numbers? Or does the proof only work for the entire number line, rather than a finite section? If that's the case, infinities are even more confusing.
There are more irrational numbers between any two irrational numbers than there are rational numbers between those two (or any other two) irrational numbers.
3:55 you forgot the ) after irrational
Hi is the Weierstrass function related to this subject?
So between any two points in the real numberline, there are infinitely many rational numbers, and infinitely many irrational numbers for each of those rational numbers. Yet between any two irrational number there exists a rational number inbetween them.
I guess the pigeon hole theorem doesn’t really hold once we have infinite holes.
2:02 not differentiable anywhere?
Weierstrass moment
I guess the merely countably infinite (actually constructivistically "existing") subset of the irrationals (that are computable by a necessarily contably infinite set of all combinatorically possible different algorithms) are sufficient here.
So is this compatible with constructivistic math?
I'm not sure.
I learned about this function in college.
That is actually so fucking genius
But what about Fourier transformation of this function?
Fourier transforms involve integrals and the rational numbers have measure zero over the real numbers. The Fourier transformation would just be zero.
afaik the point this was invented to prove is that it cannot have one.
@@farfa2937it has one if you define Fourier transform in a modern way by Lebesgue integral. Fourier transform of this function is zero since Q has Lebesgue measure 0
Wait! So does the dirichlet function even have a fourier expansion? My instinct is that it can't as it is everywhere discontinuous, and so it can't have a derivative anywhere. It also doesn't satisfy an alternate sufficiency for f-series by not being of bounded variation (Wik, '72).
Unless I am mistaken, your instinct is backed up by a theorem! Isn’t it wonderful when that happens.
If f is differentiable at a point x=c, it is continuous there.
Contrapositively, if f is discontinuous at x=c, it cannot be differentiable there.
You need Integrability for the Fourier expansion, not differentiability. But yeah, it's not Riemann-integrable. However, the Lebesgue integral of this function on any interval is 0, and it will also be the case after multiplying by a sine or cosine, so the Fourier series will have all zero coefficients and will converge to 0 (which is equal to the Dirichlet function almost everywhere, so it's not actually that bad).
In fourier theory you can ignore an arbitrary measure zero set and get the same answer. In this context, the rationals are countably infinite and so have measure zero, so we can choose to just ignore the rationals. If we do that, the function is exactly zero everywhere that remains, so the fourier transform is exactly zero everywhere.
@@timseguine2 Aren't fourier transforms reversible, though? Seems like this destroys the original function. (asking, not arguing)
@@xizar0rg Reversible in the function space L1. And the Dirichlet function is equivalent to zero in the L1 norm. As far as the L1 norm is concerned the Dirichlet function is the zero function. One way of interpreting it is that the dirichlet function has zero content at any finite frequency. Which the video hinted at.
Oh my god, I understood that...
One thing that bothers me a bit. Aren't irrational numbers suppose to be denser than rational ones since they are alef_1?
If there is at least one rational number between two irrational and one irrational between two rational it sounds like there is the same density and same amount of numbers which cannot be due to uncountable/countable Infinity
There is not just one between them; there is a coutably infinite rationals and uncountably many irrationals.
It's like there is always an integer between two numbers with difference > 1, and also a real between them, but there are more reals between.
They do have the same density. Nothing requires sets of different cardinalities to have different densities.
Even if the irrational numbers are more numerous, I'd be inclined to think that irrationals and rationals have the same *density*
@@canaDavid1You cannot have more irrationals than rationals, if you can always find a rational in between any two irrationals and vice versa.
Saying there is now more of one than the other is... Well... Logically wrong.
We can say there are more reals than integers, because there is an infinite amount of reals in between any two integers.
While I understand each of them in isolation, it makes my brain hurt to try to reconcile that a) between any two rationals there is an irrational and between any two irrationals there is a rational, and b) that rationals are countably infinite and irrationals are uncountably infinite. It feels like A should imply they have the same cardinality, even though i know that it doesn't!
You can systematize a mathematical way to represent all rationals, you can't with irrationals.
@@SioxerNikita That's just another way of saying that the rationals are countably infinite and irrationals are uncountably infinite, it doesn't really help with building an understanding or intuition for why.
@@SioxerNikita Could I not say, list out the irrationals by pairing them up with one of the rational numbers? Say I have irrationals a, b, c... and rationals A, B, C... defined such that a < b < c... and A is the rational between a and b, B is the rational between b and c, so on. Since the rationals are listable, would I not have all the irrationals listed out by pairing a with A, b with B, c with C, ect?
