2026-04-30 NYNTS Mel Nathanson

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[00:00:01]Okay. So this is a talk on um basically inspired by a preprint posted on archive in which a um AI system called Aristotle solved a problem in one of my papers and uh so I tried to figure out why it solved a problem.

[00:00:31]not so much why it solved a problem that I couldn't solve but what it did that I didn't do to solve the problem but so let me just start with some background and um okay so this is something familiar to some of you but not all of you so I just want to give a lot of background so there's what I call the sumset intersection problem and the uh products set intersection problem.

[00:01:03]Um because there's always ambiguity about this. N for me is always the positive integers and n is the non- negative integers. And we say a set is co-finite.

[00:01:16]A subset A of S is co-finite if it's complement uh in the set is finite. And if it's not co-finite, its complement is infinite.

[00:01:26]We call it co-finite. Um and if we have a sequence of sets, it's uh decreasing if uh one if they get smaller uh and strictly decreasing if each set is a proper subset of its predecessor.

[00:01:42]And then uh it's there's a a trivial argument you can show that if you have a strictly decreasing sequence and a is the intersection uh then the set A is co-finite. Its complement is infinite. U and we say that a sequence is asmmptoically strictly decreasing u if it's decreasing uh but not eventually constant. Uh so you can have the same set occurring you know two three four any finite number of times but after a while you get to a set in the sequence which is a proper subset of its uh its predecessor.

[00:02:25]So we look at semigroups. Now for me the only semigroup really is the integers and actually really the non- negative integers but that's just make believe there's some others. Um so that has to be a semigroup not necessarily a billion. So we write it multiplicatively.

[00:02:45]uh if you have subsets A1 up to AH the product of them is just all elements in the semigroup that you can write as a product A1 a2 up to ah where AI is in the subset AI for a and if all these sets are the same uh you have a to the power h that's the hold product set and similarly if the set is a billion and written additively we call these things subsets and hfold subsets.

[00:03:21]>> So let's take an index set which could be finite but for finite in the finite case uh things are often easier and different. So for simplicity, let's have our index set always be infinite. And we take a family of subsets of the set indexed by this index set q and we let a be their intersection.

[00:03:44]So a is certainly contained in a q. So the h power of a is contained in the h power of a subq. And therefore the a to the h is contained in the intersection uh of all these uh a subq's. So this inequality always holds and for additive a billion semi groupoups we have the um additive inclusion.

[00:04:25]So this is our problem. uh you take a Q to be a family of subsets of a semicroup s and a their intersection u so a to the h is always contained in the intersection of the h of the hfold products >> equal >> and the question is when do you have equality here so given a family of subsets of a semigroup this set h of aq is the product intersection um it's all h where you have equality here and I have absolutely no idea why this should be interesting.

[00:05:06]uh nor do I have any idea why well actually which may be why as far as I know no one's ever you know looked at this um but um but I happen to find it interesting so I looked at it uh so I'm interested in these are sets of positive integers u with this property that is uh for a given sequence of or family of subsets of the semigroup We want to find all the positive integers h where the intersection of the hfold subsets with sorry the intersection of the hfold subset is the uh uh uh or hfold product set is the hfold product of the intersection that's the question and additively you have the you can call it the sums set intersection problem to find this And you can just make the observation that um one is always in this uh set right because a is defined to be the intersection of a cq's. So this set of positive integers is some set of positive integers that contains the number one.

[00:06:25]Okay.

[00:06:30]and let that AQ be a family of subsets. Um let's see that's what I just said. Let me go on. Okay. So so this is this is the sum set or the products and intersection problem. You have a semi-roup S. It depends on your semigroup. You have an index set Q. It depends on your index set Q. And you want to know uh if you have a sequence of subsets when do you have this property that the intersection of the product sets is the product set the h the intersection of the hfold products is the hfold product of the intersection. Um and so for for really the most important case the set of all non- negative integers this problem is solved and the solution is very simple. Uh it says the following. If you have any decreasing sequence of sets of non- negative integers and you let a be their intersection, the sumset intersection set is every positive integer. So that if you have sets of positive integers uh if you have a sequence of sets of positive integers a subq and you let a be the intersection the hfold sum of a is exactly the hfold sum of the a the intersection of the hfold sums of q. So there is only one subset intersection set for the nine negative integers, right? Um that's actually kind of interesting.

