9 ms·
All People in Canada are the Same Age (1997)
- deleted 6y ago[deleted]
- smallnamespace 6y agoStep 9 doesn't work because it introduces a third person, and so fails to demonstrate S(2)
- coldtea 6y agoIsn't it Step 4, which uses as a premise what the whole thing is supposed to prove? "in every group of k people, everyone has the same age" You can't use your conclusion in your assumption!
- phepranto 6y agoThat's the way induction works. Assume the statement is true for n Deduce that it then must be true for n+1 Prove it for n=1 Now it's proven for every n
- NieDzejkob 6y agoYou might want to read the section "A Brief Review of the Principle of Induction".
- deleted 6y ago[deleted]
- MauranKilom 6y agoPlease take a look at the principle of induction that is explained on that page. In short: You fix k and assume that S(k) holds. If you can show that this always implies that S(k+1) holds, all you need is a base case (e.g. S(1) being true) to "recursively" prove that S(n) holds for all n >= 1 (or whatever your base case was). Here the fault is that the induction step (proving S(k+1) from S(k)) requires k >= 2 but only specifically S(1) was proven.
- azundo 6y agoNo, not quite, the conclusion is that given a group of k people where all people have the same age, then a group of k+1 people also must have the same age. Since we know that in a group of 1 all k people have the same age, then if the proof held then a group of 2 people would have the same as well, then 3, etc. This is a common tactic in induction, you make an assumption, prove that if true it implies a general result, and then give an explicit case where the assumption holds, proving the general result.
- Ericson2314 6y agoThis is easier to understand in constructive type theory. You can't pattern match on `k` to recur, but you can pattern match on `succ(k)` to recur on `k`. Even when doing classic logic, I start with my constructive intuition and the sprinkle in the continuation-passing-style spooky magic when needed to get back the classical craziness. Everyone should learn constructive first.
- kerkeslager 6y agoYou're right if we were speaking normal English, but "assuming" doesn't mean the same thing in math jargon. In math, you would say, "Assuming A, then B", to mean, "If A is true, then B is true." It's way more confusing than simply "If A, then B", and I'm not sure why the wording hasn't fallen out of favor, but that's what it is.
- OJFord 6y agoIn what way is that different from 'normal English'?
- dlkf 6y agoI don't know why you're getting downvoted, afaict 9 is the problem.
- inimino 6y agoPerhaps downvoted for spoiling the problem? Some people review the comments before looking at the article.
- deleted 6y ago[deleted]
- nathanwh 6y agoTFA will inform you of this though, it's intended to be a game of reasoning. Posting the answer is spoiling the fun for people who read the comments first
- vages 6y agoWhy isn't there a mechanism for hiding spoilers in your comments? I came to the comments to check whether my guess was right (it wasn't), and I'm very glad for the spoiler.
- Thorrez 6y agoYou don't need to come to the comments to check your guess, the website lets you check your guess yourself by clicking on the step. > See if you can figure out in which step the fallacy lies. When you think you've figured it out, click on that step and the computer will tell you whether you are correct or not, and will give an additional explanation of why that step is or isn't valid.
- deleted 6y ago[deleted]
- thaumasiotes 6y ago> it's intended to be a game of reasoning. Posting the answer is spoiling the fun for people who read the comments This is a very common example in math classes, though in my experience usually presented as a proof that "all horses are the same color".
- jhanschoo 6y agoNote that Step 9 can be read as introducing an entity that is known to exist, or introducing a variable to be universally quantified over. It isn't until Step 13 that we see that Step 9 should be read as the former.
- deleted 6y ago[deleted]
- nathanwh 6y agoFun stuff, there's a few more on this page: https://www.math.toronto.edu/mathnet/falseProofs/fallacies.html https://www.math.toronto.edu/mathnet/falseProofs/fallacies.h... The last one in particular I thought was interesting
- exmadscientist 6y agoThe ladder problem is one that I seem to remember was in our first-year classical mechanics text back when I was in grad school, and caused a lot of debate... it's a good one.