@@bebe8090 You wouldn't be able to, because irrationals are infinitely long, so by the mathematical standards of that, you can't.
@@bebe8090 Essentially, you can't list irrationals like: "I start with 0.00000...1...", because there is one that has one more zero before the 1.
While rationals, you can start systematicizing it.
Position 0,0 could be 0.0, then 0,1 would be 0.1, and so on. 1,0 could be 1.0, etc. Listable via system. You cannot do this with irrationals. You can't make a "start point".
People had a lot of free time those days.
Can quantum nonlocality/entanglement be represented by a Dirichlet function?
They’re not really directly related. Quantum entanglement comes down to interconnected probabilities, and extends far past 2 connected states (entire systems of particles can entangle together).
Also, which state is which is an arbitrary choice, which the rational/irrational inputs in the dirichlet function aren’t. (Obvious example: there are more irrational numbers than rational ones, by Cantor’s diagonal argument.) The fact we can distinguish these two means it can’t correspond to the indistinguishable entangled states.
ok but what happens with the fourier transform of that function?
If we talk about Fourier transform defines by riemman integral it doesn't exists since 1_Q is not riemman integradable so Dirichlet would probably end at that but if you define Fourier transform by Lebesgue integral Fourier series is the 0 function so Fourier series converges everywhere to 1_Q :)
Where Fourier?
I feel like with infinite precision the function would be continuous for both rational and irrational numbers
Isn’t this basically a fractal?
......so what was the fourier transform of the dirichlet function?
You said he invented it to prove a point but never said how it proves that point
Wouldn't splitting the function into two restricted functions not be continuous because there exists both rational and irrational numbers in all real numbers? For example, on the irrational restriction, wouldn't the function be discontinuous at every rational number like 0,1,2??
talking about continuity where the domain isn't an interval (or even connected) is indeed pretty weird. if we use the normal definitions of continuity for these domains, the restrictions of the dirichlet function are continuous - but so is, for example, the indicator function of x^2 < 2 on the rational numbers, which clearly has jumps
The restriction to some subset is a new function, that only sees that subset. And if on that subset the function is constant, then is continuous.
We are teaching math at high-school in the same way that mathematicians did before Cauchy: everything must be an interval, and function have always a "natural" domain. The problem is that Cauchy died in 1857
Well, I would argue, that y=Dirichlet(x) where x is from Q is dense, but it is not continuous, since Q, while dense, is only countably infinite, |Q|=|N|, i.e. does not have the cardinality of continuum, while R is uncountably infinite, thus R\Q is also uncountably infinite, |R| > |Q|, |R\Q| > |Q|, and thus y=Dirichlet(x) where x is from R\Q is really continuous subset. In other words, rational line y=1, is infinitely-times sparser than Irrational line y=0. Therefore, on average, the Dirichlet function is zero. Only occasionally, on those infinitely rare rational countable moments in the irrational continuum, is it 1.
>> "I would argue, that y=Dirichlet(x) where x is from Q is dense, but it is not continuous, since Q, while dense, is only countably infinite"
Continuity does not care about the cardinality , it only cares about the behaviour of the function in the neighbourhood of the non isolated points of the domain and the behaviour at that point.
Whenever there is a sequence that converges to a point, we can define the limit of the function at that point. Note that this point need not be in the domain. And since Q is dense in R, then every real is not an isolated point, so we can define the limit of the Dirichlet function at any real number.
This limit is determined by the behaviour of the function in the neighbourhood of the non isolated point. Since the restriction of the Dirichlet function on Q (respectively on R\Q) is constant, the behaviour of the function around any point would be constant, so any limit value would be that same constant.
Now, we recall the definition of continuity: f is continuous at a ⇔ lim(x→a) f(x) = f(a) .
In order for us to talk about the continuity at a point, that point must be both an non isolated and a point in the domain. This is already verified for the Dirichlet function restricted to Q (respectively R\Q) , all of the points of the domain are non isolated.
The function is constant, so, ∀a∈Dom(f) , f(a) = c and lim(x→a) f(x) = c , they are both equal! Thus, any constant function is continuous.
Reference: Real Analysis 26 | Limits of Functions
Just to clarify; the reference is a video on TH-cam
2:21 um... you are missing a closing parenthesis on "(irrational" and then again at 3:59.
But aren't "just the rationals" countably infinite thus not continuum in a sense
You can define continuity on discrete sets by generalizing from the real line. In fact, one can define continuity for functions from set to set, like f: R -> Q
Anyone remember when TH-cam was literally just cat videos? Yeah... I dropped out of maths at university right before the change happened.
