enter image description here

So, what's the joke in number $9$?

$9$. You understand the following joke: $\forall \forall \exists \exists$

  • 5
    $\begingroup$ Here's one interpretation: everything2.com/user/tubular/writeups/… $\endgroup$ – mavavilj Oct 1 '16 at 18:23
  • $\begingroup$ that seems pretty reasonable, and about as good of an answer as we may be able to find. $\endgroup$ – 6005 Oct 1 '16 at 18:38
  • 1
    $\begingroup$ @6005 But I don't see anything funny about it. $\endgroup$ – mavavilj Oct 1 '16 at 18:41
  • $\begingroup$ Sometimes, one does not get a joke. Without knowing what the comic artist intended, we are left with no option but to speculate and/or give up. Especially since when you do not get a joke, the most likely scenario is that we have speculated about the right interpretation, but did not find that interpretation funny in any way. $\endgroup$ – 6005 Oct 1 '16 at 18:46
  • 4
    $\begingroup$ I am really surprised that nobody said this: The phrase has as many $\forall$ signs as $\exists$ signs, which is exactly what the phrase itself is saying (except that one has to mentally add "in this phrase"). As a joke, it's not great, but it's a clever observation... $\endgroup$ – Eric Stucky Oct 2 '16 at 6:24

For a joke to convey its point, it has to allude to some nearby cultural association that many people in the audience will understand. By far the most recognizable instance of $\forall ... \exists$ statements, or any statement with more than one type of quantifier, is the epsilon-delta formalism in calculus, and the misery that is said to induce in students. It is a rite of passage in mathematics education and therefore a sign on the road to being a mathematician.

I think that the joke is alluding to the word pattern of the epsilon-delta incantations and, to some extent, its stereotypically associated emotional reaction, as a synecdoche for mathematical language, mathematicians and their habits, and higher mathematics. A student who has suffered through the calculus classes knows that as soon as the words "for every [epsilon]" appear, "there exists [delta]" will follow, with all the complications attending that.

Hence, "for every $\forall$, there is a $\exists$". There is some implied commiseration in the joke, a reference to something notorious that the audience has been through, equivalent to

  • "for every $x$ there is a $y$" (school algebra)

  • "for every superscript there is a subscript" (tensors)

  • "for every grad there is a curl" (multivariable calculus)

There is also the idea that mathematicians are those who have passed through that perplexing rite and (in addition to being able to commiserate) are the ones who are fluent and comfortable with it, who can understand the joke.

  • $\begingroup$ I think this is a fairly reasonable interpretation. $\endgroup$ – mavavilj Oct 2 '16 at 4:54
  • $\begingroup$ Excellent interpretation and convincingly argued. I want to add that there is further some irony: the statement complains that for every $\forall$ there is a $\exists$, but it is itself a statement with a $\forall$ followed by a $\exists$, thus it is complaining against itself! Perhaps it is just a peculiar aptness and not exactly irony. $\endgroup$ – 6005 Oct 3 '16 at 1:54
  • $\begingroup$ I still don't get the joke... $\endgroup$ – IAmNoOne Oct 5 '16 at 9:41
  • $\begingroup$ @Nameless Just in case, the joke reads in "For every 'for every' symbol, there exists a 'there exists' symbol". $\endgroup$ – Frenzy Li Dec 14 '17 at 22:00

I think it spoofs a common format for theorems: "For all (something) there exists (something)".


It's simple. It means literally:

For every "for every" it does exist an "exist".

Which makes the writing funny because of the symbols. It's a wonderful.. "math-pun", thanks for having shared it! xD

  • 3
    $\begingroup$ Why is it a pun though? I don't get it. $\endgroup$ – Carl Schildkraut Oct 1 '16 at 18:12
  • 2
    $\begingroup$ I think it says that "for every rule, there is an exception". (If it still bothers you, read it as For all ∀ there is an ∃.) $\endgroup$ – Cehhiro Oct 1 '16 at 18:14
  • 6
    $\begingroup$ It appears that I'm not a mathematician ... Why is this a pun? On what? Puns are usually about double meaning, but I don't see even a single meaningful meaning to either the symbol sequence or the proposed "simple" "wonderful" explanation here. $\endgroup$ – Henning Makholm Oct 1 '16 at 18:17
  • 4
    $\begingroup$ Except $\exists$ hardly means an exception, and $\exists$ and $\forall$ are both "rules." I don't buy that reading. $\endgroup$ – Thomas Andrews Oct 1 '16 at 18:19
  • 3
    $\begingroup$ @OFRBG I don't think "for every rule, there is an exception" is correct ... :) $\endgroup$ – Hagen von Eitzen Oct 1 '16 at 18:19

My guess is the joke was just not executed that well; the author of spikedmath had something in mind, but we have no idea what it was. However, here are some possibilities I can think of:

  • Very often in mathematical statements, $\forall$ and $\exists$ quantifiers alternate. $(\forall \epsilon > 0)(\exists \delta > 0) (\forall x) \ldots$, for example. It is common to transform any formula into one in prenex normal form, and sometimes when we do we additionally assume that the quantifiers alternate, for convenience. So $\forall \forall \exists \exists$ could just be the joke that the quantifiers don't alternate. It's funny in the same way as "let $\epsilon < 0$": it's not the way they are usually used.

  • You gave a link in your comment in which someone says the joke is "for every forall symbol, there is a there exists symbol nearby".

  • The joke could be simply that it is syntactic nonsense. We expect to see a variable after the $\forall$ symbol, but we do not. Instead, $\forall \forall \exists \exists$ seems to be quantifying over another forall symbol and then over another exists symbol. So it reads like syntactic nonsense that seems like it should make sense but doesn't, and may strike someone as funny in this way.


I read $$\forall\forall\exists\exists$$ as

whenever there is a statement with $\forall x: A(x)$ there is also a statement with $\exists x: A(x)$.

Mathematically this is a triviality, but its amusing that we formulate different levels of semantics using the same symbols. It's some funny kind of smearing different levels of information by means of one syntax.

  • 1
    $\begingroup$ Or rather: whenever some statement contains the symbol $\forall$ it also contains the symbol $\exists$. And even: for each $\forall$ symbol there is a $\exists$ symbol attached to it. $\endgroup$ – Did Oct 2 '16 at 21:42
  • $\begingroup$ @Did: Yes, this was my first impression! :-) $\endgroup$ – Markus Scheuer Oct 3 '16 at 18:19

What struck me (personally) as droll was:

  1. It can be construed as redundant, in that two "for every" clauses in succession can be replaced by a single "for every", and similarly for two successive "there exists" clauses.

  2. It looks like the cry of a cartoon character falling upside down from a great height.

After reading the existing responses here, I lean toward the interpretation "whenever you see a $\forall$, a $\exists$ lurks nearby", a wry comment on epsilontics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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