Very important: If there is anything about what I say that you are not sure about please ask. There are a lot of very knowledgable people roaming the comments section that you help out :) There is a very nice TEDed video about the finite version of the last of our puzzles: kzfaq.info/get/bejne/hJumfbZ_vbapp3U.html The solutions to both the infinite and the finite version are closely related. I also mention that those sets that get pushed around in the Banach-Tarski paradox are constructed using the Axiom of Choice. Vsauce in his otherwise brilliant video on the Banach-Tarski paradox actually does not mention this although this is really a big deal as far as mathematics is concerned (understandable though since his video was already very long). Here is a link to the spot in the Vsauce video where the Axiom of Choice is envoked (although you have to have a really close look to see how :) kzfaq.info/get/bejne/qZ5mYL1eqMjFeHU.htmlm2s

Want to know an anagram of banach tarski? Banach tarski banach tarski

This explanation is bogus because the assumption that Batman can be killed is incorrect.

Today - 21-012-2017 - Mathologer referenced VSauce, and VSauce referenced 3Blue1Brown. For the love of order and purity in the universe, I sincerely hope 3B1B releases a video today referencing Mathologer.

Do you really think infinite Assassins could kill Batman? You would need at least twice as many!

"it's simple. We Achilles the Batman."

It's the perfect murder because of biggest army politics. You have the biggest army with infinite assassins therefore you make the laws

Famous quote: “The Axiom of Choice is obviously true, the Well-ordering theorem is obviously false; and who can tell about Zorn’s Lemma?" Jerry Bona I like Zorn's Lemma - Algebra is so much more beautiful with it than without.

A problem may arise when each assassin realises that there is no possibility he's going to kill Batman, and can as such go home so as to not risk being caught later.

The problem is that acording to the Law of Conservation of Ninjutsu, there is always a finite amount of fighting prowess that is evenly spread out across a group of Ninjas (or assassins if you will). So your army of infinitely many assassins are each infinitesimally effective. So Batman will just defeat them all.

If one of the assassins makes a wrong choice, those after him will still know their number because they will hear him getting executed for being wrong

I'm pretty sure he would see a pile of assassin's larger than the universe.

the Mathassin's Creed: {thing} ≠ true; permit {ξ}

The second assassin whose number differs from the series knows that the last guy's number is different (too), and since the first one said there was an even number of differences, the second assassin with the wrong number knows he has to have his number wrong too, cause he doesn't see any difference with all the other assassins and the series. I know I didn't explain it very well, sorry, but it's just how it came out of my mouth... well, my fingers.

I think the axiom of choice, like almost anything else in mathematics, is a tool. It behaves in certain ways, it implies certain things about what it governs, and when you're doing math it pays to be conscious of when you are using it and to ponder what would change if you didn't.

You'll just end up with an infinite amount of knocked up and tied up assasins... because BATMAN!!!

The question is, of course, if Batman with infinite prep time could defeat an infinite number of infalible assassins.

It's not the perfect murder because an infinite number of assassins has an infinite amount of mass. So the assassins, the planet they're standing on, and Batman collapses into a black hole. From the point of view of an outside observer we know Batman can't escape the event horizon, so he's dead. But neither can the assassins, so they all got the death penalty as well!

I was just looking for videos about the axiom of choice earlier today; what a coincidence! :-D

I love this channel. Possibly my favorite maths channel. I hope there will be plenty of uploads in 2017 :D THANKS MATHOLOGER!

The Banach-Tarski paradox sounded SUPER COUNTER-INTUITIVE when I first heard about it, but after calmly comparing with the fact that it is possible to rearrange all points in [0,1] to make [0,2] (which is doubling the length and is SUPER TRIVIAL), I now think the only "weirdness" in this paradox is the "possible just by dividing the cube to finite sets and rearranging" part and not the "possible to double the volume" part. The latter part, I think actually is the origin of the (false) counter-intuitiveness for many non-experts, while it's actually not really the main point. After realizing this, I'm more "pro-axiom of choice" now.

+1 for the Michael Stevens's beard graphic at 12:00! Hilarious!