Damn, how they become continuous by themselves?
What in banach-tarski‘s two dimensional cousin is this?
Infinite decimals does not mean necessarily that the number is irrational, take for example 1/3
After watching the video I understand the problem but not the solution.
How does the formula look like for that rational/irrational function?
How doe the Fourier of it look like?
is this infinitely zoomable and if so why?
Why is it not continuous and why does every other point need to be irrational. Can't I just make up a function where every odd measurement point is 1 and every even measurement point is 2?
Like calculating if a number is irrational? That's not needed here, only the existence and properties the rationals and irrationals. Its not continuous because if you pick any two points on the real line the function will have at least discontinuity (a switch between 0 and 1, in this case) between them (in fact it will always have infinitely many). You can have a function that is 1 on the odds and 2 and the evens, yes, though that doesn't matter here are parity is only defined for integers (whole numbers) and those are sparse in the reals.
@@alexanderf8451 I don't think your answer really helped me understand the situation any better. Perhaps let's start with a simple question.
Is this a function which you can formulate in the following matter: f(x)=a*x+b
@@benrex7775 No, the function can't be written the way you describe.
@@alexanderf8451 How is it written then?
@@benrex7775 He shows you how its written in the video.
I'm pretty sure his name's pronounced Deer-ee-cleh rather than deer-ee-shleh, at least that's how Wikipedia and my professor say his name
This makes intuitive sense to me, but wouldn't this also prove that there is an equal number of rational and irrational numbers?
you can’t, it isn’t really true, look it up
I thought there were more rationals than irrationals
Isn't the number of irrationals greater than the number of rationals (different infinities), yet it appears as if they are exactly alternating (an irrational between two rationals, and a rational between two irrationals).
Are there gaps in the continuum? ;-) Insights anyone?
The irrationals are uncountable while the rationals are countable. However that doesn't matter here only their density does and both have the property of being dense in the reals. That means given any rational (or irrational) there is no meaningful "next rational" (or "next irrational"), specifically if you chose any rational (or irrational) and then declare another value to the next one there are infinitely many counterexamples. Thus they don't alternate because to alternate you'd have to be able to pick the next number.
@@alexanderf8451It's how to explain that while you can put the numbers apparently in commuting ascending order (rational, irrational, rational,..) that you{we} don't actually have what you{we} thought you{we} had, possibly because you don't 'sort' the numbers like that to actually count the rational numbers (it may also be co-related to primes and co-primes and finding them).
They’re not “alternating”, as mentioned in the video we can’t call it “oscillating” specifically because of the fact that _there’s actually infinitely many rationals and irrationals between any two (ir)rational numbers._ It’s basically impossible for our minds to conceptualize a function like this, since infinity is weird. There’s no notion of “zooming in far enough” until you see rationals and irrationals as separate, because they just aren’t ever separate.
Personally I interpret it as two vaguely ethereal “constant (line) functions” which look solid from a distance but are actually filled with holes. Kinda like matter in the universe, I guess
@@philipoakley5498 It’s provable that you can’t exhaust the real numbers using a sequence of the form {rational, irrational, rational, …}.
@@semicolumnn the question is how you explain it...
Inadvertently there is a proof in here (or at least a basis for one) that the number of rationals and number of irrationals is actually the same, and thus they are the same size of infinity.
This would contradict other proofs that show that there are more irrationals than rationals, infinitely many more making the counts different orders of infinity.
I wonder if you could find a flaw in one of these proofs using the other, or if they can both be true (which would be weird, but infinities can sometimes just be weird like that)
Between two irrationals a,b there are countably infinitely many rationals and uncountably infinitely many irrationals. (This immediately follows from the fact that (a,b) is uncountably infinite and that Q is dense in R as demonstrated in the video). While Q is dense in R, it can still be denumerated.
@@semicolumnn Ok, but according to the Dirichlet function rationals and irrationals strictly alternate. Because they strictly alternate each rational is followed by one and only one irrational, and each irrational is followed by one and only one rational. Thus according to Dirichlet and the proof started around 2:18 there is an exact 1 to 1 relationship between rationals and irrationals.
Somewhere between the different proofs exists a contradiction, but as they are describing different facets of the number line and I don't think they directly contradict each other.
