R is uncountable

[@Polarspy]

Awesome video! This is the first time I've seen the argument explained in such a clear way, great work

[@tamasvarhegyi8813]

It is a pity that Cantor's Diagonal Argument was just laid to rest for good as shown in the video : I designed 2 superb algorithms which incredibly establish 1-to-1 correspondence between positive floats and positive integers. For details please access the KZbin video “Pairing Floats and Integers” at kzbin.info/www/bejne/qpi5gKV5pMRorac You may see the step-by-step unfolding of one of the most exciting discovery, which up until now was believed to be mathematically impossible. It is amazing that a single action of digit-reversal made it possible to construct all possible floats using a pair of integers (Whole and Fractional). I predict that my discovery will withstand the combined challenges of the best set-theory experts and will earn its place in mathematics history. Let’s hope that as a result, the understanding, teaching and the relevant literature of this very important subject will considerably improve. Please leave any feedback in the COMMENT section of the video. or email to Tamas Varhegyi at [e-mail]

[@windowsforvista]

Sia's number - I thought you were talking about some interesting mathematical constant I never heard of, and then when you said it I started cracking up. Good job Dr :)

[@AbhijeetManhas]

Saw 10 videos, understood only from this one

[@geetRH]

Even I also ✌️understood by this vedio only

[@m23_vishnupreethi65]

So true

[@PrincessEev]

FINALLY! An explanation that makes sense! Today was the first time I was finally able to connect the dots and understand the proof!

[@frankkkmate4267]

this video is 1000% better than my lecturer's explanation, A HUGE Thanks for your fabulous work!!!

[@larissacollins9797]

wowwww thank you king. I learned this in lecture and it made zero sense and I watched tons of videos but this one makes the most sense to me. thank you so much. also your energy is amazing and the 13 minutes flew by :) you're killing it!

[@drpeyam]

Thanks so much!!!

[@mateocamargo7437]

I love how you have so much fun with this. Great video, I strive to be like you

[@Zonnymaka]

Sweet! Just one remark, i would have stressed more the concept of numerability, 'cause that's what real numbers lack of...and it makes them different from the rational numbers. So we start defining infinite from the "numerable" one up.

[@kelvinwong6786]

Here is the thank you and respect from a Hong Kong student. I respect your passion! I had never watch a teacher who can still be so exciting to solve a math that they already know the answer and had done so many time...

[@drpeyam]

Thank you so much!!!! 😁😁

[@shaunderoza2321]

You should have then showed (0,1) has the same cardinality as R using tan. That way with a careful construction using tan you could have snuck in another pi n joke.

[@drpeyam]

Shaun De Roza Hahaha, I should do that, maybe in a new video ;)

[@shaunderoza2321]

I can imagine this becoming a hardcore harmonic analysis channel just so you can get in you pi n and pi m jokes.

[@CliffStamp]

Super clear explanation of why the constructed number isn't in the original set.

[@grasshopperweb]

I still don't fully get it. By building a similar Cantor Diagonalization using q_1 = n_1 n_2 n_3 ... / d_1 d_2 d_3, and building a new q using the diagonals you can create new numbers. Wouldn't that mean that rational numbers are uncountable? It just really feels like a "I couldn't find the solution, so it must not exist" type non-rigorous proof to me.

[@MikeRosoftJH]

If you apply the diagonal proof to a sequence of rational numbers, then you indeed get a number not in the sequence; but if the sequence contains all rationals - and such a sequence exists - then the resulting number cannot be rational. Let's state the proof more formally, without depending on the popular visualization: * Let f be any function from natural numbers to real numbers (in other words, any infinite sequence of real numbers). * Let d(n) be, for all natural numbers, the n-th digit after the decimal point of f(n). For lack of ambiguity: if f(n) is a number with two different decimal expansions, like 0.1000...=0.0999..., then use the expansion ending with an infinite string of digits 0 rather than 9. * Let d'(n)=d(n)+5 mod 10. (In other words, d'(n) is obtained by taking the n-th digit of the n-th number in the sequence and adding 5 to it, wrapping around zero if necessary.) * Let D be a real number whose decimal expansion is d'. (The integer part of D is 0.) * By construction, the decimal expansion of D is different from the decimal expansion of f(n) for any n - the two numbers differ in the n-th digit; and the two numbers can't be equal. It can also be seen that f(n) differs from D by at least 3*10^-n. * Therefore, for no natural number n is f(n)=D. * Therefore - because I have made no assumptions about the function f - the same is true for all functions from natural numbers to real numbers. QED.

[@salmanel-farsi3744]

Very nice, easy to understand proof.. I saw another video that constructed a new number by adding 2 mod 10 to each the diagonal, but I like this proof better (more intuitive) where you can just choose any number other than 9. History tells us that Cantor was treated very poorly by other mathematicians on his work of infinite sets. This simply demonstrates that even the most logically minding humans (relying on a priori) act stupidly to other human beings simply based on belief that what you say or think is nonsense without proof.

[@rodllewellyn]

As a computer scientist, I tend to think in binary (lol!). So we can redo Cantor's argument in binary. Now we have real numbers like .00110010... etc. Our diagonalization is the same, except we flip every bit (0 => 1; 1 => 0). Naturally this does carry the risk of generating .11111... = 1, if all the diagonal digits are zero (like the 9 problem with the decimal version of the proof), but I think this is a minor issue (a match will only occur if 1 is one of the chosen real numbers). A nice thing about the binary method is it's now obvious that the cardinality of the reals is the power set of the integers.

[@piccolaradge8333]

The “Sia number” love it 😂👌

[@anthonym2499]

question: are we listing all the possible combinations of the placement of the digits without regard to their overall value? Or do we need to eliminate any duplicate values that have multiple ways of being represented (you mention if we want to be rigorous we would eliminate duplications when touting the rationals)? Also, do we have to use a base10 representation of the decimal expansions; or, is a base2 (binary) representation just as valid? For example: in base10, does 0.29999.... equal 0.3000... ? if so, do we use both representations in the list, or do we have an unused decimal expansion that can be used as the diagonalized portion? afterall, if we find a pattern that is not supposed to be on the list then we have only proven that the alteration of the diagonal of a square matrix cannot be contained within the square matrix it was derived from. Do we include the values for 0 and 1? in base10, does 0.9999... represent the same value as 1.000... If both representations are used, then we have only created a many-to-one relationship and not a bijection.

