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Examples of common false beliefs in mathematics
- jokoon 5y agoNot sure if "belief" is the right word.
- ianai 5y agoI’m getting re-asked cookie preferences every time I go to sites like SO. Is it OK for sites to ask for cookie preferences on return visits to their sites after preferences have been set and “saved”? Or maybe I’ve turned off the Cookie that would have saved preferences…
- sundarurfriend 5y agoI've just cosmetic-blocked those banners on uBlock Origin. It's a bit annoying to have to do it on every SE site, but these banners are huge and cover up a lot of the page, so it's worth it.
- leto_ii 5y agoThe weight of the names in this thread is really something. I'm constantly amazed at the quality of what is posted on Mathoverflow. Kudos to them for harbouring such a community.
- benrbray 5y agoSomething intersting about MathOverflow is that it started as an independent site, with the creators eventually agreeing to make it part of the StackExchange network. I believe part of the agreement allows the owners of MathOverflow to break off from the main site again, in the event of a major disagreement between MO and SE. I'm sure they're glad to have an escape hatch given the recent aquisition!
- tigerlily 5y agoHeh, I guess that makes them the Texas of StackExchange.
- MaxBarraclough 5y agoSomewhat related: I just learned of an Open Source rival to StackExcange by the name of Codidact. [0] Looks like they have an active math subdomain. [1] [0] https://codidact.org/ https://codidact.org/ [1] https://math.codidact.com/ https://math.codidact.com/
- throwawaysea 5y agoThese seem fairly advanced, so I don’t know if I would call them “common”. But maybe I am out of the loop. Are these concepts and very specific sounding beliefs actually widely known to math majors or CS majors?
- dwohnitmok 5y agoYes (although not for CS majors, this is very math heavy). For all of the top answers I went through, I think most math majors have believed it, know someone who has believed it, or could at least plausibly see how someone could believe it. To the question of how advanced these are, with only a few exceptions, the majority of things there are covered in undergraduate analysis, algebra, and topology courses.
- Rerarom 5y agoI have a math degree, and I could understand most of them by the end of my second year.
- qsort 5y agoMany of them are part of a standard CS curriculum. Most of them are part of a standard math curriculum. The "level of difficulty" is very uneven, though. Things like "Many students believe that 1 plus the product of the first primes is always a prime number" (a misunderstandment of Euclid's proof that there exist infinitely many prime numbers) or "If A implies B then B implies A" (difference between a necessary and a sufficient condition) are literally middle-school math. Things like "If 𝑓(𝑥,𝑦) is a polynomial with real coefficients, then the image of 𝑓 is a closed subset of ℝ" are high-school level. The top answer about dimensions of vector spaces is a well-known undergraduate-level "trap". "The Krull dimension of a noetherian integral domain is finite." is, to be brutally honest, not a sentence I understand in the slightest.
- dwohnitmok 5y agoThe Krull dimension one is one of the few that might not be found in an undergraduate program. You'd probably come across it in an algebra class that straddles the line between upper undergrad and intro graduate coursework (but Noetherian integral domain is something you would usually find in the sophomore/junior undergrad algebra class often taken as a prerequisite for the other hypothetical class). Also > "If 𝑓(𝑥,𝑦) is a polynomial with real coefficients, then the image of 𝑓 is a closed subset of ℝ" I don't think that this is high school level. Secondary school level math rarely touches anything related to multivariate polynomials (and usually doesn't have a notion of "closed sets" beyond closed intervals, but that's not as relevant apart from how the problem is stated).
- marto1 5y agoThese look so advanced to me the whole thread looks like nerds hurling dark magic spells at each other. And I consider myself technical.
- scarygliders 5y agoIt's kind of humbling to know that I can't understand any of that. I barely passed mathematics in British high school - but for some reason took to quadratic equations and other slightly more advanced maths in college, I enjoyed it back then - but simply cannot read nor understand just about all of the topics in that thread. I'd have to look up what the "/" in "M/L" means in the 5th reply (I didn't) There are so many _symbols_ used in mathematics. It's all as if they're spells written by wizened greybeard wizards and passed down to apprentices throughout the centuries.
- _lvym 5y agoQuadratic equations are now a standard part of British secondary school mathematics.
- scarygliders 5y ago"Back in mah day" - they weren't :)
- aardvark179 5y agoI think you may have wiped them from your memory. :-)
- qcoh 5y ago> I'd have to look up what the "/" in "M/L" means in the 5th reply (I didn't) M/L means that M is a field-extension over L. A concrete example would be C/R (the complex numbers over the reals). Algebraic number theory, in particular Galois theory, studies field extensions by looking at the group of symmetries: the field automorphisms of the larger field that fix the smaller field. For the concrete example above, the Galois group is a group with two elements: the identity function and the function that maps i to -i and keeps real numbers fixed. It's not a coincidence that the dimension of C as an R-vector space is the same as the size of this group (or that the degree of the polynomial that has i (and -i) as roots has degree 2).
