This is curious but remember, that function is discontinuous in the real set, so the intuition of what is continuos isn't wrong. And this show that irrational set it's bigger than rational set.
@MetaMaths4 жыл бұрын
I agree, but I wanted more people to learn about this function so I made a catchy title.
@particleonazock22463 жыл бұрын
I need to learn math to this level to appreciate popcorn in its full form.
@monalsingh68144 жыл бұрын
You really made it so simple... Thank you!!!!!!!
@radkakovarovicova65494 жыл бұрын
Hello. Can I ask you if you have also proof that this function is discontinuos at rational points? Thank you.
@MetaMaths4 жыл бұрын
Hi, this one is quite easy. In any neighbourhood around your rational point you will always have irrational points (irrationals are dense). So the difference of functional values can't be bounded by an arbitrary small number, so you can' t have continuity !
@jonarmani86544 жыл бұрын
I dig this explanation, it was helpful. Thanks!
@ruochen16584 жыл бұрын
Why is the domain [0,1]? x can be any number right?
@MetaMaths4 жыл бұрын
yes, it can be any number. I just chose to talk about [0,1] for convenience.
@krumpy82598 ай бұрын
Could the following reasoning considered to be a proof why the function is Discontinuous for rational numbers. Given that f(1)=1 and knowing that x/x+1 approaches 1 for x approaching +infinity and for the sake of it let p=x and q=x+1, don‘t we have f(x/x+1) ->0 for x->+infinity and it follows we have a jump at 1.
@SP-qi8urАй бұрын
Could u try to phrase it more formally?
@R.B.4 жыл бұрын
In integers, two (or more) consecutive numbers make sense, in that 2 and 4 adjoin 3 on both sides. Your definition of rationals makes sense as p/q where p and q are co-prime integers. And the reals are the set of irrational and rational numbers combined. I've also seen it stated that the set of irrational numbers is greater than the set of rationals. Effectively the set of rationals is countable while the irrational numbers are uncountable. This function has me thinking though. For any two irrational numbers infinitesimally close to one another, is it not possible to find a rational number which is greater than the lower bound and less than the upper bound? Wouldn't this suggest that for every irrational pair there is (at least) one rational number between them? Wouldn't this therefore mean that the set of irrational numbers is countable infinity and therefore the set of real numbers is countable infinity?
@MetaMaths4 жыл бұрын
You are right in saying that there will always be at least one rational between any two irrationals. In fact, there will always be a countably infinity of such rationals. But how do you jump to a conclusion about countability of irrationals ? In strict terms, you need a bijection with natural numbers (or rational) to prove countability. What bijection do you propose ?
@R.B.4 жыл бұрын
@@MetaMaths I'm not sure. It's a question beyond any formal education I've had, so it's a question which is probably not well formalized. I need to do more research. I can see how there could be an infinite number of rationals between two irrationals, so this seems like it's effectively the same as the set of reals between 0 and 1. Is it possible that it's actually uncountable for the same reason Cantor's Set is uncountable? As for what you'd use to bijecture, that was based on the false premise that there wouldn't be an infinite number of rationals between any two irrational numbers, but why wouldn't there be? If you can generate a rational number between two irrational numbers, there's no reason you can't find a rational number between that new rational and one of the irrationals it is between. Ad infinitum.