[@tomknapton1461]

Can you do the proof that the set of functions is bigger than the set of reals? I'd love to see that

[@drpeyam]

Tom Knapton I'll try to do that! It's tricky, but interesting :)

[@undercherries4659]

Could you please explain the mistake in Mueckenheim's argument? If all positive fractions can be enumerated, then the natural numbers of the first column of the matrix 1/1, 1/2, 1/3, 1/4, ... 2/1, 2/2, 2/3, 2/4, ... 3/1, 3/2, 3/3, 3/4, ... 4/1, 4/2, 4/3, 4/4, ... 5/1, 5/2, 5/3, 5/4, ... ... can be shuffled such that they cover the whole matrix. But by exchanging them with other fractions, never the whole matrix will be covered.

[@MuffinsAPlenty]

I think I responded to you asking this on another video, but I didn't heard back from you. The error is that the definition of "countable" is a "there exists" statement, not a "for all" statement. If *_there exists_* a way to enumerate the elements of a set, we call it countable. This means that there has to be _at least one_ way to enumerate the elements of the set; but it does *_not_* mean that every attempt to enumerate the elements of the set will work. I don't know who Mueckenheim is, but based on your comments, it seems that Mueckenheim's argument is as follows: If there is a way to enumerate the elements of the rationals, then there's also a way which doesn't. Let me give you an example. Let's say I tell you "There exists an even number less than 8." Then I say, "Look, 4 is an even number less than 8, so the statement is true!" Then Mueckenheim comes in and says, "No, there isn't an even number less than 8 because 10 is an even number bigger than 8." Are you convinced by Mueckenheim's argument here? Because it's based on the same reasoning as Mueckenheim's argument about countability.

[@undercherries4659]

@@MuffinsAPlenty Sorry, you have misunderstood. Mueckenheim has shown that Cantor's bijection cannot exist by scrutinizing Cantor's original method using his very formula k = (m + n - 1)(m + n - 2)/2 + m and his very sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...

[@MuffinsAPlenty]

@@undercherries4659 Yeah, I'm not understanding how this isn't what I suggested before. To me, it looks like Mueckenheim (still don't know who that is) takes Cantor's sequence, modifies it (thereby creating a _new, different_ sequence), and shows that this new different sequence isn't a bijection between the positive rationals and the positive integers. Let me update my analogy. Say that I want to claim that "there exists a positive even number less than 8." Then I say, "Look, 4 is a positive even number less than 8, so the statement is true!" Then Mueckenheim comes in and says, "No, there isn't an even number less than 8 because if you did have a positive even number less than 8, then you could add 8 to it. MuffinsAPlenty says that 4 is a positive even number less than 8, but 4+8 = 12, and 12 is greater than 8. See? I modified MuffinsAPlenty's example so that it no longer proves what he claimed." This is what the argument feels like to me.

[@undercherries4659]

@@MuffinsAPlenty Mueckenheim does not modify Cantor's sequence. He uses exactly what Cantor did. The only change is that he starts with indices which first are in the first column of the matrix, 1/1, 1/2, 1/3, 1/4, ... 2/1, 2/2, 2/3, 2/4, ... 3/1, 3/2, 3/3, 3/4, ... 4/1, 4/2, 4/3, 4/4, ... 5/1, 5/2, 5/3, 5/4, ... ... showing thereb that they are as many as are integer fractions. Then he applies the indices X precisely as Cantor did: XOOO... XOOO... XOOO... XOOO... ... The first index 1 remains with 1/1, the second index 2 is attached to 1/2 XXOO... OOOO... XOOO... XOOO... ... The third index 3 is attached to 2/1 XXOO... XOOO... OOOO... XOOO... ... an d so on, according to 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... . But it is obvious that the matrix never will be covered by X.

[@MuffinsAPlenty]

​@@undercherries4659 I don't understand the first part where you have X's along the first column. After reading this for a sixth time now, I finally think it's completely irrelevant to what Mueckenheim was saying. "But it is obvious that the matrix never will be covered by X." What do you mean it's obvious that the matrix will never be covered? This doesn't really seem like an "argument" to me. It seems more like looking at Cantor's argument and saying, "No, obviously not". Anyway, let me see if I can rephrase the argument to check if I'm understanding it? Basically, set up the rational numbers into an infinite array with row numbers corresponding to the numerators and column numbers corresponding to the denominators. Then sweep along the finite diagonals, one at a time. This obviously won't cover the entire grid. Is that right?

[@rahulkumar-qt9nj]

Sir your smile itself individually proof the result 😊

[@lemi11ion14]

This video is amazing. Good work!

[@imusun]

love the way he's teaching!

[@____KB]

7:17 I didn't catch the reason why xi cannot equal 9. Can someone explain this constraint?

[@drpeyam]

It’s because, for example, 0.299999... = 0.3, so we wouldn’t have a unique expansion

[@PyarMatKaro]

You assume that two numbers are equal only if their respective digits are equal. Consider 0.999... and 1. Maybe x (as constructed here) is in the list after all

[@drpeyam]

BitDancer I was emphasizing this point at the end! The only problem would be if you get an infinite expansion of 9's, but that's why I was being very careful about choosing digits other than 9 in my construction.

[@avelkm]

Dr. Peyam's Show Yeah, but you didn't prove that it's legible:) I mean that we clearly have example that two numbers can be equal even having different numbers. You avoided 9's but did not prove that other digits don't have this problem. And it might be tricky, can you actually show the proof of it?

