Comments by "eggynack" (@eggynack) on "TED-Ed" channel.

  1. There is no next free room. All the rooms are full at the beginning.
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  2. The second room was not empty. It started out with the second person inside of it. However, that room was rendered empty by the fact that he moved to the third room. The third room was not initially empty either, but the aforementioned process was used to empty it, and so on.
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  3. In spite of the fact that the hotel is infinite, all of the rooms are full. There is no higher room to put them in.
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  4. Line the infinite guests up and label each their position in line. Then, put guest one in room one, guest two in room two, and so on. What empty room is there after that?
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  5. It's alright. I'm not the biggest fan because it leaves infinitely many empty rooms. My preferred mapping is that the guest in bus m seat n goes to the nth numbered room where the mth prime is the least prime factor. The two caveats here are that the hotel counts as bus one and that the mapping of bus one is downshifted such that guest one stays in room one. This method fills every single room with exactly one guest.
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  6. There is no top. The hotel is infinite.
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  7.  @blearkob  There is no empty room. They all full.
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  8. So, if I'm getting you right here, then you're asking what's the minimum area you could enclose with an infinite quantity of fence, assuming you can't just use a single straight line. You need some caveats to even work with this in the first place. The biggest, naturally, that such a fence can't exactly be said to enclose anything, because one of the sides will be at infinity and thus not exist. That being said, we can define the area of this object as the space where you can draw a line from one piece of fence to another and go through that space, and assume for the sake of argument that the shape is convex. The straightforward answer here is infinite area. You can make the area as narrow as you want, but an instantiation of this situation will inevitably have all of that area. A more complex question is what the lower bound on the set of possible areas is. After all, you can keep reducing the area of the fence, regardless of its size. I'm inclined to say that the lower bound is infinity, because all elements of the set are greater than any finite quantity, rather than zero, because you can keep reducing the size. Notably, if you have a given finite quantity of fence, all possible areas would be positive, and the lower bound would be zero.
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  9. There is no room infinity, let alone a room infinity+1. Each room is a natural number.
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  10. Five is also a human concept, and yet I doubt you'd call people using it needlessly overcomplicated. I don't even know what your second issue is. There is no unanswerable question involved in this result. An infinite hotel absolutely can accommodate infinite guests.
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  11. Well, if only one person enters the hotel, then all our guest is doing is moving from one room to another room, and that takes a very small finite length of time. If finitely many people enter the hotel at separate times, then our guest has to do that short duration thing that many times. If infinite people enter the hotel simultaneously, then our guest might have to move to an incredibly far away room, but it is still a single move to a place finitely distant. Basically, with any of the stated scenarios, the guest will always be moving only for finite lengths of time.
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  12. No. For the sake of argument, let's say that there is one guest in the hotel for every even number. There would be infinitely many guests. Does this preclude the existence of guests assigned to the odd numbers?
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  13. Where, precisely, are you assigning each person from each bus? If I name a bus and a seat number, you should be able to identify where they're going. So, third bus fifth seat. What's their room number in your proposed system?
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  14. There is no latest room available. All the rooms are full.
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  15.  @negativepunk9638  I don't really get your question here. The problem, as stated, is that you have an infinite hotel where every room is full. Thus, there can be no available latest room. Were the hotel not full, then putting guests in an empty room would be possible. But it is, so it ain't.
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  16. A hotel with infinite rooms and infinite guests doesn't have to be full. For example, we could put a guest in every even room, leaving every odd room free. In any case, the hotel does start out full. It is by moving every guest up a room that a free room is produced. As for countable infinity, you are misunderstanding what that means. A countably infinite set is one where we can put the elements of that set in some order such that, counting by that order, we can reach any given element of the set in finite time. The natural numbers are the canonical countable infinity, and both the hotel and guests begin labelled by the natural numbers. Meanwhile, just because there are infinite guests in the hotel doesn't mean there can't be infinite guests outside the hotel. Say, for the sake of argument, that the guests in the hotel are associated with the natural numbers, and the guests outside the hotel are associated with negative numbers. Or, returning to my first point, maybe the guests in the hotel are even numbers and the guests outside are odd. It is, in fact, possible to have infinitely many countably infinite sets that are wholly disjoint from themselves and each other.
