Dope Tech Episode 2

<<@DanDart says : Yeah I knew the answer was going to be "basically everywhere that you couldn't tell what was going on">> <<@DanDart says : I'm feeling very Banach-Tarski right about now>> <<@joshhickman77 says : What point in A maps to p-1 (which is in S)?>> <<@dinkusstinkus4396 says : I refuse to believe this can't be visualized, if vsauce and veritasium can visualize the banach tarski paradox anything is possible>> <<@publiconions6313 says : Wouldn't the constant zero be in S, then subtract 1 and you get -1 which is not in S by definition?>> <<@amaarquadri says : Here's another set that you can take a subset of, rotate that by 1 radian, and get the original set back: e^(ni) for integers n ≥ 0. The set is a bunch of points on the unit circle. If you remove the first point which is at 1, then rotating everything clockwise by 1 radian makes all the other points take each other's places except for e^i which goes to 1. And we know that the chain goes on forever since there are 2*pi radians in a circle which is irrational.>> <<@5dollarfootlongcat says : Yeah when you said to write down any set of points that’s what I wrote down.>> <<@apowergso says : “Don’t think about this visually” Me, an aphantasic: I was born for this very moment>> <<@EvanG529 says : This is just a rehashing of the infinite hotel paradox, which has debatable validity>> <<@pi4313 says : I want to see the set!!!>> <<@NoIce33 says : I now have a migraine attack for half an hour after all this white on black. But interesting stuff noneteheless. Infinity is a biatch and inevitably leads to paradoxes, but we cannot do without.>> <<@rockapedra1130 says : Infinity is more trouble than it's worth.>> <<@edwardmacnab354 says : just like the quadrant where you remove the partition A and move all of infinity left , you must choose the turf carefully where this will work . So in fact you've done a bait and switch , because the subject of infinities is still not well understood , and since there are different types of infinities , you must be very careful when working with them !>> <<@unchaynd7266 says : Zach: doesn't matter, just pick a subset of the x-y plane. Me: *picks a single point* Video: 0:27 Me: 😧>> <<@TheBirthdayBros1 says : ooh ive heard of this one no i havent yes i have eeh maybe not>> <<@DavideDainoRaghnar3PCericola says : I don't agree with the example of 3:49. I understand the idea, but for many reasons I think that aleph_0 -1 is not aleph_0 (I don't axept the continuum hypotesis).>> <<@jhwheuer says : Infinities are evil>> <<@elpachanga says : I was expecting the axiom of choice but I guess that is only necessary for bounded sets>> <<@donchonealyotheoneal5456 says : I'm sorry were you speaking English but really it's pretty amazing what you can do with a graph I still don't understand it>> <<@ArionEquus says : Im not sure this is a paradox as much as it is something that just works because your selection of points is just vast and complex enough that you cover all possible places where the points could fall when you start doing the math.>> <<@tylergust8881 says : 4:10 I might be missing something here... But if S is everything outside the unit circle (or in other words, everything that is at least 1 unit away from the origin). Can't B just be a ring surrounding set S? (Like everything at least 1 unit away from the origin, but less than 2 units away from the origin) This seems too simple so I must be missing something...>> <<@bobstud3754 says : How about solving the San Francisco, Portland, Chicago, New York n Los Angeles Paradox? I will wait...>> <<@bobbob-jy5li says : Nope, got bored trying to follow this>> <<@mmenjic says : 3:56 but original S was equal to A + B where B was not contained in A and now it is so judging by that original S is not equal to new S ?>> <<@blitzar8443 says : I don't get it>> <<@D.E.P.-J. says : Too many ads.>> <<@Peaserist says : The fact that both A and S are infinite sets kind of ruins the illusion of the paradox. The same way an infinite stack of $1 bills and an infinite stack of $100 bills are worth the same, it's not exactly fair to imply that just because A is a subset of S, that A is necessarily "Smaller" than S.>> <<@totalmikie says : An infinite amount of P...>> <<@musescore1983 says : Cool video and great explanation!>> <<@TmOnlineMapper says : That's the fun part of playing with infinity. Quite a few basic conservation of properties principles just don't apply anymore.>> <<@climatebabes says : This paradox is bogus, as are the solutions. If you take the suface of S to be right of the y axis, and you remove B from it, it is not the same size. So you imagine the part of B taken off to be added, which doesn't really ever happen. Infinites are not valid they are an artifcat of mathematical process. Then the second proof is based on numbers with an imaginary component. That's one dimension, not two as it is shown as, that's just a visual aid. The problem is defined in 2 dimensions. So its not a solution to the problem. A lot of math is the result of wrong application of its methods. There is no infinity in reality, so math proofs based on it are bogus.>> <<@grandexandi says : Oh... I needed it to be visual>> <<@MrA6060 says : nah imaginary numbers aren't even real i'm falling for this one>> <<@Pseudo___ says : my set was all pairs of integers, and my subset was the pairs whos sums were even.>> <<@guepardo_6308 says : I think Zach made a mistake. All points which the constant is -1 (example: p²-1) aren't contained in A-1, because the number that when subtracted 1 gives -1 is 0 (0-1=-1) and no points in A have 0 for their constants, so, A-1≠S (example: p²-1 isn't in A-1)>> <<@TechnocratiK says : I feel as though this is, with only slight modification, entirely equivalent to Banach-Tarski if instead of using polynomials over the complex numbers we use polynomials over the unit quaternions.>> <<@forgottenetremembered says : Infinity is weird>> <<@michaeldowney2940 says : Why can you stretch A to become set S, yet rotate B and not stretch?>> <<@toniokettner4821 says : naming e^i = p and then looking at polynomials over p is a war crime>> <<@roiburshtein852 says : Wow>> <<@s1lv3rfir3 says : just make your set the set of everything. make the one that rotates a circle on the origin. then make the one that moves everything else. boom done solved. obviously this is not true, so where did I go wrong? I can't find my mistake?>> <<@wdobni says : is it a real paradox or an imaginary paradox ..... is it even possible to have an imaginary paradox>> <<@adrianmuntean8944 says : doesn't the b image, with complex numbers, when rotated, also scale at the same time?>> <<@wyboo2019 says : i finally finished the problem and the trick that i was missing seems so obvious in hindsight. to anyone else that can't figure it out: once you have one value that is definitely in A, you can figure out values in B, which are then in the union of A and B, which are then also members of A translated left by 1 and B rotated by 1 radian, which then yield new values in A and B, and etc. then its not hard to prove via induction that A and B contain the polynomials in p as described in the video. i'm not a huge fan of complex numbers and prefer a geometric algebra approach to these types of problems so i'm going to do it again tomorrow using vectors and rotors from Geometric Algebra>> <<@pulverizedpeanuts says : it's actually pronounced sherpinski mazurkyevich>> <<@micketm3 says : ...🤔 so is this... synchronicity ?>> <<@alexanderying1558 says : How about the S = entire complex plane (without 0 if it has to be proper), B = complex numbers with rational real and imaginary parts, A = everything else?>> <<@gazzaka says : Is there a reason you used B ^ S , like BS ? !>> <<@tan_k says : My brain is throwing a 404 error>> <<@nicnic1190 says : I can't pay attention long enough for this right now>>