[@ReCaptchaHeinz]

BitDancer Hi, this only happens when the number ends in periodic nines. Let's remember how periodic numbers are constructed, based on a substraction that removes the periodical part: X = 0,43333... 10X = 4,3333... 100X = 43,333... 100X - 10X = 39,000... 90X = 39 X = 39/90 = 13/30 In fact, if we divide 13/30 = 0,43333... Okey, now let's try with a number that ends in nines: X = 0.47999... 100X = 47,999... 1000X = 479,999... 1000X - 100X = 479,99.. - 47,99... 900X = 432 X = 432/900 = 48/100= 0,48 So 0,47999... = 0,48 And the same with all other numbers that end in 999... But not with other like 1 or 7, I encourage you to try an example with each one of the number from 1 to 8 and see that none of them give you a result X that is different from the original X. If you want to know why 9 causes problems, you have to know how the process works: Let's go with X=0,479999...: 100X = 47,9999... Now, we know that in the following substraction, the ",99999..." Is going to go away, so we define: Y = 47 And the same with 1000X: 1000X = 479,9999... The relation between 479 and 47 is: 479 = 10*47 + 9 The next step is the substraction: 1000X - 100X = 479,99...- 47,99... = 479 - 47 Or, in general (using Y): 1000X - 100X = [10Y + 9] - Y So: 900 X = 9*Y + 9 Removing the 9 as common factor: 100 X = Y +1 X = (Y+1)/100 And that number, since it's divided by 100 (or other multiples of 10), always is an exact number and it'll never have periodic decimals. And that's it. (Excuse my English and such a long answer)

[@PyarMatKaro]

It was not shown that d[i,i] does not have an infinite sequence of 9's. If you set the digit x[i] to differ from digit d[i,i] by 5 then the numbers x and d[i] have a finite difference, making x a number that is not in the list, and the proof is fixed

[@ReCaptchaHeinz]

Of course, that is another way of construct x. He chooses that construction because it's less restrictive, but your way is perfectly valid (I think).

[@benjaminsaalfeld4182]

So you wanted to avoid the 0.9999 case but isn't there the 0.0000 case also?

[@drpeyam]

That's a good point, but the 0.9999 is more problematic than the 0.0000 case. Ok, of course we need to avoid 0, because it's not in (0,1), but that's ok. The problem with 0.9999 is that, say, 0.519999 gives you a totally different number 0.520000, but something like 0.3400000 is just one number, namely the number 0.34. There isn't another way to write this, except with a bunch of nines!

[@taniyachaudhary4436]

Sir, why we use (0, 1) real no.?

[@mariannepunzalan3205]

There is a theorem that if a subset of a set is uncountable, then the whole set must be uncountable. The values from 0 to 1 is just a subset of Real Numbers. Thus, proving it is uncountable also proves that Real Numbers are uncountable.

[@garininderdeo8010]

Wouldn't it be you cannot construct a number that's missing X but if you can map another number on the list isn't a bijection because of horizonilization ? I'm so lost

[@mamadetaslimtorabally7363]

Oreo (cookie) for your thoughts - I like that !

[@Fematika]

Are you going to become an analysis tutoring channel?

[@drpeyam]

Fematika Probably something like analysis, calculus, linear algebra, and differential equations :) And bunnies and languages!

[@drpeyam]

But not tutoring, more like "OMG, look at this awesome math fact!!!"

[@aa561234]

x1 , x2 and x3 cant be 0 too

[@geromeponch6248]

best proof video ive seen

[@ryan_chew97]

Sia's number I fucking died

[@programmingcheatsheet]

this video is so well done

[@bullinmd]

Cantor's Pigeonhole Principle?

[@geetRH]

Thank you so much for great explain

[@amberheard2869]

Why can't we choose 9 btw i think it is for to be careful about 0.9999...=1 but by saying this i think 0 should also be avoided since 0.0000.... Just wondering. Very helpful video thank you for review.

[@drpeyam]

Because 0.999999999... = 1.000000000 so we would have a conflict of some sort

[@drpeyam]

No just one of them has to be avoided

[@doji-san]

What if I DON'T SUPPOSE we can list all the real number and just start counting... then you can't say it's "uncountable" because I am counting them... I'm confused..

[@Peter_Schluss-Mit-Lustig]

There will always be a miss

[@MikeRosoftJH]

The proof is often stated as a proof by contradiction: let a(n) be a sequence containing all real numbers; then it can be shown that D (the sequence's diagonal number) is not in the sequence - this is a contradiction, so the sequence cannot exist. But it can also be stated as a direct proof: let a(n) be any sequence of reals (any function from natural numbers to real numbers), then it can be shown that the sequence doesn't cover all real numbers (in particular, it doesn't cover its own diagonal number); and because I have made no assumption about the sequence a, it follows that the same is true for all sequences of reals.

[@alejorabirog1679]

Finally a good proof

[@aurian7238]

and therefore we're done

[@lenadubovaya190]

Awesome!- thank you

[@ilikemitchhedberg]

THE CHANDELIER! 🎵🎵😍

[@StarinaXie]

why x cannot equal 9？

[@tolga1503]

very nice of you

[@joyabrothers7316]

The man enoyed the fuck out of me. But he explained it clearly and now I understand so thank you very much.

[@matheusurbano7045]

Why can't x_i be equal to nine?

[@drpeyam]

0.99999 = 1.000000

[@SamuelAndradeGTutos]

Bit how could it be? Imagine that se have in this list all the real numbers. 0.100000... 0.110000... 0.234567... And these three is all the real numbers OK? Now, we take one number that have the digits different. 0.09999... Would work. It has all digits different, but it is the same 0.1000... So, how could you demonstrate that having all digits different means be different?

[@drpeyam]

That’s why I had the requirement that you don’t choose any 9’s, to make sure that you don’t have the trouble that 0.09999 = 0.1000. As long as you don’t have a string of 9’s, different digits means different numbers

[@SamuelAndradeGTutos]

How would you proof that different digits means different numbers? Because we have the "0,99..=1,00..", then how to proof that we don't have any other cases of "different digits / same value" numbers?

[@sahil-cb7qu]

Guy looks like Peter pettigrew

[@AdityaGhosh50]

Mind blowing 😁

[@tomificationness]

this helped a lot.

[@franchello1105]

Why can't x contain any 9s?

