978
$\begingroup$

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

$\endgroup$
27
  • 131
    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ May 6, 2010 at 0:55
  • 33
    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$
    – Unknown
    May 22, 2010 at 9:04
  • 28
    $\begingroup$ wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. $\endgroup$
    – Suvrit
    Sep 20, 2010 at 12:39
  • 25
    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
  • 27
    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

281 Answers 281

1
6 7 8 9
10
-1
$\begingroup$

Way late to the party...

"$ \mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"

$\endgroup$
5
  • 15
    $\begingroup$ What does polymod mean? $\endgroup$ Oct 20, 2010 at 11:47
  • $\begingroup$ Either the cousin needs a bit more detail if it is to be false, it is quite naive! $\endgroup$ Oct 20, 2010 at 18:25
  • 3
    $\begingroup$ Probably I understand what this means: if $f(x)=0\pmod 2$ for all $x$, then $f=0$ over $\mathbb F_2$. This is similar to my second example: mathoverflow.net/questions/23478/… $\endgroup$
    – zhoraster
    Oct 20, 2010 at 18:33
  • $\begingroup$ Consequently, there are only $4$ polynomials over $\mathbb F_2$ Isn't this convenient? :-) $\endgroup$
    – zhoraster
    Oct 20, 2010 at 18:40
  • 2
    $\begingroup$ $\mathrm{polymod}$ is "polynomial mod". Two polynomials are congruent $\mathrm{polymod} p$ iff the coefficients each power of the variable are congruent $\pmod{p}$. The equivalence classes are sets of polynomials where each coefficient ranges over an equivalence class $\pmod{p}$. For the cousin, there are many local/globals but they all seem to require additional conditions (q.v. Hensel lifting). I think the set from which $x$ was chosen was left unspecified because this "imprecise mental abbreviation" pops up at various levels of sophistication each with a different such set. $\endgroup$
    – Anonymous
    Oct 23, 2010 at 15:22
-1
$\begingroup$

From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

$\endgroup$
9
  • $\begingroup$ I followed your link, and I cannot even tell what is wrong about attaching helium balloons to both sides of a balance to model substraction on both sides of an equation. $\endgroup$
    – user11235
    Apr 10, 2011 at 20:32
  • 23
    $\begingroup$ The more I think about this "error", the less I am convinced. It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$. Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph. Also, I think it is perfectly understandable what it means to add something halfways. $\endgroup$
    – user11235
    Apr 10, 2011 at 20:50
  • 2
    $\begingroup$ It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/most-harmful-heuristic $\endgroup$ Apr 10, 2011 at 21:17
  • 3
    $\begingroup$ When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. $\endgroup$
    – roy smith
    May 9, 2011 at 3:06
  • 3
    $\begingroup$ The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. $\endgroup$ May 20, 2011 at 2:28
-1
$\begingroup$

There is a problem with the quantum logical significance of Soler's theorem.

Maria Pia Soler's theorem is a deep, difficult and above all extremely elegant purely algebraic characterization of infinite dimensional Hilbert spaces.

If $K$ is a skew field with involution, $V$ a $K$-vector space equipped with an hermitian form, then the following two conditions are equivalent:

(1) there is a infinite orthonormal sequence in $V$, and orthomodularity holds (for any subset $X$ of $V$, the orthogonal $X^\perp$ and the double orthogonal $X^{\perp\perp}$ give a algebraic direct sum decomposition of $V$)

(2) up to isomorphism (a unique one for the self-adjoint part of $K$), the $K$ with involution is one of the usual three (real, complex, quaternions, all with the usual involution) and $V$ with hermitian form is a infinite dimensional Hilbert space.

Among other things, this destroys the false belief that the concept of "Hilbert space" is something "reserved to analysts" with algebraists restricted only to more elementary cases by the lack of interaction with topology.

This mathematical achievement is not sufficiently celebrated. Above all because quantum logicians do celebrate it, but for the wrong reason. To understand why, start with a better known and simpler question.


