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I'd like to pose the community here a challenge.

It's often supposed that if we encounter intelligent alien life, mathematics will probably be how we start communicating.

We have seen Rosetta-stone-type codebreaking, which depends on assuming some common ideas (e.g. proper nouns are phonetically about the same as in another language), and Enigma-type codebreaking, which depends on our ability to analyse a given mathematical system.

But if aliens landed and immediately offered us a proof of the Riemann Hypothesis, would we recognise it?

As a fun exercise, invent a new mathematical notation and see if other people in this community can figure out which axiom/theorem/concept you are demonstrating with it.

Your notation should have an unambiguous (albeit initially secret) specification.

Kudos for

a) expressing a nontrivial axiom/theorem/concept, and

b) inventing a notation which is as dissimilar as possible to anything you'd find in an existing textbook.

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closed as off-topic by symplectomorphic, Thomas, PseudoNeo, Daniel W. Farlow, Daniel Fischer Oct 29 '15 at 13:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – symplectomorphic, Thomas, PseudoNeo, Daniel W. Farlow, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ "see if other people ... can figure out which axiom/theorem/concept you are demonstrating with it": this is a great way to write bad mathematics, not good. $\endgroup$ – symplectomorphic Oct 29 '15 at 13:04
  • $\begingroup$ OK, we start with what subject? Pythagoras thm or $ (a+b)^2$ identity, or some logical statement or what? $\endgroup$ – Narasimham Oct 29 '15 at 13:06
  • $\begingroup$ Any subject, it's an open question. The idea is that someone could decipher your notation and say "oh I get it, this proves Pythagoras" or "pfft, too easy, those are just Peano numbers". $\endgroup$ – spraff Oct 29 '15 at 13:24
  • $\begingroup$ If we try to use math to communicate with aliens, I would expect us to begin by exchanging relatively "trivial" facts until at least one side understood something of the other's notation. $\endgroup$ – David K Oct 29 '15 at 13:29
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    $\begingroup$ Interesting idea for a question, although I can see why it was closed. $\endgroup$ – Cheerful Parsnip Oct 29 '15 at 14:02
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I don't think that it is worthless to play with symbols without knowing their meaning. For instance writing programs to prove theorems require that one be familiar with the game of meaningless formalism.

From the point of view of philosophy it is especially interesting to try to imagine how one would build an intuition in (him/her/it)-self given a formal system and no clue of its meaning. Would it be necessary that two creatures like two Earthlings or two Martians or two intelligent Machines would build the same intuition or would there be a way to share this intuition among each other.

So, I've taken the freedom to give here an example:

I will use the following undefined words (concepts) "pötty", "igenyes", "közt" and I will use the language of naive set theory and logic (in English).

Axiom 1. There exist at least two pötty $A$ and $B$ such that $A\not =B$. (no plural after numbers in that martian language : ( )

Axiom 2. If $A\not = B$ are pötty then there exists a pötty $C$ such that $B$ is közt $A$ and $C$.

Axiom 3. If $B$ is közt $A$ and $C$ then $A\not=C$.

Axiom 4. If $B$ is közt $A$ and $C$ then $A$ is not közt $B$ and $C$.

Definition An igenyes of $A$ and $B$ is the set of those pötty for which either $P$ is közt $A$ and $B$ or $A$ is közt $P$ and $B$ or $B$ is közt $A$ and $P$.

Axiom 5. If the pötty $C \not = D$ and both belong to the igenyes of the pötty $A$ and $B$ then the pötty $A$ belongs to the igenyes of the pötty $C$ and $D$.

Axiom 6. There is a pötty $C$ wich does not belong to the igenyes of the pötty $A$ and $B$.

Axiom 7. If $B$ is közt $A$ and $P$ and $C$ does not belong to the igenyes of the pötty $A$ and $B$ and if there exits a pötty $F$ which is közt $B$ and $C$ then there exists a pötty such that it is közt $A$ and $C$.

Theorem For any pair of pötty $A\not =B$ there is a pötty $C$ such that $C$ is közt $A$ and $B$.

Proof ???

Intution ???

If I get enough votes up then I disclose the intuition behind.

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  • $\begingroup$ I think I get it (spoiler alert): potty are points, kozt means between, and the igenyes are lines. But it helped that you used normal English to express the logic :) Real aliens would not do this... $\endgroup$ – goblin Oct 29 '15 at 14:23
  • $\begingroup$ @goblin: You win! $\endgroup$ – zoli Oct 29 '15 at 14:23
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    $\begingroup$ Why would aliens use symbols like $\neq$ or $=$ ? Those are huge clues into your system.... $\endgroup$ – N. S. Oct 29 '15 at 14:24
  • $\begingroup$ @N.S. Don't you see that I used Google translate? The original looked like: kjhkchkbfouov7 vbcorbp1qugfuzizoboboibo. At least a character recognition program gave this blah blah. $\endgroup$ – zoli Oct 29 '15 at 14:27
  • $\begingroup$ By the way, please don't be tempted to delete this answer in future. This was an interesting and fun exercise.... $\endgroup$ – goblin Oct 29 '15 at 14:28

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