[00:08:24]So what about for integers and not necessarily non- negative integers? So the fact that every intersection set for the non- negative integers is a set of all integers isn't true for the integers. Well, sorry. First of all, it it doesn't just depend on positivity because it's not true if you take sets of non- negative rational numbers or even sets that you can construct as open sets of non- negative real numbers. So it has nothing to do with the positivity. uh and the fact that it's not true for positive rational numbers is actually uh quite interesting and more to the point or for me more important is that this problem is unsolved for sets of integers right that is uh there do exist sets of integers sequences of sets of integers a subq where a is the intersection of the aq's and this inclusion is proper All right.

[00:09:26]Um so then a natural question to ask is what sets of positive integers do occur uh as h of aq.

[00:09:41]Okay. Um let me just give an example what can happen with the integers. Um because this is a number theory seminar.

[00:09:52]So suppose we let uh for every positive integer q that a q sharp be all integers whose absolute value is at least q. So this sequence is strictly decreasing and the intersection is the empty set. And so the hfold sum of the empty set is the empty set.

[00:10:16]On the other hand, for each of these sets, a subq sharp, if you take the hfold sum for any h greater than or equal to two, you get all the integers. So each of these aq sharps is a basis of order two for the integers. And so the intersection of all the hq sharps is z.

[00:10:39]Which means for h different from two, greater than or equal to 1, for greater than or equal to two, for h different from one, these are different. And so the intersection set for this sequence is just the number one. It's not everything.

[00:10:59]And you can modify this in a nice way as follows. Suppose you let a be any set of integers whose complement is infinite. A Q sharp is that uh set we constructed a moment ago. All numbers all integers in absolute value greater than or equal to q and let aq be a union aq sharp. And because a is a co-finite set this sequence is also asmtoically strictly decreasing and a is the intersection of them.

[00:11:33]So because AQ sharp is contained in AQ AQ is a subset of AQ. H AQ sharp is everything. So H AQ is everything. AQ each of these AQs is a basis of order H for every H greater than or two. So this intersection is the integers.

[00:11:54]Recall that um we say that a set A is a basis of order H knot for Z. If the hn false sum of a is all the integers and it has the exact order hn a is a has exact order hn. If it's a basis of order h knot but not of any smaller uh integer h. So if a is a basis of order h and order for z then the hfold sum is equals z which is the intersection if and only if h is greater than or equal to h knot. So this sums set intersection set uh for this sequence a q is one plus all integers greater than or equal to hn. And in particular, if the set A is not a basis of any order for Z, then the intersection set is just the number one.

[00:12:49]So this gives an example of an intersection set which is not one and not all the integers. So other things are possible here.

[00:13:04]And let me just say you can do something similar in a a nonabelian semigroup or group. Um so uh this says if you have a semigroup s and you take as subset a whose complement is infinite. Uh then let x i be a sequence of elements in the complement of a. uh and let a q sharp be all the x i from the qth on. So this is a strictly decreasing sequence of sets and let aq be the set a union the aq sharps. The aq is asmtoically strictly decreasing and a is this intersection and you can use that construction to prove the following. If you have an infinite group, not necessarily a billion and a is any infinite proper subgroup um there's an asmtoically strictly decreasing sequence of subsets of uh g such that um where a is the intersection. The only h for which a to the h is the intersection of a q to the h is h equal 1. So in this example h of a q is one. It's not everything. That's nothing else.

[00:14:29]Okay. So, all right. Now, um the people at um Harmonic, the AI company that has Aristotle, said that um they've been feeding some of my papers into Aristotle and asking it to solve uh open problems. And in particular, I have two papers uh on archive um that have to do with these intersection problems. One from December of last year and one from according to this April of this year. So this is not that old, but uh I never remember anything. So every time I looked at this paper in the last week or so, I had to figure out what was in it. Um and there are lots of uh unsolved problems in these papers and let me just list a few of them which are uh first of all they're still unsolved and maybe more interesting than the solve problem. So, so we have a semi-group S. We have an index set Q and F Q of S is all families infinite families uh A well Q is infinite all families A sub Q of subsets of S indexed by Q. So the basic problem uh what I've just been talking about is if you have such a family of subsets of Q compute the intersection set right compute the H where the intersection of the H powers is the H powers of the intersection that's the fundamental problem and in general uh I have no idea how anyone how you solve this uh then we can specialize this a a little Right? So if you fix a subset A of your semigroup um suppose instead of considering a random sequence of subsets of S, only look at those sequences of subsets of S whose intersection is A and what is H of A Q for those sequences. In other words, if you fix the set A, what are the products intersection sets that you can get from A. So here is one uh example.