[@drpeyam]

Because then you have something like 0.5999999 = 0.600000, so two different decimal expansions correspond to the same number

[@jeffjo8732]

This video is no different than the standard presentation of the proof. So it makes all the same mistakes. None of them invalidate it as a proof, but they actually make it more complicated than it needs to be, and many of those complications are what make some think it is wrong. 1) The proposition was not that the reals are uncountable, it was that there is an uncountable set. 2) The example Cantor used in his previous proof of the proposition was the reals. It was not accepted because of properties he assumed for the set. 3) So the example set Cantor used here was EXPLICITLY not the real numbers. It was the set of all binary strings, T. This avoids the issue where, for the reals in [0,1] represented in base 2, 0.1000... and 0.01111... are the same number but different strings. 4) As presented here, the proof assumes a statement of the form "A and B", but derives the supposed contradiction from B alone. Then, the contradiction is "not A." This is not a proper proof by contradiction, which must show that a contradiction of a known truth follows from all of what is assumed ... 5) ... which is why Cantor never said it was a proof by contradiction. It is by contraposition: proving "If B then not A" also proves "If A then not B", so "A and B" can't be true. 6) Say S:X→T is a function that maps set X to a some, or all, of T. If X is the natural numbers N, then Cantor proved that there is an element of T that is not mapped by S. By contradiction, if S:X→T maps set X to all of T, then X is not N.

[@tushargupta5805]

who are these 25 peoples who disliked this..😱😱😱???

[@earthstick]

Good explanation, I finally got the the contradiction with the supposition. If you _could_ list all, then they would be finite, but you can always find one not on the list, that contradicts the supposition, therefore they are not finite, therefore they are infinite.

[@drpeyam]

That’s not true unfortunately. You can list all integers: 1, 2, 3, … but this list is not finite

[@earthstick]

@@drpeyam Can you not just add one to the last entry on the list to get another integer, thus they are also infinite?

[@ntapeknoxsiwale]

Thanks!

[@FarisSkt]

5:13

[@АндрейБойчук-м6ш]

Why would we need a Xi != 9?

[@drpeyam]

Because if not, then you’d have problems like 0.1999999... = 0.2, so it’s just to prevent redundancies!

[@АндрейБойчук-м6ш]

Ok then. Thanks!

[@tarunpurohit6522]

Love u professor

[@doji-san]

Wrong!!! Counting is a process, and thus you can count the real number... the new real number produced from the diagonal proof is added to the end of the list and "counted" ad infinitum.. :)

[@guilhermebiem582]

But there is no "end of the list". The assumption was that the list already contained all real numbers, therefore if you find a number that is not in the list, the assumption is wrong. Counting may be a process for a finite set of numbers, but not for an infinite set of numbers. In this case, you must assign every number a position all at once, otherwise you found yourself in the middle of an infinite task. If you are claiming that the set of Reals is countable, then I challenge you to come up with a function whose domain is all the Natural numbers and the Co-domain is the set of Reals. That is, for every real number r, there is a Natural number b such that f(n) = r

[@renunciant]

All of this hinges on a definition of counting, but is there a proof that counting is as it is assumed here? What if counting can be argued is not just a simple mapping or correspondence? How do you prove it then?

[@johnferenz747]

I'm a huge math geek but this diagonal thing and the uncountability of real numbers seems completely wrong. Counting between 0 and 1 is nothing but a mirror of the natural numbers. I will count: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.11, 0.21, 0.31........ on and on (Edit: The mirroring is required to preserve zeros, but those last two could be 0.12 and 0.13, which I had accidentally typed... but we will save those for the mirrors of 21 and 31). The mirror of 100 is .001, the mirror of 999 is .999 and the next pair is 1000 and .0001 (automatic labeling!). This will count every single possible combination of digits beneath 1 in the exact same way as the digits above 1, except mirrored. It's incredibly simple to do this, so I simply cannot fathom how these real numbers are considered uncountable. Thank you for the video, I loved it and I did enjoy your explanation!

[@MikeRosoftJH]

This will only cover real numbers with a finite decimal expansion; this is a strict subset of rational numbers. At what position does a number like 1/3, √2/2, or π-3 appear?

[@johnferenz7629]

@@MikeRosoftJH But those numbers you gave me aren't decimals. That's why you have to put 1/3 instead of a decimal!

[@MikeRosoftJH]

@@johnferenz7629 The decimal expansion of 1/3 is 0.333... (every digit at every position after the decimal point is 3). This is a theorem about real numbers; not about "decimals". Sure, there are countably many numbers with a finite decimal expansion - all such numbers are rational, and that there are countably many rationals has been known for over 100 years; but this is not what the theorem is about.

[@MikeRosoftJH]

@SubZee No, you can't. As I have said, this scheme only covers numbers with a finite decimal expansion; and that there are countably many such numbers is not a new result - all such numbers are rational.

[@kleoMan98]

This works if only you do not take into account the irrational numbers. R includes irrational numbers. Take as an example π or sqrt(2). With your way of proving that R is countable, how are you going to mirror π? It has infinate decimal expansion. So where do you start? π≈3,14159265.....and it keeps going, So what is the mirror of π?

[@carlosmonte9597]

I don't get this proof. Why can't you say that in a grid of rational numbers there is a number that is not on the list like here? Like here you say there is one different than the diagonal , it means the list is not complete not that the reals are uncountable no? Or the number you propose does not exist no?

[@drpeyam]

That’s the thing! Countable would mean that we write an infinite list of ALL the rational numbers, yet I’ve shown that there’s always a number that you’re gonna miss, so it’s literally impossible to list all of them, in contradiction to the definition of countability

[@aligator7181]

Actually, this is completely wrong. I have paired each such real with a unique integer. Try it, give me a real number, I will send you its corresponding integer.

[@drpeyam]

It’s correct but ok

[@MikeRosoftJH]

Okay, here you go: * 1/3 * √2 * π And don't tell me that these aren't real numbers, but "algorithms". Yes, they are real numbers; look up in some mathematical textbook, or on Wikipedia, how real numbers are actually defined. Likewise, don't give me the numbers that your function assigns to the truncations of the above numbers to finitely many decimal positions. That's not what I was asking. Plus, what you are doing is futile. Your function - no matter what it is - does not cover its own diagonal number.

