The greatest arithmeticians on this side of the Kuiper Belt had gathered together on the summit of Olympus Mons to hold the solar system's first arithmetic summit. Everyone was very excited - you could hear mutterings about sums and products in at least eight different languages (and one dwarf language), and you could see mathematicians making their enthusiastic, but nonsensical mathematician gestures with anywhere from one to twenty arms (or similar appendages)!

As Phobos set, the crowd gathered into a circle and went silent in anticipation. The keynote speaker, a distinguished Earthican mathematician, stood up, and began her address, stating various truths about the sums, products, differences, and quotients of pairs of real numbers. Unfortunately, as she continued, she realized that the crowd was bewildered - and that, every time she said "zero", the Neptunians would make a pitiful noise with their speech-glands, and every time she said "one", the Mercurians would make what is surely an obscene gesture with their noodly appendages.

She paused and thought, and it dawned on her:

  • Though every member of the audience knew that her talk would concern only the operations of addition, subtraction, multiplication, and division over the real numbers, there was no common language

  • The words "zero" and "one" are words in other languages, but they roughly translate to grievous insults about the alien's version of a "face" or "mother". She must not use them.

What series of true statements can she make such that there can be no ambiguity in translation, where each statement is of the form

$$x \text{ op } y = z$$

with $x,\,y,$ and $z$ are real numbers other than $0$ or $1$ and op one of the four elementary operations?

"Unambiguous" means that the aliens can determine the meaning of each word, only knowing that there are four operations and that the numbers are real, but not knowing which operation, for instance, "plus" refers to or which number "two" refers to (as they do not speak the language). For instance $$2+2=4$$ $$4/2=2$$ Is ambiguous because if the aliens believed "/" meant "-" (i.e. swapped the operations in all equations), they'd get the equally correct $$2+2=4$$ $$4-2=2.$$ Thus, from the first two equations, they would not be able to determine the meaning of "/", as it would be equally true if it referred to division or subtraction.

For instance, were only "+" and "*" to be used, and were $0$ and $1$ allowable words, the following three statements would be a solution: $$1*1=1$$ $$0+0=0$$ $$1+1=2$$ As the aliens would, from the first two, be able to deduce that 1 was the identity element of * and 0 was the identity element of +. The only alternate translation of those two is $$0+0=0$$ $$1*1=1$$ however, under those rules, the last statement translations to $$0*0=c$$ (where $c$ is whatever 2 translates to) - but this is false, since $0*0=0$, and the aliens can recognize that $c$ was not the same word as meant $0$. Thus, there is exactly one way to translate the first series of three statements.

  • $\begingroup$ What do you mean by a "series" of statements? Is there some desired conclusion to reach? Are we counting the number of statements, longest list wins? Is there only one finite set of statements that could possibly be made? $\endgroup$
    – jscs
    Commented Feb 12, 2015 at 0:54
  • $\begingroup$ @Josh I added an example to the question; it is more impressive to do a shorter list (since once one establishes unambiguity, adding more statements is trivial), though any solution suffices. $\endgroup$ Commented Feb 12, 2015 at 1:02
  • $\begingroup$ Okay, I see. Any requirement to provide a full "glossary", i.e., do all four operators have to be used? $\endgroup$
    – jscs
    Commented Feb 12, 2015 at 1:06
  • $\begingroup$ @Josh No requirement; our keynote speaker is just looking to communicate something (but I imagine it is harder if you don't use all the operations) $\endgroup$ Commented Feb 12, 2015 at 1:09

5 Answers 5


So you're an alien, and you go to this interplanetary talk, and you see on the board: $$w\spadesuit z=y \\ w\heartsuit w=z \\ y\clubsuit x=z \\ x\clubsuit y=w$$

You immediately conclude that there is only one possible meaning of the operators:

$(\sqrt{\frac 12})-\sqrt{2}=(-\sqrt{\frac 12})$
$ (\sqrt{\frac 12})+(\sqrt{\frac 12})=\sqrt{2}$
$(-\sqrt{\frac 12})/(-\frac 12)=\sqrt{2}$
$(-\frac 12)/(-\sqrt{\frac 12})=(\sqrt{\frac 12})$

Multiplying y, w, and z by -1 produces the only other solution.

Having identified three of the operators, you can identify the fourth by process of elimination.

Proof that there is only one possible meaning of the operators:

Haha, nope. Did you really want to read 4 paragraphs of casework? I plugged it all into a CAS.

  • 1
    $\begingroup$ Here I'm assuming that, if the Earth speaker writes four different numbers on the chalkboard, then they have four different values. That is, if you see the equation "3+7=10", you won't think that "0*0=0" is a valid interpretation. All the other answers are assuming this as well, and I think it's reasonable. $\endgroup$
    – Lopsy
    Commented Feb 14, 2015 at 15:04

Here's a presumably far-from-optimal solution providing all four operators in six statements:

Presuming the aliens know that $0$ and $1$ are not in the mix, then it's easy to disambiguate addition/multiplication from subtraction/division; there are no numbers $a,b\geq 2$ with $a+b=b$ or $a\times b=b$, and contrariwise there are no numbers with $a-a=b$. So:

  1. $\alpha\mathop{\heartsuit}\alpha = \beta$
  2. $\alpha\mathop{\spadesuit}\alpha = \beta$
  3. $\alpha\mathop{\spadesuit}\gamma = \delta$
  4. $\gamma\mathop{\heartsuit}\gamma = \epsilon$
  5. $\epsilon\mathop{\clubsuit}\gamma = \gamma$
  6. $\epsilon\mathop{\diamondsuit}\delta=\beta$

(1) and (2) establish $\alpha=2$, $\beta=4$, and $\heartsuit$ and $\spadesuit$ as the 'positive' operators. (4), (5), and (6) establish the negative operators as the inverses of the positive ones (and in particular, imply that $\clubsuit$ is inverse to $\heartsuit$ and that $\diamondsuit$ is inverse to $\spadesuit$). But now suppose that $\spadesuit$ is $\times$. Then by (3) we have $\delta=\alpha\times\gamma=2\gamma$, and by inverting (6) we get $\epsilon=\beta\times\delta=4\delta=8\gamma$. But $\gamma\heartsuit\gamma=\epsilon$ would then claim that $\gamma+\gamma=\epsilon$ or $\epsilon=2\gamma$, which is clearly a contradiction (remember that none of these can be $0$ or $1$). This gives $\spadesuit=+$ and $\heartsuit=\times$; and now lines (5) and (6) give the other two.