[00:17:01]Um suppose you take A to be the empty set. So we're only looking uh at subsets of the of the semigroup whose intersection is empty. and and we want to find out what are the sets x when does there exist a set x such that the intersection of the eighth powers of this of the sets in the sequence is empty if and only if h is an x right so that is in some sense a very very concrete problem that um I assume are well beyond AI's power to solve doesn't mean that a bright undergraduate can't solve it but I don't believe any AI can solve it.

[00:17:51]Um here we take uh a larger set of sets.

[00:17:56]This is a boldface H subq of S. So you have a semi-roup S and this is looking at HQ of A for all A and S. This just means for this given semigroup s and a given index family what are all possible intersection sets.

[00:18:17]So this is in some sense the master set of all product intersection sets for all families um uh uh of subsets of S and in particular um one can ask the following coming try to prove this if you're given a semigroup snot like the integers.

[00:18:49]So every intersection set has to contain the number one. But does there exist a set x of positive integers with one and x which can never occur for any sequence of subsets of snot? It's not the product intersection set for any sequence. So somehow for this set snot or semigroup snot, there's some x which never occurs. Now we already saw an example of that. If you let S not be the non- negative integers, the only set that does occur is the set of all positive integers. So no other set occurs. So these these like sets X which can never occur for the non-gative integers are everything except the trivial case in some sense of all non- negative integers.

[00:19:44]Now, here's an even better problem. Does there exist a set x of positive integers with one and x such that x is not an intersection set for any semigroup s? In the previous problem, we fixed a semi-group s. Now, we're looking at all semigroups in the universe. I mean, there's a there a lot of semigroups out there. And for all semi groupoups um is there some set X containing one that never occurs?

[00:20:17]Now I'm not sure when I wrote this down whether this was one of these like throwaway like off the cuff cuff remarks to say oh no one will ever solve this.

[00:20:29]This is like this is impossible.

[00:20:33]Oh, but Aristotle solved the problem. This is what Aristotle did. He said, he proved that the answer to this question is no. Right? Now, you might ask, how many hundreds of millions of dollars were spent to develop a computer system that would just say an answer to a problem is no. I mean somehow it does seem the answer should be longer. Um but no so what Aristotle proved is if you take any set of positive integers that contains one somewhere in the universe there exists a semi-group S and for every index set Q a family of subsets of S indexed by Q for which it cannot happen that the intersection set is X. Like X just never happens.

[00:21:38]Okay.

[00:21:43]Okay. So, so this is what um it was posted on archive uh a week or so ago and um so I want to explain what Aristotle did to solve the problem.

[00:22:03]And I have an advantage over someone else who wants to answer this because it's my problem in my paper where I proved a lot of stuff and my paper was fed into archive.

[00:22:21]I mean I don't know how that you feed a paper into archive sorry fed into fed into Aristotle. I mean, I don't know whether like you chop it up, turn it into a stew, pour it through a funnel into the machine. Although I I'm told by more sophisticated friends of mine that all you have to do is like uh upload a PDF of the paper. Um, so I wanted to see what I did and what Aristotle did that I did not do. All right, that's the question.

[00:22:56]All right.

[00:22:57]And then you can decide at the end whether Aristotle was really smart or I was really stupid.

[00:23:04]I mean it's it has to be one or the other. All right.

[00:23:10]>> Maybe Aristotle worked on it more than you did.

[00:23:13]>> Uh I know he did. Uh I was told the guys who uh worked for the company who who did this but I just forgot. They told me how many hours maybe it was a second I don't know I thought it was in the order of hour but who knows let's just say they did it in a microscond because it's you know okay so this is what's in my paper so u so this is kind of like a global result and uh I thought they had invented the word global until I looked at my paper and I saw that it came from a section which is a a global intersection relation for product intersection sets. So it uses I use and they use a couple of lemas. So uh the first says if you have any family of semigroups uh and each a index by a set I and for every uh index >> every index I uh you let a subi be a subset of the semigroup s subi. Um so the product of the AI is that product set is a subset of the product semi-roup and for every H this infinite product of AI to the H is just the hfold product of the just the hfold product is the hfold power of the infinite product. So in my paper I just wrote this down for I where I just has two elements in it. you know just because I was only looking at two sets but you don't change a word in the proof if you do it for any fi for i any finite set or i any infinite set. So this is just a combinatorial uh or algebraic triviality.

[00:25:02]So, so this was a theorem I proved. So that semi this is like boldface because if you read a book, if you read books in category theory, they always like to write categories like these big bold letters. So let semi be the class of all semigroups and you have to say class instead of set because people who know stuff that I don't know uh explain to me that uh it's not a set. It's it's so if you don't know what it is, you call it a class. But we know it's just all semi groupoups. So the semi be the class of all semigroups. Um so for every index set Q uh let H subq.