  • There are two ways to see why measures are real numbers.

The first: measures form a (open subset of) the positive cone in a totally ordered Archimedean abelian group with fixed unit (of measure), and any such has exactly one embedding in the real ordered additive group with 1 as unit. Either such a group is that of integers (with an arbitrary $n>0$ fixed as unit), or there is exactly one completion, the real number system. Since nothing is lost in the completion, assume real numbers.

The second: measures form a cancellative commutative semigroup, which is also totally ordered. Embed the semigroup in a group, complete it by Dedekind cuts (every poset has such a completion). You have a Dedekind complete totally ordered group, i.e. the real number system.

Everyone uses the first method (including Bourbaki, introducing real numbers essentially in this way). Why? Because the second is wrong. Yes, you can "complete" a semigroup to obtain a group (and for cancellative abelian ones one has an embedding, and a compatible total order uniquely extends), and yes, you can complete by cuts any poset (with a total order preserved as total). But the two constructions are (in general) mutually incompatible: the order completion of a totally ordered semigroup (resp. poset or total order) is again such, but when you combine the two completions (from semigroups to groups; from posets to complete lattices) you discover a incompatibility (semigroups losing cancellation of infinite elements), unless the order is Archimedean.

  • The second question is about quantum logic, formalized as a "AC irreducible orthoposet with free mobility", which corresponds to the foundations of elementary geometry.

"Irreducible AC orthoposet" means poset with a orthocomplementation whose completion by cuts (which always exist and is a orthocomplemented complete lattice) is a complete AC irreducible lattice as treated in Maeda - Maeda book "theory of symmetric lattices", including the equivalence with the ideal of finite height elements and the essentially unique coordinatization except for nonarguesian planes, lines and points: finite dimensional subspaces of a vector space equipped with an anisotropic hermitian form, normalized by the existence of a vector of "length" 1 (so that the ortholattice with a fixed atom and the vector space with such a fixed vector are really equivalent concepts, even in a categorical way). Free mobility, more generally introduced by von Neumann in quantum logic, in this context coincides with the one treated by Baer, first in a paper (solving the linear algebra version of the "Riemann - Helmholtz free mobility problem") and then in the book about linear algebra and projective geometry. The algebraic equivalent that Baer found is the generalization of the case treated by Hilbert's grundlagen (identity involution): the skew field with involution is "$*$-pythagorean $*$-formally real" (a sum $xx^*+yy^*$ has again the form $zz^*$ and is zero only when $x=0=y$), and each finite dimensional subspace has a orthonormal basis for the form. Really von Neumann had already treated a more general case involving $*$-regular rings, and this was then re-discovered in many works, the most important ones by Vidav, Hendelman, Burke, Ara.

  • There are two ways to see why an AC irreducible orthoposet with free mobility and length at least 4 (or equivalently the ortholattice of finite dimensional subspaces of a vector space of dimension at least 4 equipped with an anisotropic hermitian form with Baer condition), to be also useful in quantum logic, must be (embeddable in) a standard, i.e. Hilbertian, one.

The (historically) second: assume a complete lattice and orthomodularity, then apply Soler's theorem (in infinite length; disregard finite length).

The (historically) first: disregard points (classical logic, no need for quantum), lines (noncontextual hidden variables, no need for quantum), nonarguesian planes (not embeddable in larger irreducibly quantum logics, they have only classical interaction with other quantum components). Use von Neumann's transition probability primitive additional concepts (with relative axioms and proofs, for the easier type I case only) to have a Hilbertian embedding (without any need of von Neumann's completeness axioms). Then observe that one can always complete the resulting structure, and in presence of the completeness axioms one has a orthodox Hilbertian quantum logic.