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- Nitrolo 5y agoHere's an interesting one: "this set is open, hence it is not closed". Apparently this is wrong, and R is both open and closed. Does anyone have a clear explanation as to why this is or what this means? I can't quite understand the comment explaining it.
- akalin 5y agoIt's just unfortunate terminology -- a closed set is one whose complement is open, not one which is not open.
- prof-dr-ir 5y agoIndeed. Sets are not doors, as the saying goes, because they can be open and closed at the same time.
- zwaps 5y agoLet me try to explain it: One usually thinks of open sets as collections of numbers without a "boundary" - the boundary of the set is not part of the set but outside of it. If we look at the open set, we see that it's just "interior" points. You may have seen this when distinguishing on the number line between intervals like (-1,2) and [-1,2]. (-1,2) are all number between -1 and 2, but not including -1 and 2. [-1,2] however are the same numbers, but including -1 and 2. Slightly different sets, where the former is open and the latter is closed. Another way to think about it is for any point in (-1,2), we can find some number that is "closer to the boundary" but still in the set. This is not true for [-1,2] if we pick -1, because any number below -1 is suddenly outside the set. -1 is a boundary point, and therefore that side of the interval is not open! Why do we need this? Well, lots of calculus is about sequences that converge in smaller and smaller distances. One can for example define whether a function "jumps" by checking whether the inputs have boundary points or not (etc.). But for math people, this is not general enough, because it requires things like distances. Hence, they have a more general definition: There's a collection of stuff, and open sets are sets of this stuff that follow some rules. Now here is the important part: Closed sets do not have an extra set of rules. Instead, they are just defined as the complement (or opposite) of an open set. So R - the set of all real numbers - has a complement which is the empty set. The opposite of "all numbers" is nothing! Since R is open, by definition, the empty set is closed. Cool. But then, if we take our rules, we see the the empty set is also open. Whoops. Then, again by definition - the opposite of the empty set - which is R, so all numbers - is closed. Confusion conclusion: Both R and its complement, the empty set, are both closed and open at the same time.
- dandanua 5y agoI guess the probability theory and statistics is the field with most false beliefs. It's because human intuition doesn't really work, but everyone is trying to use it anyway. For example, the answers mention the belief that if 3 distributions are pairwise independent then they're jointly independent. Another false belief is that if 2 distributions are independent then they are conditionally independent. Also, people might think that you can pick a real number from R uniformly.
- BlueTemplar 5y ago> people might think that you can pick a real number from R uniformly This one seems to stand out though - is there a specific "non-math" situation where this is going to cause problems ?
- dandanua 5y agoI can't imagine a reasonable non-math context :)
- pfortuny 5y agoAny unspecified conceptual problem of the type “choose a number” has this issue.
- BlueTemplar 5y agoHow is "conceptual problem" different from "math-only context" ?
- coliveira 5y agoYes, people think they have some idea of what is random, when in fact the human mind is made to pickup on patterns. For example, people think that a random set of numbers is one where different numbers appear evenly distributed, which is just false.
- deleted 5y ago[deleted]
- dataflow 5y agoCan someone give a counterexample for this one? I read the comments but I still don't see how it's false: > If f is a smooth function with df = 0, then f is constant. I assume the smoothness and df = 0 criteria are said to hold on the whole domain, right? Yet the claim is that f fails to be constant over the same domain?
- flebron 5y agoConsider a disconnected domain (say, union of a few open balls in R^n), and f being constant in each connected component, but having different values in each ball. The differential is indeed everywhere 0 in the entire domain.
- deleted 5y ago[deleted]
- dataflow 5y agoD'oh, I missed the possibility of an open domain. Thanks!
- bmitc 5y agoYea, the statement is missing "locally constant". It's a poor example given the assumptions are not clearly stated.
- pfortuny 5y agoOh, these kind of problems happen time and again in real-life maths: you use a lemma until someone points out that you are assuming something which may not take place (like connectedness of the domain, here).
- bmitc 5y agoYea, but like I said, it's a poor example as a "false belief" in this context, because as stated, it isn't false for some unwritten assumptions and is false for other unwritten assumptions. In mathematics, well and actually everything, the importance of assumptions is paramount. The problem is just ill-stated. This happens in engineering and software all the time. Time and time again, issues being resolved usually revolves around clarifying assumptions.
- codeflo 5y agoI wonder what’s going on with the passive aggressive edit war in this answer: https://mathoverflow.net/a/23521 https://mathoverflow.net/a/23521 It’s about whether Euclid’s proof that there’s no finite set of primes is a proof by contradiction or not. The fact that this is disputed at all shows a certain unwillingness to use original sources — maybe each of them only looked at a different textbook’s restatement of the proof. Because no matter whether a proof of the theorem can be stated without contradiction, it took me all of 30 seconds to find a translation of the original proof to show that Euclid did in fact use one: > I say that G is not the same with any of the numbers A, B, and C. > If possible, let it be so. Now A, B, and C measure DE, therefore G also measures DE. But it also measures EF. Therefore G, being a number, measures the remainder, the unit DF, which is absurd. (http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX20.html http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX...)