- deleted 6y ago[deleted]
- toxik 6y agoHilariously I pinned down all the other problems in one try, except the everybody is the same age one.
- deleted 6y ago[deleted]
- dfee 6y agoIn 2003, I remember studying fallacies in English class. I literally had an outbreak of laughter during an exercise where the prompt was: “vote for me or admit you’re racist”. It seemed so ridiculous to teenage me that such a thing could be said. In 2020 it has been said. I’m no longer falling out of my seat laughing.
- amalcon 6y agoYou must not have been paying attention at the time. I was literally told "Vote for X or you are a traitor", which is the same fallacy, in 2003.
- kriskrunch 6y agoI know this as the Kafka trap fallacy. "A Kafka trap is a fallacy where if someone denies being x it is taken as evidence that the person is x since someone who is x would deny being x. The name is derived from the novel The Trial by the Austrian writer Franz Kafka." Source: https://debate.fandom.com/wiki/Kafka_Trap https://debate.fandom.com/wiki/Kafka_Trap
- thaumasiotes 6y ago> A Kafka trap is a fallacy where if someone denies being x it is taken as evidence that the person is x I think this concept is better known under the name "witch hunt".
- pcwelder 6y agoAlso related to the fallacy of false dichotomy. https://en.wikipedia.org/wiki/False_dilemma https://en.wikipedia.org/wiki/False_dilemma
- andrepd 6y agoWho has said it?
- microcolonel 6y agoHillary Clinton, several ways; though most famously in the Basket of Deplorables monologue.
- thoughtstheseus 6y agoThis reminds me of a Buddhist game/tradition(?) of debate where you try to get the other person to admit to a logical fallacy and sway them to your belief.
- deleted 6y ago[deleted]
- cf-d-ycom 6y agoSo this article shows that you can infer S(n+1) from S(n) for n > 1, and a base case of S(1) is true. However, you can't infer S(2) is true from assuming S(1) is true in the same way, ie. a group of 2, could be represented as two groups of S(1) and S(1). You can't claim these two S(1) groups share the same age. This means that the base case and the inductive step are not connected, which means the proof is invalid.
- cheez 6y agoNot exactly, it was saying that you assumed S(2) implicitly which is wrong.
- cf-d-ycom 6y agoYeh, that makes more sense.
- solids 6y agoWhy can’t you asume S(2)? Is it not included in the inductive hypothesis S(k)?
- cf-d-ycom 6y agoSo the base case that they prove is S(1). And the inductive step is to show: S(1) ->(implies) S(2) -> S(3) -> S(4) -> .... Or to write it more succinctly, that S(n) -> S(n + 1) assuming S(n) is true. In this particular problem, the proof that is provided can be used to show S(2) -> S(3), that S(3) -> S(4), etc... are valid and true. However, the proof could NOT be applied to show that S(1) -> S(2). So whilst you CAN assume S(2) is true in a proof, the whole inductive chain needs to be attached to a valid base case.
- cheez 6y agoThe page gives a better explanation than I could :-)
- repsilat 6y agoIt's simpler than that -- the "proof" of the inductive step is just incorrect. It wouldn't be a theorem in a sound logical system.
- solids 6y agoI heard this in the “All horses have the same color” version.
- abhgh 6y agoThe version I had come across was "all billiard balls have the same color" in Liu's Discrete Mathematics [1]. It is an exercise problem in one of the chapters. [1] https://www.amazon.com/Elements-Discrete-Mathematics-C-Liu/dp/0071005447 https://www.amazon.com/Elements-Discrete-Mathematics-C-Liu/d...
- Chrisoaks 6y agoI also think step 10 is wrong because it assumes that P ≠ Q.
- thedufer 6y agoThere's an imprecision here, but that doesn't break the proof. Note that from step 6 on, all it needs to show is "if P and Q are any members of G, then they have the same age". If P = Q, this is trivial, so we really only need to consider the P ≠ Q. This probably should have been stated, though.