Everyone uses the second method. Why? Perhaps because they avoid to read von Neumann (too long, too difficult?), but above all because they avoid to see that that the second method is wrong, for the exact same reason that the second method is wrong for the measure problem. The two axioms, orthomodularity and complete lattice, can be separately justified (any orthoposet has a completion by cuts that is again a orthoposet; any structure propositions - states has an internal logic which is orthomodular, being an orthoalgebra) but they generally clash when used together (case of pre-Hilbert spaces, incomplete infinite dimensional spaces: the orthomodular poset of splitting subspaces is not a (complete) lattice; the complete ortholattice of orthoclosed sets is not orthomodular). No completion process is generally known for an orthomodular structure to produce again a orthomodular (and complete lattice) structure. Except for the metric completion given by von Neumann's method, which needs specific assumptions.


Complementary notes.

  • [A] There is a (not so big) difference between the two cases.

For the embedding of a semigroup in the reals, the non-trivial axiom, to be added to commutative cancellative and torsion free, is clear: add the new concept of order, and postulate Archimedean (even to express the postulate "orderable" in not enough). An easy necessary and sufficient condition, and one physically "ideally falsifiable" by a "Gedankenexperiment" (with any pair of homogeneous magnitudes, one can compare multiples on one with the other; for lengths, with subatomic scale vs. cosmological scale, this is not practical but at least ideally possible. With arbitrary subsets, searching for Sup and Inf is not even ideally feasible. A equivalent form of completability is somewhat physically OK, completeness itself is not directly so).

For (the finite dimensional ideal of) a AC irreducible orthoposet with free mobility to be embedded in a Hilbert space, again von Neumann's condition (add the new concept of state, or that of transition probability, and the relative axioms) is necessary and sufficient, but this time it is not so easy to understand (and, in particular, is is not clear how much it can be formally weakened and still remain equivalent to standard embedability). But it is still a Archimedean axiom with a noncommutative twist, because probabilities take values in the Archimedean $[0,1]$, that can be replaced by a unit interval in an Archimedean totally ordered group with strong unit.

[One can also look at Chaqui work about "qualitative" foundations of probability using a binary relation "at least as probable as" between elements of a Boolean algebra; since for the type I$_n$ factors Boolean subalgebras are finite, one can apply Chaqui "countably additive" theory without philosophical problems.]

  • [B] Philosophical note.

Once Popper criticized a specific point of von Neumann - Birkhoff quantum logic, but his argument was based on a plain mathematical error (or at least a misunderstanding leading to that error). Only one review had the ardor, at the times, to say that there was something wrong with the argument. But for the rest, total silence (both from Popper and from others) in printed sources. Only many years after Popper left this world someone else has the courage to refer to the episode. https://duckduckgo.com/?q=popper+quantum+logic+birkoff Mathematics is not philosophy. One should not be afraid "by authority" to say when a mathematical argument (about quantum logic or whatever) is wrong.

  • [C] The path leading to Soler.

Before Soler's theorem there were many attempts to characterize Hilbert spaces in a "quantum logic" way.

Before purely algebraically attempts, there were partially topological ones [Zierler; Cirelli Cotta-Ramusino Novati; Szambien]. Three steps: (1) use MacLaren - Piron theorem (1963, but it is a very easy and standard extension of the finite dimensional setting of von Neumann and Birkhoff, 1936, using the fact that the ideal of finite dimensional elements completely determines the structure) to reach a vector space with an anisotropic hermitian form. (2) Use the well known (since the 1930 at least) characterizations of the three standard $*$-sfields as the only non totally disconnected T$_0$ locally (countably) compact topological rings which are sfields, and the (von Staudt) correspondence between the sfield operations and incidence geometry operations between points and line in a plane. (3) Use Amemya - Araki theorem (over the classical $*$-sfields, a infinite dimensional pre-Hilbert space with orthomodularity is Hilbert); this last step is not needed if one only wants embedability and so show that a orthomodular completion is possible.