  • 1
    $\begingroup$ I was intending that the aliens would be unaware of the fact that $0$ and $1$ are not in the mix (at best, they could deduce there are two bad numbers, but not know what they are). I think minor modification could make this answer work for that. The only problem would be that we could interpret $\heartsuit=+$, $\spadesuit=/$, $\clubsuit=-$, $\diamondsuit=*$, $\alpha=\frac{1}2$, and $\beta=1$. A statement like $\beta \spadesuit \beta = \kappa$ would suffice to eliminate that possibility. $\endgroup$ Commented Feb 12, 2015 at 1:56
  • $\begingroup$ I'd missed the 'over the real numbers' part of this; I was presuming that it was over the integers (which is also an interesting problem). $\endgroup$ Commented Feb 12, 2015 at 2:20
  • $\begingroup$ It does seem artificial to not be allowed to use 0 or 1 (which I agree is a good restriction) but for the solvers to not know that those are disallowed, but that may be a matter of taste. $\endgroup$ Commented Feb 12, 2015 at 2:21
  • $\begingroup$ (And I do think your proposed fix to my solution is a clean one and removes any ambiguity even in that case) $\endgroup$ Commented Feb 12, 2015 at 2:26

There are other ways to help identify the operations that do not depend on the arguments being natural numbers.

For example, if we say that $x\odot y=z$ and $y\odot x=z$, this tells us that the $\odot$ operator is commutative, meaning it is either addition or multiplication. Then we can do the same thing with another operator, say $\oplus$, so then we know that those two are addition and multiplication.

Then we can follow some of the same techniques: \begin{align}a\oplus a&=c\\a\odot a&=c \end{align} establishes that $a=2$ and $c=4$. Then we can state \begin{align}2\oplus4&=f\\2\odot4&=g\end{align} which tells them that $f$ and $g$ are either $6$ or $8$. Then we say $$4\odot4=f$$ This gives a lot of new information. It tells us that $\odot$ is addition and $f=8$, so then that leaves $\oplus$ as multiplication and $g=6$. From here we can establish the operations of subtraction and division with $8\oslash2=4$ and either $8\ominus2=6$ or $6\ominus2=4$. Now the aliens should be able to recognize all four elementary operations. Then we can tell them $6/2=b$, so they know $b=3$.

From here, we can add numbers to get all integers greater than $1$, and subtract to get integers less than $-1$. We can also divide to get rational numbers other than $\pm1$ and $0$.


I have a set of 12 which identifies the operators and the numbers 2-9 with no requirement that the aliens know 0 and 1 are prohibited:
Identifies 2 and 4 and tells our guests that +,* are addition and multiplication (but not which is which) and that -,/ are division and subtraction.

Identifies 6 and 8 and clarifies +,*
Then gives -,/
Identify the remaining numbers.

  • $\begingroup$ This works nicely, though it also relies on the values being whole numbers; otherwise you have the solutions involving 1/2 and 1 that Meelo mentions. $\endgroup$ Commented Feb 12, 2015 at 2:51

The challenge posed is to use the form "$x$ op $y$ = $z$" where $x,y,z$ are literals (i.e. actual numbers, not variables) other than $0$ and $1$ to unambiguously represent any aspect of arithmetic on the reals using the 4 basic operations.

Instead of establishing the identity of the numbers, it may be better to start by establishing the operators. Since Meelo (this puzzle's author) is satisfied with "saying something", this answer just has one example, the addition-multiplication relationship.

The equations $$3 + 3 = 6$$ $$3 + 6 = 9$$ $$3 * 3 = 9$$ establishes the relationship between addition and multiplication, and arguably establishes each operation on its own as well as a by-product. Naturally, the speaker can use equations involving $4$ and so on as further examples. Once these operations are established, the conference can generalize to sums and products of all integers, rationals, and reals, etc on their own. The inverses (subtraction and division) can also follow in a fairly straightforward manner.

Note that this is only unambiguous within the operation set $\{+,-,*,/\}$. If exponents were allowed, the same set of equations could still be confused with the multiplication-exponent relation.

[As an aside, it is arguable that since '3' appears 3 times in the equations with addition, and '3' features prominently in the last equation, there's also the potential of establishing the identity of that integer with this set of equations. Having this single agreed integer then allows ambiguity-free iteration to multiples of 3, as well as 'advanced' discourse of other forms (e.g. $3 + 3 = 2 + 2 + 2$) to introduce other numbers.]

  • 3
    $\begingroup$ Hello! I am from Pluto. I enjoyed your talk about the equations $\sqrt{2}*\sqrt{2}=2$, $\sqrt{2}*2=\sqrt{8}$, $\sqrt{2}+\sqrt{2}=\sqrt{8}$. $\endgroup$
    – Lopsy
    Commented Feb 14, 2015 at 13:57
  • $\begingroup$ @Lopsy Nice catch! You've snuck in the ambiguity I mentioned by introducing exponents via surds :) . $\endgroup$
    – Lawrence
    Commented Feb 14, 2015 at 22:03

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