[00:25:43]So this is this universal set I'm looking at. It's um for every semigroup s in the in the class you look at all the intersection sets that you can get from that semigroup s and you look at and you take the union over all semigroups. So this is the set of all product intersection intersection sets for all semi groupoups s and the theorem says that this set is closed under intersection right so this means if you have a family of product intersection sets for the index set Q not necessarily for the same semigroup so x1 can be a product intersection set for some semigroup s1 X2 is a product intersection set for a different semigroup S2 and so forth.

[00:26:35]Then somewhere in the universe there's another semigroup S with respect to which you can construct a family of subgroups of S indexed by Q whose product intersection set is the intersection of the X subis.

[00:26:55]Now again um so I don't I don't want to mislead anyone in my paper I was just proving that the intersection of two of these sets is in the set that is the intersection of two inter two intersection sets is an intersection set it follows that any finite intersection is an intersection set and the same proof unchanged if you need to if you want to write it down but I didn't see any reason to write it down is that this holds for infinite intersections as well as finite intersections and uh I think I um oh and then the same is true in uh HN star. So the the difference between HQ q is a is a index set and hn star is here my index set is the positive integers. So my families of subsets of a semigroup are sequences and the star means I'm only looking at sequences of semigroups sequence of subsets which are strictly decreasing or asmtoically strictly decreasing which is what I find interesting from the point of view of number theory but it doesn't matter because if you restrict your attention just to asmtoically strictly decreasing sequences of subsets of a semigroup this set is also closed under intersection.

[00:28:23]Okay. Um, I wrote down a proof, but I'm not going to uh give the proof because no one likes to see proofs in talks. And if you want to see the proof, it's on archive.

[00:28:38]Um, okay. So, here is Aristotle's theorem for both of these global inter u section sets. either you take it for every for an arbitrary indexing family Q or for strictly decreasing sequences as your families.

[00:28:59]Uh the intersection sets that can arise are everything possible namely every set of positive integers that contains one.

[00:29:08]And okay so how did Aristotle do this? Well, it starts with basically the two lemas that I used. Um, the first is that the product of h powers of sets is the h power of the product.

[00:29:31]And the second which is also sort of straightforward which says if you have a family of semigroups finite or infinite doesn't matter and you have an index set Q so in the semi groupoup SI you have a family of subsets indexed by Q and you have one family for each sub for each semi SI. So, so the intersection of AI Q as for fixed Q this is a subset of the semigroup SI and we're taking the product of those subsets. So this is something this is a set the right hand side is in the product of the semigroups si and and that's equal to what you would get if you simply took if you took the interse if you take the product of the aiq's over i. So this is something in the product of the semigroups and you take the intersection of that over q. So these two things are the same. So both of these are just you know baby algebraic identities.

[00:30:51]All right. So okay. So here's the idea that Aristotle had that I did not have. Right? You have this infinite family of semi groupoups S subi and you want to construct a new kind of semigroup.

[00:31:24]Okay. Well, sorry. Um that will come later. uh Aristotle is going to construct a very simple semigroup for every positive integer n the semigroup s subn will have this property oh so Aristotle you know these old Greek guys were smart Aristotle and you know actually Aristotle could probably have done this I don't know that it goes beyond anything that uh was known then other than this notion semigroup, but that's no big deal. So he's going to Aristotle constructs for each N a semigroup SN and a family or sequence um all right I see this should be Q and Q um all right a sequence of subsets of this semigroup and this will have the property that the intersection set of This family will be every positive integer except n. This is for n greater than or equal to two. So what this is what Aristotle is is proving and you just need one example of this and he's providing an example. It is that for every integer n there is a semi groupoup and a sequence of subsets of the semigroup whose intersection set is all integers except n. So exactly one number is missing from the intersection set and you can do this for every end.

[00:33:09]So this is Aristotle's construction right he's clever guy actually um so you look at a certain free he constructs a free semigroup he doesn't say whether it's a billion or not a billion but it makes no difference so if you want a free a billion semigroup make it a billion if you want a free nonabillion semigroup make it not a billion you look at all words so script w is all non-mpy words over the alphabet n. So you can think of this as you like is if you have palibly many variables x1, x2, x3 and just look at the free semigroup they generate. So it's all words in x1, x2 and so forth and you take two symbols alpha and beta which are not words. Okay? So your semigroup is going to consist of alpha beta and all words which are finite strings of x1 up of the x i and for every n greater than or equal to two let w strictly less than n be the subset of it's all words whose length is less than n. The length of the word is just the number of letters in the word.