[It seems to me that this partially topological approach is physically sensible, since the topological conditions can be seen to have sense looking at the need of a continuous evolution in time of the system (pure states bijectively correspond with atoms [seen as proposition answering Mackey's yes-no questions "is the system in such a pure state?"], and evolution in time gives continuous paths; (pre)compactness is required for pure states which are superpositions of a finite number of pure states). However, for reasons that I cannot understand, this method (with its merit of not requiring completeness + orthomodularity, and showing the non obvious result that a orthomodular completion exists) was not considered interesting by quantum logicians, who instead wanted a non-topological method using a complete orthomodular lattice. Note: I consider instead completely justified the mathematicians who searched for a solution of a purely mathematical problem.]

Then Morash extended Amemiya - Araki to $*$-subfields of the quaternions and some pre-Hilbert spaces over them, Gross and coauthors (Kunzi, Keller) had quite general results (among them, Keller's rightly celebrated example showing that orthomodularity alone does not force the $*$-field to be one of the three usual ones), and one of the best results for many years was Wilbur's theorem (and Holland's re-working of its methods using $*$-ordered $*$-sfields). The mayor problem with this is that only commutative and quaternional cases are covered; nothing can be said for non quaternional noncommutative $*$-sfields.

  • [D] Personal recollections.

To apply Solers's theorem in quantum logic one has not only to justify a complete orthomodular lattice (a thing that is not as obvious as quantum logicians suppose, as noted above), but also the infinite orthonormal basis. Even disregard the problem with the "infinite" part (one can dismiss logics with non-contextual hidden variables, and also non-arguesian planes since they have no embedding in any larger purely quantum logic and so they admit obly classical interactions with other parts of the system. But why can one disregard all other finite dimensional irreducible components, or equivalently say that they must have an embedding in an infinite dimensional one?) One still has the problem of justifying "orthonormal", and every attempt at this that I know boils down to a "free mobility" property of the orthogeometry, and so to von Neumann's own discussion of such a property in quantum logic (a brief discussion in his "continuous geometries with a transition probability", but a fuller discussion in other manuscripts of his reviewed in his collected papers).

Immediately after Soler's result, a (mathematical, not physical) paper appeared, with a (correct) celebration of Soler's achievement, and a final section with the application to quantum logic. A section with all the above problems (including a "ample unitary group" axiom without reference to such a "free mobility" property being relevant in quantum logic going back to von Neumann, and all other things as above). I wrote to the author that, in view of his recent paper, he should also consider von Neumann's old results.

It's better avoiding discussing his reply.

After almost 28 years I am still patiently confident that one day even quantum logicians will see von Neumann's light.

$\endgroup$
-2
$\begingroup$

For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

$\endgroup$
3
  • 6
    $\begingroup$ Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! $\endgroup$
    – Todd Trimble
    Mar 5, 2015 at 14:25
  • $\begingroup$ "$\mathbb{Z}_p$ is countable" is also a false belief for people who didn't really read the definition of $\mathbb{Z}_p$, but I don't know how much it is common. $\endgroup$ Mar 5, 2015 at 14:34
  • 10
    $\begingroup$ It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. $\endgroup$
    – Todd Trimble
    Mar 5, 2015 at 14:52
-2
$\begingroup$

In algebraic topology, I thought for a while:

  • "For $k \geq 2$, $H_k$ is the abelianization of $\pi_k$." False. True for $k = 1$. Also for all $k$ up to $n-1$ if the space is $(n-1)$-connected for $n \geq 2$ (vacuously, since this says the first $n-1$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $\pi_k \cong H_k$). See the Hurewicz theorem for more.
  • "Generically, all the $\pi_k$ are nonabelian." False. For $k \geq 2$, $\pi_k$ is abelian.