[@olavisjo]

I will try to explain where Cantor went wrong and how it IS possible to “count” the set of irrational numbers. Cantor’s goal was to find a unique natural number for every real number and vice versa, it was not to find a list with a 1 to 1 correspondence between the naturals and reals (though this would have settled the issue, this would have been a sufficient condition but it was not a necessary condition). To get to his goal, Cantor could have used the Schröder-Bernstein theorem. Which says, for two sets to be equipotent, you need only show that there exists injective functions f : A → B and g : B → A then |A| = |B| It is easy to find a unique real number given a natural number, e.g. 12 → 12.0 or 12 → 12.14159… or generally n → n.xyz… where n is a natural number and xyz… is any string of digits. Or, we can find a real number in (0,1) for every natural number. e.g. 1230 → 0.0321 or 985356295141 → 0.141592653589 But, finding a unique natural number given a real number is not as obvious, yet it can be done. Cantor assumed, in his proof, that it was necessary to list all the natural numbers, in order and on one list. This assumption is erroneous. It is like a rich man going to a small country auction, if he bids all his money ($1 million) on the first item, then he will be able to buy only that one item, but if he bids in smaller increments, he will be able to buy all the items up for bids. Cantor could have paired his purported list of ‘ALL’ real numbers in (0,1) to a small subset of the natural numbers like this. 4 → 0.5123453… 44 → 0.5125674… 444 → 0.5124195… 4444 → 0.5123676… 44444 → 0.5127295… … Then when he finds, by diagonalization, that he has not listed all the real numbers we could simply say “so what, we have not listed all the natural numbers either”. But we have satisfied our condition of finding some unique natural number for each real number in the list and we still have plenty of natural numbers left over to pair with any real number that may turn up. So, Cantors proof is thus inconclusive. We can then ask Cantor to take the diagonals of the elements in his list, change the digits and make a second infinite list of real numbers that were not on his first list. And we can then pair those real numbers to some other subset of the natural numbers like this. 14 → r1 144 → r2 1444 → r3 14444 → r4 144444 → r5 … Again, we can ask Cantor to take the diagonals of the elements in his second list, change the digits and make a new infinite list of real numbers, that were not on this second list (being careful to not duplicate any numbers from his first list). And we can then pair those real numbers to yet another subset of the natural numbers like this. 24 → rr1 244 → rr2 2444 → rr3 24444 → rr4 244444 → rr5 … And, naturally, we can keep this game going forever. The interesting thing about this is, that when the prefix of our natural numbers approaches infinity, then it will become harder to find a real number, by changing the digits of the diagonal numbers, that is not on one of the infinite quantity of lists of real numbers. Also notice that most of the natural numbers like 333, 777, 123789 etc do not appear anywhere in our pairings, but that is okay, since we have enough natural numbers to pair to our sets of irrational numbers without them. All this tells us is that it is possible that the set natural numbers is bigger than that of the set of real numbers in (0,1) or that they are equal in size. Now that we have cast doubt on Cantors diagonal proof, we can show how it is possible to find a function that does pair up the natural numbers to the irrational numbers. (the rational numbers can be paired, so we will ignore them for now) We will generate a set of random irrational numbers between 0 and 1, our list might look like this. 0.5123453… 0.5125674… 0.5124195… 0.5127676… 0.5124295… 0.05127295… … Then we can imagine a deck of cards where the first card has the number 1 on it, the second card has a 2, third a 3 and so on, a stack with all the natural numbers in order. The first number on our random list is 0.5123453… so we can pair it with the fifth card in our deck. The second number on our list is 0.5125674… since we have used the 5 card, we must now use card 15. (the first two digits in reverse) The third number on our list is 0.5124195… since we have used the 5 and 15 card, we must now use card 215. (the first three digits in reverse) The fourth number on our list is 0.5127676… since we have used the 5, 15 and 215 card, we must now use card 7215. (the first four digits in reverse) The fifth number on our list is 0.5124295… since we have used the 5, 15 and 215 card, we must now use card 4215. (the first four digits in reverse) And so on. So we have our pairing. 5 → 0.5123453… 15 → 0.5125674… 215 → 0.5124195… 7215 → 0.5127676… 4215 → 0.5124295… 50 → 0.05127295… … If you ponder on it for a while, you will see that there will never be an irrational number that we can’t find a natural number to pair with it. We could sort the natural numbers and keep the irrational that it is paired with, together with it, and then look at the nth digit of the nth real number that it is associated with. For example, 1 → 0.14159265359798… 2 → 0.23606797749979… 3 → 0.3166247903554… 4 → 0.414213562373095… 5 → 0.5123453… … Diagonal = 0.13624… then change the diagonal digits to something else and we would get an irrational number that is not in our list of irrational numbers. Anti-Diagonal = 0.34333… Since the number that we create is different from every number in our list of irrational numbers, it must be different at some finite point. (it is impossible for two irrational numbers to have the first infinite number of digits in common, then to be followed by some digits that are different.) Since this diagonal defines a finite string of digits, that has not been seen on the right side of our set, then we know that same finite string does not appear on the left side either, since the left side is simply a reflection of the digits that have appeared on the right side. So we know that there exists a natural number equivalent to that string (or a substring of that string) that will be available to pair with the irrational diagonal number that introduced that finite string. Therefore, the cardinality of the two sets is the same.