My quandary which I am not knowledgeable enough to follow through properly is if they really are in contradiction where only one could be true or if it could be solved an shown that both things are true simultaneously - that there are infinitely many more irrationals than rationals AND an exact 1 to 1 relation between rationals and irrationals
(which as I said would be weird, but sometimes infinities just do weird stuff)
@@TheFinagle A rational isnt followed by exactly one irrational. In constructing the irrational, Dr. Wood uses (a + r(b-a)) where r is an irrational number between 0 and 1. There are uncountably many irrational numbers between 0 and 1 (Because (0,1) is uncountable and Q is countable) and thus there are uncountably many irrational numbers between two rationals. The other direction is much simpler. We know that there are infinitely many rationals between any two reals, and that the rationals are only countably infinite. Thus, the number of rationals between two irrationals is countably infinite.
To recap: Between any two real numbers a and b, there exist uncountably many irrationals and countably many rationals. No paradox.
@@semicolumnn Then it should be possible to formally write that proof in a way that directly disproves that Dirichlet's function as being strictly discontinuous, ie that at some points you must have more than one irrational in a sequence because there are more irrationals than rationals. Thus forcing a continuity to exist within the function where irrationals happen in sequence.
In doing so you would break one of the 2 proofs he used to show that they strictly alternate such that the function can be discontinuous between ALL real numbers.
@@TheFinagle You can't have any real numbers in a sequence. If you could, their difference would be the smallest positive real number, but no such number exists.
The emphasis in "Dirichlet" is on "e", not on either of "i".
How can it be continuous on just the rationals if there are irrationals in between? How would such a function even be defined?
Still waiting to hear what the Fourier transform is...this was false advertising 😂
Does this have any applications for quantum mechanics and super positions?
No.
1:12 🙋♂
"Restricting it to just the rationals or just the irrationals, the function becomes continuous"...but it is not defined at every point, correct?
So if we say y=1 for x=rational, then we pick an irrational number (x = pi), then the function y becomes undefined. Therefore, the function is not continuous because it has to be defined for all x.
If you restrict it to the rationals then you can't pick an irrational number, it doesn't exist in the domain of the function.
But the restriction has a new domain, the one you are restricting on.
And if I have a function f from X to Y, I don't need to have an outside world where X lives in, for speaking about continuity.
And since our restricted function is constant on all his domain, then it is continuous.
Since, in absolute generality, a constant function g:X -> Y brween topological spaces is always continous
I clicked BECAUSE of the clever title
4:45 I think Cantor would disagree, for the set rational numbers cannot constitute a continuum. It is discrete!
It doesn't matter, euclidean metric restricted to Q induces a topology on Q and therefore we can talk about continuity
You can even talk about continuity on Z with discrete metric (but then every function is continuous so it's not as interesting). Topology, and therefore the notion of continuity, can be introduced on any nonempty set
You know that math has advanced a little since Cantor times?
Siri help
Hi
Does that mean there's an equal number of rational and irrational numbers?
No, density and cardinality are different properties of sets.
so the function is continuous but not dereivable. ¿Weierstrass?
the function is not continuous in R
No, this function is discontinuous everywhere and thus, unsurprisingly, doesn't have an derivative. Weierstrass's function is continuous everywhere but still has no derivative anywhere, which is even stranger.
what about the length of a weierstrass segment?
>> 4:38 - in fact, restricting the function to just the rationals or just the irrationals gives you a constant function
No. It gives you a function with infinite number of holes. Function does not exists in that holes or not defined. That is not a constant function.
in R, yes. but were restricting to Q, where it is a constant function. the difference is always 0 so itll always beat any epsilon regardless of the delta.
:O
This function is 0 almost everywhere, so it might as well be 0.
Imagine how useless mathematics would be if we applied your logic to everything
@@SpinDip42069 "real numbers are irrational almost everywhere so they might as well all be irrational"
what.
The engineer
@@cara-setunreal
I guess you could say Dirichlet's function is pretty much...broken
My Calculus Professor dropped this on us at the end of a Friday lecture to give something to discuss at Happy Hour.
I'm sure if the function is discovered in 1900s it would be called schrödinger function. f(x) is in superposition of both 1 and 0 until x is observed to either be rational or irrational. Upon observation f(x) collapses to either 0 or 1.
very cool link to physics here maboi
Love it. !
Probably not... unless Schrödinger were to come up with it. In reality it would probably be called the indicator function on Q, since the reluctance of mathematicians to name things after people increases with time.
No it's stupid. It is like saying f(x) = |x|/x is schrödinger function because f(x) is in superposition of both 1 and -1 until x is observed to either be negative of positive.
@@user-jn9hs5ry7h the example you’ve given seems to fit the description of Schrödinger function, your point is?