Love your videos~

mathologer, your videos are very fun and educational. i am a fan for very long time, please continue making such good videos in the future, cheers

What did Batman ever do to you?

Good news: this is math. We can take anything we wish as truth, so long as it does not contradict anything else we take to be true. Arguing about whether or not it makes physical sense is completely inconsequential until you construct a physical system where it might be applicable, which seems unlikely to happen with the types of things that make someone question the axiom of choice

I very much enjoy your videos, keep up the good work

A very interesting video. When you mentioned cheating death, it made me think of the prisoner execution problem, which has recently been puzzling me. (where the prisoner is promised to be executed without knowing when). I think that could make a good video too.

Saw the Assassin's Creed-style assassins in the thumbnail and was sold. XD

The most conterintuitive thing for me was that at 6:42 individual assasin's chances of dying are still %50, even if a finite number of an infinite number of assassins are dead no matter what happens. Though I get how this could happen, it's a shame you gloss over the fact so quickly.

This was amazing, as always. Thank you :)

Wow, I just watched the Banach-Tarski Paradox by Vsauce and then I found this. A fact that becomes quite mundane if you think about it. You mentioned and linked the Vsauce video in the description, so no wonder this got recommended.

3:42 Love the Bill Wurtz feel

Why is it 50/50 after using the close sequences strategy. Shouldn't it be more like 0 because only a finite number of them are being killed out of infinitely many?

1:45 I called it! None of our assassins can get caught for killing Batman; since, no matter, which assassin the cops choose to suspect & investigate, there’s always infinitely many infallible assassins between him and Batman, who would have done the job before the suspect. 👌🏻

9:30 Well; the 3rd guy hears: ”1”, corresponding to an odd number of differences. He sees 0 differences, which is an even number; and thus, he knows that the only difference must be on his spot. He refers to the 3rd number of the memorized sequence (1), and, knowing his hat number must be different; so, he calls out: ”0”; and, for the rest, the sequences coincide. 👌🏻

I'd say the Axiom of Choice is true as long as you don't have an Electoral College :(

Re: Axiom of Choice - It is something that is intuitively obvious - and infinite sets are anything but intuitive. Personally, I am a constructivist when it comes to proofs, but then again I am also a computer scientist: if you can't construct a thing it is generally pretty useless (with the possible exception of Turing's Oracle)

Perfect murder with infinitely many infallible assassins concentrated to a point of infinite density where Batman will be at 12:00. They can also memorize infinitely many numbers and do infinitely complex calculations. They must collectively form the entity referred to as "God" Very efficient : )

Funny, I was reading the Wikipedia article on the axiom of choice yesterday (having no prior knowledge of the concept), thinking how interesting it was and almost wishing for such a video... Great content, once again :D

In the second problem, the probability that an individual dies cannot be 50%; in fact, this probability cannot be defined at all. Indeed, assume for the sake of contradiction that the probability that an individual dies can be defined. By the fact that each individual is given a 0 or 1 hat independently and with 50% chance each, it is clear that the probability that any particular individual lives must be 50%. Thus, if I specify some set of K different individuals then the probability that all K of them live is (.5)^K. Therefore if I specify any INFINITE set of different individuals, then the probability P that all of them live must be 0: indeed, for each K, P must be less than or equal to the probability that the first K individuals in my infinite collection live, which is (.5)^K. Thus, P

The problem with this is that batman is so popular they would have to bring him back

Thx for pointing that Banach-Tarski video out, I remember doing the proof in a course. Even though I understood the math, the illustrations and thought experiments provided by that zsauce guy and of COURSE you, makes one think of it in a different light. Choice axiom is probably a bad idea. but I don't wanna live in a world without Zorns lemma, just sounds grey and boring :p Keep up the nice videos!

if the sequence doesn't fit, you must acquit

Intuitively, arriving at any mathmatical paradox implies that the problem has not been addressed in the correct manner. There is a fundamental disconnect between our perception of how things should be and reality. While we entertain notions of zero and infinity, neither absolute zero, nor infinity can exist as physical constructs. There are islands of stability in current maths but as soon as we try to pass a certain point it all decends into chaos and uncertainty in the same manner as a Mandelbrot set diagram. In order to get meaningful answers, we first have to frame the question correctly and to do that we need complete understanding of the problem. Still a long way to go.

