I am really grateful for your patient and detailed explanation about this knowledge point. I have been puzzled with this point for such a long time and I finally understand it! Thanks for your help!
when you define e as the set of positive odd integers you wrote "for some integer n" but it should be "for some positive integer n" or " for some natural number n"
Thanks a lot for this amazing video, it was really helpful! But ummm may you please explain how can i prove that the set of all even numbers (positive and negative) is countable? I mean i know it's countable but i couldn't find the right function...
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.
@@kateyepawtch I think the function is supposed to be f(n) = 2n -1. I think that the set of natural numbers actually starts from 1 and not zero...that's how I was taught anyways. The set he uses and the function he uses would make sense if the set of natural numbers was {0,1,2,3,.......}.
@@djvanschaik The function is correctly defined on N (set of positive integers, natural numbers). Also, whenever you define the function you should prove that's correctly defined, which he probably forgot to do. Values of function belong to set of E, but the function is not surjective on E. There is an element "1" which you cant get as a "2n+1", for any n (positive integer). So, even though the function is correctly defined it is not bijection and therefore it's not valid as a proof of cardinality. The function f(n) = 2n -1 is bijection and that's the function we needed.
A lot of learning packed in to this video! Bijection, cardinality and aleph naught. Love it!
man you make me love math. i struggle a lot, but your enthusiasam and love for it is rubbing off on me too. thanks
I am really grateful for your patient and detailed explanation about this knowledge point. I have been puzzled with this point for such a long time and I finally understand it! Thanks for your help!
Wonderful, clear, concise explanation!!! Thank you!
Beautiful video and great voice! Thanks for posting this!
Thank you!!! I am so happy to hear that this video helped someone:) Very cool! Made my day!!!!
sir
its just a lesson that cleared the concept of cardinality
Great!
when you define e as the set of positive odd integers you wrote "for some integer n" but it should be "for some positive integer n" or " for some natural number n"
Excellent video!
It's such a clear and on point explanation. Thank you so much!!!
You're very welcome!
Got it sir 😊 thanks for your vedio. I learned a lot
Welcome!
thank you always!
Thanks a lot for this amazing video, it was really helpful! But ummm may you please explain how can i prove that the set of all even numbers (positive and negative) is countable? I mean i know it's countable but i couldn't find the right function...
I think u can put (-1)^n in front so that u switch between positive and negative int as the value of n rises. Let me know if this helps.
Sir I am from India .. thanks 😊 sir
how do you get 1 from 2n + 1?
Little n is in Z so you can take n = 0
@@TheMathSorcerer thank you
@@lemyul you are welcome! I should make more of these:)
@@TheMathSorcerer yes. proofs are fun and you explain really well
This is a mistake he made. The function should be 2n-1 otherwise 1 will not be an image of any n in N and then f is not onto.
What a king!
Great video
Thanks
what is the cardinality of set {0,{0},{0,{0}}}
Any idea
3 elements , so card is 3
Thank you
thank you for this
beautiful
thank you Aleph null (naught) times sir!
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.
thanks
You're welcome!
Just totally lost u from red pen
the function is incorrect
explain?
@@kateyepawtch I think the function is supposed to be f(n) = 2n -1. I think that the set of natural numbers actually starts from 1 and not zero...that's how I was taught anyways. The set he uses and the function he uses would make sense if the set of natural numbers was {0,1,2,3,.......}.
@@djvanschaik The function is correctly defined on N (set of positive integers, natural numbers). Also, whenever you define the function you should prove that's correctly defined, which he probably forgot to do. Values of function belong to set of E, but the function is not surjective on E. There is an element "1" which you cant get as a "2n+1", for any n (positive integer). So, even though the function is correctly defined it is not bijection and therefore it's not valid as a proof of cardinality. The function f(n) = 2n -1 is bijection and that's the function we needed.
Word salad without content.