[@JohnRBromm]

The problem with the diagonal proof is that it does not work with infinite sets. If it did, you could prove the natural numbers are uncountable. Starting at 0 and adding 1 to each digit, you get 1. for the second number (1), adding one to its second digit, you get 11. For 2, you get 111. For 3 you get 1111 and so on. Carrying on indefinitely, your resulting number would be an infinitely repeating ...111. I guess that means either the integers are uncountable or ...111 is not an integer. No matter how many digits out you go, there are still an infinite number of digits to check. It can easily be seen that the set of all the real numbers from zero to one could be represented by the mirror reflection of the non-negative integers on the decimal point. Thus the first would be .0, then .1, then .2 and .3 and .4 and so on. The number corresponding to the integer 10 would be .01. The number corresponding to 5973 would be .3795. Applying the diagonal to this set, the first digit would be .1 then .11 then .111 and so on resulting in a number that would be .11111...., but that would be the real number that would correspond to the integer ....11111. Obviously a contradiction which would disprove the assumption that the diagonal works for an infinite series. Now you could construct a set of numbers with the first ten in the sequence being .0 through .9. For the next group of numbers to be added to the set, take the elements you currently have and add the numbers 1 through 9 to each of them giving you 1.0, 1.1, 1.2, 1.3, ... 4.0, 4.1, 4.2, .... 9.7, 9.8, and 9.9. The next group to be added to the set would be created by adding .01, .02, .03, .04, .05, .06, .07, .08, and .09 to each number already in the set. The add another group of elements to the set by adding 10, 20, 30, 40, 50, 60, 70, 80, and 90 to each element in the set. Continuing in this way adding numbers to the set by alternately adding digits to the high end and low of the numbers already in the set, you get a countable set of all he real numbers. You can then create a set of all imaginary numbers by taking the previous set and multiplying each element by the square root of -1 (i). This set is also countable. By combining this set with the set of reals we did in the previous paragraph, alternately taking an element from each set, we have a countable set of the real and imaginary numbers together. Does this pretty much do away with all the alephs except aleph null?

[@MuffinsAPlenty]

"I guess that means either the integers are uncountable or ...111 is not an integer." You are correct, and the reason is the second possibility you listed here: ...111 is not an integer. Each integer has only finitely many (nonzero) digits. This follows from the construction of the natural numbers. In non-technical terms, the construction of the natural numbers starts with 0 (or with 1) and adds 1 to get 2, then adds 1 to get 3, then adds 1 to get 4, etc. and every natural number is constructed in this way by adding 1 a finite number of times. (This may seem circular, but that's because I'm being non-technical). One of the properties of the natural numbers is that it satisfies mathematical induction. Therefore, if you can prove a property holds for 1, and if you can prove that property also holds from n+1 whenever you assume it holds for an arbitrary positive integer n, then it must hold for every natural number. We can use induction to prove that every natural number has finitely many digits. Certainly 1 does. Now, suppose n is a natural number with finitely many digits. How many digits does n+1 have? Either it has the same number of digits as n (finitely many) or has 1 more digit than n (still finitely many). Therefore, all natural numbers have finitely many (nonzero) digits. "Now you could construct a set of numbers with the first ten in the sequence being .0 through .9. For the next group of numbers to be added to the set, take the elements you currently have and add the numbers 1 through 9 to each of them giving you 1.0, 1.1, 1.2, 1.3, ... 4.0, 4.1, 4.2, .... 9.7, 9.8, and 9.9. The next group to be added to the set would be created by adding .01, .02, .03, .04, .05, .06, .07, .08, and .09 to each number already in the set. The add another group of elements to the set by adding 10, 20, 30, 40, 50, 60, 70, 80, and 90 to each element in the set. Continuing in this way adding numbers to the set by alternately adding digits to the high end and low of the numbers already in the set, you get a countable set of all he real numbers." While integers can only have finitely many (nonzero) digits, real numbers _can_ have infinitely many nonzero digits, extending to the right (after the decimal point). Your method here will miss all such real numbers, since no real number with infinitely many digits will appear in any position on your list.

[@JohnRBromm]

@@MuffinsAPlenty By integers only having a finite number of digits, does that invalidate the idea of the hotel manager with the infinite number of rooms, each occupied, then needing to find rooms for another infinite number of guests, he moves the current guests into the even number rooms leaving the odd number rooms available for the new guests? My problem with the diagonal argument is how you can get to the end of the unending list of numbers, each with an unending number of digits, Consider the subset of the reals from zero to one that only have one significant digit: .000.., .1000..., .2000..., .3000..., .4000.., .5000..., .6000..., .7000..., .8000..., and .9000... There are ten numbers, the base (B) to the power of the number of digits (D). To get the Cantor diagonal number, you need to go out to ten digits, a number that is not in the set we are considering, but is a real number that we will get to when we extend our set to ten digits. Extending our set to two significant digits, we get a set of one hundred numbers B^D. The Cantor number would need one hundred digits, again a number not in the set we are considering, but it will be in the set when we extend our set of numbers to one hundred digits. By extending our set to N digits, we now have B^N numbers, and the Cantor number needs B^N digits. Again this Cantor number will be in the set once we extend it to B^N digits. Continuing to N+1 digits, our set now has B^(N+1) numbers. The Cantor number now needs B^(N+1) digits. Again, this Cantor number will be in the set once we extend the number of digits to B^(N+1) digits. At each point of this exercise, the set of numbers with D digits as D goes from 1 to N and then to N+1 is countable and has all the numbers possible for the number of digits. The Cantor number is always further down the set of real numbers, but we will get to it once our set has that many number of digits, but by then our Cantor number will extend that much further down the list of real numbers. It seems to me that this would disprove the diagonal argument (maybe that is why Cantor's Argument is called an argument instead of a proof). On the other hand, there may be some subtleties to the diagonal argument that are not expressed in this video and any other demonstration of it that I have seen. I have tried to read Cantor's explanation of it, but I get lost with all his references to bijection and subjection and injection and rejection and ejection. Also, my explanation does not necessarily disprove that the reals are uncountable, since .3333.... as the number of digits aproaches to infinity is merely an approximation of one third, and one third is a real number. Whether the number of digits can reach infinity or not is a philosophical question. My view is that infinity is an unattainable limit, and thus any decimal representation of a number like one third or pi can only be an approximation that becomes more precise as it is extended to more digits.

[@vahagntumanyan1305]

@@JohnRBromm The difference is that you construct a particular function and then show that it is not an injection, i.e. you write down a particular ordering and show that by that ordering there are numbers that we don't take into account. Diagonalization argument claims that for ANY ordering there is a number that isn't in that set. If what you were saying was true then all rational numbers in (0,1) would also be uncountable since we can create an ordering such as 1 -> 0.1 2 -> 0.02 3 -> 0.003 . . . 45 ->0.0(x45)45 this way we cover all natural numbers but not all rational numbers. Does this mean that rational numbers have a different cardinality? Absolutely not. It just means that this particular function is not an injective. The diagonalization argument does not delve into particulars. it just says "whatever" your counting method is, I will find a number that isn't in your ordering.