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

$\endgroup$
5
  • $\begingroup$ First bullet: did you mean "True for $n=1$"? $\endgroup$
    – Michael
    Jan 15, 2019 at 23:06
  • $\begingroup$ @Michael : It's not always true for $n=1$, $\pi_1$ can be abelian, e.g. the fundamental group of the circle. For $n > 1$, $H_n \cong \pi_n$. It's easy to imagine "$\pi_n$s are (usually) nonabelian monsters and their associated homology groups are friendly abelian groups", but this difference *only* happens for $n=1$. $\endgroup$ Jan 15, 2019 at 23:31
  • $\begingroup$ I think you are confusing a few things here. Compare $H_2$ of the 2-dimensional torus with its $\pi_2$, for example. $\endgroup$
    – Michael
    Jan 15, 2019 at 23:34
  • $\begingroup$ @Michael : After actually looking up what I was talking about, I find that I have mashed together (at least) two errors to make another. Yay? $\endgroup$ Jan 16, 2019 at 4:42
  • 1
    $\begingroup$ @Michael : I think I've disentangled my mangling. I may still have a fumble-thought in the first bullet that I'm just not seeing. $\endgroup$ Jan 16, 2019 at 5:11
-3
$\begingroup$

I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

$\endgroup$
7
  • $\begingroup$ Why the downvote! I heard this from more than one student in introductory linear algebra classes and when marking. $\endgroup$
    – Benjamin
    May 12, 2015 at 22:21
  • 9
    $\begingroup$ I think this falls under $(x+y)^2=x^2+y^2$, $\endgroup$
    – Thomas Rot
    Aug 10, 2015 at 12:48
  • 7
    $\begingroup$ It always fails... But I don't think this is a common held belief. $\endgroup$
    – Thomas Rot
    Aug 10, 2015 at 21:40
  • 1
    $\begingroup$ @ThomasRot But it always fails, while $(x+y)^2=x^2+y^2$ sometimes holds, especially in characteristic 2. $\endgroup$
    – ACL
    Apr 21, 2016 at 6:37
  • 4
    $\begingroup$ I meant that $V-U$ cannot be a subspace since it doesn't contain 0. On the other hand, in any commutative ring where $1+1=0$, then the formula $(x+ y )^2=x^2+y^2$ holds. $\endgroup$
    – ACL
    Apr 21, 2016 at 10:02
-3
$\begingroup$

There is a problem with the correct statement of the second incompleteness theorem (hence also with its understanding).

No, I'm not talking of cranks.

What I report here was published two times by a reputable mathematician in 1967 and 1972.

According to S. Feferman, a summary of the bi-published assertion is

the consistency of a system S containing elementary number theory Z (directly or by translation), may be provable in S; indeed, the consistency of very strong S may even be provable in a system of primitive recursive number theory.

and Feferman remarks

this assertion seems to contradict Gödel

Since the note is short, it is possible reproduce it completely:

The best and most general version of the unprovability of consistency in the same system.[*]

Under the sole hypothesis that Z (number theory) is recursively one-to-one translatable into S, with demonstrability preserved in this direction, the consistency (in the sense of non-demonstrability of both a proposition and its negation), even of very strong systems S, may be provable in S, and even in primitive recursive number theory.

However, what can be shown to be unprovable in S is the fact that the rules of the equational calculus applied to equations demonstrable in S between primitive recursive terms yield only correct numerical equations (provided that S possesses the property which is asserted to be unprovable).

Note that it is necessary to prove this "outer" consistency of S (which for the usual systems is trivially equivalent with consistency) in order to "justify" the transfinite axioms of a system S in the sense of Hilbert's program.

("Rules of the equational calculus" in the foregoing means the two rules of substituting primitive recursive terms for variables and of substituting one such term for another one to which it has been proved equal.)

This theorem remains valid for much weaker systems than Z. With insignificant changes in the wording it even holds for any recursive translation of the primitive recursive equations into S.

[*] This has already been published as a remark to footnote 1 of the translation (1967, p. 616) of my 1931, but perhaps it has not received sufficient notice.

K. Gödel, Collected Works, II, pag. 305 (1972a, Some remarks on the undecidability results)

The commentary by S. Feferman goes from pag. 282 to pag. 287. It is very worth reading, since the bi-published note above is authored by K. Gödel.