- ogogmad 5y agoYou can read the proof as saying that for any finite set of primes, there is a prime number outside the set. And within the context of Euclid's proof, given the set {A,B,C} of primes, the proof constructs a prime G outside of it. Within constructive logic, there are different non-equivalent definitions of infiniteness (that are all equivalent in classical logic): - A set is non-finite if its finiteness leads to a contradiction. - A set is infinite if any finite subset of it can be extended. Infiniteness implies non-finiteness, but non-finiteness does not imply infiniteness. Further discussion can be found here: http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/infinite+set http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/infin... Within the context of the MathOverflow answer, the claim that Euclid's proof establishes infiniteness as opposed to non-finiteness, is correct. And the proof is therefore not "by contradiction", because that would only show non-finiteness. [edit] Edited out some impoliteness.
- codeflo 5y agoTrying to be maximally precise, the prime G isn’t really constructed, but it must exist because ABC+1 must have a prime factorization by the fundamental theorem of arithmetic. And the contraction is that G can’t be any of {A, B, C}. Edit: The above added a nit-picky side-point to your original response, which was a lot shorter. Afterwards, you edited your reply to be a lot more exhaustive and a lot less... friendly? I don't think that's the best way to use HN's edit window. But see my grandchild comment below for my response to all the points you edited in afterwards.
- IHLayman 5y agoThis article's posting on HN is possibly inspired by the recent post about the unreasonable uses of bubble sort [0] as it is a counterintuitive belief in CS that an inefficient algorithm could ever be useful. I am pretty sure that there are other beliefs in the computing world that are also common and completely false, such as "Deleting a file removes the data from the drive" or "Twice the number of cores working on a task means the process will finish twice as quickly." [0]: https://news.ycombinator.com/item?id=30112906 https://news.ycombinator.com/item?id=30112906
- AtNightWeCode 5y agoThe most common false belief I met during the years is that people think you can use math as-is in software. Reality is that you almost always need knowledge about numerical methods and for easier things you need to understand how floating-point arithmetic works.
- jdlyga 5y agoCommon false beliefs in people who know way more about math than I do.
- irrational 5y agoIt's things like this that help me to remember how very little I know about mathematics. I looked about halfway down and saw that I didn't have the slightest idea what any of them were talking about. I console myself with the thought that there are other subject matters that I am an expert in that they are not. Probably.
- BrandoElFollito 5y agoNo worries. I have a PhD in physics and understand a small fraction as well. Some I have no idea where to place (even though I went though extended math, and I quite like the subject).
- opportune 5y agoFor most of these you’d likely be able to understand the concept at a good enough level to see the contradiction if you looked up the definition of the concepts. For example I think the one about tr(AB)=tr(BA) not holding for arbitrary permutations across a set of more than two matrices is something you could try proving yourself.
- phendrenad2 5y ago> Many student believe that 1 plus the product of the first n primes is always a prime number. Wow, pretty sure I read this in New Scientist magazine in the early 2000s, quoting some "cryptography expert". Guess he wasn't such an expert.
- NieDzejkob 5y ago... or the journalist misunderstood and misquoted the expert.
- phendrenad2 5y agoI'm surprised this didn't devolve into a "0.9999 == 1" flame war. Good on the SO mods.
- astrange 5y agoMost of the proofs of that (like 1/3 * 3) feel like they're trying to trick you, but if you just write it as "0.999… == 1.000…" it feels obvious and easy to visualize to me.
- formerly_proven 5y ago> If f is a smooth function with df=0, then f is constant. Why would this be wrong? If the derivative is zero everywhere, how would the function not be constant? Or does this mean something completely different?
- pedrosorio 5y agoThink about the domain of the function EDIT: Someone already provided the answer here https://news.ycombinator.com/item?id=30146426 https://news.ycombinator.com/item?id=30146426
- z5h 5y agoGot a degree in math. For me, the repeated act of rigorously proving counter-intuitive results put me into a state of mind where I tried to avoid having "beliefs" until they were proven results.
- dhosek 5y agoI'm currently reading (almost finished, in fact), Roger Penrose's The Road to Reality. Despite having most of a master's in mathematics, I found his comment early on about trying to keep the mathematics accessible to be a bit laughable. Modern mathematics can get abstruse very quickly. It's likely that a typical HN reader could read the abstract of a PhD dissertation in just about any field except mathematics and have an idea of what the dissertation is doing. For mathematics, it's possible that even with a PhD if the dissertation is outside their specialty they might not know what it's about.
- JuanitaYoung 5y ago[dead]