- gweinberg 6y agoIf it were true that all groups of two people were the same age, then it would follow that all groups of any number of people are the same age, since any individual could be pared with any other individual to form a group of 2. But of course it is not true.
- tmabraham 6y agoFor a mathematical fallacy, see the appendix of Charles Seife's book "Zero", proving that Winston Churchill is in fact a carrot!
- sandworm101 6y agoA similar conjecture: https://en.wikipedia.org/wiki/Doomsday_argument https://en.wikipedia.org/wiki/Doomsday_argument "In other words, we could assume that we could be 95% certain that we would be within the last 95% of all the humans ever to be born."
- Dolores12 6y agoPeople is a group of more than one. So S(1) is invalid case.
- brazzy 6y agoThat's not the problem at all.
- rkagerer 6y agoReading this "proof" gave me a headache, and at first I thought it's because I'm not familiar enough with formal inductive logic to precisely follow the conversion from statements of reasoning to shorthand notation. Then I realized the aim is to trick you by playing a bit fast and loose with that convention, and hoping you don't notice. e.g. I stumbled at Step 4, and if you click on the details for it the authour admits k is ill-defined. That comes back to bite you when you hit the fallacious step. I'm fascinated how some very old works by ancient physicists and mathematicians are written using plain (if verbose) language and diagrams, and you didn't need to learn a bunch of shorthand conventions specific to the field in order to participate. Does anyone know any good books on Quantum Mechanics that don't require you to learn Dirac notation first?
- aidenn0 6y agoMany books that just give an overview of QM don't use dirac notation; look for books with "Modern Physics" or "Introduction" in their names. However, Dirac notation is used so ubiquitously in QM that your question is a bit like asking if there are linear algebra books that don't require you to learn matrix notation. Sure you can do linear algebra without matrices, but that would be a bit eccentric today.
- DarkmSparks 6y agoThey just forgot to define age as: "born after 1800"
- raverbashing 6y agoI disagree with the take here The fallacy is assuming that a group with more than 1 person has the same age. It would be ok to assume this if we were looking for a proof of contradiction, but this statement is never challenged nor contradicted This sounds very much like those "gotchas" that confuse more than help
- tsimionescu 6y agoThat is not the fallacy, that is the wrong conclusion. The fallacy is in step 9 combined with step 1. Step 9 requires at least 3 people to exist - P, Q, and R. So, step 9 only works for k>=2. So, we have proved S(1),S(k>=2) => S(k+1), but we haven't proved S(2). Of course, S(2) (in any group of 2 people, both people have the same age) is not true, so the whole conclusion is false. In inductive proofs you always need to prove some rule that says 'for any k [with some property], assuming case k is true, then case k+1 is true as well', and then you also need to prove that, for some k [with the given property], case k is actually true. Restating the proof in the article in these terms, step 9 correvtly proves that, for any k [greater than or equal to 2], if S(k) then S(k+1). But there is no proof given that there exists some k>=2 for which the statement actually holds, and in fact it can proved that NO such k exists. So overall the proof doesn't hold.
- raverbashing 6y agoI see your point, but I kinda disagree, and again, that's why this is more confusing than helpful. You're taking a false premise and running with it, then tripping far ahead and saying that's the fallacy. > but we haven't proved S(2). Well, not surprising you haven't proven it, because you're already deep down in the mud on steps 7 and 8 > Step 7: Consider everybody in G except P. These people form a group of k people, so they must all have the same age > Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age. Given that > Let G be an arbitrary group of k+1 people This is already false Saying that > Let R be someone else in G other than P or Q. Is something completely natural for a group with k+1 elements (with the exception of k < 3), but the "proof" is so deep down in its absurdity at this point calling this the fallacy is almost a technicality
- jmchuster 6y agoThe link to the answer: https://www.math.toronto.edu/mathnet/falseProofs/guess27.html https://www.math.toronto.edu/mathnet/falseProofs/guess27.htm... It took me a couple visits to realize that each `Step 1` `Step 2` heading is a link to an explanation for why that step holds true.