[@JohnRBromm]

@@vahagntumanyan1305 I don't understand your response. My simple argument is that you cannot take a list of an infinite number of numbers each with an infinite number of digits and change one digit in each of these numbers. By being infinite, there is no end to the list, so there is no way to get to the end. Thus you can not change one digit from each of the numbers. This in no way invalidates the possibility that there may be uncountable sets or there may be different orders of infinity. It just means that the Diagonal Argument is fundamentally flawed as it is usually presented. If there is another way it can be presented so I can understand it and accept it, I would sure like to see it. This has been haunting me since the mid 1960s when my high school algebra teacher showed it to the class, and it was brought up again when I was studying for my BS in Math. None of the instructors I have been able to talk to about the Diagonal Argument has been able to explain it beyond what you did in this video. Here is an infinite list of numbers, dot, dot, dot, each with an infinite number of digits, dot, dot, dot... Now create a new number where the first digit is different than the first digit of the first number and the second digit is different than the second digit of the second number, dot, dot, dot, and you come up with a number that isn't there. It reminds me of the Math class where the instructor writes a problem on the board, steps back and looks at it, and then writes 27 as the answer. One of the students raises his hand and says he doesn't understand how the instructor got that answer. The instructor erases the 27 from the board, steps back, looks at the problem again, and then goes back to the board and again writes 27. The same student again raises his hand and says he still doesn't understand how the instructor came to that answer. The instructor says, "OK," again erases the 27 from the board, steps back and looks at the problem again, and again goes back to the board and writes 27. The student again raises his hand and says he still doesn't understand how the instructor came to that answer. The instructor turns to the student with a look of frustration and says, "Come on, I've solved it for you three different ways."

[@vahagntumanyan1305]

@@JohnRBromm Then I misunderstood your problem with the proof. The problem you seem to have is with visualizing infinities themselves. I am no math professor, so I may be out of my depth trying to "explain" this and making assumptions on which parts you seem to be uncertain about (sorry for that). If I am getting you right this time you would also have problems with a statement such as "The sum of an infinite series is EXACTLY equal to its limit", because after all the sum only tends to that number, and you can never actually reach it. (sorry for ranting) Think of the diagonalization argument not as finding a particular "diagonal number", but showing a process that if continued infinitely would produce such a number. Suppose the set of rational numbers IS NOT in fact infinite. Then you obviously can diagonalize it and find a number that isn't in that set right? Now add that number into the set, and repeat this process infinitely, so regardless of how large the set is a diagonal number will always exist, and by doing this process infinitely you will end up with actual infinity.

[@vettypasanga9955]

Who doesn't know that real numbers are uncountable

[@wildtiga07]

1.) You know how to teach. Thank you. 2.) That is one giant chalkboard eraser lmao

[@briancabana6380]

As far as I can see, this does not prove that the reals are uncountable. It just proves that whatever system you used to count them is inadequate to count them. Similarly, I could count the rationals with the denominator of 3, and then cite numbers that don't have a denominator of three, and claim that the rationals were uncountable.

[@MikeRosoftJH]

But if any enumeration of real numbers (infinite sequence, or function from natural numbers to real numbers) is incomplete, that's precisely what it means that real numbers are uncountable. Given any function from natural numbers to real numbers, there is some real number (and, by extension, uncountably many real numbers) which the function does not cover. (Sure, there exists an enumeration of rational numbers which is incomplete; but there exists another one which does cover all rational numbers. And that's no more surprising than that the function n->n+1 - as in Hilbert's Hotel - does not cover all natural numbers, but there exists another function - such as the identity function - on natural numbers that does. An infinite set can be mapped one-to-one with its strict superset or subset.)

[@briancabana6380]

@@MikeRosoftJH I understand this. My point here is that the proof as shown, so far as I can tell, does not show that *any* enumeration of reals is incomplete, just that one is, namely one that chooses a random real number. It's possible that one can show that the very randomness of the function means that it can stand in for "any ennumeration", but it does not strike me as obvious that this is the case, or with no further need of justification. Example, why can't I take the first diagonalization, that where n'ij = nij +1 mod 10, and call it 1d1, and then I reorder my reals to count 1d1, 1d2, 1d3, and so on up to 1 d9. Now I have accounted for the additional diagonalized numbers. I can reorder my list such that 1d0 through 1d9 (1d0 is 1 in my original list) are new elements 1-10. 1d0-2d9 are my new elements 11-20, and so on. Doubtless there are infinite permutations of diagonalizations that I can't account for this way, but doubtless does not mean proven.

[@MikeRosoftJH]

@@briancabana6380 The proof does not assume anything about the sequence. Let f(n) be any arbitrary sequence of real numbers (function from natural numbers to real numbers). Then there exists a diagonal number, which differs from every number in the sequence; given any natural number n, f(n) (the n-th number in the sequence) differs from the diagonal number in the n-th digit after the decimal point. And if you believe that the sequence could cover its own diagonal number, at which position is it? (No matter what natural number you propose, I can say: no, that's not it.) Sure, you can add the diagonal number to the sequence. You can't add it to the end, because there is no end; but you can add it to the beginning (or, if you want, to the millionth position), shifting all numbers after it one position forward. You can even do this infinitely many times, by interspersing the two sequences. You can even do *that* infinitely many times - observe that the set of all natural numbers can be mapped one-to-one with the set of all pairs of natural numbers (or, equivalently, the set of all natural numbers can be split into a countably infinite collection of countably infinite sets). (Map first sequence to the positions 1, 3, 5, ... - numbers that are not divisible by 2; second to positions 2, 6, 10, ... - numbers that are divisible by 2 but not by 4; third to positions 4, 12, 20 - numbers that are divisible by 4 but not by 8; and so on.) But what you get is still a sequence of real numbers, and the diagonal theorem applies to it (it has its own diagonal number, different from the original, which it does not cover). And that's not all: just about how many diagonal numbers are there? The video (somewhat arbitrarily) shows one possible way of constructing it for a given sequence. But it can be seen that for every position there are at least 7 different digits to choose from. (For position n, you can't choose the digit which the n-th number in the sequence already has at the same position; and you may want to avoid using the digits 0 or 9 in order to avoid a problem with numbers which have two different decimal expansions: 0.1000...=0.0999... .) This yields uncountably many diagonal numbers (you can re-interpret these choices - 7 for every decimal position - as base-7 representations of real numbers), and none of these appear in the sequence.

