# What are some interesting sole exceptions or counterexamples? [duplicate]

Many theorems assert that a particular property holds for all objects in a class except those in a given list of exceptions. Examples of rules that admit precisely one exception include:

• All primes are odd, except for $2$
• All automorphisms of $S_n$ are inner for all $n$ except $6$
• All simple Lie algebras have abelian outer automorphism group except for $D_4$ ($\text{Out}(D_4) \cong S_3$, which leads to the exceptional phenomenon of triality)

What are some other interesting examples of results that admit (essentially) one exception?

Edit (modified from a comment below): Read strictly this question is subordinate to the one someone suggested it duplicated, but that question asks (more or less) about classifications of exceptional objects more generally, and not specifically about the case of a single exception. Partly because of this, very few of the answers to that question are admissible answers to this one (and two of them are actually already given in the question here).

## marked as duplicate by mrf, user7530, beep-boop, Grigory M, Aditya HaseDec 21 '14 at 7:03

• All primes are $\not \equiv 0$ mod $3$, except $3$ ;) – Bruno Joyal Dec 19 '14 at 3:48
• @JimmyK4542 Thanks, I didn't find that in my search for earlier related questions. Read strictly the present question is subordinate to the one you link, but that question asks (more or less) about classifications of exceptional objects more generally, and not specifically about the case of a single exception. Partly because of this, very few of the answers to that question are admissible answers to this one (and two of them are already given in the question here). – Travis Dec 19 '14 at 4:17
• Related: Accidents of small $n$ – MJD Dec 19 '14 at 9:15
• Community Wiki? – Jemmy Dec 20 '14 at 18:54
• How do you define "interesting"? – Paul Dec 21 '14 at 4:57

The smooth structure on $\mathbb{R}^n$ is unique up to diffeomorphism, except if $n = 4$.

• And the exception is particularly interesting because for n=4 there are uncountably many different smooth structures. That is a serious counter-example. – Francis Davey Dec 20 '14 at 22:58

If the $n$th Fibonacci number is prime then $n$ is prime, except that $F_n=3$ when $n=4$.

• I'ld be interested to see the proof of this. – Lyndon White Dec 20 '14 at 3:17
• @oxinabox First show that if $k\mid n$ then $F_k\mid F_n$. (Write $n=km$ and use induction on $m$.) Since $F_1=F_2=1$, this means that if $F_n$ is prime then $n$ can have no factors except $1,2$ and $n$. The only such numbers are $1$, primes and $4$. – David Dec 20 '14 at 8:38

A generalization of the Four Color Theorem says that the chromatic number of a closed surface with Euler characteristic $\chi$ (the number of colors needed to color any map on the surface) is bounded above (sharply) by $$\left\lfloor \frac{7 + \sqrt{49 - 24 \chi}}{2} \right\rfloor,$$ except for the Klein bottle, which has Euler Characteristic zero, so that the above formula gives a count of $7$ colors, but which only requires $6$. (See Ringel, G. and Youngs, J. W. T. Solution of the Heawood Map-Coloring Problem. Proc. Nat. Acad. Sci. USA 60, 438-445, 1968.)

(This is borrowed from this answer to the question JimmyK4542 references in the comments above.)

• I'm not a mathematician, so maybe I'm missing something; but if that formula provides an upper bound and not an exact answer, how is the Klein bottle an exception? 7 would seem to be an accurate upper bound. – Chris Hayes Dec 19 '14 at 5:29
• @ChrisHayes That's what's meant by the term "sharp". Put more explicitly, the for each surface (except the Klein bottle), there is some map for which you need the number of colors given in the formula, i.e., you can't improve the formula for that shape. For the Klein bottle, however, the formula gives seven colors, even though there's no map that requires that many, but there is a map that requires six. – Travis Dec 19 '14 at 5:33
• Ahh, I see. Thanks! – Chris Hayes Dec 19 '14 at 5:34

The group of units of $\mathbb Z/p^n \mathbb Z$ is cyclic for every prime power $p^n$, except when $p=2$; then it's cyclic only for $n=1,2$.

The duals of $L^p$ spaces: $$1\le p<\infty,\ \frac1p+\frac1q = 1\implies (L^p)^*=L^q,$$ but $$(L^\infty)^*\ne L^1.$$

$x^a-y^b=1$ has no solution in prime numbers, except $3^2-2^3$.

• It's Catalan's conjecture (now proved). The important you've missed is that $x,y,a,b>1$ since we have infinitely many natural number solutions to $x-y=1$. – user26486 Dec 19 '14 at 18:03
• Using the same trick as for Fermat, I replaced naturals by primes. – Yves Daoust Dec 20 '14 at 11:26
• Please stop weakening problems. It makes them a lot less elegant and useful. – user26486 Dec 20 '14 at 14:42

All spheres $S^n$ are simply-connected except $S^1$.