Besides that comment you can also read some posts on mathoverflow:

Clarification of Gödel's second incompleteness theorem

the choice of representing formulas and Gödel's second incompleteness theorem

Relationship between first and second incompleteness theorems

Before that second volume was published in 1990 (and so I could read it), I found no warnings in logic textbook (but after I discovered that E. Mendelson had a small warning-footnote in at least one edition of his textbook), I never found a paper that when using the result checks (or at least warns that one should check) for the hypotheses (like the derivability conditions), I never found (among the professors, all of them not specialized in logic, then at my reach, in the pre-internet-for-all era) an answer to my questions:

(A) why call that arithmetical proposition "the" formalization in S of the consistency? Are not there many possible formalizations, why should they be equivalent?

(B) How can one exclude that completely different ways to formalize the consistency exist, and give even more inequivalent formalizations, and some of them would be provable?

At the time, I did not have other obvious questions:

(1) why bother with a S-proof of a S-proposition expressing S-consistency? If S is inconsistent it can prove anything, including that proposition, so existence of a proof would mean nothing.

(2) why call "formalization of consistency" a proposition $\forall n\phi(n)$ [where, for each fixed numeral $n$, $\phi(n)$ is evidently decidable and it is true iff the $n$-th proof in S is not a proof of contradiction]? The assertion of consistency is the infinite conjunction of those propositions (and is outside first-order language), and the first incompleteness theorem exactly shows that for propositions of that form the universal quantifier is strictly stronger than the infinite conjunction (over the standard numerals, but the first order quantifier requires truth in all models, including nonstandard $n$).

[Note that (2) is related to the next (3) and to the sometime proposed counter-examples to Church-Turing thesis, like the one that Tarski possibly believed at the end of his career.]

(3) consider the following pseudo-argument. ZF (with infinity, with or without choice), or ZF with the negation of infinity (which is essentially the same as first order Peano arithmetic) are known to be "reflexive": they prove the consistency of their own finitely axiomatized subtheories. But a theory is consistent iff does not prove, using only a finite number of axioms, a contradiction. So ZF proves its own consistency.

Finally, a very good (in my opinion) way to avoid to check derivability conditions when using the theorem is this. Work only in finitely axiomatized set theories (like NBG, or bounded Zermelo, or the intersection of the axioms for bounded Zermelo and NFU, or NFU, or extensions of the above with a finite number of axioms). Note that for a finitely axiomatized first order theory T, the proposition "M is a model of T" is a formal set theoretic proposition without the need of any syntactical tricks. [This corresponds exactly to what Feferman noted in 1960 about a "best" way to (try to) express consistency of a finitely axiomatized T with a arithmetic proposition of the form $\forall n\phi(n)$ as above]. Now use the form of the theorem (T does not prove "there exists a model of T", unless T is inconsistent) as proved by Jech (a proof that at that time I misunderstood, conflating "models of T including the well-foundation axiom", "transitive models of T including well-foundation", "well founded models of T"). See also Godel's Second Incompleteness theorem and Models and https://mathoverflow.net/a/51786/33039

Bottom line: the second incompleteness theorem is way, way subtle[*], and many non-logicians in mathematics do not realize this.

[*] As is was once said (perhaps also by the person who at the end of his career did go to his office only to have the privilege of walking home with Gödel): make things as simple as possible, but not simpler. The statement "a sufficiently strong consistent theory cannot prove its own consistency" is unfortunately of the "simpler" class.

[PS: the answer "all Gödel sentences are true because they say of themselves they are unprovable" is instead related to the first incompleteness theorem. The answer about "incompleteness, AI, formalization" is again unrelated.]