[@briancabana6380]

@@MikeRosoftJH I don't doubt the claim. I continue to argue that the argument *as presented* wasn't airtight. Look at all you had to say to solidify it. The claim presented was "*This* diagonal number is not on the list, therefore R is uncountable". The rebuttal to that is I can simply shift the list to include *that* diagonalization. Therefore, *proving* rather than claiming that R is uncountable requires more work than the proof *as presented*. I don't doubt there is a more rigorous formulation of this proof that solidifies it. The crux of the uncountability of R, so far as I can see, has little to do with the diagonalization argument (which I argue begs the question) and much more to do with the absence of any closed form for uncountable elements in R. You can't count these elements because you can't describe them As soon as any element is given a closed form, it can be counted, rational or irrational. Examples include the algebraic numbers and logarithms with rational bases. And as soon as you count an element, you necessarily close its form. So Dr. Peyam never counted a single "uncountable element". Hence the diagonalization argument "works" because R is uncountable, not the other way around. It only "works" because each number in the original list has no closure. If it did, any defined list (say 10X10) would yield various functions containing all variants of every element of that list, and then you'd move on to 11X11, 12X12, counting away.

[@wernerhartl2069]

x doesn’t exist because it is the last member of an endless list.

[@MikeRosoftJH]

No, it isn't. Please look up how real numbers are actually defined in mathematics. Sure, the diagonal number is a limit of an infinite sequence of real numbers. But a limit is *not* an "infinity-th" element of a sequence; there's no such thing. Again, please look up how limit is defined.

[@wernerhartl2069]

@@MikeRosoftJH Seems to me a proof of “R is uncountable” should begin with a definition of R.

[@wernerhartl2069]

@@MikeRosoftJHSuppose by chance Cantor’s construction generates the sequence: .3, .33, .333,…..This has the limit 1/3, but the sequence doesn’t end. (Assuming you could list undefined numbers in the first place.)

[@MikeRosoftJH]

@@wernerhartl2069 Let's state the proof more formally. * Let f be any function from natural numbers to real numbers (in other words, any infinite sequence of real numbers). * Let d(n) be, for all natural numbers, the n-th digit after the decimal point of f(n). For lack of ambiguity: if f(n) is a number with two different decimal expansions, like 0.1000...=0.0999..., then use the expansion ending with an infinite string of digits 0 rather than 9. * Let d'(n)=d(n)+5 mod 10. (In other words, d'(n) is obtained by taking the n-th digit of the n-th number in the sequence and adding 5 to it, wrapping around zero if necessary.) * Let D be a real number whose decimal expansion is d'. (The integer part of D is 0.) * By construction, the decimal expansion of D is different from the decimal expansion of f(n) for any n - the two numbers differ in the n-th digit; and the two numbers can't be equal. It can also be seen that f(n) differs from D by at least 3*10^-n. * Therefore, for no natural number n is f(n)=D. * Therefore - because I have made no assumptions about the function f - the same is true for all functions from natural numbers to real numbers. QED. Observe that I am not constructing the decimal expansion in steps - I define it as a whole. And that the various sequences (functions whose domain is the set of all natural numbers) exist follows from set theory axioms, namely the axiom of powerset and axiom of separation. You wanted to know what is a limit. Okay: let a(n) be a sequence of real numbers. Suppose that there exists a real number L with the following property: given any (arbitrarily small) positive real number ε, there exists some natural number n, so that for no m>n does a(n) differ from L by more than ε. Then L is the limit of sequence a(n). (Observe that a sequence can have only one limit; for the contrary, let ε=|L1-L2|/4.) Now it can be seen that the sequence 0.3, 0.33, 0.333, ... has limit 1/3. And - as I have told you - the value of the decimal expansion 0.333... (real number whose all digits after the decimal point are 3) *is* the limit. It's not any of the elements of the sequence, let alone the "infinity-th" element - there's no such thing. Finally, what are real numbers? You can look it up in any math textbook, or on Wikipedia. A full formal definition is slightly technical; but the essence is the following: It's a mathematical structure which is closed on the operation of addition, subtraction, multiplication, and division by non-zero values (where the operations work as you would expect from them), which is ordered (again, the ordering relation < has the expected properties), and which has the property of supremum: Let A be any set of real numbers which is non-empty and bounded from above (there exists some real number l such that no element of A is greater than l). Then there exists the minimum number L with the same property: 1) it is an upper bound in the previously defined sense; and 2) no number less than L is an upper bound. We then call the number L the supremum of the set. From this it follows that every sequence which is monotonous and bounded has a limit, which is equal to its supremum. It's also not a priori given that real numbers have a decimal expansion. To the contrary, that every real number has some decimal expansion, and every decimal expansion defines a real number, is something that would need to be proven.

[@wernerhartl2069]

@@MikeRosoftJH You have to show that an “infinite” sequence of digits is a real number. Step 1: Define “infinite.” Step 2: Show that they are a Dedekind-complete ordered field.

[@varalakshmi2824]

Very bad , I had have the idea about it . You did not try to say something new .

[@MiroslavMakaveli]

Number theory is so boring...with all these numbers

[@MiroslavMakaveli]

It's not, sorry.

[@tamasvarhegyi8813]

Hi, Is this a joke or what. At the end of the video he attempts to prove that 0.99999... equals 1. He does that by first claiming that 0.99999.... = 1- 0.00000...1 which of course he does not prove. Then he states without proof that 0.00000...1 equals 0. An attempt to prove one claim by substituting 2 unproven claims in its place is called kicking the can down the road. Of course the main claim of the video about the uncountability of floating point numbers is also wrong for one inconvenient fact. They are countable in the sense that there exists a one-to-one mapping between them and positive integers.