N.J.Wildberger definitely doesn't ! In fact I thought it one of his most compelling arguments against 'real' numbers. Famous Problems 19d - modelling the continuum ?

For me the axiom of choice is a no brainer. Just because we can't explicitly point to any choice function it doesn't mean it doesn't exist. And it seems to me very intuitive that there's at least one,so I'm ok with using it. It has lot of very beautiful consequences and I can't imagine myself doing algebra without assuming there are maximal ideals in any ring.

Equivalence portioning can be used in software testing to reduce the number of test cases needing to be done when different values in a set are functionally equivalent when related to the logic being applied to each value.

does it matter if a finite or infinite/2 number of assassins die, since there's always infinitely more to replace them?

it's high noon

Thank you Burkard. It's good to finally have an awesome Australian mathematician! Happy Australia day btw :)

The hat problem was equivalent to a problem I had on one of my problem sheets during a countability course! Took me a while to get that one.

I'm glad I do engineering, we just let you guys do the tricky stuff XD

3:51 You probably shouldn't say "list"! The set of infinite sequences of 0's and 1's is not listable!

Excellent video! You really are a fantastic teacher! I’m not sure why the directional assassin problem requires all the complexity of memorizing all infinite binary sequences and placing them in an infinite number of boxes. My solution is to have the first assassin simply call out the mod 2 sum of all of the assassins he can see. The next assassin also computes the mod 2 sum of all the assassins he sees and lastly performs a mod 2 addition with the previous assassin’s called out number. In this manner (and with no memorization or need for the Axiom of Choice) it seems to accomplish the same result - 50% survival for the first assassin and 100% survival of all others. As I thought more about this, I guessed that you might say that the sum of an infinite number of binary digits (mod 2 or otherwise) does not converge (I really loved your videos on infinite sums!). But I would argue that if you don’t think the infinite sum converges, why do you think an assassin could memorize (and bundle) an infinite number of infinite sequences?

Nothing clever say; just thank you for a really entertaining video.

I am not sure how I feel about the new format, I think that the videos were more engaging when you are on screen. :)

Isn't the first one just Zeno's paradox about turtle but formulated in terms of assassins and batman?

I was thinking about how there are uncountably many boxes of sequences in this puzzle, and in light of your "Transcendental numbers powered by Cantor's infinities" video, I was wondering if I could prove this using Cantor's diagonal argument. The natural approach would be assume there are countably many boxes, choose a representative from each box, list those in order, and then construct a new sequence by making the sequence differ in the nth place from the nth sequence. Of course, this technique does not work here, since it would only guarantee that the newly constructed sequence differs from any sequence in 1 place. It could still be in the same box. So I came up with the following modification. Assume there are countably many boxes. For each box, choose a representative sequence. List these sequences. For each sequence in the list, we may associate it with a prime number p. We do this by associating the nth sequence in the list with the nth prime number. (By Euclid's argument that there are infinitely many primes, we are guaranteed that every sequence in the list has a prime number associated to it.) Now, we construct a new sequence according to the following algorithm: To determine whether a 0 or 1 goes in position n: 1. Consider the prime factoization of n. If n is _not_ a pure power of a prime, place a 0 in position n. 2. If n _is_ a pure power of a prime, i.e., n = p^k for some prime number p, then consider _which_ prime number p it is. In other words, it is the mth prime number for some positive integer m. 3. Consider the mth sequence on your list. Look at the number in position n. If it is a 0, make position n in your constructed sequence a 1; if is a 1, make position n in your constructed sequence a 0. Now, our constructed sequence must differ from each sequence in the list in infinitely many places. This is because for each positive integer m, there are infinitely many powers of the mth prime. Hence, the constructed sequence differs from the mth sequence in the list at all (infinitely many) positions given by powers of the mth prime. Therefore, the constructed sequence is not in the same box as any of the sequences on the list. Hence, we have demonstrated that we missed a box. Therefore, there are uncountably many boxes.