[00:34:31]And the set and sn that's going to be a semigroup. It's all words of length less than n together with alpha and beta. Now this is a set.

[00:34:44]All right? It's not yet a semigroup. So I have to make it a semigroup. So you have to define a multiplication of these words. So if you take two words, so sn is words of length less than n together with these symbols alpha and beta. So if you take two words less than n u and v u and v u * v is just their concatenation.

[00:35:12]And we define a multiplication in in this set sn as follows. So this is first of all alpha is like a zero. Alpha x and X time alpha are alpha for everything.

[00:35:24]Beta* X and X time beta is equal to alpha for everything. So that describes what happens when you multiply alpha or beta by something you always get alpha.

[00:35:37]And if you take two words whose length is less than n, what is the product in this semigroup? So if the sum of the length is less than n the product of uv is just the word product uv.

[00:35:51]If the length is equal to n the product is beta and if the u plus v is is greater than n then the product is alpha. Okay. So that is a well- definfined multiplication and the only question is is it associative? Um, so it is I mean it it takes a couple of minutes to check, but that's an associative multiplication.

[00:36:24]All right. Um, let me just go back.

[00:36:28]So, so this is clever, right? Um, so you're looking at words in countably infinitely many symbols. You're looking at all words whose length is strictly less than n for any integer n at least two. You throw in two mysterious symbols alpha and beta. And this is an infinite set and you define a multiplication like this. All right, that's you have to think of it but once you think of it it's not hard to check and you can check and it is associative. All right. Okay.

[00:37:04]So this n is a semigroup and in this semigroup I need to construct a sequence of subsets.

[00:37:16]So in the semigroup sn the qth subset will be alpha that one symbol and I'm just using the notation that's in Aristotle's paper. So instead of calling the M variable X subm, he just writes M in square brackets. So this is just all letters uh M for M greater than or equal to Q.

[00:37:42]All right. So this is the Q set in the semigroup SN. And and when you go from Q to Q + one, you're throwing away one element. So this sequence is strictly decreasing. and eventually you thrown away everything. So the intersection is just alpha. All right.

[00:38:04]So so my set capital A which is the intersection of these things is just alpha. And then you can check. So if you take h strictly less than n and you take the c set and raise it to the eth power, you get alpha together with all words whose letters are greater or all all words of length h whose letters um are at least q.

[00:38:49]If h is equal to n, you get alpha and beta. If h is greater than n, you just get alpha. Right? So this is a calculation.

[00:39:03]So when you're looking at the h power of these sets, for h less than n, it's this. For h equal to n, it's alpha and beta. For h greater than n, it's just alpha.

[00:39:16]And you do the calculation you see and and what you get right what you get is for this particular sequence the intersection set is all positive integers except n right see in one case when h is equal to n a subn q to the h is alpha and beta And if you take powers of alpha, you're never going to get beta.

[00:39:51]Okay.

[00:39:53]All right. So, so Aristotle had two ideas that I did not have. The first was to construct these semigroups in which the intersection set is everything except one particular number.

[00:40:15]But then if you're clever, you realize that's all you need because a moment ago we proved I proved and Aristotle proved that the intersection of these uh eighth power intersection sets is another eighth power intersection set. So uh if you take any set X of integers that contains one. Its complement is some set of integers that doesn't contain one. Call that set I. For every N in this set I, let Sn be the building block you constructed above. Let S be the product of so SN is a semi groupoup and let SN be the product indexed by the set I of the sets Sn. So this is a product semi-roup and you let aq be the product of the corresponding sequences for each sn and this is strictly decreasing. This is a strictly decreasing sequence because all these aq and q's are an and a which is the intersection.

[00:41:40]Well, so a is the product of the intersection of the a and q's. Those are all equal to alpha and a to the h is just alpha to the h which is just alpha because alpha times anything is alpha.

[00:41:59]So maybe I should just say this in words because it's easier.

[00:42:04]Given a set of integers positive integers that contains one I want to construct a set a semigroup and a sequence of subsets in that semigroup whose intersection set is equal to x.

[00:42:23]So how can I do that? Look at all the numbers that are not in x. Suppose n is not an x. Then I have a semigroup sn and a sequence of subsets of sn whose intersection set is everything except n.

[00:42:40]So I take the product over all the ends that are not in x and that will be and that will be the intersection of the intersection sets. That will be a set whose intersection set is everything except the numbers that are not in X.

[00:43:04]That means X you just get X.

[00:43:07]I mean I don't think I'm saying it so clearly but um but I'm not Aristotle. Of course Aristotle doesn't say it either.