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  17. If every even room of an infinite hotel is full, and no odd rooms, how many guests, exactly, are there?
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  18. Infinity/2 is still infinity. That's the point. The set of all even numbers is just as much infinity as is the set of all integers, so you can trivially fill only part of the hotel with infinite guests. What, precisely, is stopping me from only putting guests in even rooms? You can't just say, "Because that's impossible." There has to be a specific reason. I don't know what you mean with a condition to apply to infinity. I'm not stipulating conditions. I'm just putting people in some rooms and not others. You're defining properties of infinity that really aren't a thing.
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  19. It's very much possible to half fill in spite of infinity/2 equaling infinity. It's just that the number of guests when the hotel is half full and the number of guests when the hotel is completely full are equal. We haven't set any boundaries here. We're just putting guests into some set of rooms, and then counting those guests. Let's reframe the problem a bit. How many elements are in the set of all even numbers?
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  20. And yet, the set of numbers between one and two would not fit in the hotel
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  21.  @MikeRosoftJH  Yeah, I think that's a caveat that showed up in the original proof, the one about numbers being written in a particular form.
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  22. "Uncomputable numbers", also known as transcendental numbers, are real numbers. They are contained within that set, and therefore are accounted for by simply saying, "All real numbers." The reals are uncountably infinite, and the video said they were using the reals, so they were using an uncountably infinite set. As a side note, e^pi is actually transcendental. So, it cannot be finitely computed.
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  23. It means that every room has a person inside it. By moving guests around, it is possible to make the hotel be not full.
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  24. Simply put, not every existing guest is already in. Only infinitely many guests are required to fill the hotel, not all possible guests. For the sake of argument, let's say that every person in existence has some unique number assigned to them, and that the people inside the hotel are the natural numbers. Well, we could still add people by using the negative number people. Notably, the natural numbers contain infinitely many non-overlapping infinite sets.
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  25. An infinite hotel can absolutely be full if you have infinite people. Consider taking all the hotel's guests and lining them up. Give each guest a numeric label equivalent to their position in line. Now, put person one into room one, person two into room two, person three into room three, and so on. If you name any room, it is filled with a guest I can identify immediately by label. Thus, all rooms are full, and so the whole hotel is full.
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  26. Still a resolvable problem. Let's assume that the guests can reach any given room in any desired finite length of time, and that they do indeed need to enter one at a time (which, honestly, not sure why that'd be the case. The lining up thing was just to assign rooms, not a method of entrance). The first person reaches their room in a second, the second person reaches theirs in half a second, the third person in a quarter second, the fourth person in an eighth second, and, generally speaking, the people reach their room in 1/2(n-1) seconds, where n is their place in line. After two seconds, every person will be inside their room.
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  27. The person in room one is able to go to room two because room two is unoccupied because the person who was in room two went to room three. Each room is being made temporarily empty by the fact that the person who was in it moved up a room. Except for the first room, which is made permanently empty, because no one from beneath it is entering, allowing for a new guest.
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  28. There is no last room. The hotel is infinite.
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  29. There is no "room n+1". Room n+1 refers to a room one higher relative to any particular guest. So, for the person in room 4, room n+1 is room 5. Room 5 starts out full though, and is thus not a valid place to put the new guest.
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  30. Simply put, for each room put a person in it.
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  31. Foreign language usage is pretty typical for various things. Cause, y'know, it's distinct from the regular letters you're using. Greek letters are also pretty common.
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  32. There's an infinite amount of rooms, but there's also an infinite amount of people, one present in every room. Someone, at some point, has to move somewhere, in order to admit a new guest. And given that the person moving has to move to one of the infinite full rooms, you're going to have to have infinite moves of some variety.
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  33. You can. You just can't empty a single bus before moving on. A good method is the abacaba pattern. Every time a number is listed, that's the bus you take a guest from: 121312141213121... If you continue this pattern, you hit every single number arbitrarily many times. Alternatively, you can just have everyone go to their rooms simultaneously. Have the algorithm be posted at the front of the bus, and everyone just heads to their mathematically determined room.
