# Theorems that disappointed mathematicians

I was looking for a list of theorems that disappointed mathematicians, in the sense that some disgruntlement regarding the fact that the theorem in consideration being true/false was notable. An example could be a theorem whose negation was notably hoped to be true or a theorem which ended up implying the negation of statements notably hoped to be true. The only examples I know are Gödel’s incompleteness theorems, and I was hoping to know if there are any other similar examples.

Thank you.

• I personally still find the unsolvability of the Word Problem for groups to be deeply disappointing. I think we've collectively gotten over the unsolvability of the quintic, but perhaps at the time people were distressed by it.
– lulu
Commented Aug 9, 2023 at 17:42
• The pythagoreans could not even accept the irrationality of $\sqrt{2}$ Commented Aug 9, 2023 at 17:52
• Hilbert's vision is the best example , but not at all the only one. Very disappointing was that the opriginal version of set theory (which was very very strongly believed to be consistent) turned out to be inconsistent. Commented Aug 9, 2023 at 17:54
• For programmers , Rice's theorem is probably disappointing. Commented Aug 9, 2023 at 17:57
• Although this is a very broad question I think it is of historical interest. Please don't close this, a little variety on MSE would be good for the site. Many of the more interesting posts are unfortunately the older posts since the current atmosphere is less accepting Commented Aug 9, 2023 at 18:11

Just to collect some of the comments into an answer:

• Unsolvability of the word problem for groups, that is, there exist groups for which there is no algorithm determining when two elements are equal
• Unsolvability of the general quintic (and unsolvability of particular quintics such as $$x^5-x+1=0$$) in terms of radicals
• To the Pythagoreans, existence of irrationals (in particular, irrationality of $$\sqrt2$$)
• Russell's paradox, and in particular, the inconsistency of Frege's set theory
• Rice's theorem that, roughly, all "interesting" questions about programs are undecidable. (The halting problem, the mother of all undecidable problems, is a special case.)
• The category $$\sf Top$$, whose objects are topological spaces and whose morphisms are continuous functions, is not Cartesian closed, in particular it does not have exponential objects. This motivates the search for "nicer" categories of topological spaces
• Hilbert's tenth problem : there is no algorithm for polynomial Diophantine equations.
• Noneuclidean geometry. In one way, mathematicians were disappointed by the proof that the parallel postulate did not follow from the other more intuitive axioms. That disappointment was clearly tempered by the pleasure in expanding the idea of a geometry.
• The existence of non-measurable sets in set theory with the axiom of choice. (Partly tempered by the Solovay model, a model of set theory without the (full) axiom of choice in which every set is measurable, though I don't know how much working measure theorists care about it.)
• The independence of the continuum hypothesis from ZFC set theory, on the existence of a set of intermediate cardinality between the naturals and the reals. Georg Cantor tried to prove it for many years, and it was Hilbert's first problem in his 1900 address. Gödel proved its consistency with ZFC in 1940, and Cohen proved the consistency of its negation in 1963.
• The Weierstrass function, the first continuous nowhere-differentiable function, got some pretty nasty comments thrown its way
• Arrow's impossibility theorem that, when there are at least three alternatives, there is no ranked voting system (other than dictatorship!) that satisfies two reasonable-sounding conditions. Informally, every ranked voting system sacrifices some form of "fairness".
• Community wiki'd of course Commented Aug 10, 2023 at 2:58
• It occurs to me that the irrationality of $\sqrt2$ is essentially "unsolvability of the quadratic by the field operations ($+$, $-$, $\times$, $\div$)" Commented Aug 10, 2023 at 5:35
• What about Proofs that Disappointed Mathematicians?. Like the four colour problem (the proof was a great relief of course but far from satisfying) or The global nonlinear stability of the Minkowski space, which I heard was a 100+pp proof.
– NBF
Commented Aug 10, 2023 at 17:12
• Probably anything from mathoverflow.net/questions/11540/… qualifies. And probably worth adding The Halting Problem. Commented Aug 10, 2023 at 20:05
• @OmnipotentEntity Nothing but love for our spiky boy Commented Aug 11, 2023 at 18:54

I'm not sure if this theorem disappointed mathematicians, but it certainly let down sailors, the military, and anyone hoping for a perfect flawless map of the Earth. When someone creates a flat map, they aim for it to accurately represent the Earth by preserving certain physical properties like distance, directions, or area. In each case, the induced mapping between the Earth and the map is known as isometric, conformal, or equiareal respectively.

• The Mercator projection preserves angles and shapes but distorts areas as you move away from the equator.
• The Peters projection preserves areas but distorts shapes.
• The Robinson projection aims to balance various properties but doesn't preserve any of them perfectly.

However, Carl Friedrich Gauss demonstrated that there is no hope for distance-preserving maps, even on a small scale. In modern mathematical terminology, his celebrated theorem (Theorema Egregium) can be stated as: The Gaussian curvature $$K$$ of a surface is invariant under local isometry.

• Surely the existence of triangles with straight (i.e. length-minimizing) sides whose angles add up to more than 180 degrees gives a short proof of this fact, without using all the machinery of Gauss's theorem. (Also, it might be useful to give a short definition of Gaussian curvature, namely that it's the product of the principal curvatures. Notably, the principal curvatures themselves are not invariant under local isometry.) Commented Aug 11, 2023 at 3:44

There's a collection of results which most will learn in their first course on non-commutative rings which I recall being distinctly disappointed by and then heard that I was far from alone in this opinion.

When it comes to studying non-commutative rings it initially seems theoretically very interesting! When you work over fields, modules (in this case, vector spaces) have a very simple and lovely theory based around bases but lots of results from that context (even the existence of a basis) can fail when working over general non-commutative rings and it can be great fun to try and work out different theoretical requirements you could put on your ring to try and fix this (e.g. being a division ring, or semisimple or being an algebra over a field etc.).