[00:43:18]Aristotle just does it. Uh but so the incredibly clever idea that Aristotle had was to use the fact that any intersection of these intersection sets is another intersection set and then to construct for every integer n an intersection set which is everything except n and then by taking an appropriate choice intersection of these like one element missing intersection sets you get every intersection set. So for any set X not containing one you can construct a semi-group for which that X is an intersection set right that was what Aristotle did now that's that he did this where the index set is just the positive integers we actually looking at sequences and actually looking at asmtoically strictly decreasing sequences uh but what about when you have an arbitrary index that Q. Um, in that case he does essentially he has the same idea but he he constructs a different set of semigroups and let me just show the semigroup that Aristotle constructed.

[00:44:31]So we have this index set Q which is infinite. So in particular it contains of these two elements. So that q1 and q2 be two distinguished elements of the set q. It's like when we for n for in the previous case we chose an alpha and the beta but now we let sn is going to be a semigroup. So for each n sn is going to be a semigroup of all non- negative integers from zero up to n cub plus n squ right. So if you like number theory these are really numbers.

[00:45:08]This is good. Right? And we defined an operation on this set to make it into a semigroup which is x times y in this set is going to be the minimum of x + y or n cub + n 2 which is truly a stupid definition right I mean what a ridicul I mean right um but as they say sometimes something stupid is brilliant So um so you can check uh that this is a semigroup. It's a computer of monoid with the identity zero. And um so in particular if you multiply h things they're they're or their product whatever you want to call it it's the minimum of the sum of the numbers just adding integers and n cub plus n ^2 right every addition is truncated n cub plus n 2 and that's a semigroup um and let me All right. And and for this semigroup they construct sequences uh of subsets.

[00:46:29]Well, they they turn this into a semigroup in which uh the product intersection set is except end. Um and I would really encourage you to look at the Aristotle paper on um archive to see what they do. Uh since I just have a few minutes left, I just want to um just mention something else which is interesting. So so I'm done with I'm done with harmonic and Aristotle, right? That's one AI machine. I don't know what you call it. Machine software device.

[00:47:14]uh it's one thing that was I think it was developed with the specific purpose of doing mathematics whereas other AIs tend to be much more um uh general in scope. So one which uh I had never heard of until yesterday or so was chat GPT 5.5 grow.

[00:47:36]So I know I have chat GPT on my telephone. is just there. Uh doesn't cost anything and presumably that's it's not worth very much but chat GPT 5.5 Pro is maybe like top of the line if you like.

[00:47:53]So um in that same paper about some set intersection sets uh that paper was fed into the chat GPT 5.5 pro program and it solved a different problem that was in that same paper. So not the one I just discussed a completely different problem but it solved it.

[00:48:16]uh and just you know two days ago or so I think and that was interesting it was very interesting and it was beautiful proof but I thought it looked somewhat familiar and uh uh since uh I can never remember anything I start digging around uh at piles on my desk and I have a collaborator in Brazil with whom we've already written one paper on this subject and we have another paper coming out and he had sent me an email with approve oh two weeks ago I mean two weeks ago before this was done on chat 5.5 chat GPT 5.5 pro uh in which he also solved a problem and his problem was almost identical in its ideas and what it did to the one that came out of chat GPT 5.5 pro so I thought there are several I mean how Can this happen?

[00:49:16]Right? So, of course, it could happen in several ways. One, it could be that I wasn't smart enough to see the solution, but my friend in Brazil was and chat GPT was, and they came up with exactly the same solution. Another is that um these huge, you know, uh computer systems that suck up data everywhere somehow found the email this guy had sent to me uh and plagiarized it.

[00:49:49]>> Right. Or another possibility was that my friend wasn't as thought as smart as I thought. He had also asked chat GPT to solve the problem and it solved the problem. So what do you do? So I sent him an email. I said, "Diego, um, did you use chat GPT to solve this problem?" Because he came up with the same solution. And he said, "No, I I I" He's like me. He's like, you think of mathematics is like, you know, poetry, like you you write poetry, you write mathematics, you don't want a machine to do it. But he said, "My English is bad."

[00:50:28]So after I wrote up my solution, I put it into chat GPT just to improve the English. So maybe chat GPT already had this in his memory. You know, in any case, >> the answer is I have no idea what happened, nor uh uh maybe it's not possible to know what happened, but it's just really interesting that in this case, Chachi PT came up with uh exactly the same solution as this guy. Um and I guess it's because they're both very smart. Um and uh and actually there was one other instance with my friend in Brazil who had sent me something else which which again he had done but he had run it through chat GPT to fix the English or to make it grammatical and whatever and that also popped up somewhere else in something chat GPT did in solving uh but chat GPT 5.5 I don't know anything about these machines, but um Tim Gowers, who's a right guy at Cambridge in England, uh evidently uh was interested in one of my problems about the sizes of some sets and asked Chat GPT to try to solve it. Um and Chat GPT didn't. But it's somehow like the two of them are working collaboratively, right?