@tonaxysam3 жыл бұрын
@@R.B. Two things that I want you to consider: 1. The Cantor's Set is uncountable; meaning, there is NOT a way to match the Natural numbers (0, 1, 2, ...) with all the elements of the Cantor's Set such that: Is a one to one matching, meaning unique pairs, one of the natural is not going to be matched with two or more of the Cantor's set, and viseversa; and all the elements of the Cantor's set have one pair. 2. There is actually a countable infinity (meaning, the same amount of elements as the Natural numbers) amount of racional between any two numbers... But that doesn't have anything to do with the fact that the irrationals are uncountable infinity. You can think of this like so: Imagine that there is a countable infinity amount of irrational numbers; every single one of them can be written as A + 0.abcde... Where A is the integer part of the irrational, and 0.abcde... is the decimal part; a, b, c, d and e are the digits of the number. Now, i'm not going to proof this thing; but, all the real numbers, and all the numbers in [0, 1] have the same amount of elements. So let's imagine that there is a list of all the irrational numbers between 0 and 1 (a list is another way of thinking about the match between the Natural Numbers and the irrationals): 1. 0.10100100010... 2. 0.14156813783... 3. 0. 3191502857... ... Even if you use all the infinitely many Natural Numbers on your list; I can show you that there exists some irrational number that is not in your list, and with that, that there is an infinite amount of irrational numbers that are not on your list: If x and y are irrational numbers, then: x = y if and only if they have the same decimals on the same place, ej. 0.10100100010... is not equal to 0.20100100010... Because of that different decimal. Now with the previous thing, I will create a number that is not in the list: 0. We look at the first decimal of the first irrational number 0.10100100010..., and we add 1 to it 0.2 We look at the second decimal of the second irrational number 0.14156813783..., and we add 1 to it 0.25 We look at the third decimal of the third irrational number 0. 3191502857... and if it is nine, we take 1 from it: 0.258 Continuinng like this, we will create some irrational number, let's call it x x = 0.258... The big question: Is this number on our list? Well it can't be the first irrational number, because it's first decimal is different Well it can't be the second irrational number, because it's second decimal is different Well... So even if we used all the infinitely many natural numbers, there is still one irrational number that doesn't have a pair... You could think, it doesn't matter, we can create a new list and add x to it, right? 1. 0.258... 2. 0.10100100010... 3. 0.14156813783... 4. 0. 3191502857... ... Nice, but now I will create an irrational number that is not in our new list... Can you guess how? Therefore, we can conclude that, yes there is an infinitely many amount of irrational numbers between 0 and 1... But there are A lot more than all the rational numbers in the real numbers. So lot in fact that it's even hard to compare to something, but you can think of it like this: Compare 1 apple to a countable Infinity amount of apple's (meaning, you have as many apples as Natural Numbers it can be) That little 1 apple is basicaly nothing compare to the Infinity amount of apple's. Well that little 1 apple is our infinity amount of apple's compare to all the UNcountable infinity amount of apple's (one apple of every irrational number). Is basicaly nothing :v I do recomend to you this videos if you want to keep searching on the topic: The Cantor Set: kzbin.info/www/bejne/m4TKoJqgjs-Sq9k Banach-Tarski Paradox (a.k.a from 1 sphere you can create 2 spheres exactly equal to the first, same size, same areas, same everything): kzbin.info/www/bejne/qWmZXo1jeMeUfqM But... From where the Natural numbers came in the first place?: kzbin.info/www/bejne/lWiqhmV7o96lb6c That's all I got uwu
@yoavmor90022 жыл бұрын
The functions you suggested isn't one-to one, many irrational pairs would share an arbitrarily chosen rational number that is in the regions of both pairs
@schweinmachtbree10132 жыл бұрын
The whole message presented in this video is just plain wrong, because there is a misunderstanding of what a continuous function is! (which is very ironic considering the title). 'Continuous function' means an _everywhere-continuous_ function, so there is nothing wrong with the intuition that "a continuous function can be graphed (on each connected component of its domain) without lifting ones pen from the paper" - this video wrongly supposes that this intuition would continue to apply to not-everywhere-continuous functions, i.e. discontinuous functions! (and it also confuses being continuous at a point and being locally continuous at a point). If you're going to make educational videos about analysis then you really need to make sure you understand what you're talking about, because otherwise you're just going to end up causing your viewers even more confusion about one of the most technical and difficult-to-grapple-with branches of math.
@MetaMaths2 жыл бұрын
Fully agree. I went for a broader title here. How else could you advertise Thomae’ s function so that more people learn it ?
@schweinmachtbree10132 жыл бұрын
@@MetaMaths You could title the video something like "Fractal popcorn (Thomae's function) - a function which is continuous almost everywhere but discontinuous everywhere in between" :)
@toniokettner48214 жыл бұрын
this guy assumed that everyone confused continuity with differentiability. i doubt that most people made this mistake
@tonaxysam3 жыл бұрын
No, he meant continuity on a point; think about the |x| function, it is not diferentiable at x = 0 but you can draw it without lifting your pen of the paper; but this function is continuous at x = 1 / e, for example; and it can't be drawn without lifting your pen of the paper; and if you are thinking about a function continuous everywhere/differenciable nowhere (meaning, some Weierstrass function) even if that function is like a fractal; it can be drawn without lifting your pen of the paper.... However, here we have a function continuous that can't be drawn like that... So no, is not directly related to differentiability; because there exists functions that can be drawn without lifting your pen of the paper that aren't differenctiable.
@toniokettner48213 жыл бұрын
@@tonaxysam that has nothing to do with my comment