Because the mail can't handle all the letters lawyers would write and no-one could afford to pay for them anyway! Ah, so this is just the old mathematician vs. engineer joke, with the naked lady/man/tasty treat/whatever - they can only move half way from their current position to the prize each second, the mathematician just storms off because he can't ever reach the prize, the engineer proceeds anyway claiming he'll be close enough for all 'practical purposes'.

But the real question is how does one convince an infinite assassins to kill one person

but infinite number of asassins will fill the entire universe , even if the asassins is smaller than a quark....

When I saw this in my recommandation and I looked at the picture and the title I was like, is this some gametheory video? Then I see mathloger... So here I am

So there are two important lemmas here: 1. Closeness (equality modulo finely many positions) is an equivalence relation. That is, each sequence belongs to exactly one putative "equivalence class". 2. Each infinite sequence can be associated with its equivalence class even if we lack knowledge of one of those places. We already know the sequence differs in at most a finite number of places, and finite + 1 is still finite, so this is trivial. One thing that's left implicit is that each assassin knows his position in the sequence. This follows from hearing N prior assassins say 0 or 1. 2 also implies that every assassin knows the correct equivalence class since they are lacking knowledge of no more than N+1 digits where N is their position in the sequence (starting at 0, of course). This means that they know the class even if the assassins before him in the sequence mess up and say the wrong number. However, each assassin after the 0th will say the wrong number unless either they also know whether each prior assassin lives or dies (in which case they know which number was really on that hat) OR that set of assassins makes an even number of mistakes.

If we don't accept the Axiom of Choice, then we are forced to conclude that not every vector space has a basis. This seems very, very bad.

This is my guess for the assassins lined up facing to one side that can guess their number one at a time. Haven’t watched the answer yet. edit: this isn’t the answer in the video, so no spoilers :) The first assassin can give some type of parity bit, like saying “one!” if the number of assassins with ones on their heads ahead of him are odd (thus making it an even parity). That way, when it comes to a certain assassins turn, he can just add up all the previous assassins one’s with all the one’s ahead of him and yell one or zero to maintain that even parity. This should work flawlessly for a finite number of assassins, but i’m not so sure for infinite because I wonder if it could ever be even or odd?

It seems odd to have so much explanation on the technical strategies the assassins would use at every step of the execution, but hand-wave past the underlying math, at least considering that the video is only 12 minutes. It's a really interesting logical result that you can achieve these low finite numbers of deaths, but you just have to take the interesting math that leads to it(dividing up infinitely many numbers into these classes, the choosing, and problems with that choosing) for granted. Even the titular Axiom of Choice is only really mentioned at the very end of the video, not in particularly much detail, and the relation between Banach-Tarski and it isn't explained.

The 2nd assassin whose number is different heard that there's an even number of differences, but he also heard that the person behind him said that he was different. Because of this, he knows that there's an odd amount of differences in front of that person. Since he only sees an even amount of differences (in this case, 0), he must be one of those differences.

I don't understand the idea behind the boxes at all. First of all, they still have 50% chance of dying. You seem to be saying that 50% of infinity is somehow suddenly finite. I don't see how you could get away with that. Further; the details of how they could divide their sequences without ambigiouity don't make sense to me: Let's assume that they all get together and agree that a certain (randomly chosen) sequence, A, will make up the backbone of the first box (which they write "A" on the lid): if they see any "close" series to this one, they all agree that this is the one they will shout. Imagine then that they find another sequence, B, that is not close to A. They put this into another box marked "B". Then I, being a fan of Batman, come along. I'm feeling a bit trollish, so I decide to "help" the assassins: I offer a hand in sorting out all sequences into their correct box. (They first answer that they are infinitely many so my contribution will be too small to matter, but I argue that the task is so large that they need all the help they can get.) They agree, and so I start with sequence A1: it is the same as sequence A, except at the first place that A has a difference from B, I swap that number for what is in B. This is still close to A, so it goes into the same box. I then take A1 and pick the first number where it is different from B and swap that entry for what is in B, and call the new sequence A2. Then I take A2 and pick the first number differing from B and swap it out for what is in that place in B and call it A3. All of these are of course close to A, so they go in the box labled "A". I continue with this exercise an infinite amount of times until eventually I end up with An = B. But: as I create them, each is just one number removed from entries already existing in box "A". And because of the identiy that if X=Y and Y=Z then X=Z, all of the entries must be put into box "A". Eventually I get close to B, and by similar argument can lump B in with A. I continue working through the night, building simliar bridges to all other of the asssassins' boxes. Now the assassins will have the choice that they either have to 1) shout the same sequence no matter what they see around themselves (essentially they have just all agreed beforehand what to shout), or 2) Since any member of A is also a member of B, they could not agree on which box a particular sequece would belong to and end up shoulting numbers from different sequences with equally depressing results. But as I watch infinitely many assassins get executed, I wonder: how could they possibly have made any meaningful partition to begin with?