However to actually see this difference in any meaningful way, you want a good supply of genuinely non-commutative rings which are somewhat reasonable in their properties:

• One example is that you might look at finite rings and we know that there is exactly one finite field of size equal to each prime power for each prime, so why not relax that condition to being just an integral domain but still commutative? It's not very difficult to argue that the finite integral domains are all fields...
• OK, so let's instead say that the rings we look at are division rings (i.e. every nonzero element has an inverse but the ring is not necessarily commutative). Unfortunately, a little more work gets you Wedderburn's Little Theorem which tells you that the finite division rings are all fields.
• OK let's even relax associativity! We'll look at finite alternative division rings, i.e. ones where $$x(xy) = (xx)y$$ and $$y(xx) = (yx)x$$ but all $$x,y$$ in the ring but $$a(bc)$$ doesn't have to equal $$(ab)c$$ for all $$a,b,c$$ in the ring. Well the Artin-Zorn Theorem tells you that in this case they are still fields.
• Another place you might look is real division algebras (which we suppose are associative but not commutative). $$\mathbb{R}$$ and $$\mathbb{C}$$ are rather simple examples, so you might try and apply the same kind of construction you performed on $$\mathbb{R}$$ to get $$\mathbb{C}$$ on $$\mathbb{C}$$ instead (this is the Cayley-Dickson construction). This works and gives you the quaternions which are very interesting and can help with understanding all kinds of things in group theory, number theory and physics.
• Let's try that again! Well actually you only get the octonions, which are still very interesting but they aren't actually distributive so don't really classify as what we're looking for. OK, are there any other finite-dimensional real division algebras? Frobenius' Theorem says that $$\mathbb{R}, \mathbb{C}, \mathbb{H}$$ are the only examples unfortunately.
• How about allowing potentially countably-infinite dimensional division algebras? Unfortunately this still only gives $$\mathbb{R}, \mathbb{C},\mathbb{H}$$ due to application of a result which has a number of names (in books I've read it goes variously by the non-commutative nullstellensatz or the Amitsur-Schur Lemma etc.).
• There are lots of similarly restrictive requirements on real division algebras even if we loosen the definition further. Suppose we allow them to only be alternative? Zorn proved you only get $$\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$$.
• What if we allow it to not even be alternative but it must be a normed vector space with a norm which satisfies $$\| ab\| = \|a\| \|b \|$$? Hurwitz tells us you only get $$\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$$.
• What if we require it to be associative and have a norm which satisfies $$\|ab\| \leq \|a\| \|b\|$$ and be complete (a Banach algebra)? Gelfand-Mazur tells you it's just $$\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}$$.
• The most general possible situation in this area is any real division algebra which isn't necessarily associative or alternative and no other requirements. Then Kervaire-Milnor tells you (by applying some deep results from algebraic topology) that they can still only be of dimensions 1,2,4 or 8 (and I think people feel there are more stringent requirements on such rings but this is still an area of research).

The outcome is that you struggle to prove things about these much more complex non-commutative rings but the moment you want to find some kind of family of them which is vaguely comprehensible (as opposed to, say, general matrix rings which are tough for a first course) you find none! I and a number of others felt that Ring Theory might be a lot richer if there had been finite, non-field division rings for example...

• "every nonzero element has a right and left inverse but they aren't necessarily the same" - I thought left and right inverses must be equal in rings if both exist? Commented Aug 11, 2023 at 3:47
• @AkivaWeinberger That's right: it's only after we relax associativity that left and right inverses may not be equal. Commented Aug 11, 2023 at 3:54
• Note that the situation gets much better if you replace $\mathbb{R}$ with (say) $\mathbb{Q}$: the Brauer group of $\mathbb{Q}$ (which counts finite-dimensional central division algebras over $\mathbb{Q}$) is quite large, as opposed to $\mathbb{R}$'s puny Brauer group $C_2$ (whose elements correspond to $\mathbb{R}$ itself and $\mathbb{H}$ - note that $\mathbb{C}$ isn't central as an $\mathbb{R}$-algebra). Commented Aug 12, 2023 at 3:47
• @AkivaWeinberger yes, sorry I guess the major difference is non-commutativity. I'll edit that out. Commented Aug 14, 2023 at 11:50

Almost every reasonable conjecture in the theory of infinite dimensional Banach spaces, which were resolved by very difficult counter examples. The one I am most familiar with is the Gowers-Maurey space: https://arxiv.org/abs/math/9205204.

(I am told that several complex variables suffers from a similar problem, but I don't know any details. Also, I believe that people will be disappointed if the Navier-Stokes is shown to not necessarily have regular solutions.)

One such example which I can think of is that Mertens conjecture was proven false. This states that Mertens function which is given by $$M(x)=\sum_{n\leq x}\mu(n)$$ satisfies $$\vert M(x)\vert\leq \sqrt{x}$$. I should note that $$\mu(n)$$ is the Möbius function in number theory which is $$0$$ unless $$n$$ is square-free in that case it is given by $$\mu(p_1p_2...p_k)=(-1)^k$$ One reason that the disproof of this conjecture is disappointing is that Mertens conjecture implies the Riemann hypothesis, RH is equivalent to the statement that $$M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$$. Thus, we expect the square root cancellation in the summation; however, the disappointing part of Mertens’ conjecture being false is that as much as we expect RH to be true, if one were to attempt to prove RH by showing $$M(x)=O\left(x^{\frac{1}{2}+\epsilon}\right)$$, then the implied constant will not be $$1$$ (and more likely than not must depend upon $$\epsilon$$).