The professor turned to me and said, "I believe that all numbers in mathematics can be expressed with only 6 symbols."

"A bold claim," I replied. "I suppose the you want all numbers to be encoded in base 6."

"Not at all. I am referring to Base 10."

"Then how is such a thing possible?" I questioned. "There are 10 digits, and even those aren't enough to let you express decimals!"

"On the contrary. See, the symbols I have chosen are not limited to numbers. They could include things like an addition sign, or possibly an exponentation sign."

"But even if you choose your symbols carefully, could you express every rational number? What about e? What about i or pi? What about sines, roots, and logarithms? Math is too complicated to be expressed in only six symbols!"

"You misjudge my work. Everything you have just mentioned can be expressed with my 6 symbols!"

What were the six symbols the professor chose?


  • We will assume Base 10 and standard notation for this question.

  • The symbols are all fairly simple, and do not include things like summation or randomness.

  • The following can be expressed with the symbols:

    • Any rational number

    • e, i, and pi.

  • You can perform the following operations on the above numbers:

    • Addition

    • Subtraction

    • Negation

    • Multiplication

    • Division

    • Exponentation

    • Rooting

    • Sine, cosine, and tangent

    • Log with any base

  • This means you can combine any of these. For example, you could express √(2*pi).

  • The answer is very simple, although not very obvious. If your answer seems like a cheat, it's probably not correct.

  • The final goal is to be able to express all combinations of the above numbers and operations. Infinity is not included, and infinitely long sequences are neither.

  • 2
    $\begingroup$ Can we use a countably or uncountably infinite number of symbols to express a number? $\endgroup$ Jan 24, 2017 at 0:43
  • $\begingroup$ So 1+1 counts as 3 symbols for the number 2? I have to be able to show every single numbers with the exact same 6 symbols? $\endgroup$ Jan 24, 2017 at 0:47
  • $\begingroup$ @stackreader I believe the OP means 6 different symbols, and however many of each. 1, + settles the naturals, adding - (minus) settles all integers, / settles rationals, and ^ might be enough to get irrationals, and even the entire complex plane, but I can't prove it. $\endgroup$ Jan 24, 2017 at 0:52
  • $\begingroup$ @greenturtle3141 if you use these 4 symbols, you only have 2 left for digits right? $\endgroup$ Jan 24, 2017 at 0:57
  • $\begingroup$ Yes. Just 1 (1) should do the trick though, unless it doesn't cover irrationals. $\endgroup$ Jan 24, 2017 at 1:00

2 Answers 2


I believe we can do everything described in the question with the following symbols:

1, -, exp, log, and parentheses.

So, here's how it goes. First of all,

0 is 1-1; then x+y is x-(0-y); now xy and x/y are exp(log x +- log y). So we have all the usual arithmetic operations and can therefore form all the rational numbers.


e is exp(1), i is exp(1/2 log(-1)), and i pi is log(-1).

From this we get

cos x in (exp(ix)+exp(-ix))/2, sin x = (exp(ix)-exp(-ix))/2i. Tan is the ratio of these two.

For exponentiation and rooting

we have x^y = exp(y log x) and y'th root of x = exp(1/y log x).

For log to arbitrary base


I'm pretty sure this gives

the right branch cuts for exponentiation and rooting

but I'm a little uneasy about

multiplication and division in the presence of zeros: we need log to give "$-\infty" in that case, and exp to accept that as input and yield 0.

  • $\begingroup$ Very close to my answer, but not the same. Keep in mind parenthesis count as symbols, so you are at 7 symbols. I just verified that your symbols can solve the problem, but I am waiting for a six symbol answer. $\endgroup$ Jan 24, 2017 at 1:49
  • $\begingroup$ Ah yes, I wasn't counting parens (not least because we don't really need them; we could adopt Polish notation or something). Anyway, I added them and compensated with an "obvious" optimization. $\endgroup$
    – Gareth McCaughan
    Jan 24, 2017 at 1:51
  • $\begingroup$ Can't you just change a/0 to 1-1 anyway? $\endgroup$
    – boboquack
    Jan 24, 2017 at 2:26
  • $\begingroup$ Yeah, I guess. But if you want to do actual mathematics rather than just constructing lots of individual numbers, you need a way of writing down products that doesn't break down when one multiplicand happens to be zero. $\endgroup$
    – Gareth McCaughan
    Jan 24, 2017 at 3:02
  • $\begingroup$ I see someone downvoted this answer. I would (note: this is a sincere statement of fact, not snark) be very interested to know why. $\endgroup$
    – Gareth McCaughan
    Jan 25, 2017 at 0:39

There is no solution if infinite strings are not allowed.

There are a countable number of finite strings on an alphabet of 6 symbols, and an uncountable number of complex numbers. There is no surjection from a countable set to an uncountable one.

  • $\begingroup$ This isn't really the point of the problem. Given any of the numbers listed, you could express it with only 6 symbols. We aren't counting infinity here, so you can assume that the number given doesn't incorporate it in any way, e.g. (1/2) × √(1/3) x √(√(1/4))... To infinity. $\endgroup$ Jan 24, 2017 at 2:14
  • $\begingroup$ Reading the question, seems that rational numbers and e, pi, i are all we need to express. The OP doesn't seem to be looking for every complex number. $\endgroup$
    – Ankoganit
    Jan 24, 2017 at 2:25
  • $\begingroup$ @AnonymousApple: Oh, any of the numbers listed. But could you express any rational or irrational number? If you mean that you could, then that's not possible. $\endgroup$
    – Deusovi
    Jan 24, 2017 at 2:29
  • $\begingroup$ The original question has been edited, which should hopefully clarify my intent. If you are still confused, please ask $\endgroup$ Jan 24, 2017 at 2:32
  • $\begingroup$ @AnonymousApple: So there are irrational numbers that cannot be written with the six symbols? $\endgroup$
    – Deusovi
    Jan 24, 2017 at 3:14

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