life lessons from mathologer: it is a bad idea to try to be intimitate with someone who has nothing in common with you because "being CLOSE is an equivalence relation."

What software do you use to make animations? Love your videos by the way.

By the time the final question got asked I believe I had the correct idea in my mind. You would want to use some form of encoding where a change in bits and a continuation of bits is equal to the bit you are currently on. You can't just say the same that's infront of you, as you die. But you need to learn your own hat from somewhere behind and also give away the next hat while your own call. Not quite sure if just the first guy as to gamble this way tho. It's not the full solution - but I will watch it now

How about a video on the relationship between the Axiom of Choice and The Continuum Hypothesis - I remember that being interesting but can't remember what it is (I know, I know, easy to look up these days, but it's more fun to watch Mathologer videos...)

The axiom of choice was accepted, not because it seemed reasonable or necessary, but only because they'd have to throw out a bunch of theorems if they rejected it

I'd like to see one million assassins standing on a pin.

I think that if you try to convict a gang of *infinitely many* infallible assassins, you will have other problems than proving which one killed Batman.

Practical application, of course I have to make some adjustments but this is actually kind of helpful.

I'm a simple mathematician, i read axiom of choice, i see the video

Hello there! First of all, I'd like to thank you for your videos; they've been always really enjoyable to watch. I'd like to ask you about your thoughts on intuitionistim. I take it that you don't accept all its premises since you declared your subscription to the axiom of choice; still, I'd really think it would be nice to hear (or read) what are your main opinions about this take on logic and math foundation. Thanks in advance, Best regards, Matheus

I love these puzzles, superb content! However, I think the strategy in both puzzle 2 and 3 should be made a bit more precise if the tails of the two sequences are shifted. For instance (1,0,1,0,1,0,1,0,1,0,1,.....) and (0,1,0,1,0,1,0,1,0,1.....) have the same tail, but they do not match in any entry. In puzzle 2 this can be solved, if before calling out the numbers, the last assassin simply shifts everybody so that the tails match. In puzzle 3, however, the shift might result in the following: For instance let (1,0,1,1,|0,1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,0,......) be the hat sequence and (|0,1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,0,......) the representative sequence, then after matching the tails, the first 4 terms 1,0,1,1 of the hat sequence aren't matched. In this case one can extend the "shorter" sequence by zeroes so that the lengths of the two sequences are the same and the strategy works again.

What resources do you recommend if one would want to read further into the Axiom of Choice? That aside, the only solution that my high school educated brain can come up with is that in a problem with an infinite amount of solutions, the predictability of those solutions (at least to the assassins) can only be as predictable as those willing to die for the cause. For example, if we reached assassin #2045, and he sees that there are an odd number of pattern changes as far as he can see, what isn't to say that from his position that the solution of the 1's and 0's has changed and assassin #2045 would get killed and those assassins who previously had their pattern change now had the correct hats?

I would love to see an introduction to the proof of Fermat's Last Theorem by Andrew Wiles from Mathologer!

Im guessing that the if a one gets it wrong he will cause the person infront of him to have it wrong as well. So every mistake will only cause two of them to die. But after that the mistake will have been corrected so noone after that (have to) have it wrong. Kinda because negative times negative is possitve. Is that correct?