$\endgroup$
-5
$\begingroup$

I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k $, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

$\endgroup$
3
  • $\begingroup$ I don't understand your third paragraph. Are you saying that under the assumptions in the 2nd paragraph, the limiting distribution (rescaling if necessary) is always Levy-stable? $\endgroup$
    – Yemon Choi
    Apr 12, 2011 at 1:28
  • $\begingroup$ @Yemon, Yes, this is what I was implying. Perhaps I was a little too cavalier? Certainly the sum of (well enough behaved) i.i.d. r.v.'s with power law tails converge to a Levy-Stable distribution... $\endgroup$ Apr 12, 2011 at 23:53
  • 6
    $\begingroup$ Generally such a limiting distribution doesn't exist. Perhaps you need to divide your sum by the square root of $n$? $\endgroup$ Dec 29, 2011 at 13:56
-5
$\begingroup$

One common misconception is that Category Theory is abstract. It is in fact, no more abstract than set theory or calculus. This common complaint is the usual complaint that people have when they first meet set theory or category theory. Since people interested in category theory are usually already interested in mathematics, it is strange - but perhaps not that strange - that they voice this complaint. It's underlying reason is that it's notions are unfamiliar, rather than abstract. Amd it doesn't help that there isn't a bestiary of easy examples that we wpuld have for algebra and calculus.

This needs to be developed.

$\endgroup$
4
  • 6
    $\begingroup$ Set theory is considered more abstract than calculus. $\endgroup$
    – Asaf Karagila
    Apr 3, 2021 at 10:29
  • $\begingroup$ @Asaf Karaglia: From the point of view of the informed everyman, they are all as abstract as each other. The point I'm making, is that in the usual pedagogy of mathematics, geometry, algebra and logic are threaded through each other and inform each other. However, this is not as true for Category Theory, which is seen as the new kid on the block and hence its unfamiliarity to mathematicians and mathematically informed disciplines like physics. $\endgroup$ Apr 3, 2021 at 10:49
  • 1
    $\begingroup$ I think I qualify as an 'informed everyman' on these subjects (I took undergraduate calculus up through real analysis out of Baby Rudin, and taught myself set theory and category theory). I would say there was about a 3-4 month period of intense headaches and confusion moving from real analysis in Rudin to set theory in Monk, because the level of abstraction was so much higher. You work directly with primitive notions and axioms, as opposed to 'concrete mathematical entities' like the reals together with algebraic operations etc. (I didn't downvote, but I did upvote Asaf's comment) $\endgroup$
    – Alec Rhea
    Apr 4, 2021 at 0:23
  • $\begingroup$ @Alec Rhea: I wouldn't count you as an informed everyman but as a mathematician. Especially as you know enough to discriminate between 'baby Rudin' and 'adult Rudin'. $\endgroup$ Apr 4, 2021 at 6:48
-11
$\begingroup$

When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characteristic two and I didn't get a good grade that day!

$\endgroup$
2
  • 19
    $\begingroup$ Quoth the question, "The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed)". $\endgroup$
    – JBL
    Dec 1, 2010 at 23:39
  • 3
    $\begingroup$ Also, this is of course just a special case of the more general “law of universal linearity”, which iirc was mentioned in earlier answers… $\endgroup$ Dec 2, 2010 at 0:40
-14
$\begingroup$

I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, $x^2$, and the fact that its graph does not look like a line at any value of $x$.

$\endgroup$
3
  • 8
    $\begingroup$ I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do). $\endgroup$ May 5, 2010 at 4:53
  • $\begingroup$ Yes, I did not read the question very carefully. I realize it is not a good comment, and, yes, it is more of a abd heuristic than anything else. $\endgroup$
    – Herb
    May 25, 2010 at 23:59
  • 1
    $\begingroup$ it is also a comment on the imprecision of the words locally, infinitesimally,.... This once led Oort-Steenbrink to give some careful restatements of results previously called as "local Torelli theorems"... $\endgroup$
    – roy smith
    Apr 14, 2011 at 19:02
1
6 7 8 9
10

Not the answer you're looking for? Browse other questions tagged or ask your own question.