- trabant00 6y agoI don't understand some things regarding the resoning: If you assume S(n) is true, n being any natural number, what good does it to to prove that S(n+1) is also true since it is included in the initial assumption imho. If you define "P and Q are any members of G" then "everybody in G except P" can only mean to me an empty group. Also "Let R be someone else in G other than P or Q" can only mean R must be outside the group. Can somebody explain these to me?
- zwaps 6y agoEdit: Let me try to be as clear as possible. Your confusion comes from the language in the question of "ANY SPECIFIC N" versus "ANY, as in ALL N". Say, n=3. Then, we assume S is true for the value n=3, but we don't yet know if S holds for any other n (1,2,4,100 etc.), which we have not assumed. However, we show that IF S is true for n=3, THEN that alone implies it's also true for 4 (n+1). Of course then, we could probably show the same process for SOME value of n, 3 or otherwise, and the corresponding n+1! So instead of writing n=3, we just write n and save ourselves the task of having to check all those numbers. Nevertheless, our assumption was that S was true for one specific n, and not all n at the same time! So we do two things, and it makes sense to think of it in reverse order. First, if S is true for any specific n we pick (and not necessarily ALL other n), is it THEN also true for n+1? At this stage, we have identified two "n" for which S holds of all possible n. More precisely, we have picked one and assumed S holds, and then found a second one. However, note that the choice of which n we assume to be true was "free" among ALL n. Think of "dynamics" instead of "state" if you are an engineer. But we started with the assumption that S is true for at least one n, the n we chose. Maybe we can not find such an n, in which case it doesn't matter that it would also be true for n+1. Therefore, the second question is: can we find an n for which S is true? Then it's true for ALL n! But we need the two components. We need to know the relationship of n->n+1, and we need at least one "real" value of n where S does in fact hold. Your second question is similar. The author means: Pick any two members of G, but you gotta pick two specific members. Since the proof works independently of whom you pick, it goes through for all others. However, it does not mean that you "pick all members" of G! The "gotcha" moment you had, seeing that the choice of the two members was arbitrary, is usually what completes such a proof. You show the thing you want to show for two specific members of G, but then you circle back and state proudly: "But see, I could have picked ANY member of G. Hence, this must holds for all "dyads" of two people in G!" Edit: And to understand why the proof fails - the author does not actually proof the n->n+1 step for any arbitrary n, only for n>=2. The specific "real" value he finds, however, is n=1. Therefore, the two steps are disconnected, the n->n+1 does not hold if n=1. And this is precisely where the "ANY n" versus "ANY as in ALL n" comes into play again. n=1 is qualitatively very different to n=2. S is obviously true for n=1 (one person), but it's obviously not always true for n=2 (two people). While the author proofs that n=2 implies S for all n, this is not true if we start with n=1. However, n=1 is the only thing we can actually show to be really true. In that sense, for the purpose of showing that all Canadians are of the same age, the asserted step of n->n+1 (given n>1) is irrelevant, because we can not find such an n>1 where we can show that this is true! Here you can see that while we do the proof for any arbitrary but specific n, not assuming it is true for any other value of n, we still need to ensure that we do in fact mean ANY of the n we can pick, including n=1!
- zwaps 6y agoAnother simple way to see this is to reverse the usual order of steps, starting with the induction step. The author shows that S(n)->S(n+1) and that proof is correct if n is at least larger than 1. The hidden assumption of n>1 is the "gotcha" moment, but that part of the proof is still valid as induction step. Imagine starting with this and amending the instructions with "given that n>1". However, for the purpose of showing that all Canadians are the same age, we now need to find a base case - the first (and now second) step of induction. And here, we see that while n->n+1 holds (for n>1), there simply isn't any n>1 for which S is true! The case of S(1) is irrelevant, since it's not included in the assumption of the induction step. If we would have started with the induction step, and concluded that our argument holding for n>1 is good enough, we would have then clearly realized that there is no base case and therefore we can not complete the induction proof.