Integers & Rationals are both infinite but is it the SAME infinity?

[@jsnmad]

Excellent break down. He did not take anything for granted. Gradually building by using clear visual presentations. Concepts in mathematics can be easier understood when visualized. Nice work.

[@MyOneFiftiethOfADollar]

I wonder if Cantor had anyone he could hug when he first noticed this. What a lonely moment if not! He would shake your hand if he could on your rendering of his famous diagonal argument

[@henriquenunes7196]

History says that Cantor wrote to a friend when he first found out that R is aleph1 " I see but I dont believe....."

[@standupstandout]

Since the number 1 appears multiple times in the rational number grid (as 1/1, 2/2 etc.) have you actually shown that the set of rational numbers is smaller because there is not a strictly one to one mapping with the integers?

[@itskathan7940]

Just nailed it, Solved by countability doubt in Theory of Computation. Love from India 💕

[@RuichenZhao]

The diagram with a zigzag path is so illustrative

[@alainrogez8485]

I like also the Stern-Brocot tree. It is quite elegant.

[@half-soul8393]

Great video. However, I didn't understand how would you biject the negative and positive together..?

[@DrTrefor]

Alternate. 0,1,-1,2,-2,3,-3… that’s a list of positive and negatives

[@shayhan6227]

I don't get why that method of elimination is justified isn't the 1/1, 2/2 being the same just matter of encoding the numbers in that format?

[@rebelgames2546]

You would assign 5 to the next value the “snake” would come across which is 3/1, skipping over 2/2. You can do this every time this situation arises and you still come to same conclusion as they are still one-one.

[@dominicellis1867]

You can generate an infinite set of rational numbers with a single numerator with natural number p. You can then create a similar infinite set of rational numbers for each denominator q. If there is a correspondence for each combination p/q with a natural number p, then why does this fail when going from rationals to reals?

[@MisterrLi]

Yes, you can show that the set of Rationals is listable. It's like taking the full set of Natural numbers, and add more natural numbers to it. Sure, all the positions of the Natural numbers are filled up, but you can always find more empty positions by moving values around. For example, you can relabel all Natural numbers to the position plus 1 to fit in one extra number, and move numbers to position times 2 to fit in an infinity of new numbers, and so on. You could also pair up the Natural numbers with the same natural number values in the Rational number set. That way you can see that the actual infinite number of rational numbers is bigger, since it contains infinitely many values that are not found in the set of natural numbers. This method works when the two sets have the same elements, so that these can be paired off, and then you simply compare the rest of the elements. For Rationals, they then get a different number of infinite elements, about N^2, depending on the definition of them, and if you remove numbers with similar values.

[@starfishsystems]

It's amazing that after so much elaboration you still got it wrong. The natural numbers, the integers, and the rationals, are all COUNTABLY infinite sets. They are the SAME size. Go back and think about this some more. en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

[@MisterrLi]

@@starfishsystems Well, you could say that they are in the same "size group", just as all different infinite sets are all bigger than all finite sets. It all comes down to how much information you have, and the precision you allow in the system you use. If you allow for more precision than just "more than finite" you could see that you can allow for size differences between sets of countable infinite elements, if you know that identical elements can be paired off. And you do here.

[@aenimosity7]

Great video, you're a great teacher. Thanks!

[@DrTrefor]

I appreciate that!

[@BlaqueT]

This was an excellent video. I struggled with this proof for so long, but you made it clear in a very short amount of time. I do have a question though, can you map from the naturals to the negative integers, since the naturals have no negative elements by definition?

[@DrTrefor]

Indeed! The function f(x)=-x works.

[@NURULISLAM-bq7br]

Excellent explanation. Thank you

[@eppssilon]

ngl, it almost feels like our math is flawed if we consider them the same size, one is an infinite line and the other is an infinite square that can be zigzagged into a line Especially in my case, I am trying to find a function from Q (rationals) to N (naturals or positive integers) and it needs to be injective, meaning each element from the starting group Q should have a unique image in the landing group N, no two elements should share the same image, (if f(a) = f(b), it should mean that a = b, because no two have the same image), if we consider the two groups the same size, we should be able to assign them in this way, but I just can't seem to visualize it at all (abstractly or graphically), the explanation of the zigzag is pretty cool to visualize that its countable or "listable", but there is no real logic beyond visualization, there is no way you could tell me what's the Nth number is unless you count by hand, which yeah, isn't really practical if you ask me maybe I am missing something but this feels like a paradox and that there is no such function

[@MyOneFiftiethOfADollar]

The zigzag or alternate diagonal graphical way does place the rational numbers into a 1-1 correspondence with the naturals making it “countable” even with redundancy. I have seen the function you are interested in in a real analysis text somewhere and will let you know if I find it. I understand there are many who doubt the validity of “proof by picture” but you can be sure that Cantor noticed the zig zag in that grid of numbers before he produced an injection/bijection if he ever did ?