I don't get this on a such a level that I can't think of what questions to ask.

You could even extend the last puzzle to one where the assassins are numbered A(1), A(2), A(3), ... , the A(i) being arbitrary reals from the half-open interval [0,1). By the axiom of choice all the assassins could agree on the representative sequence R(1), R(2), R(3), ... where A(i) - R(i) is nonzero for a finite set S of i. The first assassin would call out C(1) := (sum A(i) - R(i) over i >1 in S) mod 1 and for each n > 1 the n-th assassin would call out C(n) := (sum A(i) - R(i) over i > n in S) - C(n-1) mod 1, so that C(n) = A(n) would be guaranteed for all n>1.

Batman will not be killed. He can only be killed by the first assassin he meets, no subsequent assassin will get the chance. As there is no first assassin, he will not be killed.

Part 3 Cheat Death does not require any memorization of infinite sets - only the ability to count them. The first man counts the quantity of 1s in front of him, and he says 0 if that quantity is even and 1 if it's odd. Having both heard those before and seeing those to come, every other participant has the information necessary to call the number on their head.

great information for any system to consider

Proving continuum hypothesis , proving inconsistency in ZFC , constructing ZFC from naive set specification , resolving Russell's paradox , constructing infinite number system , construct and ensure overall consistent mathematical universe and developing arithmetic system - edition 8 May 2024 DOI: 10.13140/RG.2.2.21713.75361 LicenseCC BY-NC-ND 4.0

Hi Mathologer. Could you make a video explaining John Conway's surreal numbers? How can it "create new numbers"?

As a sound guy I have to say that the "pause and ponder" music is a bit too loud in relation to the rest of the show. Awesome videos! Thank you :)

Hey Mathologer, I challenge you a similar problem. The setup is just like part 3 (7:00) but the hats are a little different. Hats can be any existing color. (There are infinite numbers of colors). The assassins are mathassins so they can memorize all existing color or memorize infinite sequence of numbers. Remember, the assassins have to say the color but nothing. I made this problem ::) (If you really want the answer I'll give you the answer.)

Could you do a video about "game numbers" (number-like objects that are actually games of infinite lengths between 2 players)? I find it interesting how we can define a whole complete lot of "winning strategies for player 1" and "winning strategies for player 2" - and then screw them both simultaneously by never subscribing to either of strategies, always making moves that break out of either strategy until we have an (infinte) collection of moves that are neither 1-winning or 2-winning strategy, since our both sets of winning strategies are complete. It is an interesting theme, but the math for this is a bit hard, even in Wikipedia, so basically it is "mind = blown" stuff for me. Could you explain this theme in layman terms?

as i understand theres a logic that we dont use (paraconsistent) that doesnt need the not(not a and a) rule, that logic works by itself afaik but its not the one that we base our general knowledge on, i was wondering is the math without the axiom of choice still powerful enough to be used on science? also does that math doesnt have the axiom or has an axiom that negates the axiom of choice?

fantastic video

Are the points in space in the example of the Banach-Tarski paradox zero dimensional or 1 dimensional or higher?

(1) Remark: The set of possible hat types does not have to consist of two elements. They can try to guess a real number, or a picture (set of points), and the solution works as well. (2) I know a similar AC puzzle: Consider 100 students and infinitely many boxes in a room. Every box contain a real number. The task for each of the students is following: Come to the room when all boxes are closed, open all but one box, and try to guess the number in the remaining closed box. The process of opening boxes can be consecutive and transfinite, for example, student can open infinite number of boxes leaving infinite number of other boxes closed, then make a decision and continue in the opening process. The overall task is to make a strategy in which at most one student fail. The solution uses dividing into equivalence classes described in the video. I can write it down if you are interested.

It's not the fault of the Axiom of Choice, but ours for measuring weird sets.

for the first question, i said that for the iTh assassin, P(assasin_i is guilty) = 0, Is that a correct solution also?

I don't even understand how it is possible to choose a particular number from the set of non-constructible numbers. Wouldn't that choice procedure amount to a construction of the chosen number?