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  34. You're the one implicitly numbering the seats in the setup. You can just personally label the first seat of each bus one, or, if you want to have a seat zero, then you can modify the setup with that in mind.
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  35. It's quite possible for an infinite hotel to be full if you have infinite people in it.
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  36. A never-ending hotel absolutely can be filled with never-ending guests. Consider that you wouldn't likely raise an objection to the idea that every room has a bed, a bathroom, or walls. Well, it is equivalent to say that every room has a person.
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  37. Infinity/infinity isn't "mathematically impossible". It's just an indeterminate. This means that it can take on basically any value. One of those possible values is one, which represents the situation where there is a single person in each room.
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  38.  @tomasantunes6789  It is full in the sense that every room has a person in it.
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  39. That doesn't really generate a clean mapping though. For example, where does the guest that was originally in room five go?
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  40. Consider that you wouldn't take issue with the rooms having beds, bathrooms, or floors. It's equivalent to say that the rooms have people.
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  41. Sure they can. You just need infinite people.
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  42. There's no division involved. Put all the people in a line, and label each person with a number based on their position in the line. Send person one to room one, send person two to room two, send person three to room three, and so on. In this fashion, for each room there will exist one person occupying that room. The impossibility of division here doesn't imply that it's impossible to assign people to rooms such that this specific thing can happen. What it implies, if it implies anything, is that you can assign people to rooms such that anything you want can happen. There are assignments that put infinite people in each room, assignments that give infinite rooms to each person, assignments that leave infinite rooms empty, and assignments that leave any finite quantity of rooms you desire empty.
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  43. Again, you're trying to abstract out something that need not be abstracted out. I'm not subtracting the number of people from the number of rooms, getting a value, and using that value to determine how full the hotel is. I'm telling you exactly where every guest is going, and then telling you that, once every guest has reached their room, every room has a guest in it. I could also tell you where every guest is going and have exactly one room not have a guest in it. Not every guest mapping scheme will have this hotel filling quality. But, in order for the hotel to be full, there need only be one such mapping.
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  44. Eh, as long as we're working in a weird infinite space, I don't see much reason to not make these people capable of reaching their room in any finite time. So, person one takes a second to reach their room, person two takes half a second, person three takes a quarter of a second, then an eighth, and so on. After two seconds, every room would be filled.
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  45. They do need time. Two seconds. Ask yourself this. We've lined up all the people, assigned them rooms, and then had each go to their room in the timespan I stated. Which room is empty after three seconds? If you can't name a single room that is empty (or, rather, if such a room cannot be named), then no room is empty, which means all the rooms are full.
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  46. Where'd all the posts and such go?
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  47. That's silly. Infinite doesn't mean all. Take out all the occupants of even numbered rooms and have them stand outside and you have infinite people inside the hotel and infinite people outside. In point of fact, it is possible to find infinitely many completely disjoint infinite sets of people inside the hotel. Such is the way of infinity.
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  48. If each of the infinite buses has a countable infinity of people, then both my method and the video's method will successfully house any arbitrary person in any arbitrary bus after finite time without any room ending up with more than one guest. My method will also not have any room with less than one guest. I don't know where all this stats stuff could possibly fit in. There's no randomness involved in this process. Every person, even when infinite people are present, is assigned to a very specific room. Name any room and I can tell you exactly which guest my method will assign that guest to, and, name any passenger, and I can tell you where my method assigns that passenger. The video only accounts for some of the rooms, so for that mapping I can tell you whether a room has a guest, and, if it does, where that guest came from. Room 128, in both mappings, contains exactly one guest. In the video's mapping, it has the guest that was originally in room 7. In mine, assuming you assign the guest in room one to room one, it has the guest that was originally in room 65.
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  49. Yep.
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  50. Because there are already guests in those rooms. You can toss the new guests into whatever room you want, but whereever you put them it's going to create some infinite chain of person movement. Any solution is necessarily going to require infinite movement of some sort, or else we'd be forced to conclude the patently absurd result that this claim also holds true for finitetly many rooms.
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