[@shortpianocovers713]

You are such a great teacher 😊.Thank you so much...❤

[@henriquenunes7196]

Dear Dr. Bazett. Topologically speaking the interior of Q (rationals - an aleph0 set) is empty. Lots of people tell me that the interior of R\Q is also an empty set. I believe that the interior of R\Q is at least an aleph0 set. I hope this is not a Godel/Cohen issue, but it just may.......could you share some thoughts please? Thanks very much.

[@CrazyGo3]

this video saved my life

[@rryan916]

Anyway I could reach out to you to show you a method I came up with that would suggest all real numbers are countable? It’s based off this model with some tweaks!

[@JavedAlam24]

What is it?

[@jingyiwang5113]

Thank you so much for this amazing and patient explanation! It is really helpful to me!

[@03abdulmateen99]

I think irrationals have bigger 'size' than integers

[@DarkKnightLives]

Wow!! Just Wow!! Eye Opening!!

[@imad9948]

thank you so much, sir

[@3attaaroo]

WONDERFUL.. many THANKS

[@swarnendumaity6363]

I think irrational no. sets is bigger than rational no. because we can never line them up...there are infinite no. of digits in a single irrational no. itself.

[@TheCanvaStudio]

What about negative rational numbers??

[@DrTrefor]

This can be done similarly, same pattern but just do the negative of each number before moving on

[@TheCanvaStudio]

Okay thanks @@DrTrefor

[@jimmyandtheband9849]

You are such an inspiring teacher. Thank you for your passion and time to break things down!! This was eye-opening!!! Love the vids!!! :D

[@SachinAggarwal-p3g]

Well thats not the case that every positive infinite set has the same size,as in case of set of real number we cannot map it with integers..

[@ShinyLP]

7:13 Bruh you just spent 7 mins proving that this is not true

[@jairoselin5119]

Answer for the final question is No.

[@ilovetiananmen]

Genius!!

[@Creationweek]

How can you just throw away 2/2? That's something thats always bothered me about this method. You don't hit every number exactly 1 time some numbers get hit more then 1 time. This in my mind feels like a cheat and that should suggest the rational numbers are smaller then countable numbers.

[@ValoriYT]

You don’t technically need to show that it maps “perfectly” both ways. I would highly recommend reading chapters 1.1-1.3 (the first three chapters) of “Bartle and Sherbert: Introduction to Real Analysis”. There is a nice proof in 1.3 that shows why this is OK to do :)

[@ayeshaafzal2716]

Is the same order we will find 22/7 in this table of rational numbers which is equal to pi an irrational number ......so how this method is true... Plzzzz tell me

[@ayeshaafzal2716]

@@DrTrefor ok i got it thank you 😊

[@MyOneFiftiethOfADollar]

22/7 does not equal pi. 22/7 is a rational approximation of pi. 355/113 even closer I think.

[@LinaLina-gr2br]

🧡

[@alphainfinitum3445]

This along with all other proofs of countable sets work only because of ONE REASON; we humans don't have a grasp of infinity. It feels intuitively dishonest that all the natural numbers can be put into 1-1 correspondence with all the natural numbers, and at the same time all the even numbers can be put into 1-1 correspondence with the same natural numbers. I strongly think that all those proofs rely or lean on our ignorance of the infinite, because they don't work for finite sets.(which we humans have a firm understanding of). It's the same reason that Hilbert's hotel works, simply because infinity is a black box, or the magician's trick of choice. It's like anything you don't understand, you can just pass it through that black magic box, and it works for an incomprehensible reason, but the results line up with our logic.

[@stevencraeynest7729]

I agree

[@goman9998]

You forgot zero✌️

[@cameronmyron5776]

And negatives

[@MyOneFiftiethOfADollar]

OK, make the bottom row 0/1 , 0/2, 0/3, 0/4, ...... :)

[@willjackson5885]

Literally makes no sense, for every integer there’s an infinite amount of rational numbers (between 0-1, 1-2…) So how can there be the same amount?

[@Just_Jude100]

That's what I said. He even says there are more rational numbers than integers. But YET.... I don't get that.

[@christam7847]

The notion countability has been disproved. 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 used to index all fractions (including those of the first column). In short, there is a permutation such that the X's of the first column XOOOO... XOOOO... XOOOO... XOOOO... XOOOO... ... after exchanging them with the O's cover all matrix positions. But this is obviously impossible.

[@أحمدالدسوقي-ت9س]

Hello professor, your videos are great and helpful but please accept my critique. I don't know about other people but I find your speech to be of high volume and really quick which makes not enjoying when I am listening. Try to make it more soothing and peaceful. Other than that, your are really great.

[@farmerjohn6526]

cantors argument is flawed....once you create the infinite set...its by definition bijective with natural numbers..the diagonal exists too. the compliment of the diagonal existed too. if you rearrange the real numbers then there is no contradictions that occur with the original complimented diagonal line. however there is a contradiction with the compliment of the new diagonal...however its a different number..so basically the impossible or contradicting numbers include all real numbers. that is illogical..thus the original argument is faulty.

[@farmerjohn6526]

both sizes are infinite...so there is no such thing as a size...that contradicts the meaning of infinite.

[@leemoore9527]

What a waste of time