[00:51:58]Like Chat GPT does something is a good idea. Tim does something I think feeds it back in. I think something like that is going on. Um anyway, uh we seem to be entering a world where um this AI stuff uh is almost as useful as u talking to a guy over coffee down the hall. So, uh it's maybe not as bad as I thought. Uh, in any case, I thank you all for your attention.

[00:52:36]I enjoyed that.

[00:52:41]That's the story.

[00:52:52]So, Mel, uh, what was the query that you put to chat GPT to come up with a with a with the answer?

[00:53:01]>> What did you ask it?

[00:53:02]>> Are you asking me?

[00:53:04]>> Yes.

[00:53:05]>> I've never asked Chad GPT for anything.

[00:53:08]>> But you said you you entered it like >> No, no, no.

[00:53:14]um someone who's using the AI entered it. I have not used AI um for this at all. At all. Um though I will say that um some friends of mine got a very lovely lema out of uh chat GPT 5.5 Pro >> with a beautiful proof and I went to my 20 bucks a month Claude AI and I typed in the lema and I asked Claude AI to prove it and Claude thought AI came back with exactly the same proof and it wasn't maybe the proof was sort of obvious somehow but in each of them there was like an idea but they came up with the same ideas so that's why for all I know all of these AI systems really just depend on a couple of guys locked up in a penitentiary somewhere with a math library and whenever a query comes in they solve it and they type it out real fast >> maybe They used to do chess they used to do chess that way. They would hide a human in the chess machine and that's why they won. They don't do that anymore. But uh but but there are stories like that.

[00:54:37]>> Yeah, I suspect to be to be obviously that's not what happen. That's not what's happening here. But it is kind of curious. Uh >> may maybe someone uh sometimes uh do a wrong proof on purpose and feed into the uh AI chat GPT to see what happened.

[00:55:00]I I have asked Claude I mean I I ask Claude AI stuff like if I'm you know like lazy I say you know find this paper for me or what's the definition of this and once in a while Claude AI will come back with the answer I don't know and sometimes we'll come back with the answer this might be correct but I'm not sure. Um, okay.

[00:55:23]>> And I'm I'm impressed that it has that degree of human um that sorry, it was programmed with that or it taught itself because it's going to take over the world soon how to imitate a human.

[00:55:36]>> They're not overtaking the world.

[00:55:40]>> Well, you and I probably won't live long enough to see it, but I'm sure my children will.

[00:55:46]>> No. No.

[00:55:48]You're not convinced me.

[00:55:59]questions or comments or uh I would absolutely encourage people if uh who's anyone who's interested that some of the problems and that I mentioned are really nice and too nice for uh a computer to solve, right?

[00:56:15]>> Better if a human does it.

[00:56:20]very surprised.

[00:56:23]>> So, I've been um I'm teaching a course in set theory uh semester and and the difference between a set and a class.

[00:56:32]>> Well, yeah.

[00:56:34]Um so, so um I'm out of the textbook and I like to have my let my students have have lecture notes and my handwritten notes are just terrible. So I I've I've been mainly sort of quickly before like the day before I I teach a class, I dictate what I'd like to have in some lecture notes for the students. I basically dictate an outline of my lecture to to whatever AI we've got that >> uh my university subscribes to Gemini.

[00:57:06]So I tend to use that one uh for these things though it's terrible for mathematics. Um and it's mainly been good. But I've been talking about a thing called Goodstein's theorem. I wanted to give a lecture on a thing called Goodstein's theorem which I don't know if you're familiar with it. Um it's it's about a sequence of na it's proving that a certain sequence of natural numbers converges but the sequence can grow very very quickly. It involves putting things in hereditary number bases and then bumping up the number base. And so I I I dictated the outline of this to I tried this on three different AIs to the AI to get a lecture and they all sort of crashed. And the reason is that rather than work with the formal sort of um breakdown of the integer examples that I gave it, it actually tried to compute the sequence starting with some of those numbers and it very quickly grew so quickly. I mean, the proofs that this thing converges can't be done in piano arithmetic, even though it's a normal number theoretic sequence. So, it it it it broke all of the AIS I tried it on in creating creating lecture notes for my students.

