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There are many "strange" functions to choose from and the deeper you get involved with math the more you encounter. I consciously don't mention any for reasons of bias. I am just curious what you consider strange and especially like.

Please also give a reason why you find this function strange and why you like it. Perhaps you could also give some kind of reference where to find further information.

As usually: Please only mention one function per post - and let the votes decide :-)

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    $\begingroup$ m.reddit.com/r/math/comments/9txzv/… $\endgroup$
    – Regenbogen
    Commented Apr 22, 2010 at 14:21
  • $\begingroup$ books.google.com/… $\endgroup$
    – Regenbogen
    Commented Apr 22, 2010 at 14:23
  • $\begingroup$ thank you... would like to see this community's answers and votes! $\endgroup$
    – vonjd
    Commented Apr 22, 2010 at 14:24
  • $\begingroup$ What restrictions, if any, are you placing on the functions? Do they have to be defined (or valued) on the reals or complex numbers, or are functions between arbitrary topological spaces, algebraic objects, unstructured sets etc. acceptable? What about functors between categories? $\endgroup$
    – Robin Saunders
    Commented Apr 22, 2010 at 14:39
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    $\begingroup$ Counterexamples in analysis has some nice ones: books.google.nl/… $\endgroup$
    – skupers
    Commented Aug 21, 2010 at 18:40

37 Answers 37

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Any of the isomorphisms $\mathbb{C}'\to S^{1}$, where $S^{1}$ is the unit circle and $\mathbb{C}'$ is the non-zero complex numbers, with the group operation for both being multiplication.

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  • $\begingroup$ Could you please give an example of such an isomorphism, or at least an argument to prove that these groups are isomorphic? I find it quite surprising. $\endgroup$
    – Bernikov
    Commented Jul 17, 2012 at 7:31
  • $\begingroup$ The proof of their existence comes from the structure theorem for divisible abelian groups. Both are isomorphic to $\mathbb{Z}/\mathbb{Q}\times \bigoplus_\omega \mathbb{Q}$. $\endgroup$
    – Daniel Miller
    Commented Jul 17, 2012 at 11:56
  • $\begingroup$ @DanielMiller perhaps I'm being dense, but surely $S^1$ is not isomorphic to $\mathbb{Z}/\mathbb{Q}\times\bigoplus_\omega\mathbb{Q}$? the former is uncountable, but the latter is a cartesian product of two countable sets and is hence countable. ($\bigoplus_\omega\mathbb{Q}$ is the set of functions from $\omega$ to $\mathbb{Q}$ with finite support) $\endgroup$
    – user147820
    Commented Dec 11, 2020 at 2:48
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    $\begingroup$ You're right. The subscript should be $\aleph_1$. $\endgroup$
    – Daniel Miller
    Commented Dec 12, 2020 at 18:59
  • $\begingroup$ @DanielMiller now I'm being a bit pedantic, for which I apologize, but I think it should be $2^{\aleph_0}$, unless we're assuming the continuum hypothesis. anyway +1, I agree this isomorphism is a very bizarre function indeed! $\endgroup$
    – user147820
    Commented Apr 29, 2021 at 15:37
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Characteristic p commutative algebra leads naturally to the construction of various continuous functions on [0,1]^m that have beautiful self-similarity properties; for explication and some pictures see:

Pedro Teixeira, Syzygy gap fractals--I, arXiv 1008.0583
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How about a function f: f(f(x)) = exp(x).

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  • $\begingroup$ Could you give a reference to understand more about it - thanks $\endgroup$
    – vonjd
    Commented Jul 17, 2012 at 7:46
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How about the function given by the Banach-Tarski paradox? This maps a ball into two copies of the same size ball, and is composed of isometries on subsets of $\mathbb{R}^3$.

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Just a simple construction to illustrate Nate Eldredge's answer about functions with dense graphs. Pick any $\mathbb{R}$-vector space E with a norm. On E, choose a non-continuous linear form $L: E \to \mathbb{R}$; now this can only be done if $\dim(E)=\infty$, of course.

Then, pick y such that $L(y)=1$, and let $T: E \to E$ be defined by $Tx=x-L(x)y$. Then obviously T maps E onto the kernel of L; it is not difficult to prove that $\ker (L)$ must be dense in E for any non-continuous L (the two conditions are even equivalent), and thus the graph of T must be dense in $E \times E$.

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The formula for the nth term in the Fibonacci sequence

$F_{n} = \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$

This is interesting because it is a non-recursive expression for the Fibonacci sequence and also because it involves the golden ratio.

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  • $\begingroup$ Yes, but isn't $\,x^n\,$ defined recursively? You still don't avoid recursion. Also, the golden ratio is not rational. If you define $\,\phi\,$ as the positive root of $\,x^2=x+1\,$ then $\,\phi^n = F_{n-1}+F_n\phi\,$ so $\,\phi^n\,$ is defined by $\,F_n.$ $\endgroup$
    – Somos
    Commented Jan 2, 2019 at 22:31
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I like the beauty and mysticism of Euler's identity:

$$f(\theta) = e^{i\theta} = \cos\theta + i \sin\theta$$

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