• and $S^0$! :)  – Benjamin Dec 19 '14 at 11:19
• @Benjamin: How does $S^0$ violate simply-connectedness? – fishlips Dec 19 '14 at 19:36
• @fishlips it's not path-connected, although the fundamental group will be trivial, regardless of basepoint. – Benjamin Dec 19 '14 at 20:38

$$\forall x,y \in \mathbb{Z_+}$$ $$|y^3-x^2| \ne 2$$ except for $x = 5$ and $y=3$

Ie 26 is the only number between a square and a cube.

• although $2^3 - 3^2 = 1$ – rbp Dec 20 '14 at 20:30

A lot of answers to this question can be given by taking unique answers and negating them. The one that came to mind just now was "the number of critical points of a Morse function on an $n$-dimensional manifold is not equal to 2 unless the manifold is an $n$-sphere." We also have things like "a division algebra over the reals of any finite dimension except 8 is associative."

For any prime $n$ except $2$, $x^n+y^n=z^n$ has no solution in nonzero integers.

• $n\in\mathbb N_{>2}$ is sufficient; you don't need $n\in\mathbb P_{>2}$. FLT says it's for positive integers $x,y,z$, but it's easily expandable to integers by checking cases: $n$ is odd and $1)$ $x,y<0$ (multiply both sides by $-1$), $2)$ $x>0,y<0$ ($\implies x^n=z^n+a^n$, where $y=-a$) and when $n$ is even (the signs of $x,y,z$ don't matter). – user26486 Dec 19 '14 at 17:56
• I avoided the naturals on purpose, as $\mathbb N_{>2}$ amounts to considering two exceptions ($1$ and $2$). With primes, $1$ vanishes. – Yves Daoust Dec 20 '14 at 11:24

All compact orientable two dimensional surfaces admit a metric with a constant non-positive curvature, except the sphere $S^2$.

For all $n$, the braid group on $n$-strings $B_n(M)$ of a closed orientable surface $M$ is torsion free unless $M=S^2$, the $2$-sphere.

The reason the same proof for other surfaces can't be used for the sphere is that the sphere is the only closed surface with non-trivial higher homotopy (because all other surfaces have a contractible universal cover, but the sphere is its own universal cover). This means that the Fadell-Neuwirth fibration sequence

$$F_{m+r,n−r} S^2 \to F_{m,n} S^2 \to F_{m,r} S^2$$

(here $F_{m,n}M$ is the ordered configuration space of $m$ point in $M\setminus\{n\mbox{ points}\}$) induces a long exact sequence in homotopy which isn't bounded at $\pi_1$, but has non-trivial groups appearing for all higher $\pi_k$ and so importantly, $F_{n,0}(S^2)$ is not an Eilenberg-Maclane space. This is crucial in the proof for other surfaces because the fundamental group of a $K(G,1)$ with the homotopy type of a finite CW-complex is torsion-free by an old theorem of P.A. Smith.

The reason such groups actually do have torsion is a consequence of the well-known 'Dirac string trick'. Essentially, if you apply a full twist to the trivial braid, the new braid it is not trivial, but if you do two full twists, then you can isotope the braid to the trivial braid by 'pulling the strings around the sphere'

Just an observation from $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$:

For any prime $p$ such that $p\neq 2$, $$\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}<p$$ and $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2.$$

• Why would this be restricted to primes? This seems like a rather simple function that just happens to cross y=x at x=2. – John Dvorak Dec 19 '14 at 17:12
• Yeah, this too seems true. The graph plotted shows y=x and y=(x+y)^(1/2) intersect at $x=0$ and $x=2$. But as the case is, out of all numbers primes are of prime importance. Might be thats why I checked only for primes. Anyways thanks for this observation. – supremum Dec 20 '14 at 1:38
• More specifically: The function is defined as f x=sqrt(x+f x) => (f x)^2 = x + f x => f x = 1/2 +- sqrt(4*x + 1)/2 (where the upper branch is used here). This function's derivative converges to zero monotonically, which means this inequality is true for any reals > 2. x=0 is the other intersection with x=y and also a vertex of the parabola. – John Dvorak Dec 20 '14 at 2:06

Here is one I found. The sum of two complex sinusoids with the same frequency is another sinusoid of that frequency, with 1 exception:

For amplitudes $A_1,A_2 \in \mathbb C$ and phases $\varphi_1, \varphi_2 \in \mathbb R$ and frequency $f \in \mathbb R$:

If $f_1(x) = A_1\cos(2\pi f~x + \phi_1)$ and $f_2(x) = A_2 \cos(2\pi f ~x + \phi_2)$

Then there is an amplitude $A_3 \in \mathbb C$ and phase $\varphi_3 \in \mathbb R$ such that

$f_1(x) + f_2(x) = A_3\cos(2\pi f~x + \varphi_3)$

with the exception:

$f_1(x) + f_2(x) = \cos(x) \pm i~\sin(x)$

...and all transforms of the above exception by moving/scaling the coordinate system.