[00:58:17]>> What is the proof? where where does the proof lie if not >> so so the way it works is so the idea is that if you have a natural number you can you can write it in base three for example but some of the exponents might be bigger than three so then you've got to write those in base three as well so hereditarily base n means that when you've written it out everything including the exponents is sort of in that in that number base. So the sequence is you take a number, you write it in hereditarily base 2, you bump all the exponents up to three, but you also subtract one from the number. And then you sort of iterate that next time, three to four, but you subtract one.

[00:58:59]Four to five, you subtract one. Um, and it eventually converges. Uh, this is uh Julius Goodstein from like 1944. So it's a very venerable result. But the way the proof works is you replace all the natural numbers with the ordinal omega.

[00:59:19]And so you track it. So you corresponding to each number written in this hereditarily base n form. You've got an equivalent form but written using using omega. And if you so it's an ordinal. And if you track what happens to the ordinals, even though the natural the natural numbers in the sequence might be growing, the ordinals are decreasing. and you can't have a an infinite decreasing set of ordinals. So eventually it has to converge.

[00:59:47]Um so it's it's um it's kind of a nice example if you're teaching undergraduate set theory of of of how you could actually use ordinals for something that you can understand which is this sequence of of natural numbers. But but yeah, no, you can pro it's provable that that that there's no no proof of it uh of convergence using piano arithmetic, which is also kind of kind of neat. That's that's much newer.

[01:00:18]But um yeah, you know, so formally in in one's head when one's writing out examples, one's just thinking of everything in terms of, you know, 2 to the 2 to the 2 + 1 plus, you know, things like that. But chat GPT and and Gemini were just trying to compute those integers first and then subtracting one and then reincverting them into hereditarily bas internally. And by doing that it just it just broke the broke the systems.

[01:00:51]>> That's beautiful David.

[01:00:53]>> Yeah.

[01:00:54]>> Thank you for sharing that story. That's score one for the humans. Yep.

[01:01:01]>> Someone asked u how do you write a paper if AI did all the proofs or sorry how do you Yeah. Like who are the co-authors?

[01:01:13]>> Exact. Yeah. That's a good question.

[01:01:15]>> You write that the proof was done by AI with prompts by you. You're a co-author.

[01:01:20]You you prompted it.

[01:01:22]>> Well, suppose you didn't prompt it. Just asked a question.

[01:01:25]>> That's a prompt.

[01:01:27]>> Sorry. If you say uh um Aristotle prove RH and it comes back with a valid proof, that's not a prompt. You just asked it to solve the problem.

[01:01:39]You know, if you if you have a suggestion like you know, you know, try using complex analysis. That's that's like maybe a prompt. But >> no, anything you ask it as a prompt, you can write the paper and say this was done with the following very simple prompt, which is just the question. No, that's perfectly acceptable.

[01:01:59]>> Uh, the integers journal forbids all papers that use AI for anything including code.

[01:02:11]>> Seems uh rather draconian.

[01:02:14]>> How about other computer assisted proofs like four color theorem?

[01:02:19]Would would integers I mean You're allowed to use computers, just not AI.

[01:02:28]>> Well, but AI is just a computer.

[01:02:31]>> Yeah, you would think >> it's not transparent. It's not transparent. It's not reproducible what you're doing.

[01:02:38]>> So, if I if I tell a computer, >> uh, I want you to look at all configurations uh, color configurations of this graph or something like that, that is something that's reproducible. I could read it and do it. If I say prove something and it comes back to me that that by the way happens in insurance a lot um there's a whole problem with predictive analytics. Regulators will not put up with something unless they understand what it means. So you can come to someone and say well they predicted we need this much reserves and they'll say can I see the outcome they say we don't know it. A regulator will not accept that you're playing with people's bank accounts and everything else. So there's a lot of logic to it.

[01:03:23]Uh you know I work at CMS on the side and we had a we had a problem like this.

[01:03:29]So they created a >> what is CMS?

[01:03:33]>> The center for Medicare Medicaid services. That's my daytime job. Today uh something I was working on which had a deadline crashed. So I said let me go back and enjoy myself with number theory. And thank you Mel very much.

[01:03:46]>> But uh I'll tell you what they did. They created something called Atlas. And what Atlas is good for, you cannot access it.

[01:03:54]Is if you want to know if the innovation center at CMS produced a model on something, it will do it. But they did it very cleverly. They said, "I want you to tell me what's known about this. Then I want you to give me the references.

[01:04:11]Then I want you to evaluate how confident uh someone could be in this."

[01:04:17]So when you ask them, did someone test this model once? They'll tell you yes or no.