• $A_1\cos(2\pi f x + \phi_1) + A_2 \cos(2 \pi f x + \phi_2)$ can be written as $A_3 \cos (2\pi f x + \varphi_3)$ if and only if $\phi_1 = \phi_2 + \pi k$ or $A_1 / A_2$ is a real number (or $A_2$ is zero); consider for example the real zeros of $A_3 \cos (2\pi f x + \varphi_3)$. I don't see how "$A_1 / A_2$ is not a real number" is "only 1 exception". – JiK Dec 20 '14 at 12:52

If $a,b\in\mathbb Z, n\in\mathbb N_{\ge 2}, (a,b)=1, a>b$, then $$\exists p\in\mathbb P( \forall k\in\mathbb N\cap[1;n)(p\mid a^n+b^n\wedge p\not\mid a^k+b^k))$$

The only exception to this is $(a,b,n)=(2,1,3)$.

The following variation of it

If $a,b\in\mathbb Z, n\in\mathbb N_{\ge 2}, (a,b)=1, a>b$, then $$\exists p\in\mathbb P( \forall k\in\mathbb N\cap [1;n)(p\mid a^n-b^n\wedge p\not\mid a^k-b^k))$$

has more than one exception, namely $(a,b,n)=(2,1,6)$ and any pair $(a,b,n)=(k,l,2)$ with $k+l=2^m$ for some $m\in\mathbb N$.

• Symbolic logic is a useful tool but it's pretty hard to read. – Matt Samuel Dec 19 '14 at 21:02
• @MattSamuel I thought my way of using it was the standard way. I always clearly put parantheses to the "such that" part of "exists" and to the "holds" (or "we have that") part of "for all" (i.e. "$\exists p(\ldots)$"$\equiv$"$\exists p$ such that $\ldots$", "$\forall x(\ldots)$"$\equiv$"$\forall x$ holds $\ldots$"). Therefore the symbols in my answer translate to "Exists $p\in\mathbb P$ such that for all $k\in\mathbb N, k<n$ holds $p\mid a^n+b^n$ and $p\not\mid a^k+b^k$". – user26486 Dec 19 '14 at 21:51
• I think the point Matt was making is that it's usually better to use words than symbols (unless you are treating formulas as objects, or you are in a context requiring more precision than usual.) Anyway, I think your notation is not quite standard: you use commas for too many things. Probably you should make the conjunction explicit in your hypothesis, and in the conclusion you should say something like $\forall k \in \mathbb{N}\,(k < n \implies ...)$. As it stands, it's not clear to me what the "$k<n$" is doing. – Trevor Wilson Dec 19 '14 at 22:56
• @TrevorWilson Alright, I've changed it to $k\in\mathbb N\cap [1;n]$. – user26486 Dec 19 '14 at 23:54
• That's arguably worse. I agree with the comments - this could as easily be expressed with words and would be much clearer into the bargain: 'for all $a, b\in \mathbb{Z}$ and all $n\geq 2$ there's a prime $p$ such that $p$ divides $a^n+b^n$ but doesn't divide $a^k+b^k$ for any $k\lt n$'. – Steven Stadnicki Dec 20 '14 at 0:00

There are Graeco-Latin squares of all orders greater than two except of order six.

Here are a couple involving zero, that students often forget to their chagrin:

The square of every real number is positive, except for $0$.

Every real number has a multiplicative inverse, except for $0$.

• Of course, the second is simply one of the field axioms. – Travis Dec 19 '14 at 5:17
• True, but students still forget it regularly. – paw88789 Dec 19 '14 at 5:18
• @Travis Is it really an axiom and not a theorem proven by $0^{-1}(0)=0^{-1}(0-0)=0^{-1}\cdot 0-0^{-1}\cdot 0=1-1=0\neq 1$? – user26486 Dec 19 '14 at 16:30
• @mathh It is usually taken as such, anyway, see en.wikipedia.org/wiki/… . – Travis Dec 19 '14 at 16:42
• @mathh The argument in your comment shows that $0$ has no inverse, but it does not show that every other real number does have an inverse. – David Dec 19 '14 at 18:42

Fermat's last theorem: There are no solutions in $\mathbb{Z}$ in the equation $a^n+b^n=c^n$, except for $n=2$

$0$ properties:

The only number that does not have a multiplicative inverse is $0$.

The only number that satisfies $-x=x$ is $0$.

$\frac xx=1, x \not \in \{0\}$

$\frac 0x=0, x \not \in \{0\}$

$x^x\in \mathbb{C}, x \not \in \{0\}$

• there are solutions for $n=1$ ... – greggo Dec 20 '14 at 2:34

No $3$ consecutive odd integers are prime except $3,5,7$.